NCERT So1utions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

 

NCERT So1utions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

Question 1.
Determine which of the following polynomials has (x +1) a factor.
(i) x³+x²+x +1
(ii) x⁴ + x³ + x² + x + 1
(iii) x⁴ + 3x³ + 3x² + x + 1
(iv) x³ – x² – (2 +√2 )x + √2
Solution:
The zero of x + 1 is -1.
(i) Let p (x) = x³ + x² + x + 1
∴ p (-1) = (-1)³ + (-1)² + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p (- 1) = 0
So, (x+ 1) is a factor of x³ + x² + x + 1.

(ii) Let p (x) = x⁴ + x³ + x² + x + 1
∴ P(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1)+1
= 1 – 1 + 1 – 1 + 1
⇒ P (-1) ≠ 1
So, (x + 1) is not a factor of x⁴ + x³ + x² + x+ 1.

(iii) Let p (x) = x⁴ + 3x³ + 3x² + x + 1 .
∴ p (-1)= (-1)⁴ + 3 (-1)³ + 3 (-1)² + (- 1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x⁴ + 3x³ + 3x² + x+ 1.

(iv) Let p (x) = x³ – x² – (2 + √2) x + √2
∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2
= -1 – 1 + 2 + √2 + √2
= 2√2
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x³ – x² – (2 + √2) x + √2.

Question 2.
Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases
(i) p (x)= 2x³ + x² – 2x – 1, g (x) = x + 1
(ii) p(x)= x³ + 3x² + 3x + 1, g (x) = x + 2
(iii) p (x) = x³ – 4x² + x + 6, g (x) = x – 3
Solution:
(i) We have, p (x)= 2x³ + x² – 2x – 1 and g (x) = x + 1
∴ p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
= 2(-1) + 1 + 2 – 1
= -2 + 1 + 2 -1 = 0
⇒ p(-1) = 0, so g(x) is a factor of p(x).

(ii) We have, p(x) x³ + 3x² + 3x + 1 and g(x) = x + 2
∴ p(-2) = (-2)³ + 3(-2)²+ 3(-2) + 1
= -8 + 12 – 6 + 1
= -14 + 13
= -1
⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

(iii) We have, = x³ – 4x² + x + 6 and g (x) = x – 3
∴ p(3) = (3)³ – 4(3)² + 3 + 6
= 27 – 4(9) + 3 + 6
= 27 – 36 + 3 + 6 = 0
⇒ p(3) = 0, so g(x) is a factor of p(x).

Question 3.
Find the value of k, if x – 1 is a factor of p (x) in each of the following cases
(i) p (x) = x² + x + k
(ii) p (x) = 2x² + kx + √2
(iii) p (x) = kx² – √2 x + 1
(iv) p (x) = kx² – 3x + k
Solution:
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x² + x + k
Since, p(1) = (1)² +1 + k
⇒ p(1) = k + 2 = 0
⇒ k = -2.

(ii) Here, p (x) = 2x² + kx + √2
Since, p(1) = 2(1)² + k(1) + √2
= 2 + k + √2 =0
k = -2 – √2 = -(2 + √2)

(iii) Here, p (x) = kx² – √2 x + 1
Since, p(1) = k(1)² – (1) + 1
= k – √2 + 1 = 0
⇒ k = √2 -1

(iv) Here, p(x) = kx² – 3x + k
p(1) = k(1)² – 3(1) + k
= k – 3 + k
= 2k – 3 = 0
⇒ k = [latex]\frac { 3 }{ 4 }[/latex]

Question 4.
Factorise
(i) 12x² – 7x +1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x – 6
(iv) 3x² – x – 4
Solution:
(i) We have,
12x² – 7x + 1 = 12x² – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x² -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x² + 7x + 3 = 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

(iii) We have, 6x² + 5x – 6 = 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
Thus, 6x² + 5x – 6 = (2x + 3)(3x – 2)

(iv) We have, 3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
Thus, 3x² – x – 4 = (3x – 4)(x + 1)

Question 5.
Factorise
(i) x³ – 2x² – x + 2
(ii) x³ – 3x² – 9x – 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² – 2y – 1
Solution:
(i) We have, x³ – 2x² – x + 2
Rearranging the terms, we have x³ – x – 2x² + 2
= x(x² – 1) – 2(x² -1) = (x² – 1)(x – 2)
= [(x)² – (1)²](x – 2)
= (x – 1)(x + 1)(x – 2)
[∵ (a² – b²) = (a + b)(a-b)]
Thus, x³ – 2x² – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) We have, x³ – 3x² – 9x – 5
= x³ + x² – 4x² – 4x – 5x – 5 ,
= x² (x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x² – 4x – 5)
= (x + 1)(x² – 5x + x – 5)
= (x + 1)[x(x – 5) + 1(x – 5)]
= (x + 1)(x – 5)(x + 1)
Thus, x³ – 3x² – 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) We have, x³ + 13x² + 32x + 20
= x³ + x² + 12x² + 12x + 20x + 20
= x²(x + 1) + 12x(x +1) + 20(x + 1)
= (x + 1)(x² + 12x + 20)
= (x + 1)(x² + 2x + 10x + 20)
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)(x + 2)(x + 10)
Thus, x³ + 13x² + 32x + 20
= (x + 1)(x + 2)(x + 10)

(iv) We have, 2y³ + y² – 2y – 1
= 2y³ – 2y² + 3y² – 3y + y – 1
= 2y²(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)(2y² + 3y + 1)
= (y – 1)(2y² + 2y + y + 1)
= (y – 1)[2y(y + 1) + 1(y + 1)]
= (y – 1)(y + 1)(2y + 1)
Thus, 2y³ + y² – 2y – 1
= (y – 1)(y + 1)(2y +1)

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NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

 

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5

Question 1.
Use suitable identities to find the following products
(i) (x + 4)(x + 10)
(ii) (x+8) (x -10)
(iii) (3x + 4) (3x – 5)
(iv) (y²+ [latex]\frac { 3 }{ 2 }[/latex]) (y²– [latex]\frac { 3 }{ 2 }[/latex])
(v) (3 – 2x) (3 + 2x)
Solution:
(i) We have, (x+ 4) (x + 10)
Using identity,
(x+ a) (x+ b) = x² + (a + b) x+ ab.
We have, (x + 4) (x + 10) = x²+(4 + 10) x + (4 x 10)
= x² + 14x+40

(ii) We have, (x+ 8) (x -10)
Using identity,
(x + a) (x + b) = x² + (a + b) x + ab
We have, (x + 8) (x – 10) = x² + [8 + (-10)] x + (8) (- 10)
= x² – 2x – 80

(iii) We have, (3x + 4) (3x – 5)
Using identity,
(x + a) (x + b) = x² + (a + b) x + ab
We have, (3x + 4) (3x – 5) = (3x)² + (4 – 5) x + (4) (- 5)
= 9x² – x – 20

Question 2.
Evaluate the following products without multiplying directly
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
Solution:
(i)We have, 103 x 107 = (100 + 3) (100 + 7)
= ( 100)² + (3 + 7) (100)+ (3 x 7)
[Using (x + a)(x + b) = x² + (a + b)x + ab]
= 10000 + (10) x 100 + 21
= 10000 + 1000 + 21=11021

(ii) We have, 95 x 96 = (100 – 5) (100 – 4)
= ( 100)² + [(- 5) + (- 4)] 100 + (- 5 x – 4)
[Using (x + a)(x + b) = x² + (a + b)x + ab]
= 10000 + (-9) + 20 = 9120
= 10000 + (-900) + 20 = 9120

(iii) We have 104 x 96 = (100 + 4) (100 – 4)
= (100)²-4²
[Using (a + b)(a -b) = a²– b²]
= 10000 – 16 = 9984

Question 3.
Factorise the following using appropriate identities
(i) 9x² + 6xy + y²
(ii) 4y²-4y + 1
(iii) x² – [latex]\frac { { y }^{ 2 } }{ 100 }[/latex]
Solution:
(i) We have, 9x² + 6xy + y²
= (3x)² + 2(3x)(y) + (y)²
= (3x + y)²
[Using a² + 2ab + b² = (a + b)²]
= (3x + y)(3x + y)

(ii) We have, 4y² – 4y + 1²
= (2y)² + 2(2y)(1) + (1)²
= (2y -1)²
[Using a² – 2ab + b² = (a- b)²]
= (2y – 1)(2y – 1 )

Question 4.
Expand each of the following, using suitable identity
(i) (x+2y+ 4z)²
(ii) (2x – y + z)²
(iii) (- 2x + 3y + 2z)²
(iv) (3a -7b – c)^z
(v) (- 2x + 5y – 3z)²
(vi) [ [latex]\frac { 1 }{ 4 }[/latex]a -[latex]\frac { 1 }{ 4 }[/latex]b + 1] ²
Solution:
We know that
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(i) (x + 2y + 4z)²
= x² + (2y)² + (4z)² + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x)
= x² + 4y² + 16z² + 4xy + 16yz + 8 zx

(ii) (2x – y + z)² = (2x)² + (- y)² + z² + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x)
= 4x² + y² + z² – 4xy – 2yz + 4zx

(iii) (- 2x + 3y + 2z)² = (- 2x)² + (3y)² + (2z)² + 2 (- 2x) (3y)+ 2 (3y) (2z) + 2 (2z) (- 2x)
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx

(iv) (3a -7b- c)² = (3a)² + (- 7b)² + (- c)² + 2 (3a) (- 7b) + 2 (- 7b) (- c) + 2 (- c) (3a)
= 9a² + 49b² + c² – 42ab + 14bc – 6ac

(v)(- 2x + 5y- 3z)² = (- 2x)² + (5y)² + (- 3z)² + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

Question 5.
Factorise
(i) 4 x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
Solution:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
= (2x)² + (3y)² + (- 4z)² + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x)
= (2x + 3y – 4z)² = (2x + 3y + 4z) (2x + 3y – 4z)

(ii) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
= (- √2x)² + (y)² + (2 √2z)²y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x)
= (- √2x + y + 2 √2z)²
= (- √2x + y + 2 √2z) (- √2x + y + 2 √2z)

Question 6.
Write the following cubes in expanded form

Solution:
We have, (x + y)³ = x³ + y³ + 3xy(x + y) …(1)
and (x – y)³ = x³ – y³ – 3xy(x – y) …(2)

(i) (2x + 1)³ = (2x)³ + (1)³ + 3(2x)(1)(2x + 1) [By (1)]
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 12x² + 6x + 1

(ii) (2a – 3b)³ = (2a)³ – (3b)³ – 3(2a)(3b)(2a – 3b) [By (2)]
= 8a³ – 27b³ – 18ab(2a – 3b)
= 8a³ – 27b³ – 36a²b + 54ab²

Question 7.
Evaluate the following using suitable identities
(i) (99)³
(ii) (102)³
(iii) (998)³
Solution:
(i) We have, 99 = (100 -1)
∴ 99³ = (100 – 1)³
= (100)³ – 1³ – 3(100)(1)(100 -1)
[Using (a – b)³ = a³ – b³ – 3ab (a – b)]
= 1000000 – 1 – 300(100 – 1)
= 1000000 -1 – 30000 + 300
= 1000300 – 30001 = 970299

(ii) We have, 102 =100 + 2
∴ 102³ = (100 + 2)³
= (100)³ + (2)³ + 3(100)(2)(100 + 2)
[Using (a + b)³ = a³ + b³ + 3ab (a + b)]
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200 = 1061208

(iii) We have, 998 = 1000 – 2
∴ (998)³ = (1000-2)³
= (1000)³– (2)³ – 3(1000)(2)(1000 – 2)
[Using (a – b)³ = a³ – b³ – 3ab (a – b)]
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 +12000
= 994011992

Question 8.
Factorise each of the following
(i) 8a³ +b³ + 12a²b+6ab²
(ii) 8a³ -b³-12a²b+6ab²
(iii) 27-125a³ -135a+225a²
(iv) 64a³ -27b³ -144a²b + 108ab²

Solution:
(i) 8a³ +b³ +12a²b+6ab²
= (2a)³ + (b)³ + 6ab(2a + b)
= (2a)³ + (b)³ + 3(2a)(b)(2a + b)
= (2 a + b)³
[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (2a + b)(2a + b)(2a + b)

(ii) 8a³ – b³ – 12o²b + 6ab²
= (2a)³ – (b)³ – 3(2a)(b)(2a – b)
= (2a – b)³
[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a³ – 135a + 225a²
= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)
= (3 – 5a)³
[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a³ -27b³ -144a²b + 108ab²
= (4a)³ – (3b)³ – 3(4a)(3b)(4a – 3b)
= (4a – 3b)³
[Using a³ – b³ – 3 ab(a – b) = (a – b)³]
= (4a – 3b)(4a – 3b)(4a – 3b)

Question 9.
Verify
(i) x³ + y³ = (x + y)-(x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²)
Solution:
(i) ∵ (x + y)³ = x³ + y³ + 3xy(x + y)
⇒ (x + y)³ – 3(x + y)(xy) = x³ + y³
⇒ (x + y)[(x + y)2-3xy] = x³ + y³
⇒ (x + y)(x² + y² – xy) = x³ + y³
Hence, verified.

(ii) ∵ (x – y)³ = x³ – y³ – 3xy(x – y)
⇒ (x – y)³ + 3xy(x – y) = x³ – y³
⇒ (x – y)[(x – y)² + 3xy)] = x³ – y³
⇒ (x – y)(x² + y² + xy) = x³ – y³
Hence, verified.

Question 10.
Factorise each of the following
(i) 27y³ + 125z³
(ii) 64m³ – 343n³
[Hint See question 9]
Solution:
(i) We know that
x³ + y³ = (x + y)(x² – xy + y²)
We have, 27y³ + 125z³ = (3y)³ + (5z)³
= (3y + 5z)[(3y)² – (3y)(5z) + (5z)²]
= (3y + 5z)(9y² – 15yz + 25z²)

(ii) We know that
x³ – y³ = (x – y)(x² + xy + y²)
We have, 64m³ – 343n³ = (4m)³ – (7n)³
= (4m – 7n)[(4m)² + (4m)(7n) + (7n)²]
= (4m – 7n)(16m² + 28mn + 49n²)

Question 11.
Factorise 27x³ +y³ +z³ -9xyz.
Solution:
We have,
27x³ + y³ + z³ – 9xyz = (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
Using the identity,
x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
We have, (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
= (3x + y + z)[(3x)³ + y³ + z³ – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x² + y² + z² – 3xy – yz – 3zx)

Question 12.
Verify that
x³ +y³ +z³ – 3xyz = [latex]\frac { 1 }{ 2 }[/latex] (x + y+z)[(x-y)² + (y – z)² +(z – x)²]
Solution:
R.H.S
= [latex]\frac { 1 }{ 2 }[/latex](x + y + z)[(x – y)²+(y – z)²+(z – x)²]
= [latex]\frac { 1 }{ 2 }[/latex] (x + y + 2)[(x² + y² – 2xy) + (y² + z² – 2yz) + (z² + x² – 2zx)]
= [latex]\frac { 1 }{ 2 }[/latex] (x + y + 2)(x² + y² + y² + z² + z² + x² – 2xy – 2yz – 2zx)
= [latex]\frac { 1 }{ 2 }[/latex] (x + y + z)[2(x² + y² + z² – xy – yz – zx)]
= 2 x [latex]\frac { 1 }{ 2 }[/latex] x (x + y + z)(x² + y² + z² – xy – yz – zx)
= (x + y + z)(x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz = L.H.S.
Hence, verified.

Question 13.
If x + y + z = 0, show that x³ + y³ + z³ = 3 xyz.
Solution:
Since, x + y + z = 0
⇒ x + y = -z (x + y)³ = (-z)³
⇒ x³ + y³ + 3xy(x + y) = -z³
⇒ x³ + y³ + 3xy(-z) = -z³ [∵ x + y = -z]
⇒ x³ + y³ – 3xyz = -z³
⇒ x³ + y³ + z³ = 3xyz
Hence, if x + y + z = 0, then
x³ + y³ + z³ = 3xyz

Question 14.
Without actually calculating the cubes, find the value of each of the following
(i) (- 12)³ + (7)³ + (5)³
(ii) (28)³ + (- 15)³ + (- 13)³
Solution:
(i) We have, (-12)³ + (7)³ + (5)³
Let x = -12, y = 7 and z = 5.
Then, x + y + z = -12 + 7 + 5 = 0
We know that if x + y + z = 0, then, x³ + y³ + z³ = 3xyz
∴ (-12)³ + (7)³ + (5)³ = 3[(-12)(7)(5)]
= 3[-420] = -1260

(ii) We have, (28)³ + (-15)³ + (-13)³
Let x = 28, y = -15 and z = -13.
Then, x + y + z = 28 – 15 – 13 = 0
We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz
∴ (28)³ + (-15)³ + (-13)³ = 3(28)(-15)(-13)
= 3(5460) = 16380

Question 15.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given
(i) Area 25a² – 35a + 12
(ii) Area 35y² + 13y – 12
Solution:
Area of a rectangle = (Length) x (Breadth)
(i) 25a² – 35a + 12 = 25a² – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)
Thus, the possible length and breadth are (5a – 3) and (5a – 4).

(ii) 35y²+ 13y -12 = 35y² + 28y – 15y -12
= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).

Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume 3x² – 12x
(ii) Volume 12ky² + 8ky – 20k
Solution:
Volume of a cuboid = (Length) x (Breadth) x (Height)
(i) We have, 3x² – 12x = 3(x² – 4x)
= 3 x x x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) We have, 12ky² + 8ky – 20k
= 4[3ky² + 2ky – 5k] = 4[k(3y² + 2y – 5)]
= 4 x k x (3y² + 2y – 5)
= 4k[3y² – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).

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CBSE Class 12 Maths Notes Chapter 1 Relations and Functions

 

CBSE Class 12 Maths Notes Chapter 1 Relations and Functions

Relation: A relation R from set X to a set Y is defined as a subset of the cartesian product X × Y. We can also write it as R ⊆ {(x, y) ∈ X × Y : xRy}.

Note: If n(A) = p and n(B) = q from set A to set B, then n(A × B) = pq and number of relations = 2^pq.

Types of Relation
Empty Relation: A relation R in a set X, is called an empty relation, if no element of X is related to any element of X,
i.e. R = Φ ⊂ X × X

Universal Relation: A relation R in a set X, is called universal relation, if each element of X is related to every element of X,
i.e. R = X × X

Reflexive Relation: A relation R defined on a set A is said to be reflexive, if
(x, x) ∈ R, ∀ x ∈ A or
xRx, ∀ x ∈ R

Symmetric Relation: A relation R defined on a set A is said to be symmetric, if
(x, y) ∈ R ⇒ (y, x) ∈ R, ∀ x, y ∈ A or
xRy ⇒ yRx, ∀ x, y ∈ R.

Transitive Relation: A relation R defined on a set A is said to be transitive, if
(x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R, ∀ x, y, z ∈ A
or xRy, yRz ⇒ xRz, ∀ x, y,z ∈ R.

Equivalence Relation: A relation R defined on a set A is said to be an equivalence relation if R is reflexive, symmetric and transitive.

Equivalence Classes: Given an arbitrary equivalence relation R in an arbitrary set X, R divides X into mutually disjoint subsets A, called partitions or sub-divisions of X satisfying

• all elements of A_i are related to each other, for all i.
• no element of A_i is related to any element of A_j, i ≠ j
• A_i ∪ A_j = X and A_i ∩ A_j = 0, i ≠ j. The subsets A_i and A_j are called equivalence classes.

Function: Let X and Y be two non-empty sets. A function or mapping f from X into Y written as f : X → Y is a rule by which each element x ∈ X is associated to a unique element y ∈ Y. Then, f is said to be a function from X to Y.
The elements of X are called the domain of f and the elements of Y are called the codomain of f. The image of the element of X is called the range of X which is a subset of Y.
Note: Every function is a relation but every relation is not a function.

Types of Functions
One-one Function or Injective Function: A function f : X → Y is said to be a one-one function, if the images of distinct elements of x under f are distinct, i.e. f(x₁) = f(x₂ ) ⇔ x₁ = x₂, ∀ x₁, x₂ ∈ X
A function which is not one-one, is known as many-one function.

Onto Function or Surjective Function: A function f : X → Y is said to be onto function or a surjective function, if every element of Y is image of some element of set X under f, i.e. for every y ∈ y, there exists an element X in x such that f(x) = y.
In other words, a function is called an onto function, if its range is equal to the codomain.

Bijective or One-one and Onto Function: A function f : X → Y is said to be a bijective function if it is both one-one and onto.

Composition of Functions: Let f : X → Y and g : Y → Z be two functions. Then, composition of functions f and g is a function from X to Z and is denoted by fog and given by (fog) (x) = f[g(x)], ∀ x ∈ X.
Note
(i) In general, fog(x) ≠ gof(x).
(ii) In general, gof is one-one implies that f is one-one and gof is onto implies that g is onto.
(iii) If f : X → Y, g : Y → Z and h : Z → S are functions, then ho(gof) = (hog)of.

Invertible Function: A function f : X → Y is said to be invertible, if there exists a function g : Y → X such that gof = I_x and fog = I_y. The function g is called inverse of function f and is denoted by f^-1.
Note
(i) To prove a function invertible, one should prove that, it is both one-one or onto, i.e. bijective.
(ii) If f : X → V and g : Y → Z are two invertible functions, then gof is also invertible with (gof)^-1 = f^-1og^-1

Domain and Range of Some Useful Functions

Binary Operation: A binary operation * on set X is a function * : X × X → X. It is denoted by a * b.

Commutative Binary Operation: A binary operation * on set X is said to be commutative, if a * b = b * a, ∀ a, b ∈ X.

Associative Binary Operation: A binary operation * on set X is said to be associative, if a * (b * c) = (a * b) * c, ∀ a, b, c ∈ X.
Note: For a binary operation, we can neglect the bracket in an associative property. But in the absence of associative property, we cannot neglect the bracket.

Identity Element: An element e ∈ X is said to be the identity element of a binary operation * on set X, if a * e = e * a = a, ∀ a ∈ X. Identity element is unique.
Note: Zero is an identity for the addition operation on R and one is an identity for the multiplication operation on R.

Invertible Element or Inverse: Let * : X × X → X be a binary operation and let e ∈ X be its identity element. An element a ∈ X is said to be invertible with respect to the operation *, if there exists an element b ∈ X such that a * b = b * a = e, ∀ b ∈ X. Element b is called inverse of element a and is denoted by a^-1.
Note: Inverse of an element, if it exists, is unique.

Operation Table: When the number of elements in a set is small, then we can express a binary operation on the set through a table, called the operation table.

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NCERT Solutions for Class 10 Maths Unit 2

NCERT Solutions for Class 10 Maths Unit 2

Polynomials Class 10

Exercise 2.1

Q1 :

The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the

number of zeroes ofp(x), in each case.

Answer :

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

Exercise 2.2 

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4

Solution:
(i) x2 – 2x – 8 = x2 – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
Either x + 2 = 0 or x – 4 = 0
⇒  x = -2 or x = 4
Hence, zeroes of this polynomial are -2 and 4.
Verification:
Sum of the zeroes = (-2) + (4) = 2
=  = 
Product of zeroes = (-2) (4) = -8 =  =  
Hence verified.

(ii) 4s2 – 4s + 1 = (2s – l)2 = (2s – l)(2s – 1)
Either 2s – 1 = 0 or 2s – 1 = 0
i.e. , s =   , 
Hence, the two Zeroes are  and 
Verification:

Hence verified.

(iii) 6x2 – 3 – 7x = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x (2x – 3) + 1(2x – 3)
= (2x – 3) (3x + 1)
Either 2x – 3 = 0 or 3x+1 = 0

Hence verified.

(iv) 4u2 + 8u  ⇒  4u(u + 2)
Either 4u = 0 or u + 2 = 0
⇒ u = 0 or u = -2
Hence, the two zeroes are 0 and -2.
Verification:
Sum of the zeroes = 0 + (-2) = -2
=  = 
Product of zeroes = 0 x (-2) = 0 =  
Hence verified.

(v) t2 – 15 = t2 – ()2
= (t + () (t- ()
Either t + ( = 0 or t – ( = 0
⇒ t = -( or t = (
Hence, the two zeroes are -( and + .
Verification:
Sum of the zeroes = -( +  = 0
= 
Product of zeroes = – x  = -15
= 
Hence verified.

(vi) 3x2 – x – 4 = 3x2 – 4x + 3x – 4
=  x(3x – 4) + l(3x 4)
= (x + 1) (3x – 4)
Either x + 1 = 0 or 3x-4 = 0
⇒  x = -1 or x =  
Verification:
Sum of the zeroes = -1 +  =   =  
Product of zeroes = -1 x  =   =  
Hence verified.

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively:

Solution:
(i) Let the zeroes of polynomial be α and β.
Then, α + β =  and αβ = -1
∴ Required polynomial is given by,
x2 – (α + β)x + αβ = x2 – x + (-1)
= x2 – x – 1
= 4x2 – x – 4

(ii) Let the zeroes of polynomial be α and β.
Then, α + β= √2 and αβ = 
∴  Required polynomial is:
x2 – (α + β)x + αβ = x2 – √2x + 
= 3x2 – 3√2x + 1

(iii) Let the zeroes of the polynomial be α and β.
Then, α + β = 0 and αβ = √5
∴ Required polynomial
= x2 – (α + β)x + αβ
= x2– 0 x x + √5 = x2 + √5

(iv) Let the zeroes of the polynomial be α and β.
Then, α + β = 1 and αβ = 1.
∴  Required polynomial
= x2 – (α + β)x + αβ
= x2 – x + 1

(v) Let the zeroes of the polynomial be α and β.
Then, α + β = –  and αβ = 
∴ Required polynomial
= x2 – (α + β)x + αβ
= x2 – (-  ) + 
=  4x2 + x + 1 = 0

(vi) Let the zeroes of the polynomial be α and β.
Then, α + β = 4 and αβ = 1.
∴ Required polynomial = x2 -(α + β)x + αβ
= x2 – 4x + 1

                   Exercise 2.3

Question 1.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) p(x) = x4– 5x + 6, g(x) = 2 – x2
Solution:

 Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.
(i) t2 – 3, 2t4 + 3t3 – 2t2– 9t – 12
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x2 + 3x + 1, x5 – 4x3 + x2 + 3x + 1
Solution:

∴  Remainder is 0, therefore, t2 – 3 is a factor of polynomial 2t4 + 3t3 – 2t2 -9t – 12.

∴ Remainder is 0, therefore, x2 + 3x + 1 is a factor of polynomial 3x4 + 5x3 – 7x2 + 2x + 2.

∴ Remainder = 2 ≠ 0, therefore, x3 – 3x + 1 is not a factor of polynomial x5 – 4x3 + x2+ 3x + 1.

 Question 3.
 Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are  and  and –
Solution:

=  x (3x2– 5).Since both  and(3x2– 5)are the factors, therefore 3x2 – 5 is a factor of the given polynomial.
Now, we divide the given polynomial by 3x2 – 5.

Hence, the other zeroes of the given polynomial are -1 and –1.

 Question 4.
On dividing x3 – 3x2 + x + 2bya polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).
Solution:
x3 – 3x2 + x + 2  = g(x) x (x – 2) + (-2x + 4) [By division algorithm]

 Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and:
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
(i) p(x) = 2X2 + 2x + 8,
q(x) = x2 + x + 4,
g(x) = 2 and r(x) = 0

(ii) p(x) = x3 + x2 + x + 1,
q(x) = x + 1,
g(x) = x2 – 1 and r(x) = 2x + 2

(iii) p(x) = x3 – x2 + 2x + 3,
g(x) = x2 + 2,
q(x) = x – 1 and r(x) = 5

                                      Exercise 2.4

 Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3 + x2 – 5x + 2;  , 1, -2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Solution:
(i) Comparing the given polynomial with ax3 + bx2 + cx + d, we get:
a = 2, b – 1, c = -5 and d = 2.
∴  p(x) = 2x3 + x2 – 5x + 2


(ii) Compearing the given polynomial with ax3 + bx2 + cx + d, we get:
a = 1, b = -4, c = 5 and d = – 2.
∴ p (x) = x3 – 4x2 + 5x – 2
⇒  p(2) = (2)3 – 4(2)2 + 5 x 2 – 2
= 8 – 16+ 10 – 2 = 0
p(1) = (1)3 – 4(1)2 + 5 x 1-2
= 1 – 4 + 1 – 2
= 6-6 = 0
Hence, 2, 1 and 1 are the zeroes of x3 – 4x2 + 5x – 2.
Hence verified.
Now we take α = 2, β = 1 and γ = 1.
α + β + γ = 2 + 1 + 1 =  = 
αβ + βγ + γα = 2 + 1 + 2 =  = 
αβγ = 2 x 1 x 1  =  = .
Hence verified.

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Solution:
Let α , β and  γ be the zeroes of the required polynomial.
Then α + β + γ = 2, αβ + βγ + γα = -7 and αβγ = -14.
∴ Cubic polynomial
= x3 – (α + β + γ)x2 + (αβ + βγ + γα)x – αβγ
= x3 – 2x2 – 1x + 14
Hence, the required cubic polynomial is x3 – 2x2 – 7x + 14.

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a-b, a, a + b, find a and b.
Solution:
Let α , β and  γ be the zeroes of polynomial x3 – 3x2 + x + 1.
Then α =  a-b, β = a and γ = a + b.
∴ Sum of zeroes = α + β + γ
⇒   3 = (a – b) + a + (a + b)
⇒  (a – b) + a + (a + b) = 3
⇒  a-b + a + a + b = 3
⇒       3a = 3
⇒ a =   = 1 …(i)
Product of zeroes = αβγ
⇒ -1 = (a – b) a (a + b)
⇒ (a – b) a (a + b) = -1
⇒   (a2 – b2)a = -1
⇒  a3 – ab2 = -1   … (ii)
Putting the value of a from equation (i) in equation (ii), we get:
(1)3-(1)b2 = -1
⇒ 1 – b2 = -1
⇒ – b2 = -1 – 1
⇒  b2 = 2
⇒ b = ±√2
Hence, a = 1 and b = ±√2.

 Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± √3, finnd other zeroes.
Solution:
Since two zeroes are 2 + √3 and 2 – √3,
∴  [x-(2 + √3)] [x- (2 – √3)]
= (x-2- √3)(x-2 + √3)
= (x-2)2– (√3)2
x2 – 4x + 1 is a factor of the given polynomial.
Now, we divide the given polynomial by x2 – 4x + 1.

So, x4 – 6x3 – 26x2 + 138x – 35
= (x2 – 4x + 1) (x2 – 2x – 35)
= (x2 – 4x + 1) (x2 – 7x + 5x – 35)
= (x2-4x + 1) [x(x- 7) + 5 (x-7)]
= (x2 – 4x + 1) (x – 7) (x + 5)
Hence, the other zeroes of the given polynomial are 7 and -5.

Question 5.
If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
We have
p(x) = x4 – 6x3 + 16x2 – 25x + 10
Remainder = x + a   … (i)
Now, we divide the given polynomial 6x3 + 16x2 – 25x + 10 by x2 – 2x + k.

Using equation (i), we get:
(-9 + 2k)x + 10-8 k + k2 = x + a
On comparing the like coefficients, we have:
-9 + 2k = 1
⇒ 2k = 10
⇒ k =  = 5  ….(ii)
and 10 -8k + k2– a   ….(iii)
Substituting the value of k = 5, we get:
10 – 8(5) + (5)2 = a
⇒   10 – 40 + 25 = a
⇒  35 – 40 =   a
⇒   a =   -5
Hence, k = 5 and a = -5.

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