Ex 2.2 NCERT Solutions Class 9 Chapter 2
The solutions to exercise 2.2 (from class 9th's maths NCERT book) are given below. The exercise has 4 questions which have been explained thoroughly with proper reasoning.
(i) p(x)= 5x – 4x2 + 3
Putting x=0
p(0) = 5(0) -4(0)2 + 3 => 3
(ii) p(x)= 5x – 4x2 + 3
Putting x= -1
p(-1) = 5(-1) -4(-1)2 + 3
= -5 -4 (1) + 3 {square of -1 is +1}
= -5 -4 +3
= -9 + 3
= -6
(iii) p(x)= 5x – 4x2 + 3
Putting x= 2
p(2) = 5(2) -4(2)2 + 3
= 10 -4 (4) + 3 {square of -1 is +1}
= 10 -16 + 3
= -6 + 3
= -3
(i) p(y) = y2 – y + 1
p(0) = (0)2 – 0 + 1
= 1
p(1) = (1)2 – 1 + 1
= 1
p(2) = (2)2 – 2 + 1
= 4 -1
= 3
(ii) p(t) = 2 + t + 2t2 – t3
p(0) = 2 + 0 + 2(0)2 – (0)3
= 2
p(1) = 2 + 1 + 2(1)2 – (1)3
= 3 + 2 -1
= 4
p(2) = 2 + 2 + 2(2)2 – (2)3
= 2 + 2 + 8 – 8
= 4
(iii) p(x) = x3
p(0) = (0)3
p(0) = 0
p(1) = (1)3
p(1) = 1
p(x) = x3
p(2) = (2)3
p(2) = 8
(iv) p(x) = (x – 1) (x + 1)
p(0) = (0 – 1) (0 + 1)
= -1
p(1) = (1 – 1) (1 + 1)
= 0 x 2 => 0
p(2) = (2 – 1) (2 + 1)
= 3
p(- ⅓) = 3(- 1/3) +1
= -1 + 1
= 0
(ii) p(x) = 5x – Ï€
p(⅘) = 5(4/5) – Ï€
⇒ 4 – Ï€
(iii) p(x) = x2 – 1, x = 1, –1
p(1) =(1)2 – 1
= 1 -1
⇒ 0
p(-1) =(-1)2 – 1
= 1 -1
⇒ 0
(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
p(-1) = (-1 + 1) (-1 -2)
⇒ 0 x (-3)
⇒ 0
p(2) = (2 + 1) (2 -2)
⇒ 3 x 0
⇒ 0
(v) p(x) = x2, x = 0
p(0) = (0)2
⇒ 0
(vi) p(x) = lx + m, x = – m/l
p(- m/l) = l(-m/l) + m
⇒ -m + m
⇒ 0
(vii) p(x) = 3x2 – 1; x = -1/3, 2/3
p( -1/3) = 3(-1/3)2 – 1
⇒ (3 x 1/9) - 1
= 1/3 -1 ⇒ -2/3
p( 2/3) = 3(2/3)2 – 1
⇒ (3 x 4/9) - 1
= 4/3 - 1 ⇒ 1/3
(viii) p(x) = 2x + 1; x=1/2
p(1/2) = 2(1/2) + 1
⇒ 1 + 1
⇒ 2
(i)p(x) = x + 5
0= x + 5
x= -5
So, -5 is the zero of polynomial p(x) = x + 5
(ii)p(x) = x – 5
0= x - 5
x= 5
5 is the zero of polynomial p(x) = x - 5
(iii) p(x) = 2x + 5
0 = 2x + 5
2x = -5
x= -5/2
-5/2 is the zero of polynomial p(x) = 2x + 5
(iv) p(x) = 3x – 2
0 = 3x -2
2 = 3x
x= 2/3
-2/3 is the zero of polynomial p(x) = 3x – 2
(v) p(x) = 3x
0 = 3x
x=0
0 is the zero of polynomial p(x) = 3x
(vi) p(x) = ax, a≠ 0
0 = ax
x=0
0 is the zero of polynomial p(x) = ax, where a≠ 0
(vii)p(x) = cx + d, c ≠ 0, c, d are real numbers
0 = cx + d
cx = -d
x= -d/c
-d/c is the zero of polynomial p(x) = cx + d, where c ≠ 0, c, d are real numbers
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