NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2.

• NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.1
• NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.2
• NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.3
• NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.4
• NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.5

Ex 2.2 Class 9 Maths Question 1.
Find the value of the polynomial 5x – 4x² + 3 at:
(i) x = 0                    (ii) x = –1                 (iii) x = 2

Solution:
Let p(x) = 5x – 4x² + 3
(i) Putting x = 0, we get, p(0) = 5(0) – 4(0)² + 3 = 0 – 0 + 3 = 3
Thus, the value of 5x – 4x² + 3 at x = 0 is 3.
(ii) Putting x = –1, we get, p(–1) = 5(–1) – 4(–1)² + 3 = – 5 – 4 + 3 = –9 + 3 = –6
Thus, the value of 5x – 4x² + 3 at x = –1 is –6.
(iii) Putting x = 2, we get, p(2) = 5(2) – 4(2)² + 3 = 10 – 4(4) + 3 = 10 – 16 + 3 = –3
Thus, the value of 5x – 4x² + 3 at x = 2 is –3.

 

Ex 2.2 Class 9 Maths Question 2.

Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y² – y + 1                              (ii) p(t) = 2 + t + 2t² – t³
(iii) p(x) = x³                                                              (iv) p(x) = (x – 1) (x + 1)

Solution:
(i) Given that p(y) = y² – y + 1.
∴ P(0) = (0)² – 0 + 1 = 0 – 0 + 1 = 1
p(1) = (1)² – 1 + 1 = 1 – 1 + 1 = 1
p(2) = (2)² – 2 + 1 = 4 – 2 + 1 = 3

(ii) Given that p(t) = 2 + t + 2t² – t³
∴ p(0) = 2 + 0 + 2(0)² – (0)³
            = 2 + 0 + 0 – 0 = 2
P(1) = 2 + 1 + 2(1)² – (1)³
        = 2 + 1 + 2 – 1 = 4
p(2) = 2 + 2 + 2(2)² – (2)³
        = 2 + 2 + 8 – 8 = 4

(iii) Given that p(x) = x³
∴ p(0) = (0)³ = 0, p(1) = (1)³ = 1
p(2) = (2)³ = 8

(iv) Given that p(x) = (x – 1)(x + 1)
∴ p(0) = (0 – 1)(0 + 1) = (–1)(1) = –1
p(1) = (1 – 1)(1 + 1) = (0)(2) = 0
P(2) = (2 – 1)(2 + 1) = (1)(3) = 3

 

Ex 2.2 Class 9 Maths Question 3.
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = –1/3
(ii) p(x) = 5x – π, x = 4/5
(iii) p(x) = x² – 1, x = x – 1
(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
(v) p(x) = x², x = 0
(vi) p(x) = lx + m, x = –m/l
(vii) P(x) = 3x² – 1, x = –1/√3, 2/√3
(viii) p(x) = 2x + 1, x = 12

Solution:
(i) We have, p(x) = 3x + 1

p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0

Since, p(-1/3) = 0, therefore, x = -1/3 is a zero of 3x + 1.

(ii) We have, p(x) = 5x – π

∴ p(4/5) = 5(4/5) – π = 4 – π

Therefore, x = 4/5 is not a zero of 5x – π.

(iii) We have, p(x) = x² – 1
∴ p(1) = (1)² – 1 = 1 – 1 = 0
Since, p(1) = 0, so x = 1 is a zero of x² – 1.
Also, p(-1) = (-1)² – 1 = 1 – 1 = 0
Since p(-1) = 0, so, x = -1 is also a zero of x² – 1.

(iv) We have, p(x) = (x + 1)(x – 2)
∴ p(-1) = (-1 + 1) (-1 – 2) = (0)(-3) = 0
Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).
Also, p(2) = (2 + 1)(2 – 2) = (3)(0) = 0
Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x²
∴ p(0) = (0)² = 0
Since, p(0) = 0, so, x = 0 is a zero of x².

(vi) We have, p(x) = lx + m

∴ p(-m/l) = l(-m/l) + m = -m + m = 0

Since, p(-m/l) = 0, so, x = -m/l is a zero of lx + m.

(vii) We have, p(x) = 3x² – 1

(viii) We have, p(x) = 2x + 1
∴ p(1/2) = 2(1/2) + 1 = 1 + 1 = 2
Since, p(1/2) ≠ 0, so, x = 1/2 is not a zero of 2x + 1.

 

Ex 2.2 Class 9 Maths Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x)= ax, a≠0
(vii) p(x) = cx + d, c ≠ 0 where c and d are real numbers.

Solution:
(i) We have, p(x) = x + 5. Since, p(x) = 0
⇒ x + 5 = 0
⇒ x = -5.
Thus, zero of x + 5 is -5.

(ii) We have, p(x) = x – 5.
Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = 5
Thus, zero of x – 5 is 5.

(iii) We have, p(x) = 2x + 5. Since, p(x) = 0
⇒ 2x + 5 = 0
⇒ 2x = -5
⇒ x = −5/2
Thus, zero of 2x + 5 is −5/2.

(iv) We have, p(x) = 3x – 2. Since, p(x) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
⇒ x = 2/3
Thus, zero of 3x – 2 is 2/3.

(v) We have, p(x) = 3x. Since, p(x) = 0
⇒ 3x = 0 ⇒ x = 0
Thus, zero of 3x is 0.

(vi) We have, p(x) = ax, a ≠ 0.
Since, p(x) = 0 ⇒ ax = 0 ⇒ x = 0
Thus, zero of ax is 0.

(vii) We have, p(x) = cx + d. Since, p(x) = 0
⇒ cx + d = 0 ⇒ cx = -d ⇒ x = −d/c
Thus, zero of cx + d is −d/c.

 NCERT Solutions for Maths Class 10

NCERT Solutions for Maths Class 12