- 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 Polynomials Ex 2.2
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 – Ï€, 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 = 1/2
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 – Ï€
∴ p(4/5) = 5(4/5) – Ï€ = 4 – Ï€
Therefore,
x = 4/5 is not a zero of 5x – Ï€.
(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.
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