Class 10

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 4 Quadratic Equations Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 4 Quadratic Equations

Short/Long Answer Type Questions

Question 1.
If x = 3 is one root of quadratic equation x² – 2kx – 6 = 0, then find the value of k.
Solution :
Since, x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0. So, we put x = 3 in the given equation, we get
(3)² – 2k × 3 – 6 = 0
⇒ 9 – 6k – 6 = 0
⇒ – 6k + 3 = 0
⇒ k = \(\frac {3}{6}\)
⇒ k = \(\frac {1}{2}\)

Question 2.
If one root of the quadratic equation 6x² – x – k = 0 is \(\frac {2}{3}\) then find the value of k.
Solution :
Since, \(\frac {2}{3}\) is one root of the quadratic equation 6x² – x – k = 0.
So, we put x = \(\frac {2}{3}\) in the given equation, we get
6 × (\(\frac {2}{3}\))² – \(\frac {2}{3}\) – k = 0
⇒ 6 × \(\frac{4}{9}-\frac{2}{3}\) – k = 0
⇒ \(\frac{8}{3}-\frac{2}{3}\) – k = 0
⇒ \(\frac {6}{3}\) – k = 0
⇒ 2 – k = 0
⇒ k = 2

Question 3.
Find the possible root of \(\sqrt{3 x^2+6}\) = 9.
Solution :
The given equation is
\(\sqrt{3 x^2+6}\) = 9
Squaring on both sides, we get
3x² + 6 = 81
⇒ 3x² = 81 – 6
⇒ 3x² = 75
⇒ x² = \(\frac {75}{3}\)
⇒ x² = 25
Taking square root both sides, we get
\(\sqrt{x^2}\) = \(\sqrt{25}\)
x = ± 5

Question 4.
Solve : \(\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}=\frac{1}{x}\) where a + b ≠ 0.
Solution :
The given equation is :

⇒ – (a + b) × ab = (a + b)x (a + b + x)
⇒ (a + b)x (a + b + x) + ab (a + b) = 0
⇒ (a + b) [x(a + b + x) + ab] = 0
⇒ x (a + b + x) + ab = 0
⇒ ax + bx + x² + ab = 0
⇒ x² + ax + bx + ab = 0
⇒ x(x + a) + b (x + a) = 0
⇒ (x + a) (x + b) = 0
⇒ (x + a) = 0 and (x + b) = 0
⇒ x = – a and x = – b

Question 5.
A train travels 300 km at a uniform speed. If the speed had been 10 km/hr more, it would have 1 hr less for same journey. Find the speed of train.
Solution :
Let the speed of train be x km/hr
Distance = 300 km (given)
Time taken by train = \(\frac {300}{x}\)hrs.
[∵ Time = Distance / Speed]
If speed had been 10 km/hr more then new speed = (x + 10) km/hr.
Time taken = \(\frac{360}{x+10}\) hrs.
According to question

Question 6.
The diagonal of a rectangular field is 25 metres more than the shorter side. If longer side is 23 metres more than the shorter side, find the sides of the field.
Solution :
Let the shorter side be x m. Then diagonal = (x + 25)m and longer side = (x + 23)m.
We known that each angle of a rectangle is 90°. In a right ΔABC, we have (x + 25)² = (x + 23)² + x².

⇒ x² + 50x + 625 = x² + 46x + 529 + x²
⇒ 2x² + 46x + 529 – x² – 50x – 625 = 0
⇒ x² – 4x – 96 = 0
⇒ x² – (12 – 8)x – 96 = 0
⇒ x² – 12x + 8x – 96 = 0
⇒ (x² – 12x) + (8x – 96) = 0
⇒ x(x – 12) + 8 (x – 12) = 0
⇒ (x – 12) (x + 8) = 0
⇒ x – 12 = 0 and x + 8 = 0
⇒ x = 12 and x = – 8 (Reject)
Hence, longer side 12 + 23 i.e., 35 m and shorter side = 12 m.

Question 7.
In a flight of 600 km, an discraft was slowed down due to bad weather. The average speed of the strip was reduced by 200 km/hr and time of flight increased by 30 minutes. Find the duration of flight.
Solution :
Distance = 600 km.
Let the speed of discraft be x km/hr. Time taken by discraft (T₁) = \(\frac {600}{x}\) hrs.
It speed of is craft was slow down by 200 km/ hr then new speed = (x – 200) km/hr.
Time taken by discraft (T₂) = \(\frac{600}{x-200}\) hrs
According to question,

⇒ 2,40,000 = x² – 200x
⇒ x² – 200x – 2,40,000 = 0
⇒ x² – (600 – 400)x – 2,40,000 = 0
⇒ x² – 600x + 400x – 2,40,000 = 0
⇒ x(x – 600) + 400(x – 600) = 0
⇒ (x – 600) (x + 400) = 0
⇒ x – 600 = 0 and x + 400 = 0
⇒ x = 600 and x = – 400 (Reject)
So, speed of discraft = 600 km/hr and duration of flight = \(\frac{600}{600-200}\) hrs
= \(\frac{600}{400}=\frac{3}{2}\) = 1\(\frac {1}{2}\)hrs.

Question 8.
The speed of a boat in still water is 18 km/hr. It takes \(\frac {1}{2}\) an hours extra in going 12 km upstream instread of going the same distance down stream. Find the speed of the stream.
Solution :
Let the speed of stream be x km/hr. Speed of boat in still water = 18 km/hr (given)
Speed of boat in down stream = (18 + x) km/hr
Time taken in going down stream (T₁) = \(\frac{12}{(18+x)}\) hrs
Speed of boat in upstream (18 – x) km/hr.
Time taken in going upstream (T₂) = \(\frac{12}{(18-x)}\) hrs
According to question,

⇒ 48x = – x² + 324
⇒ x² + 48x – 324 = 0
⇒ x² + (54 – 6)x – 324 = 0
⇒ x² + 54x – 6x – 324 = 0
⇒ x(x + 54) – 6(x + 54) = 0
⇒ (x + 54) (x – 6) = 0
⇒ x + 54 = 0 and x – 6 = 0
⇒ x = – 54 (Reject) and x = 6
Hence, speed of stream = 6 km/hr.

Question 9.
I can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately ?
Solution :
Let the time taken by larger pipe be x hours and time taken by smaller pipe be y hours
The portion of tank filled by larger pipe in 1 hour = \(\frac {1}{x}\)part
The portion of tank filled by smaller pipe in 1 hour = \(\frac {1}{y}\)part
Two pipe can fill the tank in 12 hours.
∴ The portion of tank filled by both pipe in 1 hr = \(\frac {1}{12}\)part
According to condition 1st
\(\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\) ……………(1)
If larger pipe used for 4 hours, then the portion of tank filled by larger pipe in 4 hours = \(\frac {4}{x}\)part
And smaller pipe used for 9 hours, then. Portion of tank filled by smaller pipe in 9 hours = \(\frac {9}{y}\)part
According to condition 2nd
\(\frac{4}{x}+\frac{9}{y}=\frac{1}{2}\) ………..(2)
Let \(\frac{1}{x}\) be a and \(\frac{1}{y}\) be b, from equations (1) and (2) we get
a + b = \(\frac {1}{12}\) ………(3)
4a + 9b = \(\frac {1}{2}\) ………….(4)
By cross-multiplication, we get

Putting the values of a and b, we get
\(\frac{1}{x}=\frac{1}{20}\) and \(\frac{1}{y}=\frac{1}{30}\)
⇒ x = 20 and y = 30
Hence, time taken by larger pipe is 20 hours and smaller pipe is 30 hours.

Question 10.
Find the value of k for which are roots of the equation 3x² – 10x + k = 0 are reciprocal of each other.
Solution :
The given equation is :
3x² – 10x + k = 0
Let one root of the equation be a since roots are reciprocal each other so, other root is \(\frac {1}{α}\)
Product of two roots = \(\frac {c}{a}\)
⇒ α × \(\frac {1}{α}\) = \(\frac {k}{3}\)
⇒ 1 = \(\frac {k}{3}\)
⇒ k = 3

Question 11.
For what value of k, the given quadratic equation kx² – 6x – 1 = 0 has no real roots ?
Solution :
The given equation is :
kx² – 6x – 1 = 0
The condition for no real roots is :
D < 0
⇒ b² – 4ac < 0
⇒ (-6)² – 4 × k × -1 < 0
⇒ 36 + 4k < 0
⇒ 4k < – 36
⇒ k < –\(\frac {36}{4}\)
⇒ k < – 9
Hence, k should be less than – 9 i.e., (-10, – 11, …..).

Question 12.
For what values of k, the roots of the equation kx² + 4x + k = 0 are real ?
Solution :
The given equation is :
x² + 4x + k = 0
∵ The given equation has real roots
∴ D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ (4)² – 4 × 1 × k ≥ 0
⇒ 16 – 4k ≥ 0
⇒ – 4k ≥ – 16
⇒ k ≤ \(\frac {-16}{-4}\)
⇒ k ≤ + 4
Therefore, k should be ≤ 4 i.e., (4, 3, 2, 1, ….)

Question 13.
Find the k so that the quadratie equation (k + 1)x² – 2 (k + 1)x + 1 = 0 has equal roots.
Solution :
The given equation is :
(k + 1)x² – 2(k + 1) x + 1 = 0
Since, equation has equal roots
∴ D = 0
⇒ b² – 4ac = 0
⇒ [-2(k + 1)]² – 4 × (k + 1) × 1 = 0
⇒ 4(k² + 2k + 1) – 4k – 4 = 0
⇒ 4k² + 8k + 4 – 4k – 4 = 0
⇒ 4k² + 4k = 0
⇒ 4k(k + 1) = 0 and k + 1 = 9
⇒ k = – 1
k = – 1 (Reject since if we put the k = – 1 in the equation (-1 + 1)x will be zero
k = 0

Question 14.
If – 3 is a root of the quadratic equation 2x² + Px – 15 = 0 while the quadratic equation x² – 4Px + k = 0 has equal root. Find the value of k.
Solution :
Since, – 3 is a root of the equation
2x² + Px – 15 = 0.
So, we put x = – 3 in this equation, we get
2(-3)² + P(-3) – 15 = 0
⇒ 18 – 3P – 15 = 0
⇒ 3 – 3P = 0
⇒ P = \(\frac {-3}{-3}\) = 1
Putting the value of P in the equation, we get
x² – 4 × 1 + k = 0
⇒ x² – 4x + k = 0
The equation x² – 4x + k = 0 has equal roots.
So, D = 0
⇒ b² – 4ac = 0
⇒ (-4)² – 4 × 1 × k = 0
⇒ 16 – 4k = 0
⇒ k = \(\frac {-16}{-4}\)
⇒ k = 4

Question 15.
If the roots of the quadratic equation (x – a) (x – b) + (x – b) (x – c) + (x – c)(x – a) = 0 are equal, then show that a = b = c.
Solution :
The given equation is :
(x – a) (x – b) + (x – b) (x – c) + (x – c)(x – a) = 0
x² – ax – bx + ab + x² – bx – xc + bc + x² – cx – ax + ac = 0
⇒ 3x² – 2ax – 2x – 2x + ab + bc + ac = 0
⇒ 3x² – 2 (a + b + c) x + (ab + bc + ac) = 0
Since, roots are equal, then
D = 0
⇒ b² – 4ac = 0
⇒ [- 2 (a + b + c)]² – 4 × 3 × (ab + bc + ac) = 0
⇒ 4(a² + b² + c² + 2ab + 2bc + 2ca) – 12 (ab + bc + ac) = 0
⇒ 4[a² + b² + c² + 2ab + 2bc + 2ca – 3ab – 3bc – 3ac] = 0
⇒ a² + b² + c² – ab – bc – ac = 0
⇒ \(\frac {1}{2}\)[2a² + 2b² + 2c² – 2ab – 2bc – 2ac] = 0
⇒ \(\frac {1}{2}\)[(a² + b² – 2ab) + (b² + c² – 2bc) + (a² + c² – 2ac)] = 0
⇒ (a – b)² + (b – c)² + (a – c)² = 0
⇒ (a – b)² = 0, (b – c)² = 0, (a – c)² = 0
⇒ a = b, b = c, a = c
∴ a = b = c
Hence Proved.

Question 16.
If the root of the quadratic equation (c² – ab)x² – 2 (a² – bc)x + b² – ac = 0 in x are equal, then show that either a = 0 or a³ + b³ + c³ = 3abc.
Solution :
The given equation is:
(c² – ab)x² – 2(a² – bc)x + (b² – ac) = 0
Since, roots are equal, then
D = 0
⇒ b² – 4ac = 0
⇒ [- 2(a² – bc)]² – 4 × (c² – ab) (b² – ac) = 0
⇒ 4(a⁴ + b²c² – 2abc) – 4(b²c² – ac³ – ab³ + a²bc) = 0
⇒ 4[a⁴ + b²c² – 2a²bc – b²c² + ac³ + ab³ – a²bc) = 0
⇒ [a⁴ – 3a²bc + ac³ + ab³] = 0
⇒ a[a³ + b³ + c³ – 3abc] = 0
⇒ a = 0 or a³ + b³ + c³ – 3abc = 0
⇒ a = 0 or a³ + b³ + c³ = 3abc
Hence proved.

Question 17.
A pole has to erected at a point on the boundary of a circular park of diameter 17 metres in such a way that the differences of its distances from two dimetrically opposite fixed gates A and B on the boundary is 7 metres. It is possible to do so? If yes, at what distances from the two gates should the pole be erected ?
Solution :
Let us first draw the diagram. Let P be location of the pole. Let the distance of pole from the gate B be x m ie., BP = x m. Now distances of the pole from the two gates = AP – BP (or BP – AP) = 7 m.
Therefore AP = (x + 7) m.
Now,
AB = 17 m.

∠APB = 90° (angle in a pamicircle is 90°) In a right triangle APB, we have
AB² = AP² + BP²
[By Pythagoaes theorem]
⇒ 17² = (x + 7)² + x²
⇒ 289 = x² + 49 + 14x + x²
⇒ 289 = 2x² + 14x + 49
⇒ 2x² + 14x + 49 – 289 = 0
⇒ x² + 7x – 120 = 0
⇒ x² + (15 – 8) x – 120 = 0
⇒ x² + 15x – 8x – 120 = 0
⇒ x(x + 15) – 8(x + 15) = 0
⇒ (x – 8) (x + 15) = 0
⇒ x – 8 = 0 or x + 15 = 0
⇒ x = 8 or x = – 15 (Reject)
Hence, the pole has to be erected on the boundary of the park at a distance of 8 m from gate Band 8 + 7 i.e., 15 m from gate A.

Fill in the Blanks

Question 1.
A quardratic equation ax² + bx + c = 0 has two………..real roots, if b² – 4ac > 0.
Solution :
Distinct

Question 2.
A quadratic equation ax² + bx + c = 0 has no ………… roots, if b² – 4ac < 0.
Solution :
Real

Question 3.
If we can factorise ax² + bx + c, a ≠ 0, into a product of two linear factors, then the roots of the quadratic equation ax² + bx + c = 0 can be found by equating each ………. to zero.
Solution :
Factor

Question 4.
A quadratic equation can also be solved by the method of completing the …………..
Solution :
Square

Question 5.
ar² + bx + c = 0, a ≠ 0 is called the standard form in a quadratic ………..
Solution :
Equation

Question 6.
A quadratic equation has atmost ……….. roots.
Solution :
Two.

Multiple Choice Questions

Question 1.
The root of the quadratic equation x² – 0.04 = 0 are:
(a) ± 0.2
(b) ± 0.02
(c) 0.4
(d) 2
Solution :
(a) ± 0.2

x² – 0.04 = 0
⇒ x² – (0.2)² = 0
⇒ (x + 0.2) (x – 0.2) = 0
⇒ x + 0.2 = 0
and x – 0.2 = 0
⇒ x = – 0.2
and x = 0.2
⇒ x = ± 0.2

Question 2.
If x² + 2kx + 4 = 0 has a root x = 2, then value of k is:
(a) – 1
(b) – 2
(c) 2
(d) – 4.
Solution :
(b) – 2

x² + 2kx + 4 = 0
one root x = 2 then
(2)² + 2k(2) + 4 = 0
4k + 8 = 0
⇒ k = – 2
So correct choice is (b).

Question 3.
(x² + 1)² – x² = 0 has
[NCERT Exemplar Problems]
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real roots.
Solution :
(c) no real roots

(x² + 1)² – x² = 0 get x² = y, then
(y + 1)² – y = 0
y² + 1 + 2y – y = 0
y² + y + 1 = 0
∵ b² < 4ac
So, no real root.
Hence correct choice is (c).

Question 4.
Which of the following equations has two distinct real roots ?
[NCERT Exemplar Problems]
(a) 2x² – 3\(\sqrt{2}\)x + \(\frac {9}{4}\)
(b) x² + x – 5 = 0
(c) x² + 3x + 2\(\sqrt{2}\) = 0
(d) 5x² – 3x + 1 = 0.
Solution :
(b) x² + x – 5 = 0

(a) 2x² – 3\(\sqrt{2}\)x + \(\frac {9}{4}\) = 0
D = (-3\(\sqrt{2}\))² – 4 × 2 × \(\frac {9}{4}\) = 0

(b) x² + x – 5 = 0
D = (1)² – (4) (1) (-5) = 21
D > 1
Hence correct choice is (b).

Question 5.
The roots of the equation 3x² – 4x + 3 = 0 are :
(a) real and unequal
(b) real and equal
(c) imaginary
(d) none of these.
Solution :
(c) imaginary

3x² – 4x + 3 = 0
a = 3, b = – 4, c = 3
b² = (-4)² = 16
4ac = 4 × 3 × 3 = 36
∵ b² < 4ac
So roots are imaginary Hence correct choice (c).

Haryana State Board HBSE 10th Class Maths Notes Chapter 15 प्रायिकता Notes.

Haryana Board 10th Class Maths Notes Chapter 15 प्रायिकता

→ घटना E की सैद्धांतिक (या परंपरागत) प्रायिकता P(E) को निम्नलिखित रूप में परिभाषित किया जाता है-

जहाँ हम कल्पना करते हैं कि प्रयोग के सभी परिणाम समप्रायिक हैं।

→ एक निश्चित या निर्धारित घटना की प्रायिकता 1 होती है।

→ एक असंभव घटना की प्रायिकता 0 होती है।

→ घटना E की प्रायिकता एक ऐसी संख्या P(E) है कि 0 ≤ P(E) ≤ 1

→ वह घटना जिसका केवल एक ही परिणाम हो एक प्रारंभिक घटना कहलाती है। किसी प्रयोग की सभी प्रारंभिक घटनाओं की प्रायिकता का योग 1 होता है।

→ किसी भी घटना E के लिए P(E) + P(\(\bar{E}\)) = 1 होता है, जहाँ E घटना ‘E नहीं’ को व्यक्त करता है। E और \(\bar{E}\) पूरक घटनाएँ कहलाती हैं।

Haryana State Board HBSE 10th Class Maths Notes Chapter 14 सांख्यिकी Notes.

Haryana Board 10th Class Maths Notes Chapter 14 सांख्यिकी

→ सांख्यिकी- सांख्यिकी वह विज्ञान है, जो संख्यात्मक आंकड़ों के संग्रह, प्रस्तुतीकरण एवं विश्लेषण की उपयोगी विधियों एवं तकनीकों का अध्ययन करती है तथा उन पर आधारित निष्कर्ष निकालती है।

→ सांख्यिकी आंकड़ों का निरूपण- साठिाकी आंकड़ों को निरूपित करने के लिए निम्नलिखित प्रकार के आलेखों या आरेखों का उपयोग किया जाता है-

• आयतचित्र,
• वारंवारता बहुभुज,
• वारंवारता बक,
• दंड आरेख,
• चित्रालेख,
• पाई चार्ट या वृत्तीय आरेख।

→ अवर्गीकृत आंकड़ों का माध्य (समांतर माण)-सांख्यिकी आंकड़ों का आदर्श मापक ‘माध्य’ होता है क्योंकि माध्य अथवा औसत ज्ञात करने के लिए सभी आंकड़ों को निरूपित किया जाता है। इसे \(\bar{x}\) द्वारा प्रकट किया जाता है तथा प्राप्त आंकड़ों का माध्य सभी प्रेक्षणों के योग को प्रेक्षणों की संख्या से भाग देकर ज्ञात किया जाता है।
यदि n प्रेक्षण x₁, x₂, x₃, …….., x_n हो तो,

→ वर्गीकृत आंकड़ों के माध्य का परिकतन-इसकी निम्नलिखित दो विधियों हैं-
(a) प्रत्यक्ष विपि : यदि घर x के x₁, x₂, x₃, ………, x_n अलग-अलग मान हों तथा इनकी संगत वारंवारता f₁, f₂, f₃, …….., f_n हो तो,

(b) कथित माध्य विधि : यदि बारंबारताएँ अधिक हों तो गणना कठिन हो जाती है और तब इस विधि का प्रयोग करते हैं। इस विधि में हम एक खेच्छ अचर मान ‘a’ को लेते हैं (ध्यान रहे कि ‘a’ का गान x का वह मान लेना चाहिए, जो बंटन के मध्य भाग में हो)। ‘a’ के मान को x_i में से घटाते जाते हैं। पटाने पर प्राप्त मान (x – a) को विषलन मान ‘d’ कहते हैं अर्थात d_i = x_i – a तो माध्य \(\bar{x}\) = a + \(\bar{d}\), जहाँ \(\bar{d}\) = \(\frac{\sum f_i d_i}{\sum f_i}\)
(c) पग-विचलन विपि । कभी-कभी कल्पित माध्य विधि में प्राप्त विचलनों di को वर्ग-माप h से भाग देते हैं तथा u_i प्राप्त करते हैं अर्थात् u_i = \(\frac{x_i-a}{h}\) तव माध्य (\(\bar{x}\)) = \(a+\left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h\)

→ किसी वर्ग-अंतराल का मध्य बिंदु (वर्ग चिहून) उसकी उपरि और निचली वर्ग सीमाओं का औसत होता है। अर्थात

→ बहुलक : बहुलक (Mode) दिए हुए प्रेक्षणों में वह मान है जो सबसे अधिक बार आता है अर्थात उस प्रेक्षण का मान जिसकी वारंवारता अधिकतम है।

→ बहुसक वर्ग : एक वर्गीकृत वारंवारता बंटन में, वारंवारताओं को देखकर बहुलक ज्ञात करना संभव नहीं है। यहाँ, हम केवल वह वर्ग (class) ज्ञात कर सकते हैं जिसकी वारंवारता अधिकतम है। इस वर्ग को बहुतक वर्ग (modal class) कहते हैं।

→ वर्गीकृत जाँकड़ों का बहुतक निम्नलिखित सूत्र द्वारा ज्ञात किया जाता है-
बहुलक = \(l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h\)
जहाँ l = बहुलक वर्ग की निम्न (निचली) सीमा
h = वर्ग अंतराल की माप (यह मानते हुए कि सभी अंतराल बराबर मापों के हैं)
f₁ = बहुलक वर्ग की वारंवारता
f₀ = बहुलक वर्ग से ठीक पहले वर्ग की बारंबारता तथा
f₂ = बहुलक वर्ग के ठीक बाद में आने वाले वर्ग की वारंवारता है।

→ संचवी बारंबारता : किसी पारंवारता चंटन में किसी वर्ग की संचयी वारंवारता उस वर्ग से पहले वाले सभी वर्गों की बारंबारताओं का योग होता है।

→ माध्यक माध्यक (median) बेटीय प्रवृत्ति का ऐसा मापक है, जो आंकड़ों में सबसे बीच के प्रेक्षण का मान देता है। अवर्गीकृत आँकड़ों का मायक ज्ञात करने के लिए, पहले हम प्रेक्षणों के मानों को आरोही क्रम में अवस्थित करते हैं। अब, यदि विषम है, तो माध्यक \(\left(\frac{n+1}{2}\right)\) प्रेक्षण का मान होता है, यदि n सम है, तो माध्यक \(\frac{n}{2}\) और \(\left(\frac{n}{2}+1\right)\) ये प्रेक्षणों के मानों का औसत (माध्य) होता है।

→ माध्यक वर्ग : हम दिए गए सभी वर्गों की संचयी बारंबारताएँ और \(\frac{n}{2}\) शास करते हैं। अब, हम वह वर्ग खोजते हैं जिसकी संचयी वारंवारता से अधिक और उसके निकटतम है। इस वर्ग को माध्यक वर्ग (median class) कहते हैं।

→ वर्गीकृत ओंकड़ों का माध्यक निम्नलिखित सूत्र द्वारा ज्ञात किया जाता है-
माध्यक = \(l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h\)
जहाँ l = माध्यक वर्ग की निम्न सीमा
n = प्रेक्षणों की संख्या
cf = माध्यक वर्ग से ठीक पहले वाले वर्ग की संचयी बारंबारता
f = माध्यक वर्ग की वारंवारता
h = वर्ग-माप (यह मानते हुए कि वर्ग-माप बराबर है)

→ तीनों केंद्रीय प्रवृत्ति के मापकों में संबंध : 3 माध्यक = बहुतक + 2 माध्य

→ संचयी वारंवारता बंटनों को आतेखीय रूप से संचयी बारंबारता वकों या से कम प्रकार के’ या ‘से अधिक प्रकार के तोरण द्वारा निरूपण।

→ वर्गीकृत ओंकड़ों का माध्यक इनके दोनों प्रकार के तोरणों के प्रतिच्छेद बिंदु से सैतिज अक्ष पर तंब डालकर संब और कैतिज अक्ष के प्रतिच्छेद बिंदु के संगत मान से प्राप्त हो जाता है।

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 5 Arithmetic Progressions Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 5 Arithmetic Progressions

Short/Long Answer Type Questions

Question 1.
For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an AP?
Solution :
The consecutive terms of an AP are k + 9, 2k – 1 and 2k + 7, then
2k – 1 = \(\frac{k+9+2 k+7}{2}\)
⇒ 4k – 2 = 3k + 16
⇒ 4k – 3k = 16 + 2
⇒ k = 18.

Question 2.
Find how many integers between 200 and 500 are divisible by 8.
Solution :
The numbers divisible by 8 between 200 and 500 are : 208, 216, 224,…..496, which an AP.
Here, a = 208, d = 8 and a_n = l = 496
We know that an = a + (n – 1)d
⇒ 496 = 208 + (n – 1) × 8
⇒ 496 – 208 = 8n – 8
⇒ 288 = 8n – 8
⇒ 288 + 8 = 8n
⇒ \(\frac {296}{8}\) = n
⇒ n = 37
Hence, required integer divisible by 8 = 37.

Question 3.
Find the middle term of an AP: 213, 205, 197……..
Solution :
The given sequence of an AP is 213, 205, 197, ………37
Here, a = 213, d = 205 – 213 = -8, l = 37
Let the number of terms ben
a_n = l = a + (n – 1)d
⇒ 37 = 213 + (n – 1) × (- 8)
⇒ 37 = 213 – 8n + 8
⇒ 8n = 221 – 37
⇒ n = \(\frac {184}{8}\) = 23
The middle term will be \(\frac{23+1}{2}\) i.e 12^th
So, a₁₂ = 213 + (12 – 1) × (-8)
= 213 – 88 = 125
Hence, middle term os AP is 125.

Question 4.
For what value of n are the nth terms of two APs : 63, 65, 67 …..and 3, 10, 17……..
Solution :
The two APs are 63, 65, 67,…. and 3, 10, 17……
Let a, d and A, D be Ist term and common difference of two given APs
Here, a = 63, d = 65 – 63 = 2 And
A = 3, D = 10 – 3 = 7
n^th terms of two APs are eqal
aⁿ = Aⁿ
⇒ a + (n – 1)d = A + (n – 1)D
⇒ 63 + (n – 1) × 2 = 3 + (n – 1) × 7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 2n + 61 = 7n – 4
⇒ 61 + 4 = 7n – 2n
⇒ 65 = 5n
⇒ n = \(\frac {65}{5}\) = 13
Hence, 13^th term of two given APs are equal.

Question 5.
Write the nth terms of the AP:
\(\frac{1}{m}, \frac{1+m}{m}, \frac{1+2 m}{m}\) (CBSE-Comptt 2017)
Solution :
The given sequence of an APs is

Question 6.
Which term of the AP. 20, 19\(\frac {1}{4}\), 18\(\frac {1}{2}\), 17\(\frac {3}{4}\), is the first negative term. (CBSE 2020)
Solution :
The given sequence of AP is
20, 19\(\frac {1}{4}\), 18\(\frac {1}{2}\), 17\(\frac {3}{4}\), …………
Here, a = 20, d = 19\(\frac {1}{4}\) – 20 = \(\frac {-3}{4}\)
Let n^th term of an AP be the first negative term, then,

Hence, 28^th term of the AP is the first negative term.

Question 7.
If m times the m^th term of an AP is equal to n times its n^th term and m ≠ n, show the (m + n)^th term of AP is zero. [CBSE 2019]
Solution :
Let a be first term and d be common difference of AP.
We have, ma_m = na_n
⇒ m[a + (m – 1)d] = n[a + (n – 1)d]
⇒ ma + m²d – md = na + n²d – nd
⇒ m²d – n²d – md + nd = na – ma
⇒ d(m² – n²) – d(m – n) = a(n – m)
⇒ d[m² – n² – (m – n)] = a(n – m)
⇒ d[(m + n) (m – n) – (m – n)] = – a(m – n)
⇒ d[(m – n) (m + n – 1)] = – a(m – n)
⇒ d(m + n – 1) = \(\frac{-a(m-n)}{(m-n)}\)
⇒ d(m + n – 1) = – a
⇒ d = – \(\frac{a}{m+n-1}\)
Now, a_(m+n) = a + (m + n – 1)d
⇒ a_(m+n) = a + (m + n – 1) × – \(\frac{a}{(m+n-1)}\)
⇒ a_(m+n) = a – a = 0
Hence, (m + n)^th term = 0.

Question 8.
A manufactures of TV set produce 720 sets in fourth year and 1080 set in the sixth year, Assuming that the production increase uniformly by a fixed number every year, then find total production in first 9 year.
Solution :
Production of TV sets in 4th year = 720,
Production of TV sets in 6th year = 1080
Let a and d be first term and common difference respectively then,
a₄ = a + (4 – 1)d
⇒ 720 = a + 3d
⇒ a + 3d = 720 …….(1)
And a₆ = a + (6 – 1)d
⇒ 1080 = a + 5d
⇒ a + 5d = 1080 …….(2)
Substracting equ. (2) from equ. (1), We get

⇒ d = \(\frac {-360}{-2}\) = 180
Substituting the value of d in equ. (1), Weget
a + 3 × 180 = 720
⇒ a + 540 = 720
⇒ a = 720 – 540
⇒ a = 180
Production of TV sets increases uniformly by a fixed numbers.
So, Total productions of TV sets in first 9 year = a₉
a₉ = a + (9 – 1) 180
= 180 + 8 × 180
= 180 + 1440 = 1620
Hence, production in first 9^th year = 1620.

Question 9.
Find the sum of the first 20 terms of the following AP : 1, 4, 7, 10,……..
Solution :
The given sequence of AP is 1, 4, 7, 10 …………
Here, a = 1, d = 4 – 1 = 3, n = 20
We know that
S_n = \(\frac {n}{2}\)[2a + (n – 1)d]
= \(\frac {20}{2}\)(2 × 1 + (20 – 1) × 3]
= 10 (2 + 57)
= 10 × 59 = 590
Hence, sum of first 20 term = 590

Question 10.
Find the sum of all two digit natural numbers which are divisible by 4.
Solution :
The all two digit numbers which are divisible by 4 are :
12, 16, 20,…………..96
Here, a = 12, a_n = l =96, d = 16 – 12 = 4
∴ a_n = a + (n – 1)d
⇒ 96 = a + (n – 1)d
⇒ 96 = 12 + (n – 1) × 4
⇒ 96 = 12 + 4n – 4
⇒ 96 – 8 = 4n
⇒ 88 = 4n
⇒ n = \(\frac {88}{4}\) = 22
Sum of first 22 terms = S₂₂
∴ S₂₂ = \(\frac {22}{2}\)[12 + 96]
= 11 × 108 = 1188
Hence, sum of all two digit numbers divisible by 4 = 1188

Question 11.
How many terms of AP – 6, \(\frac {-11}{2}\), – 5, \(\frac {-9}{2}\), ………. are needed to give their sum zero.
Solution :
The given sequence of AP is
– 6, \(\frac {-11}{2}\), – 5, \(\frac {-9}{2}\), ……….
Here, a = – 6, d = \(\frac {-11}{2}\) – (-6) = \(\frac{-11}{2}+\frac{12}{2}=\frac{1}{2}\)
and S_n = 0 (given)
Let AP has n terms
S_n = 0
⇒ \(\frac {n}{2}\)[2a + (n – 1)d] = 0
⇒ \(\frac {n}{2}\)[2 × (-6) + (n – 1) × \(\frac {1}{2}\)] = 0
⇒ \(\frac {n}{2}\)[- 12 + \(\frac{n}{2}-\frac{1}{2}\)] = 0
⇒ \(\frac {n}{4}\)[- 24 + n – 1] = 0
⇒ n[- 25 + n] = 0
⇒ n² – 25n = 0
⇒ n(n – 25) = 0
⇒ n = 0 or n – 25 = 0
⇒ n = 0 or = n = 25
Reject n = 0
Hence, terms needed = 25

Question 12.
How many terms of the AP : 45, 39, 33………must be taken so that their sum is 180 ? explain the double answer.
Solution :
The given sequence of AP is 45, 39, 33, ………….
Here, a = 45, d = 39 – 45 = – 6 and S_n = 180 (given)
Let AP has n terms
S_n = 180
⇒ \(\frac {n}{2}\)[2a + (n – 1)d] = 180
⇒ \(\frac {n}{2}\)[2 × 45 + (n – 1)(-6)] = 180
⇒ \(\frac {n}{2}\)[90 – 6n + 6] = 180
⇒ \(\frac {n}{2}\)[96 – 6n] = 180
⇒ 96n – 6n² = 360
⇒ 6n² – 96n + 360 = 0
⇒ n² -16n + 60 = 0
⇒ n² – (10 + 6)n + 60 = 0
⇒ n – 10n – 6n + 60 = 0
⇒ n(n – 10) – 6 (n – 10) = 0
⇒ (n – 10)(n – 6) = 0
⇒ n = 10, or n = 6
Hence, 10 or 6 terms must be taken

Question 13.
Solve the equation : 1 + 4 + 7 + 10 + …… +x = 287
Solution :
The given equation is 1 + 4 + 7 + 10 +……+ x = 287
Here, a = 1, d = 4 – 1 = 3, a_n = x and S_n = 287
∴ a_n = a + (n – 1)d
⇒ x = 1 + (n – 1) × 3
⇒ x = 1 + 3n – 3
⇒ x = 3n – 2 ………….(i)
Also S_n = \(\frac {n}{2}\)[1 + l]
⇒ 287 = \(\frac {n}{2}\)[1 + x]
⇒ 287 = \(\frac {n}{2}\)[1 + 3n – 2]
⇒ 287 = \(\frac {n}{2}\)[3n – 1]
⇒ 574 = 3n² – n
⇒ 3n² – n – 574 = 0
By quadratic formula, we have
n = \(\frac{1 \pm \sqrt{(-1)^2-4 \times 3 \times(-574)}}{2 \times 3}\)
= \(\frac{1 \pm \sqrt{1+6888}}{6}=\frac{1 \pm \sqrt{6889}}{6}=\frac{1 \pm 83}{6}\)
⇒ n = \(\frac{1+83}{6}\) or n = \(\frac{1-83}{6}\)
⇒ n = 14 or n = \(\frac {- 41}{3}\) (Reject)
So, putting the value of n in equation (1), we get
x = 3 × 14 – 2 = 42 – 2 = 40
Hence, x = 40

Question 14.
If second and third terms of an AP are 3 and 5 respectvely, then find the sum of first 20 terms of it.
Solution :
Let a be first term and common difference bed of an AP
a₂ = 3
⇒ a + d = 3 ………..(1)
And a₃ = 5
⇒ a + 2d = 5 ……….(2)
Subtracting equ. (2) from equ. (1), we get

⇒ d = 2
Substituting the value of d in equ (1), we get
a + 2 = 3
⇒ a = 1
S₂₀ = \(\frac {20}{2}\)[2 × 1 + (20 – 1) × 2]
= 10 [2 + 38]
= 10 × 40 = 400
Hence,
S₂₀ = 400

Question 15.
If the sum of the first n terms of an AP is \(\frac {1}{2}\)(3n² + 7n), then find its nth terma. Hence write its 20^th terms.
Solution :
We have
S_n = \(\frac {1}{2}\)[3n² + 7n]
S₁ = \(\frac {1}{2}\)[3 × 1² + 7 × 1]
= \(\frac {1}{2}\) × 10 = 5
S₂ = \(\frac {1}{2}\)[3 × 2² + 7 × 2]
= \(\frac {1}{2}\)[12 + 14]
= \(\frac {1}{2}\) × 26 = 13
∴ a₁ = S₁ = 5, a₂ = S₂ – S₁ = 13 – 5 = 8
d = 8 – 5 = 3
So, AP is 5, 8, 11, ……………….
N^th term (a_n) =a + (n – 1)d
= 5 + (n – 1) × 3
= 5 + 3n – 3
= 3n + 2
And a₂₀ = 5 + (20 – 1) × 3
= 5 + 57 = 62.
Hence, AP is, 5, 8, 11, ………… and a₂₀ = 62

Question 16.
If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of its n terms.
Solution :
Let first terms be a and common difference be d
S₄ = 40 (given)
⇒ \(\frac {4}{2}\)[2a + (4 – 1) × d] = 40
⇒ 2[2a + 3d] = 40
⇒ 2a + 3d = 20 …. (1)
And S₁₄ = 280
⇒ \(\frac {14}{2}\)[2a + (14 – 1)d] = 280
⇒ 2a + 13d = \(\frac {280}{7}\)
⇒ 2a + 13d = 40 ………(2)
Subtracting the equ (2) from equ. (1), we get

⇒ d = \(\frac {- 20}{- 10}\) = 2
Substituting the value of d in the equ. (1), we get
2a + 3 × 2 = 20
⇒ 2a + 6 = 20
⇒ 2a = 20 – 6 = 14
⇒ a = \(\frac {14}{2}\) = 7
Its sum of n^th term (S_n) = \(\frac {n}{2}\)[2a + (n – 1)d]
= \(\frac {n}{2}\)[2 × 7 + (n – 1) × 2]
= \(\frac {n}{2}\)[14 + 2n – 2]
= \(\frac {n}{2}\)[12 + 2n]
= n (6 + n) = n² + 6n
Hence, sum of its nth = n² + 6n.

Question 17.
Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\).
Solution :
We have, first term = a
second term = b
Then, common difference (d)
= b – a
last term (l) = c
Let n be the number of terms in the AP
∴ a_n = l = c
⇒ a + (n – 1) × (b – a) = c
⇒ (n – 1) (b – a) = c – a

Hence Proved.

Question 18.
Find the sum of all 11 terms of an A.P. whose middle terms is 30.
Solution :
Let first term be a and common difference be d of, AP consisting of 11 terms, (\(\frac{11+1}{2}\))^th i.e. 6^th terms is middle terms. But it is given that middle most term is 30.
∴ a₆ = 30
⇒ a + (6 – 1)d = 30
⇒ a + 5d = 30
Sum of 11 terms (S₁₁) = \(\frac {11}{2}\)[2a + (11 – 1)d]
= \(\frac {11}{2}\)[2a + 10d]
= 11 (a + 5d)
= 11 × 30 [using eq. (1)]
= 330
Hence, S₁₁ = 330.

Question 19.
If in an A.P. the sum of first m times is n and the sum of its first n terms is m, then prove that the sum of its first (m + n) terms is -( m + n)
Solution :
Let first term be a and common difference be d of given A.P. then.
S_m = n (given)
⇒ \(\frac {m}{2}\)[2a + (m – 1)d] = n
⇒ 2ma + m (m – 1)d = 2n ………….(1)
And S_n = m (given)
⇒ \(\frac {n}{2}\)[2a + (n – 1)d] = m
⇒ 2na + n(n – 1)d = 2m ………….(2)
Substracting eq. (2) from eq. (1), we get

2a(m – n) + [m (m – 1) – n(n – 1)]d = 2(n – m)
⇒ 2a(m – n) + [(m² – m – n² + n)]d = – 2(m – n)
⇒ 2a(m – n) + [(m² – n²) – (m – n)]d = – 2(m – n)
⇒ 2a(m – n) + [(m + n) (m – n) – (m – n)d)] = 2(m – n)
(m – n)[2a + (m + n – 1)d] = – 2(m – n)
⇒ 2a + (m + n – 1)d = \(\frac{-2(m-n)}{(m-n)}\)
2a + (m + n – 1)d = – 2 ……..(3)
Now, S (m+n) = \(\frac{(m+n)}{2}\)[2a + (m + n – 1)d]
= \(\frac{(m+n)}{2}\) × (-2)
(using equation) ….(3)
= – (m + n)
Hence, S (m + n) = – (m + n)
Hence proved.

Question 19.
If m^th term of AP is \(\frac {1}{n}\) and n^th term is , find the sum of first mn terms.
Solution :
Let the first term of an AP is a and its common difference be d. them.
a_m = \(\frac {1}{n}\) …………(1)
⇒ a + (m – 1)d = \(\frac {1}{n}\)
And a_n = \(\frac {1}{m}\)
⇒ a + (n – 1)d = \(\frac {1}{m}\) …………(2)
Subtracting the equation (2) from (1), We get

Question 20.
A Motor car travels 175 km distance from place A to B, at a uniform speed 70km/ hr, passes through all ten green traffic signals. Due to heavy traffic it stops for one minute at first signal 3 minutes at second signal, 5 minutes at third signal and so on stops for 19 minutes at tenth signal How much total time to reach at the place B. Solve by suitable mathematical method.
Solution :
Distance travelled by car (s) = 175km.
Speed of a car (s) = 70km/hr.
Time taken by a car = \(\frac {D}{S}\)
to reach place AtoB = \(\frac {175}{70}\) hrs = \(\frac {5}{2}\)hours.
Due to heavy traffic a car stops at signal Iat to signal s 10 Time taken by a car at signal Ist to signal 10 = 1 min + 3 min + 5 min + ——- 19
Here a₁ = – 1, a₂ = – 3, a₃ = 5 and a_m = 19, n = 10.
Since, a₂ – a₁ = 3 – 1 = 2, a₃ – a₂ = 5 – 3 = 2,
a₃ – a₂ = a₂ – a₁
So, the given sequence form an AP
∴ S₁₀ = \(\frac {n}{2}\)[a + l]
= \(\frac {10}{2}\)[1 + 19]
= 5 × 20 = 100 min.
So, car stops at ten signal = 100 min
= \(\frac {100}{60}\)hrs
= 1\(\frac {40}{60}\)hrs
= 1\(\frac {2}{3}\) = \(\frac {5}{3}\)hrs
Total time taken by car = \(\frac{5}{2}+\frac{5}{3}=\frac{15+10}{6}\)
= \(\frac {25}{6}\) = 4\(\frac {1}{6}\)hrs

Question 21.
A Car travels 260km distance from place A to place B, at a uniform speed 65 km/hr passes through all thirreen green traffic signals, 4 minutes at first signal, 7 minutes at second signal 10 minutes at third signal and soon stops for 40 minutes at thirteen signal. How much total it takes to reach at the place B? Solve by suitable mathematical method.
Solution :
Distance covered from place A to B = 260 km/hr.
Speed of a car = 65km/hr
Time taken by a car = \(\frac{260}{65}=\frac{20}{5}\) = 4 hrs.
Car stops at thirteen green signals So, time taken by car at 13 green signals in
min = 4 + 7 + 10 + …….. 40
Since, a₂ – a₁ = 7 – 4 = 3, a₃ – a₂ = 10 – 7 = 3
a₃ – a₂ = a₂ – a₁ =
So, the given sequence form a_n AP
Here, a = 4, d = 3, a_n = 40 and n = 13
∴ S₁₃ = \(\frac {n}{2}\)[a + l]
= \(\frac {13}{2}\)[4 + 40]
= \(\frac {13}{2}\) × 44
= 13 × 12 = 156 min.
So, time taken (extra) by a car = 156 min
= \(\frac {156}{60}\) hrs
2\(\frac {36}{60}\)hrs = 2\(\frac {3}{5}\)hrs = \(\frac {13}{5}\)hrs
Total time taken by a car
= 4 + 2\(\frac {3}{5}\)hrs
= 6\(\frac {2}{5}\)hrs

Fill in the Blanks

Question 1.
If a, b, c are in AP, then ………… = \(\frac{a+c}{2}\) and b is called the airthmetic mean.
Solution :
b

Question 2.
The constant difference between any two consecutive terms of an AP is called the ……………. of the AP.
Solution :
commondifference

Question 3.
= \(\frac {x}{2}\)[2a + (n – 1)d]
Solution :
S_n

Question 4.
The common differeence of an AP can be positive, negative or ………..
Solution :
zero

Question 5.
An arithmetic progression is a list of numbers in which each term is obtained by adding or substracting a fixed number of the …………. term expect the first term.
Solution :
preceding

Question 6.
A ……….. may be difined as an arrangement of numbers in some definite order and according to some rule.
Solution :
sequence.

Multiple Choice Questions

Question 1.
The common difference of the AP.
\(\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}\) is :
(a) 1
(b) \(\frac {1}{p}\)
(c) – 1
(d) – \(\frac {1}{p}\)
Solution :
(c) The sequence of given AP is
\(\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}\), ………..
Common difference (d)
=\(\frac{1-\mathrm{P}}{\mathrm{P}}-\frac{1}{P}=\frac{1-\mathrm{P}-1}{\mathrm{P}}=\frac{-\mathrm{P}}{\mathrm{P}}\) = – 1

Question 2.
The n^th term of the AP. a, 3a, 5a, ……….
(a) na
(b) (2n – 1)a
(c) 2(2n + 1)
(d) 2na
Solution :
(b) The sequence of given AP is a, 3a, 5a …….
Here,
a₁ = a, d = 3a – a = 2a
n^th term (a_n) = a + (n – 1)d .
= a + (n – 1) × 2a
= a + 2na – 2a
= 2na – a
= a (2n – 1)

Question 3.
Value of x for which 2x, (x + 10) and (3x + 2) are there consecutive terms of an AP are :
(a) 6
(b) – 6
(c) 18
(d) – 18
Solution :
(a) Three consecutive terms of AP are 2x, (x + 10) and (3x + 2) Then,
x + 10 = \(\frac{2 x+3 x+2}{2}\)
⇒ 2x + 20 = 5x + 2
⇒ 20 – 2 = 5x – 2x
⇒ 18 = 3x
⇒ x = \(\frac {18}{3}\) = 6

Question 4.
The first term of AP is p and common difference is q, then its 10th term is :
(a) q + qp
(b) p – 9q
(c) p + 9q
(d) 2p + 9q.
Solution :
(c) p + 9q

a = p, d = q
a_n = a + (n – 1)d
⇒ a₁₀ = p + (10 – 1) = p + 9q

Question 5.
The 5th terms of the sequence defined by t₁ = 2, t₂ = 3 and t_n = t_n-1 + t_n-2 for n ≥ 3 :
(a) 16
(b) 13
(c) 15
(d) 2.
Solution :
(b) t₁ = 2, t₂ = 3

and t_n = t_n-1 + t_n-2
t₃ = t₂ + t₁ = 3 + 2 = 5
t₄ = t₃ + t₂ = 5 + 3 = 8
t₅ = t₄ + t₃ = 8 + 5 = 13
Hence correct choice is (b).

Question 6.
The sum of first 20 odd natural numbers is : ……………..
(a) 400
(b) 200
(c) 500
(d) 800.
Solution :
(a) Odd number is 1, 3, 5 ……
a = 1, d = 2, n = 20
S_n = \(\frac {n}{2}\)[2a + (n – 1)d]
= \(\frac {20}{2}\)[2 × 1 + 19 × 2]
= 10 × 40
= 400
So, correct choice is (a).

Haryana State Board HBSE 10th Class Science Solutions Chapter 13 Magnetic Effects of Electric Current Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 13 Magnetic Effects of Electric Current

HBSE 10th Class Science Magnetic Effects of Electric Current Textbook Questions and Answers

Question 1.
Which of the following correctly describes the magnetic field near a long straight wire?
(a) The field consists of straight lines perpendicular to the wire.
(b) The field consists of straight lines parallel to the wire.
(c) The field consists of radial lines originating from the wire.
(d) The field consists of concentric circles centered on the wire.
Answer:
(d) The field consists of concentric circles centered on the wire.
Reason: As per right hand thumb rule.

Question 2.
The phenomenon of electromagnetic induction is ………………
(a) the process of charging a body
(b) the process of generating magnetic field due to a current passing through a coil
(c) producing induced current in a coil due to relative motion between a magnet and the coil
(d) the process of rotating a coil of an electric motor
Answer:
(c) producing induced current in a coil due to relative motion between a magnet and the coil

Question 3.
The device used for producing electric current is called a …………….
(a) generator
(b) galvanometer
(c) ammeter
(d) motor
Answer:
(a) generator
Reason: Generator converts mechanical energy to produce electrical energy.

Question 4.
The essential difference between an AC generator and a DC generator is that
(a) AC generator has an electromagnet while a DC generator has a permanent magnet.
(b) DC generator will generate a higher voltage.
(c) AC generator will generate a higher voltage.
(d) AC generator has slip rings while the DC generator has a commutator.
Answer:
(d) AC generator has slip rings while the DC generator has a commutator.

Question 5.
At the time of short circuit, the current in the circuit ………………..
(a) reduces substantially
(b) does not change
(c) increases heavily
(d) vary continuously
Answer:
(c) increases heavily

Question 6.
State whether the following statements are true or false.
(a) An electric motor converts mechanical energy into electrical energy
(b) An electric generator works on the principle of electromagnetic induction
(c) The field at the centre of a long circular coil carrying current will be parallel straight lines
(d) A wire with a green insulation is usually the live wire of an electric supply
Answer:
(a) False
Reason: An electric motor converts electrical energy into mechanical energy.
(b) True
(c) True
(d) False

Question 7.
List two methods of producing magnetic fields.
Answer:
Magnetic fields can be produced by passing an electric current through –

• A straight conductor
• A circular loop and
•  A solenoid.

Question 8.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current-carry solenoid with the help of a bar magnet? Explain.
Answer:
(i) A current-carrying solenoid behaves like a bar magnet because the magnetic fields of both solenoid and bar magnet are almost similar. Moreover, just like the bar magnet, one end of solenoid has N-polarity and the other end has S-polarity.

(ii) Yes, the bar magnet can be used to determine the north and south poles of a current-carrying solenoid.

• Tie the solenoid in a brass hook and suspend it with a long thread so that it can move freely.
• Bring the north pole of bar magnet near one of the ends of solenoid. If that end of solenoid moves towards the bar magnet then it would be south pole, if it repels, then it would be north pole.

Question 9.
When is the force experienced by a current-carrying conductor placed in a magnetic field largest?
Answer:
When the current in the conductor is perpendicular to the magnetic field, the force experienced by the conductor will be the largest.

Question 10.
Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?
Answer:
According to Fleming’s left hand rule, the magnetic field acts in vertically downward direction.
It should be noted that the direction of current will be opposite to that of electron beam.

Question 11.
Draw a labeled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor?
Answer:
Electric generator:
1. An electric generator is a device which converts mechanical energy into electric energy. Principle:
2. The electric generator works on the principle of electromagnetic induction.

Construction:
(Note: The construction and working shown here is of an Alternating Current (AC) generator.)

1. An AC electric generator, as shown in the figure consists of a rotating rectangular coil ABCD placed between the two poles of a permanent magnet.

2. The two ends of this coil are connected to the two rings R₁ and R₂. The inner side of these rings are insulated.

3. The two conducting stationary brushes B₁ and B₂ are kept pressed separately on the rings R₁ and R₂ respectively.

4. The two rings R₁ and R₁ are internally attached to an axle. The axle may be mechanically rotated from outside to rotate the coil inside the magnetic field.

5. Outer ends of the two brushes are connected to the galvanometer to show the flow of current in the given external circuit.

Working:

1. Suppose the axle attached to the two rings is rotated such that the arm AB moves up (and the arm CD moves down) in the magnetic field produced by the permanent magnet. Then the coil ABCD gets rotated clockwise.

2. By applying Fleming’s right-hand rule, we can determine that the induced currents are set up in these arms along the directions AB and CD. Thus an induced current flows in the direction ABCD or say from B₂ to B₁.

(Note: If there are larger numbers of turns in the coil, the current generated in each turn adds up to give a large current through the coil.)

3. After half a rotation, arm CD starts moving up and AB, down. As a result, the directions of the induced currents in both the arms change. This gives rise to the net induced current In the direction DCBA. The current in the external circuit now flows from B₁ to B₂.

4. Thus after every half rotation, the polarity of the current in the respective arms changes. Such a current, which changes direction after equal intervals of time, ‘s called an alternating current (abbreviated as AC).

Function of split ring:
The function of split ring is to work as a commutator for reversing the direction of current in the coil after every half rotation.

Question 12.
Name some devices in which electric motors are used.
Answer:
Electric motors are used in fan, air conditioner, mixer and grinder, washing machine, water-pumps, etc.

Question 13.
A coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is
(i) pushed into the coil,
(ii) withdrawn from inside the coil,
(iii) held stationary inside the coil?
Answer:
(i) Electric current will be induced in the coil and the galvanometer will show a deflection.
(ii) Electric current will be induced in the coil but in opposite direction. The galvanometer will show a deflection in reverse direction.
(iii) No current will be induced in the coil. The galvanometer will not show any deflection.

Question 14.
Two circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.
Answer:
Yes. Current will be induced in coil B.
Reason: When the current in coil A is changed the magnetic field around it changes.
As the two circular coils are placed close to each other, the magnetic field lines linked with coil is also changed. Therefore, sohrie current is induced in coil B.

Question 15.
State the rule to determine the direction of a
(i) magnetic field produced around a straight conductor-carrying current,
(ii) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and
(iii) current induced in a coil due to its rotation in a magnetic field.
Answer:
(i) Direction of magnetic field:

1. The direction of magnetic field associated with the electric current can be found with the help of the ‘Right Hand Thumb Rule’.

2. For knowing the direction of magnetic field, imagine that you are holding the conducting wire in your right hand so that your thumb points in the direction of the current (I), then the direction in which your fingers encircle the wire will give the direction of magnetic field around the wire.

3. The wrapped fingers demonstrate closed loops.

(ii) Fleming’s left hand rule:

1. Arrange your left hand such that the fore finger, the center finger and thumb remain at right angle to each other.

2. Adjust your hand in such a way that the forefinger points in the direction of magnetic field and the centre finger points in the direction of current, then the direction in which the thumb points will be the direction of magnetic force.

(iii) Fleming’s right hand rule:

1. The direction of electric current can be found out with the help of Fleming’s right hand rule.
2. Arrange the forefinger, middle finger and thumb of a right hand at right angle to one another.
3. Adjust the forefinger in the direction of magnetic field, and thumb pointing in the direction of motion of conductor.
4. The direction of middle finger will then indicate the direction of induced electric current.

Question 16.
Explain the underlying principle and working of an electric generator by drawing a labeled diagram. What is the function of brushes?
Answer:
Electric generator:
1. An electric generator is a device which converts mechanical energy into electric energy. Principle:
2. The electric generator works on the principle of electromagnetic induction.

Construction:
(Note: The construction and working shown here is of an Alternating Current (AC) generator.)

1. An AC electric generator, as shown in the figure consists of a rotating rectangular coil ABCD placed between the two poles of a permanent magnet.

2. The two ends of this coil are connected to the two rings R₁ and R₂. The inner side of these rings are insulated.

3. The two conducting stationary brushes B₁ and B₂ are kept pressed separately on the rings R₁ and R₂ respectively.

4. The two rings R₁ and R₁ are internally attached to an axle. The axle may be mechanically rotated from outside to rotate the coil inside the magnetic field.

5. Outer ends of the two brushes are connected to the galvanometer to show the flow of current in the given external circuit.

Working:

1. Suppose the axle attached to the two rings Is rotated such that the arm AB moves up (and the arm CD moves down) in the magnetic field produced by the permanent magnet. Then the coil ABCD gets rotated clockwise.

2. By applying Fleming’s right-hand rule, we can determine that the induced currents are set up in these arms along the directions AB and CD. Thus an induced current flows in the direction ABCD or say from B₂ to B₁.

(Note: If there are larger numbers of turns in the coil, the current generated in each turn adds up to give a large current through the coil.)

3. After half a rotation, arm CD starts moving up and AB, down. As a result, the directions of the induced currents in both the arms change. This gives rise to the net induced current In the direction DCBA. The current in the external circuit now flows from B₁ to B₂.

4. Thus after every half rotation, the polarity of the current in the respective arms changes. Such a current, which changes direction after equal intervals of time, ‘s called an alternating current (abbreviated as AC).

Question 17.
When does an electric short circuit occur?
Answer:
1. An electric short circuit takes place when the live wire and neutral wire of the electric supply line touch each other directly or indirectly via. a conducting wire.
2. This occurs when the plastic insulation of the wires gets torn or there is a fault in the electrical appliance.

Question 18.
What is the function of an earth wire? Why is it necessary to earth metallic appliances?
Answer:
1. The earthing wire is a green colour wire connected to a metal plate. The metal plate is buried deep in the earth near our house.

2. In case, if the live wire touches the metal case of the appliance (i.e. in case of leakage of current), then due to earthing, the current will directly pass into the earth without causing electrical shock to the user.

3. To avoid electrical shocks, metal appliances such as toaster, electric iron, refrigerator, table fan, etc. are connected to the earth wire. Thus, a person is saved from any electrical shocks occurring due to leakage of current.

HBSE 10th Class Science Magnetic Effects of Electric Current InText Activity Questions and Answers

Textbook Page no – 224

Question 1.
Why does a compass needle get deflected when brought near a bar magnet?
Answer:
1. When you bring the magnetic compass near the bar magnet, the magnetic field of bar magnet exerts force on both the poles of the compass needle.
2. The force experienced by two poles is equal and opposite. So, these two forces form a couple and hence deflect the compass needle.

Textbook Page no – 228

Question 1.
Draw magnetic field lines around a bar magnet.
Answer:

Question 2.
List the properties of magnetic field lines.
Answer:
Characteristics of magnetic field lines:
(1) The magnetic field lines outside the magnet starts from the north pole (N) and reach the south pole (S), whereas, inside the magnet, these lines start from south pole (S) and end at north pole (N).

• The inner and outer lines together form closed loops.

(2) The region where the field lines are closer has a stronger magnetic field compared to the region where the field lines are farther.

• Thus the magnetic field is strong near the poles where the lines are quite close to each other.

(3) Magnetic field is a vector quantity. This means it has magnitude as well as direction.

• The tangent drawn at any point of a magnetic field line (the direction of magnetic needle at that point) shows the direction of magnetic field at that point.

(4) Magnetic field lines do not intersect each other.

Question 3.
Why don’t two magnetic field lines Intersect each other?
Answer:
No, the two magnetic lines of force cannot intersect.
Reason:
If the two magnetic field-lines intersect then it would mean that at the point of intersection, the compass needle would point towards two directions which is not possible.

Textbook Page no – 229

Question 1.
Consider a circular loop of wire lying In the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.
Answer:
1. By applying right hand thumb rule, we can find the direction of the magnetic field inside and outside the circular loop of wire carrying electric current.

2. The magnetic field lines (dotted) are found to be perpendicular to the plane of the paper.

3. The front side of the loop behaves as the south pole where as the back side i.e. the face touching the plane of the table behaves as the north pole.

Question 2.
The magnetic field in a given region is uniform. Draw a diagram to represent it.
Answer:
A uniform magnetic field in a given region can be represented by straight, parallel equally spaced field lines.

3. Choose the correct option:

The magnetic field inside a long straight solenoid-carrying current
(a) Is zero
(b) decreases as we move towards its end
(c) increases as we move towards its end
(d) Is the same at all points
Answer:
(d) Is same at all points
Reason: Since the magnetic field line is uniform, represented by parallel equally spaced lines, the current is same at all points.

Textbook Page no – 231.

Question 1.
Which of the following properties of a proton can change while It moves freely in a magnetic field? (There may be more than one correct answer.)
(a) mass
(b) speed
(c) velocity
(d) momentum
Answer:
The correct options are (c) and (d).
Reason: The magnetic force acts perpendicular to the direction of motion of the proton. It does not change its mass and speed but changes its direction of motion. So, both velocity and momentum get changed.

Question 2.
In Activity 13.7, how do we think the displacement of rod AB will be affected if ……………
(i) current in rod AB is increased;
(ii) a stronger horse-shoe magnet is used; and
(iii) length of the rod AB is Increased?
Answer:
(i) When the current in the rod AB is increased, force exerted on the conductor increases.
∴ The displacement of the rod increases.

(ii) When a stronger horse shoe magnet is used, the magnitude of the magnetic field increases. This increases the force exerted on the rod and the displacement of the rod.

(iv) The displacement of the rod will increase, because F α L.

Question 3.
A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is ……………
(a) towards south
(b) towards east
(c) downward
(d) upward
Answer:
(d) Upward
Reason: As per Fleming’s left hand rule.

Textbook Page no – 233.

Question 1.
State Fleming’s left-hand rule.
Answer:

1. Arrange your left hand such that the fore finger, the center finger and thumb remain at right angle to each other.
2. Adjust your hand in such a way that the forefinger points in the direction of magnetic field and the centre finger points in the direction of current, then the direction in which the thumb points will be the direction of magnetic force.

Question 2.
What is the principle of an electric motor?
Answer:
Electric motor: An electric motor is a device which converts electric energy into mechanical energy.

Principle :
1. When a current carrying conductor (or a coil) is placed in a magnetic field, the conductor experiences force and it rotates,
2. Electric motor works on the same principle.

Construction :

1. An electric motor, as shown in figure, consists of a rectangular coil ABCD made from insulated copper wire.
2. The coil is placed between the two poles of a magnetic field such that the arm AB and CD are perpendicular to the direction of the magnetic field.
3. The ends of the coil are connected to the two halves P and Q of a split ring.
4. The inner sides of these halves are insulated and attached to an axle such that they can rotate easily.
5. The external conducting edges of P and Q touch two conducting stationary brushes (i.e.carbon strips) X and Y respectively.
6. Finally, the terminals of the battery are connected with brushes ‘X’ and ‘Y’.

Working : Current in the coil ABCD enters from the source battery through conducting brush X and flows back to the battery through brush Y.
1. The current in arm AB of the coil flows from A to B, whereas in arm CD it flows from C to D, i.e. opposite to the direction of current through arm AB.

2. On applying Fleming’s left hand rule, we find that the force acting on arm AB pushes it downwards while the force acting on arm CD pushes it upwards. As a result, the coil and the axle O, mounted free to turn about an axis, rotate anti-clockwise. At half rotation, Q makes contact with the brush X and P with brush Y. So the current in the coil gets reversed and flows along the path DCBA.

3. The reversal of current also reverses the direction of force acting on the two arms AB and CD. Thus the arm AB of the coil that was earlier pushed down is now pushed up and the arm CD previously pushed up is now pushed down.

4. Hence, the coil and the axle rotate half a turn more in the same direction. The reversing of the current is repeated at each half rotation, giving rise to a continuous rotation of the coil and to the axle.

Uses: The electric motor is used in appliances such as electric fan, mixer, washing machine, CD player, etc.

Question 3.
What is the role of the split ring in an electric motor?
Answer:
The split ring acts as a commutator. Its function is to reverse the direction of the current flowing through the coil after every half rotation of the coil.

Textbook Page no – 236.

Question 1.
Explain different ways to induce current in a coil.
Answer:
(1) Current can be induced in a coil by moving a magnet towards or away from the coil or vice-versa.
(2) Changing the current in the coil placed near it.
(3) Moving a coil in a non-uniform magnetic field or by changing a magnetic field around steady coil.
(4) Rotating it in a magnetic field or by rotating a magnet placed near the coil.

Textbook Page no – 237.

Question 1.
State the principle of an electric generator.
Answer:
1. A device that converts mechanical energy into electrical energy is called an electric generator.
2. The electric generator works on the principle of electromagnetic induction.

Question 2.
Name some sources of direct current.
Answer:
Electrochemical dry cells, batteries, DC generators, solar cells, etc. are the sources of direct current (DC).

Question 3.
Which sources produce alternating current?
Answer:
Alternating current is produced at AC generators (thermal power plants), car alternators, hydroelectric power plant, etc.

Question 4.
Choose the correct option.
A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once In each
(a) two revolutions
(b) one revolution
(c) half rotation
(d) one-fourth revolution
Answer:
(c) half rotation

Textbook Page no – 238.

Question 1.
Name two safety measures commonly used In electric circuits and appliances.
Answer:
1. Installing safety fuse of proper rating
2. Connecting earth wire to all devices that have a heavy power rating.

Question 2.
An electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5A. What result do you expect? Explain.
Answer:
1. The current drawn by the oven
\(I=\frac{P}{V}=\frac{2 K W}{220 \mathrm{~V}}=\frac{2000 \mathrm{~W}}{220 \mathrm{~V}}=9.09 \mathrm{~A}\)
2. The current of 9.09 A that the oven draws is much more than the current rating of 5A. As a result, the circuit will get overloaded, heat a lot and eventually trip the fuse.

Question 3.
What precaution should be taken to avoid the overloading (domestic electric circuits?
Answer:
Following precautions should be taken to avoid overloading of domestic circuits:

• Wires and fuse used should be of proper rating.
• Smaller appliances such as TV, fan, bulb, etc. should run on a 5A circuit whereas big appliances like AC, water motor, oven, etc. should run on 15A circuit.
• Too many appliances that have high power rating should not be run simultaneously.
• One should not operate too many electrical appliances on a single socket simultaneously.

Activities

Activity 1.

Aim: To demonstrate that a magnetic field is produced around a current carrying wire.
Material required: A thick copper wire, a compass needle, a plug-key, a resistance wire (Resistor R) and a battery of 6 V.

Procedure:
1. Take a straight thick copper wire and place it between the points X and Y in an electric circuit, as shown in figure.
2. Horizontally place a small compass near to this copper wire. See the position of its needle.
3. Pass the current through the circuit by inserting the key into the plug.
4. Observe the change in the position of the compass needle.

Observation and conclusion:

On passing current through the copper wire XY, the compass needle gets deflected from its position of rest. Since a magnetic needle can be deflected only by a magnetic field, we conclude that the current carrying wire produces a magnetic field around it or it behaves like a magnet.

Activity 2.

Aim : To obtain magnetic field lines around a bar magnet.

Procedure:
1. Fix a sheet of white paper on a drawing board using some adhesive material.
2. Place a bar magnet in the centre of it
3. Sprinkle some iron filings uniformly around the bar magnet Figure. A salt-sprinkler may be used for this purpose.
4. Now tap the board gently.
5. What do you observe?

Observation and conclusion

1. The iron fillings arrange themselves in a pattern as can be seen in the figure.
2. The magnet exerts a force on the iron-fillings scattered around it. This force arranges the filling along definite lines extending from one end to another. The region upto which the fillings got arranged is the region upto where the magnet exerted its force. This region is called magnetic field and the lines are called magnetic field lines.

Activity 3.

Aim: To draw magnetic field lines of a bar magnet using a compass.
Material: A sheet of white paper, a drawing board, adhesive tape, a bar magnet, a compass and a pen/pencile.

Procedure:
1. Take a small compass and a bar magnet.

2. Place the magnet on a sheet of white paper fixed on a drawing board, using some adhesive material.

3. Mark the boundary of the magnet.

4. Place the compass near the north pole of the magnet. How does it behave? The south pole of the needle points towards the north pole of the magnet. The north pole of the compass is directed away from the north pole of the magnet.

5. Mark the position of two ends of the needle.

6. Now move the needle to a new position such that its south pole occupies the position previously occupied by Its north pole.

7. In this way, proceed step by step till you reach the south pole of the magnet as shown in Figure 9.

8. Join the points marked on the paper by a smooth curve. This curve represents a field line.

9. Repeat the above procedure and draw as many lines as you can. You will get a pattern shown in Figure10. These lines represent the magnetic field around the magnet. These are known as magnetic field lines.

Observation and conclusion:

1. As we move the compass needle towards the poles of bar magnet, the deflection increases. Reason for this is that the magnetic field is stronger near the two poles and so the field exerts large force on the compass needle.

2. Now, magnetic field is a vector quantity and it has both direction and magnitude. So, the direction of the magnetic field will be the direction in which the north pole of the compass needle moves. Thus, we consider that the field lines emerge from the north poles and merge at the south poles. On the other hand the direction of field line is from its south pole to its north pole inside the magnet. The magnetic field lines form closed curves.

Activity 4.

Aim: To show that the direction of magnetic field due to current depends on the direction of the current.

Procedure:
1. Take a long straight copper wire, two or three cells of 1.5V each, and a plug key. Connect all of them in series as shown in Figure (a).

2. Place the straight wire parallel to and over a compass needle.

3. Plug the key in the circuit.

4. Observe the direction of deflection of the north pole of the needle. If the current flows from north to south, as shown in Figure (a), the north pole of the compass needle would move towards the east.

5. Replace the cell connections in the circuit as shown in Figure (b). This would result in the change of the direction of current through the copper wire, that is, from south to north.

6. Observe the change in the direction of deflection of the needle. You will see that now the needle moves in opposite direction, that is, towards the west [Figure (b)]. It means that the direction of magnetic field produced by the electric current is also reversed.

Observation and conclusion:

1. When the current flows from north to south (as shown in figure(a)), the north pole of the compass deflects towards east. On reversing the current the compass needle also deflects in opposite side i.e. towards west.

2. We conclude that the direction of magnetic field due to the current depends on the direction of the current. So, the direction of magnetic field due to the current gets reversed when the direction of current is reversed.

Activity 5.

Aim: To study the pattern of magnetic field formed when current passes through a straight conductor.

Procedure:

1. Take a battery (12 V), a variable resistance (or a rheostat), an ammeter (0-5A), a plug key, connecting wires and a long straight thick copper wire.

2. Insert the thick wire through the centre, normal to the plane of a rectangular cardboard. Take care that the cardboard is fixed and does not slide up or down.

3. Connect the copper wire vertically between the points X and Y, as shown in Figure (a), in series with battery, a plug and key.

4. Sprinkle some iron filings uniformly on the cardboard. (You may use a salt sprinkler for this purpose.)

5. Keep the variable of the rheostat at a fixed position and note the current through the ammeter.

6. Close the key so that a current flows through the wire. Ensure that the copper wire placed between the points X and Y remains vertically straight.

7. Gently tap the cardboard a few times. Observe the pattern of the iron filings. You would find that the iron filings align themselves showing a pattern of concentric circles around the copper wire.

8. What do these concentric circles represent? They represent the magnetic field lines.

9. How can the direction of the magnetic field be found? Place a compass at a point (say P) over a circle. Observe the direction of the needle. The direction of the north pole of the compass needle would give the direction of the field lines produced by the electric current through the straight wire at point P. Show the direction by an arrow.

10. Does the direction of magnetic field lines get reversed if the direction of current through the straight copper wire is reversed? Check it.

Observation and conclusion:

(1) It can be observed that the iron fillings align themselves forming a pattern of concentric circles around the copper wire. These circles represent the magnetic field lines.
(i) When we reverse the direction of current through the straight wire, the direction of magnetic field also gets reversed. This can be seen by the deflection of the needle.

(2) Also note that if we increase the current, the deflection increases and if we decrease it, the deflection decrease. This means the deflection of the needle changes when the intensity of current is charged.

(3) When we move the compass away from the wire, the concentric circles around the current carrying conductor becomes large. i.e. the radius of the circle increases.

Activity 6.

Aim: To study the pattern of the magnetic field lines produced by a current carrying circular coil.

Procedure:

1. Take a rectangular cardboard having two holes. Insert a circular coil having large number of turns through them, normal to the plane of the cardboard.
2. Connect the ends of the coil in series with a battery, a key and a rheostat, as shown in Fig 9.
3. Sprinkle iron filings uniformly on the cardboard.
4. Plug the key.
5. Tap the cardboard gently a few times. Note the pattern of the iron filings that emerges on the cardboard.

Observation and conclusion:
1. As can be seen in the figure, the iron fillings get arranged in a circular pattern.
2. The lines of force around the wire are in the form of concentric circles.
3. At the centre of the loop, the lines of force are almost in the form of parallel straight lines. At the centre the magnetic field is nearly uniform.

Activity 7.

Aim: To demonstrate the existence of force on current carrying conductor in a magnetic field.

Procedure:
1. Take a small aluminium rod AB (of about 5 cm). Using two connecting wires suspend it horizontally from a stand, as shown in Fig. 12.

2. Place a strong horse-shoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. For this put the north pole of the magnet vertically below and south pole vertically above the aluminium rod (Fig.12).

3. Connect the aluminium rod in series with a battery, a key and a rheostat.

4. Now pass a current through the aluminium rod from end B to end A.

5. What do you observe? It is observed that the rod is displaced towards the left. You will notice that the rod gets displaced.

6. Reverse the direction of current flowing through the rod and observe the direction of its displacement. It is now towards the right. Why does the rod get displaced?

Observation and conclusion:

1. The aluminium rod gets displaced because a force is exerted on the current-carrying rod when it is placed in the magnetic field.

2. When we reverse the direction of current, the direction of displacement of rod i.e. the direction of force on the rod also gets reversed.

3. Suppose we reverse the direction of magnetic field by interchanging the poles of the magnet. Then we observe that the direction of the force on the rod also gets reversed. This means that the direction of force on the conductor depends upon the direction of current and the direction of magnetic field.

4. The rod gets displaced maximum i.e. the magnitude of the force is maximum when the direction of current is perpendicular to the direction of magnetic field.

We conclude the following:

• A current-carrying conductor experiences a force when placed in a magnetic field.
• The direction of the force and that of the displacement of the conductor depends upon the direction of the current and the direction of the magnetic field.

Activity 8.

Aim: To study the phenomenon of electromagnetic induction.

Procedure:

1. Take a coil of wire AB having a large number of turns.

2. Connect the ends of the coil to a galvanometer as shown in Figure.

3. Take a strong bar magnet and move its north pole towards the end B of the coil. Do you find any change in the galvanometer needle?

4. There is a momentary deflection in the needle of the galvanometer, say to the right. This indicates the presence of a current in the coil AB. The deflection becomes zero the moment the motion of the magnet stops.

5. Now withdraw the north pole of the magnet away from the coil. Now the galvanometer is deflected toward the left, showing that the current is now set up in the direction opposite to the first.

6. Place the magnet stationary at a point near to the coil, keeping its north pole towards the end B of the coil. We see that the galvanometer needle deflects toward the right when the coil is moved towards the north pole of the magnet. Similarly the needle moves toward left when the coil is moved away.

7. When the coil is kept stationary with respect to the magnet, the deflection of the galvanometer drops to zero. What do you conclude from this activity?

Observation and conclusion:

1. When we move the magnet towards away from the coil or when we move the coil towards/away from the magnet, the magnetic lines of force passing through the closed coil gets charged.
2. This induces potential difference in the coil which in turn induces current in the circuit.
3. We can observe the presence of current in the galvanometer needle.
4. The needle shows no movement when both, the coil and the magnet remain stationary.
5. We conclude that when there is no change in the magnetic lines of force passing through the coil, no potential difference is induced and hence no current is induced.

Activity 9.

Aim: To study the phenomenon of electromagnetic induction.

Procedure:
1. Take two different coils of copper wire having large number of turns (say 50 and loo turns respectively). Insert them over a non-conducting cylindrical roll, as shown in Figure. (You may use a thick paper roll for this purpose.)

2. Connect the coil-i, having larger number of turns, in Figure. Current is induced in cou-2 when current in coil-i is changed series with a battery and a plug key. Also connect the other coil-2 with a galvanometer as shown.

3. Plug in the key. Observe the galvanometer. Is there a deflection in its needle? You will observe that the needle of the galvanometer instantly jumps to one side and just as quickly returns to zero, indicating a momentary current in cou-2.

4. Disconnect coil-i from the battery. You will observe that the needle momentarily moves, but to the opposite side. It means that now the current flows in the opposite direction in coil-2.

Observation and conclusion:

1. Coil 1 is called primary coil whereas coil 2 is called secondary coil.

2. When the key is plugged, the needle of the gets galvanometer deflected to one side and then quickly returns of zero. This indicates a momentary current in coil 2.

3. When coil 1 is disconnected from the battery, the needle of the galvanometer moves to the opposite side. This means, now the current flows in the opposite direction i.e. in coil 2.

4. In short in this activity, we observe that as soon as the current in coil 1 reaches either a steady value or becomes zero, the galvanometer connected to coil 2 shows no deflection.

Conclusion:

(a) A potential difference is induced in coil 2 whenever the electric current through the coil 1 changes.
(b) As the current in coil 1 changes, the magnetic field associated with it also changes. At that time, magnetic field lines around coil 2 also change. Thus, the change in magnetic field lines associated with the secondary coil is the cause of induced electric current in it.

Haryana State Board HBSE 10th Class Maths Notes Chapter 4 Quadratic Equations Notes.

Haryana Board 10th Class Maths Notes Chapter 4 Quadratic Equations

Introduction
You are already familiar with linear equations in one and two variables and their solutions. In chapter 2, you have studied different types of polynomials. Recall that polynomials of degree 2 are called quadratic polynomials. The standard form of quadratic polynomial is ax² + bx + c, where a, b, c are real numbers and a ¹ 0. When we equate this polynomial to zero, we get a quadratic equation. Quadratic equations come up when we deal with many real-life situations. In this chapter, we shall study about quadratic equations and solving them by various methods. We shall also discuss some applications of quadratic equations in daily life situations.

1. Equation : An equation is an algebraic expression obtained by equating a polynomial to zero. It has two parts separated by an equal (=) sign. Left part is called LHS (Left Hand Side) and right part is called RHS (Right Hand Side).
For example:
p(x) = x + 5 [Linear Polynomial]
put p(x) = 0
x + 5 = 0 [Linear Equation]
p(y) = 2y² + 3y + 4 [Quadratic Polynomial]
put p(y) = 0
2y² + 3y + 4 = 0 [Quadratic Equation]
2. Roots of an equation : A real number a is said to be a root of an equation p(x) = 0, if p(a) = 0.
3. The zeroes of a polynomial p(x) and roots of the respective equation p(x) = 0 are the same.
4. Imaginary roots : If an equation does not have real roots then it is said to be having imaginary roots.
5. \(\text { Speed }=\frac{\text { Distance }}{\text { Time }}\)
6. Distance = Speed × Time
7. \(\text { Time }=\frac{\text { Distance }}{\text { Speed }}\)

Quadratic Equation
If p(x) is a quadratic polynomial, then p(x) = 0 is called a quadratic equation.
The general form of a quadratic equation is ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. It is also known as the standard form of a quadratic equation.
For example: 2x² + 5x + 7 = 0. -7x² + 3x – 4 = 0, x² + 5x + 6 = 0 are quadratic equations in variable x.
But x² + \(\frac{3}{x}\) + 4 = 0, x² + \(2 \sqrt{x}\) – 3 = 0, x² + \(\frac{1}{x^2}\) + 1 = 0 are not quadratic equations.

Roots of a quadratic equation : Let p(x) = 0 be a quadratie equation, then the zeroes of the polynomial p(x) are called the roots of the equation p(x) = 0.
Thus, α is a root of quadratic equation p(x) = 0 if p(α) = 0, and x = α is the solution of the given equation.
In chapter 2, we studied that quadratic polynomial have at mont two zeroes. It follows from this that a quadratic equation has at most two real roots.
Let us consider some examples.

Solution of a Quadratic Equation by Factorisation Method
Let ax² + bx + c = 0, a ≠ 0 be a quadratic equation and let quadratic polynomial ax² + bx + c be expressible as a product of two linear factors, say (mx + n) and (qx + p), where m, n, q, p are real numbers such that m ≠ 0, q ≠ 0.
Then ax² + bx + c = 0
⇒ (mx + n)(qx + p) = 0
⇒ mx + n = 0 or qx + p = 0
⇒ x = \(-\frac{n}{m}\) or x = \(-\frac{p}{q}\)
Hence, x = \(-\frac{n}{m}\) and x = \(-\frac{p}{q}\) are the possible roots of the quadrntic equation ax² + bx + c = 0.

Solution of a Quadratic Equation by Completing the Square
It is possible to take a quadratic equation whose left hand side is not a perfect square and change it into one that is, then we solve the resulting equation by extracting root. The technique of solving quadratic equations in this manner is called completing the square. To solve the quadratic equation by the method of completing the square we may use the following steps:
Step I: Consider the quadratic, equation
ax² + bx + c = 0, a ≠ 0, a b, c ∈R
Step II: Make the coefficient of x unity. If it is not, i.e., we get
\(x^2+\frac{b}{a}+\frac{c}{a}=0\)
Step III: Transposing the constant term \(\frac{c}{a}\) to R.H.S., we get
\(x^2+\frac{b}{a} x=-\frac{c}{a}\)
Step IV: Add(\(\frac{1}{2}\) × coefficient of x)² on both sides, we get
\(x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=\left(\frac{b}{2 a}\right)^2-\frac{c}{a}\)
Step V: Express L.H.S. as the perfect square of a suitable binomial expression and simplify R.H.S. we get
\(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\)
Step VI: Take square root of both sides, we get
\(x+\frac{b}{2 a}=\pm \sqrt{\frac{b^2-4 a c}{4 a^2}}\)
Step VII: Obtain the values of shifting the constant term \(\frac{b}{2 a}\) to R.H.S.

Solution of a Quadratic equation by Quadratic Formula
[Shreedhara Charya’s Rule]
Consider the quadratic equation ax² + bx + c = 0, a ≠ 0, a, b, c ∈R.
ax² + bx + c = 0


where D = b² – 4ac and D is called discriminant.
Thus, If D = b² – 4ac ≥ 0, then the quadratic equation ax² + bx + 0 = 0 has two real roots α and β given by :

Remarks: This is a formula for the roots of a quadratic equation expressed in terms of the coefficienta a, b, c. This is given by an ancient Indian Mathematician Shreedhara Charya around 1025 A.D. and is known as Shreedhara Charya’s formula for determining the roots of the quadratic equation ax² + bx + c = 0.

Nature of Roots
In previous section, you have seen that roots of the equation ax² + bx + c = 0, a ≠ 0 are given by :
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
There are three different cases:
Case I: If b² – 4ac > 0
In this case, the roots are real and distinct.
These roots are given by
\(x=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\)
and \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

Case II: If b² – 4ac = 0
In this case, the roots are real and equal
Then \(x=\frac{-b \pm \sqrt{0}}{2 a}\)
\(x=-\frac{b}{2 a}\) and x = \(-\frac{b}{2 a}\)

Case III: If b² – 4ac < 0
In this case there is no real number whose square is b² – 4ac. Therefore, there are no real roots for the given quadratic equation. Since, b² – 4ac determines whether the quadratic equation ax² + bx + c has real roots or not, b² – 4ac is called the discriminant of this quadratic equation. It is generally denoted by D.
So, a quadratic equation ax² + bx + c = 0 has:
(1) Two distinct real roots, if b² – 4ac > 0 i.e, D > 0.
(2) Two equal and real roots (ie., coincident roots) if b² – 4ac = 0 i.e., D = 0.
(3) No real roots, if b² – 4ac < 0 i.e, D < 0.

Haryana State Board HBSE 10th Class Maths Notes Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Notes.

Haryana Board 10th Class Maths Notes Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन

→ घनाभ का आयतन (ν) = लं० × चौ० × ॐ०

→ घनाभ का संपूर्ण पृष्ठ तल का क्षेत्रफल = 2 (लं० × चौ० + चौ० × ॐ० + ऊँ० × लं०)

→

→ धन क आयतन = (भुजा)³

→ धन का पृष्ठ तल क्षेत्रफल = 6 (भुजा)²

→ धन का विकर्ण = भुजा \(\sqrt{3}\)

→ r त्रिज्या तथा h ऊँचाई वाले बेलन का-
(i) वक्र पृष्ठीय क्षेत्रफल = 2πrh
(ii) संपूर्ण पृष्ठीय क्षेत्रफल = 2πr(r + h)
(iii) आयतन = πr²h

→ त्रिज्या तथा h ऊँचाई वाले शंकु के लिए-
(i) तिर्यक ऊँचाई (l) = \(\sqrt{r^2+h^2}\)
(ii) वक्र पृष्ठीय क्षेत्रफल = πrl
(iii) संपूर्ण पृष्ठीय क्षेत्रफल = πr (r + l)
(iv) आयतन = \(\frac{1}{3}\)πr²h

→ r त्रिज्या वाले गोले के लिए-
(i) पृष्टीय क्षेत्रफल = 4πr²
(ii) आयतन = \(\frac{4}{3}\)πr³

→ r त्रिज्या वाले अर्धगोले के लिए-
(i) वक्र पृष्ठीय क्षेत्रफल = 2πr²
(ii) संपूर्ण पृष्ठीय क्षेत्रफल = 3πr²
(iii) आयतन = \(\frac{2}{3}\)πr³

→ (i) शंकु के छिन्नक का आयतन = \(\frac{1}{3}\)πh(r₁² + r₁r₂ + r₂²)
(ii) शंकु के छिन्नक का वक्र पृष्ठीय क्षेत्रफल = πl(r₁ + r₂)
जहाँ l = \(\sqrt{h^2+\left(r_1-r_2\right)^2}\)
(iii) शंकु के छिन्नक का संपूर्ण पृष्ठीय क्षेत्रफल = πl(r₁ + r₂) + πr₁² + πr₂²

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 2 Polynomials Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 2 Polynomials

Short/Long Answer Type Questions

Question 1.
Find the zeroes of the quadrate polynomial x² + x – 2, and verify the relationship between the zeroes and coefficients.
Solution :
Let f(x) = x² + x – 2
= x² + (2 – 1)x – 2
= x² + 2x – x – 2
= x(x + 2) – 1(x + 2)
= (x + 2) (x – 1)
To find the zeroes of the polynomial f(x)
Put
f(x) = 0
⇒ (x + 2) (x – 1) = 0
⇒ x + 2 = 0
⇒ x = – 2 And
(x – 1) = 0
⇒ x = 1
Therefore, zeroes of the polynomial f(x) are :
α = – 2, β = 1
Now, sum of the zeroes = α + β = – 2 + 1 = – \(\frac {1}{1}\)
= – Coefficient of x / Coefficient of x²
And product of zeroes α × β = – 2 × 1 = – \(\frac {2}{1}\)
= Constant term / Coefficient of x²
∴ Zeroes of the polynomial f(x) are – 2, 1.

Question 2.
Find the zeroes of the quadratic polynomial 7y² – \(\frac {11}{3}\)y – \(\frac {2}{3}\) and verify the relationship between the zeroes and the coefficients.
Solution :
Let p(x) = 7y² – \(\frac {11}{3}\)y – \(\frac {2}{3}\)
= \(\frac {1}{3}\) (21y² – 11y – 2)
= \(\frac {1}{3}\) [12y² – (14 – 3)y – 2]
= \(\frac {1}{3}\) [12y² – 14y + 3y – 2]
= \(\frac {1}{3}\) [7y (3y – 2) + 1 (3y – 2)]
= \(\frac {1}{3}\) (3y – 2) (7y + 1)
To find the zeroes of the polynomial p(x).
Put
p(x) = 0
⇒ \(\frac {1}{3}\) (3y – 2)(7y + 1) = 0
⇒ (3y – 2)(7y + 1) = 0
⇒ 3y – 2 = 0 and 7y + 1 = 0
⇒ y = \(\frac {2}{3}\) and y = – \(\frac {1}{7}\)
Therefore, zeroes of the polynomial P(x) are:
α = \(\frac {2}{3}\) and β = – \(\frac {1}{7}\)
Now, sum of the zeroes = α + β = \(\frac{2}{3}-\frac{1}{7}\)
= \(\frac{14-3}{21}=\frac{11}{21}=\frac{11}{7 \times 3}=\frac{\frac{11}{3}}{7}\)
= – Coefficient of x / Coefficient of x²
And product of zeroes = α × β = \(\frac{2}{3} \times \frac{-1}{7}=\frac{-\frac{2}{3}}{7}\)
= Constant term / Coefficient of x²
Hence, zeroes of p(x) = \(\frac{2}{3}, \frac{-1}{7}\)

Question 3.
Find the zeroes of a quadratic polynomial x² – 2x – 15 and verify the relationship between the zeroes and coefficient of this polynomial.
Solution :
Let p(x) = x² – 2x – 15
= x² – (5 – 3)x – 15
= x² – 5x + 3x – 15
= x(x – 5) + 3(x – 5)
= (x – 5) (x + 3)
To find the zeroes of polynomial p(x),
Put
p(x) = 0
⇒ (x – 5) (x + 3) = 0
⇒ x – 5 = 0 and x + 3 = 0
⇒ x = 5 and x = – 3
Therefore, zeroes of the polynomial p(x) are :
α = 5 and β = – 3
Sum of zeroes = α + β = 5 + (-3)
= 2 = \(\frac {2}{1}\) = \(\frac{-(-2)}{1}\)
= – Coefficient of x / Coefficient of x²
And product of zeroes = α + β = 5 × (-3) = \(\frac {-15}{1}\)
= Constant term / Coefficient of x²
Hence, zeroes of the p(x) = 5, – 3.

Question 4.
Find the value of k such that the polynomial x² – (k + 6)x + 2(2k – 1) has sum of its zeroes equal to half to their product.
Solution :
Let
p(x) = x² – (k + 6)x + 2(2k – 1) and let α and β be zeroes of p(x)
Then sum of zeroes = α + β = – \(\frac{-(k+6)}{1}\)
= + (k + 6)
And product of zeroes = α × β = \(\frac{2(2 k-1)}{1}\)
= 2(2k – 1)
According to question
Sum of zeroes of p(x) = \(\frac {1}{2}\)Product of zeroes of it
⇒ (k + 6) = \(\frac {1}{2}\) × 2 (2k – 1)
⇒ k + 6 = 2k – 1
⇒ 6 + 1 = 2k – k
⇒ k = 7

Question 5.
Find the quadratic polynomial whose sum and product of zeroes are and respectvely.
Solution :
Sum of zeroes = α + β = \(\frac {21}{8}\)
And product of zeroes = α × β = \(\frac {5}{16}\)
So, the required quadratic polynomial is :
= x² – (α + β)x + αβ
= x² – \(\frac {21}{8}\)x + \(\frac {5}{16}\)
= \(\frac {1}{16}\) (16x² – 42x + 5)

Question 6.
Find a quadratic polynomial whose zeroes are reciprocal of the zeroes of the polynomial f(x) = ax² + bx + c, a ≠ 0, c ≠ 0.
Solution :
f(x) = ax² + bx + c (given)
Let α and β be the zeroes of f(x)
Then,
α + β = – \(\frac {b}{a}\) And αβ = \(\frac {c}{a}\)
Since, zeroes of required polynomial are reciprocal of zeroes of polynomial f(x).
Therefore zeroes of required polynomial are \(\frac {1}{α}\) and \(\frac {1}{β}\)
∴ Sum of zeroes = \(\frac{1}{a}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-\frac{b}{a}}{\frac{c}{a}}\)
= \(\frac{-b}{a} \times \frac{a}{c}=\frac{-b}{c}\)
And product of zeroes = \(\frac{1}{\alpha} \times \frac{1}{\beta}=\frac{1}{\alpha \beta}=\frac{1}{\frac{c}{a}}\)
= \(\frac {a}{c}\)
Required polynomial = x² – (sum of zeroes)x + product of zeroes
= x² + – (- \(\frac {b}{c}\))x + \(\frac {a}{c}\)
= x² + \(\frac{b x}{c}+\frac{a}{c}=\frac{1}{c}\)(cx² + bx + a)

Question 7.
If α and β are zeroes of a polynomial x² – 4\(\sqrt{3}\)x + 3, then find the value of α + β – αβ.
Solution :
Let p(x) = x² – 4\(\sqrt{3}\)x + 3
Since, α and β are zeroes of p(x), then α + β = sum of zeroes = \(\frac{-(-4 \sqrt{3})}{1}\) = 4\(\sqrt{3}\)
And α × β = product zeroes = \(\frac {3}{1}\) = 3
Now, α + β – αβ = 4\(\sqrt{3}\) – 3.

Question 8.
Divide x³ – 6x² + 11x – 6 by x – 2 and verify the division algorithm.
Solution :
Let f(x) = x³ – 6x² + 11x – 6 and g(x) = x – 2
Now, we divide f(x) by g(x), as follows

Hence, q(x) = x² – 4x + 3 and r(x) = 0.
According to division algorithm of polynomials.
f(x) = g(x) × q(x) + r(x)
⇒ x³ – 6x² + 11x – 6 = (x – 2) (x² – 4x + 3) + 0
⇒ x³ – 6x² + 11x – 6 = x³ – 4x² + 3x – 2x² + 8x – 6
⇒ x³ – 6x² + 11x – 6 = x³ – 6x² + 11x – 6
∴ LHS = RHS
Hence, the division algorithm is verified.

Question 9.
Verify g(x) = x³ – 3x + 1 is a factor of P(x) = x⁵ – 4x³ + x² + 3x + 1 or not.
Solution :
If g(x) = x³ – 3x + 1 is a factor of p(x) = x⁵ – 4x³ + x² + 3x + 1, then p(x) is divisible by g(x) completely i.e., obtained remainder is zero.
Now we divide p(x) by g(x) as follows

Since, obtained remainder is not zero. So, g(x) is not a factor of p(x).

Question 10.
If 4 is zero of the cubic polynomial x³ – 3x² – 10x + 24, find its other two zeroes.
Solution :
Let f(x) = x³ – 3x² – 10x + 24
We know that if a is a zero of f(x) then (x – α) is a factor of f(x)
Since 4 is zero of f(x) then (x – 4) is a factor of f(x)
Now, we divide f(x) by (x – 4) as follows

According to division algorithm of polynomials
x³ – 3x² – 10x + 24 = (x – 4) (x² + x – 6)
= (x – 4) [x² + (3 – 2)x – 6]
= (x-4) [x² + 3x – 2x – 6]
= (x – 4) [x(x +3) – 2(x + 3)]
= (x – 4) (x + 3) (x – 2)
For zeroes of the polynomial f(x) = 0
∴ (x – 4)(x + 3)(x – 2) = 0
⇒ x = 4, x = – 3, x = 2
Hence, other two zeroes of the polynomial are : – 3, 2

Question 11.
Find all zeroes of the polynomial 2x⁴ + – 9x³ + 5x² + 3x – 1, if two of its zeroes are (2 + \(\sqrt{3}\)) and (2 – \(\sqrt{3}\)).
Solution :
Since, (2 + \(\sqrt{3}\)) and (2 – \(\sqrt{3}\)) i.e., (2 ± \(\sqrt{3}\)) are zeroes of polynomial 2x⁴ – 9x³ + 5x² + 3x – 1 then (x – 2 – \(\sqrt{3}\))(x – 2 + \(\sqrt{3}\)) or (x – 2)² – (\(\sqrt{3}\))² or x² – 4x + 1 is a factor of f(x).
Now, we divide f(x) by x² – 4x + 1 as follows

According to division algorithm of polynomials
2x⁴ – 9x³ + 5x² + 3x – 1
= (x² – 4x + 1)(2x² – x – 1) + 0
= [(x – 2)² – (\(\sqrt{3}\))²][(2x² – 2x + x – 1]
= (x – 2 – \(\sqrt{3}\))(x – 2 + \(\sqrt{3}\)) [2x (x – 1) + 1(x – 1)]
= (x – 2 – \(\sqrt{3}\))(x – 2 + \(\sqrt{3}\))(x – 1) (2x + 1)
For the zeroes of the polynomial f(x) = 0
(x – 2 – \(\sqrt{3}\))(x – 2 + \(\sqrt{3}\))(x – 1) (2x + 1) = 0
⇒ x = 2 + \(\sqrt{3}\), 2 – \(\sqrt{3}\), 1, – \(\frac {1}{2}\)
Hence, all zeroes of polynomial are:
2 + \(\sqrt{3}\), 2 – \(\sqrt{3}\), 1, –\(\frac {1}{2}\)

Question 12.
For what value of k, is the polynomial f(x) = 3x⁴ – 9x³ + x² + 15x + k completely divisible by 3x² = 5,
Solution :

Fill in the Blanks

Question 1.
The zeroes of a polynomial p(x) are the…………of the points, where the graph of y = p(x) intersects the x-axis.
Solution :
x-coordinates

Question 2.
A cubic polynomial has at most ………….. zeroes.
Solution :
three

Question 3.
If α and β are the zeroes of the quadratic polynomial ax² + bx + c = 0, a ≠ 0 . then:
α + β = …………..
αβ = ……………
Solution :
\(\frac{-b}{a}, \frac{c}{a}\)

Question 4.
If α, β and γ are the zeroes of the cubic polynomial ax² + bx² + cx + d = 0, a ≠ 0, then:
αβ + βγ + γα = ………….
αβγ = ………….
Solution :
\(\frac{c}{a}, \frac{-d}{a}\)

Question 5.
A ………….. polynomial has atmost two zeroes.
Solution :
quadratic

Question 6.
A linear polynomial has only ……… zero.
Solution :
one.

Multiple Choice Questions

Question 1.
If one zero of the quadratic polynomial x² + 3x + k is 2, then value of k is :
(a) 10
(b) – 10
(c) – 7
(d) – 2
Solution :
(b) – 10

Let p(x) = x² + 3x + k and let other zero be β, then
α + β = – \(\frac {b}{a}\)
⇒ 2 + β = \(\frac {- 3}{1}\)
⇒ β = – 3 – 2 = – 5
And α × β = \(\frac {c}{a}\)
⇒ 2 – 5 = \(\frac {k}{1}\)
⇒ – 10 = k

Question 2.
The quadratic polynomial, the sum of whose zeroes is – 5 and their product is 6, is :
(a) x² + 5x + 6
(b) x² – 5x + 6
(c) x² – 5x – 6
(d) – x² + 5x + 6
Solution :
(a) x² + 5x + 6

∵ Sum of zeroes = (α + β) = – 5
And product of zeroes = α × β = 6
So, the required quadratic polynomial is
f(x) = x² – (α + β)x + αβ
= x² – (-5)x + 6
= x² + 5x + 6.

Question 3.
The zeroes of the polynomial x² – 3x – m (m + 3) are :
(a) m, (m + 3)
(b) – m, (m + 3)
(c) m, – (m + 3)
(d) – m, – (m + 3)
Solution :
(b) – m, (m + 3)

∵ Let f(x) = x² – 3x – m (m + 3)
Putting x = – m, we get
⇒ f(x) = (m)² – 3 × – m – m (m + 3)
= m² + 3m – m² – 3m = 0
And putting
x = (m + 3), we get
f(x) = (m + 3)² – 3(m + 3) – m (m + 3)
= (m + 3) [m + 3 – 3 – m]
= (m + 3) (0) = 0.
Hence, – m and (m + 3) are zeroes of the given polynomial

Question 4.
If 2 is factor of polynomial f(x) = x⁴ – x³ – 4x² + kx + 10
(a) 2
(b) – 2
(c) – 1
(d) 1
Solution :
(c) – 1

f(2) = 2⁴ – 2³ – 4 × 2² + k × 2 + 10
⇒ 0 = 16 – 8 – 16 + 2k + 10
⇒ 0 = 2k + 2
⇒ 2k = -2
∴ k = – 1
Hence option (c) is correct.

Question 5.
Find that polynomial whose zeros are – 5 and 4 :
(a) x² – x – 20
(b) x² + x – 20
(c) x² – x + 20
(d) x² + x + 20
Solution :
(b) x² + x – 20

Let β
α = – 5 and β = 4
Then
α + β – 5 + 4 = 1 and
αβ = – 5 × 4 = – 20
Quadratic polynomial x² – (α + β)x + αβ = 0
x² – (-1)x + (-20) = 0
x² + x – 20 = 0
Hence, option (b) is correct.

Question 6.
Expression (x – 3) will be a factor of polynomial f(x) = x³ + x² – 17x + 15 if:
(a) f(3) = 0
(b) f(-3) = 0
(c) f(2) = 0
(d) f(- 2) = 0
Solution :
(a) f(3) = 0

f(x) = x³ + x² – 17x + 15
f(3) = (3)³ + (3)² – 17 (3) + 15
= 27 + 9 – 51 + 15 = 0
f(3) = 0
Hence, option (a) is correct.

Haryana State Board HBSE 10th Class Maths Notes Chapter 1 Real Numbers Notes.

Haryana Board 10th Class Maths Notes Chapter 1 Real Numbers

Introduction
We have studied about natural numbers, whole numbers, rational numbers, irrational numbers, and real numbers in earlier classes. We have also studied their properties and basic fundamental operations upon them. Recall that real number is the collection of rational and irrational numbers. It is denoted by R.

In this chapter, we shall learn about prove of irrationality of \(\sqrt{2}\), \(\sqrt{3}\)……, decimal expansion of rational numbers as terminating or non-terminating repeating decimal expansions, Euclid’s division lemma and fundamental theorem of arithmetic Euclid was the first Greek mathematician who suggests, the divisibility of two positive integers. He says that any positive integer a can be divided by another positive integer bin such a way that it leaves a remainder r such that 0 ≤ r ≤ b, and we compute HCF of two positive integers by the technique of Euclid’s division algorithm.

Fundamental theorem of arithmetic tells us that every composite number can be expressed as a product of primes in a unique way. Both of these results have wide and significant applications in the field of mathematics However, we shall discuss their use only in a few areas.

1. Real numbers: A collection of all rational and irrational numbers is called real numbers.
Example:- 1, 0, 1\(\frac{1}{2}\), 2.25, 4.010010001, \(\sqrt{3}\), π ….., etc.

2. Rational numbers: The numbers which can be written in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0, are called rational numbers.
Example : –\(\frac{4}{5}\), \(\frac{3}{7}\), \(\frac{9}{11}\) etc.

3. Irrational numbers: The numbers which cannot be written in the form \(\frac{p}{q}\), where p and q are integers and q ≠ 0, are called irrational numbers.
Example: \(\sqrt{3}\), \(\sqrt{19}\), 0.04004000400004…., etc.

4. Integers: The collection of whole numbers and their negatives is called integers.
Integers are of three type:
(a) Positive integers [1, 2, 3, …];
(b) Zero and
(c) Negative integers [- 1, -2, -3, …]

5. Positive integers/Natural numbers: The numbers which are used for counting objects are called positive integers or natural numbers.
Esrample: 1, 2, 3, 4, … etc.

6. Non-positive integers: Zero and the negative integers are collectively called non-positive integers.
Example : 0, -1, -2, -3, … etc.

7. Prime numbers: Numbers having exactly two factors (1 and the numbers itself) are called prime numbers.
Example : 2, 3, 5, 7, 11, 13, 17, ….etc.

8. Composite numbers: Numbers having more than two factors are called composite numbers. Composite numbers can be expressed as the product of primes.
Example : 4, 6, 12, 35, …. etc.

9. Co-primes: Two numbers are said to be coprimes, if they do not have a common factor other than 1. HCF of co-primes is 1.
Example : 2 and 3, 7 and 11.

10. Lemma : A lemma is a proven statement used for proving another statement.

11. Algorithm: An algorithm is a series of well defined steps which provides a procedure for solving a type of problem.

Euclid’s Division Lemma

Theorem 1.1.
For any two given positive integers a and b there exist unique whole numbers q and r such that
a = bq + r
where 0 ≤ r < b.
here a = dividend, b = divisor, q = quotient and r = remainder.
Dividend = divisor × quotient + remainder

Euclid’s Division Algorithm:
To find the HCF of two positive integers, say a and b, with a > b, the following steps:
Step I: Apply Euclid’s division lemma to a and b. So, we find whole numbers q and r, such that a = bq + r, where 0 ≤ r < b.
Step II: If r = 0, b is the HCF of a and b. If r ≠ 0, apply the division lemma to b and r.
Step III: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

Fundamental Theorem of Arithmetic
Theorem 1.2
Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factor recur. Let be a composite number. Then factorization of can be written as x = P₁ × P₂ × P₃ × …..P_x, where P₁, P₂, …. P_x are primes and written in ascending order.
For example:
25410 = 2 × 3 × 6 × 7 × 11 × 11
= 2 × 3 × 6 × 7 × 112
Note: This method is also called the prime factorination method.
1. For any two positive integers a and b.
HCP (a, b) × LCM (a, b) = a × b
We can find other relations with the help of this formula.
(I) HCF (a, b) = \(\frac{a \times b}{\mathrm{LCM}(a, b)}\)
(II) LCM (a, b) = \(\frac{a \times b}{\mathrm{HCF}(a, b)}\)
(III) a = \(\frac{\mathrm{HCF}(a, b) \times \mathrm{LCM}(a, b)}{b}\)
(IV) \(b=\frac{\mathrm{HCF}(a, b) \times \mathrm{LCM}(a, b)}{a}\)

2. For any three positive integens x, y and z.
(a) HCF (x, y, z) × LCM (x, y, z) ≠ x × y × z
(b) LCM (x, y, z)
= \(\frac{x \times y \times z \times \mathrm{HCF}(x, y, z)}{\mathrm{HCF}(x, y) \times \mathrm{HCF}(y, z) \times \mathrm{HCF}(z, x)}\)
(c) HCF (x, y, z)
= \(\frac{x \times y \times z \times \mathrm{LCM}(x, y, z)}{\mathrm{LCM}(x, y) \times \mathrm{LCM}(y, z) \times \mathrm{LCM}(z, x)}\)

Revisiting Irrational Numbers
Irrational Numbers : An irrational number is a non-terminating and non-recurring decimal, that is, it cannot be in the form \(\frac{a}{b}\), where p and q are both integers and q ≠ 0.
Example: (i) 0.12119111211112….
(ii) \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\)……….. 3\(\sqrt{2}\), 5\(\sqrt{7}\), ……
(iii) 3\(\sqrt{2}\), 3\(\sqrt{5}\)…. and π.

Theorem 1.3
Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.

Theorem 1.4
Prove that \(\sqrt{2}\) is an irrational number.
Solution:
Let us assume that \(\sqrt{2}\) is a rational number. It can be expressed in the form of \(\frac{a}{b}\), where a, b are coprime positive integers and b ≠ 0.
∴ \(\sqrt{2}\) = \(\frac{a}{b}\) [HCF of a and b is 1]
[because a and b are coprime.]
⇒ 2 = \(\frac{a^2}{b^2}\) (Squaring both sides)
⇒ 2b² = a² ……(i)
Therefore, 2 divides a², it follows that 2 divides a [By Theorem 1.3]
Let a = 2c [Where c is any integer]
Put a = 2c in equation (i), we get
⇒ \(b^2=\frac{4 c^2}{2}=2 c^2\) ……(ii)
It means bis divided by 2 and so b is divided by 2 [By Theorem 1.3]
From (i) and (ii) we say that 2 is a common factor of both a and b. But this contradict the fact that a and b are coprime, so they have no common factor. So our assumption that \(\sqrt{2}\) is a rational number is wrong.
Therefore, \(\sqrt{2}\) is an irrational number. Proved

Revisiting Rational Numbers And Their Decimal Expansions
1. Rational Numbers: Rational numbers are of the form of \(\frac{p}{q}\), where p and q are integers
and q ≠ 0.

2. Rational numbers in decimal form: A rational number has either a terminating or nonterminating but recurring decimal.
For example: (i) \(\frac{2}{5}\) = 0.4 [Terminating decimal expansion].
(ii) \(\frac{1}{3}\) = 0.333 …… [Non-terminating but recurring decimal expansion]

3. Theorem 1.5 : Let x be a rational number whose decimal expension terminates. Then x can be expressed in the form \(\frac{p}{q}\), where p and q are coprime, and the prime factorisation of q in of the form 2ⁿ5^m, where n, m are non negative integers.
For example:
(i) 1.04 = \(\frac{104}{100}=\frac{104}{10^2}=\frac{104}{2^2 \times 5^2}\)
(ii) 0.125 = \(\frac{125}{1000}=\frac{125}{10^3}=\frac{125}{2^3 \times 5^3}\)

4. Theorem 1.6 : Let x = \(\frac{p}{q}\) be a rational number, such that p and q are co-prime and the prime factorisation of q is of the form 2^m5ⁿ or 2ⁿ5^m, where m, n, are non-negative integers. Then has terminating decimal expansion.
For Example : \(\frac{23}{40}=\frac{23}{2^3 \times 5^1}\)
Hence \(\frac{23}{40}\) has a terminating decimal.

5. Theorem 1.7: Let x = \(\frac{p}{q}\) be a rational number, such that p and q are coprime and the prime factorisation of q is not of the form 2^m5ⁿ where m, n are non-negative integers. Then x has a non terminating repeating (recurring) decimal expansion.
For example : \(\frac{53}{343}\)
∵ 343 = 73 [Not of the form 2^m5ⁿ, where m and n are non-negative integers]
∴ \(\frac{53}{343}\) has non terminating repeating decimal.

Haryana State Board HBSE 10th Class Maths Notes Chapter 3 Pair of Linear Equations in Two Variables Notes.

Haryana Board 10th Class Maths Notes Chapter 3 Pair of Linear Equations in Two Variables

Introduction
We have studied in previous classes about linear equations in one variable. Its general form is ax + b = 0; where a and b are real numbers, a ≠ 0 and x is a variable. If ax + b = 0 ⇒ x = \(-\frac{b}{a}\) is the solution of the equation.

In class IX, we have studied about a linear equation in two variables and its graphical representation The general form of linear equation in two variables is ax + by + c = 0; where a, b, c are real numbers, a ≠ 0, b ≠ 0 and x, y both are variables. Any pair of values of andy which satisfies the equation or ax + by + c = 0, is called its solution. Recall that, the graph of a linear equation in two variables is a straight line. The coordinates of every point on a graph satisfies the given equation and coordinates of every point of a graph is the solution of the equation. In this chapter we shall study about a pair of linear equations in two variables and various methods of solving them as algebraic and graphical. We shall also study about some applications of linear equations in two variables in solving simple problems from different areas.

Pair of linear equations in two variables and their graphical representation
A pair of linear equations in two variables is said to form a system of simultaneous linear equations in two variables.
The general form of a pair of linear equations in two variables x and y is :
a₁x + b₁y + c₁ = 0
and a₂x + b₂y + c₂ = 0.
Where a₁, a₂, b₁, b₂, c₁, c₂ are all real numbers and a₁ ≠ 0, b₁ ≠ 0, a₂ ≠ 0, b₂ ≠ 0, aa₁² + b₁² ≠ 0, and a₂² + b₂² ≠ 0. A pair of values of x and y satisfying each one of the given equations is called a solution of the system.
For example: x = 2, y = -2 is the solution of the simultaneous equations in two variables 2x + y = 2 and x – y = 4.

2. The system of simultaneous linear equations in two variables have either a unique solution, or infinitely many solutions or no solution. A system of simultaneous linear equation in two variables could be either consistent or inconsistent.
Consistent System : A system of simultaneous linear equations is said to be consistent, if it has at least one solution.
For example : x = 2, y = 1 is the solution of linear equations 2x + 3y = 7 and 2x – y = 3. So, these linear equations are consistent
Inconsistent System : A system of simultaneous linear equations is said to be inconsistent if it has no solution.
For example. The system of linear equations x + 3y = 4 and 3x + 9y = 10 has no solution. So, these linear equations are inconsistent.

Graphical Representation of Linear Equations in two Variables
We know that the graph of a linear equations in two variables is a straight line. We have studied in class IX that for the given two lines in a plane, only one of the following three possibilities are there :

1. The two lines will intersect at one point.
2. The two lines will not intersect, i.e., they are parallel.
3. The two lines will be coincident, i.e., one line overlaps the other line.

Thus, the graphical representation of a pair of simultaneous linear equations in two variables will be in one of the ahead forms:


Fig. Parallel Lines Coincident Lines i.e. one line overlap other line

Graphical Method of Solution of a pair of Linear Equations
Let a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are two linear equations and on comparing the efficients of x, y and c₁, c₂, there are three possible cases :
Case I : If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), then graphs of the equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 intersect each other at a point P(α, β) as shown in the figure. Then a = α, y = β is the unique solution of the given linear equations. If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), then the pair of linear equation is consistent.

Case II: If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), then graphs of equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel lines. Therefore, there is no solutions of given equations and the pair of linear equations is inconsistent.

Case III : If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), then the graphs of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, are coincident lines. Hence, the linear equations have infinitely many solutions. In this case the pair of linear equations is consistent.

We show these cases by the ahead table :

Algebraic Methods for Solving a Pair of Linear Equations
In the previous section we discussed to solve a pair of linear equations in two variables by graphical method. This method is not convenient in such cases when the point representing the solution of the linear equations has non-integral values like (-2.37, 4.19), (\(\frac{1}{9}, \frac{1}{17}\)), etc. It gives us approximate values of x and y. Algebraic methods provide us exact and correct solution of every pair of linear equations.

The most commonly used algebraic methods of solving a pair of linear equations in two variables are as follows:
(a) Substitution Method
(b) Elimination Method
(c) Cres Multiplication Method.

(a) Substitution Method
Working Rule :
Step-1: Find out the value of any one of the two variable in terms of the other from any one of the two given equations.
Step-2: Sabattute this value in other equation This gives us a linear equation in one variable.
Step-3: Solve the linear equation and get the value of variable, which obtained in Step 2.
Step-4: Put this value in any one of the given equations and get the value of other variable.
Step-5: The values of and y obtained from Step-3 and 4 respectively, in the solution of the given two linear equations

(b) Elimination Method by Equating Coefficients:
Working rule :
Step-1: Multiply the given equations by suitable non zero real numbers so as to make the coeficients of any of the one variable numerically equal in both equations.
Step-2: Add or wubtract new equations so obtained to eliminate the one variable of both equntions whose coefficients are numerically equal.
Step-3: Solve the simple equation in one variable thus obtained.
Step-4: Substitute the value found in step 8 in any one of the given equations and find the value of other variable.

(c) Cross Multiplication Method
Working Rule:
Consider a system of linear equations in two variables x and y.
a₁x + b₁y + c₁ = 0; a₁ ≠ 0, b₁ ≠ 0 ……(1)
a₂x + b₂y + c₂ = 0; a₂ ≠ 0, b₂ ≠ 0 ……(2)
Let us solve the system of equations by elimination method as follows:
Step 1 : Multiplying equation (1) by b₂, and equation (2) by b₁, we get
b₂a₁x + b₂b₁y + b₂c₁ = 0 ……(3)
b₁a₂x + b₁b₂y + b₁c₂ = 0 ……(4)
Step 2: Subtracting equation (4) from equation (3), we get

Step 3: Substituting the value of in equation (1), we get


Thus, x and y can be obtained from equations (5), (6) as formula.
The following diagram helps in writing the above solution :

The downward arrows indicate the first product while the upward arrow indicate the second product. The second product is to be subtracted from the first product. Since the denominators have been obtained by cross multiplying the coefficients, this method is known as the method of cross multiplication for solving a system of linear equation.
Remarks: If we write the system of linear equations as :
a₁x + b₁y = c₁
and a₂x + b₂y = c₂
Then the following diagram helps in writing the solution:

Conditions for solvability of linear equations : Consider the system of a pair of linear equations :
a₁x + b₁y + c₁ = 0
and a₂x + b₂y + c₂ = 0
(1) If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Then equations has unique solution i.e, system is consistent.
(2) If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Then equations has infinitely many solutions Le system is consistent.
(3) If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Then equations has no solution i.e. system is inconsistent.

Equations Reducible to a pair of Linear Equations in Two Variables
In this process, we shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. The process of solving the system of equations involving such cases has been explained through some examples. We solve obtained linear equations by any algebraic method.

Haryana State Board HBSE 10th Class Maths Notes Chapter 12 वृतों से संबंधित क्षेत्रफल Notes.

Haryana Board 10th Class Maths Notes Chapter 12 वृतों से संबंधित क्षेत्रफल

→ वृत्त की परिधि-वृत्त के अनुदिश एक बार चलने में तय की गई दूरी उसका परिमाप होता है, जिसे प्रायः परिधि (circumference) कहा जाता है। त्रिज्या r वाले वृत्त की परिधि 2πr होती है।

→ पाई (π)-वृत्त की परिधि और उसके व्यास के अचर अनुपात को ‘यूनानी’ अक्षर π (पाई) से व्यक्त किया जाता है।

→ π एक अपरिमेय संख्या है और इसका दशमलव प्रसार अनवसानी और अनावर्ती होता है। व्यावहारिक कार्यों के लिए इसका मान लगभग \(\frac{22}{7}\) या 3.14 लिया जाता है।

→ r त्रिज्या वाले वृत्त का क्षेत्रफल = πr²

→ r त्रिज्या वाले अर्धवृत्त का क्षेत्रफल = \(\frac{\pi r^2}{2}\)

→ r त्रिज्या वाले चतुर्थांश का क्षेत्रफल = \(\frac{\pi r^2}{4}\)

→ त्रिज्या : वाले वृत्त के एक त्रिज्यखंड, जिसका कोण अंशों में θ है, के संगत चाप की लंबाई \(\frac{\theta}{360}\) × 2πr होती है।

→ त्रिज्या : वाले वृत्त के एक त्रिज्यखंड, जिसका कोण अंशों में θ है, का क्षेत्रफल \(\frac{\theta}{360}\) × πr² होता है।

→ एक वृत्तखंड का क्षेत्रफल = संगत त्रिज्यखंड का क्षेत्रफल – संगत त्रिभुज का क्षेत्रफल

→ समबाहु त्रिभुज का क्षेत्रफल = \(\frac{\sqrt{3}}{4}\) (भुजा)²