# Ncert Solutions For Class 9 Maths Ex 2.2

## Ncert Solutions For Class 9 Maths Chapter 2 Ex 2.2

Q.1. Find the value of the polynomial atf(x)=5a4a2+3$f(x)= 5a -4a^{2}+ 3$ at

(i) a= 0

(ii) a = – 1

(iii) a = 2

Let f(x)=5a4a2+3$f(x)= 5a -4a^{2}+ 3$

(i) When a=0

f(0)=5(0)+4(0)2+3=3$f(0) = 5(0) + 4(0)^{2} + 3= 3$

(ii) When a=-1

f(a)=5a+4a2+3f(1)=5(1)+4(1)2+3=54+3=6$f(a) = 5a + 4a^{2}+ 3\\ f(-1) = 5(-1) + 4(-1)^{2} + 3 = -5 – 4 + 3 = -6$

(iii) When a=2

f(a)=5a+4a2+3f(2)=5(2)+4(2)2+3=1016+3=3$f(a) = 5a + 4a^{2} + 3\\ f(2) = 5(2) + 4(2)^{2} + 3 = 10 – 16 + 3 = -3$

Q.2. Find p(0), p(1) and p(2) for each of the following polynomials:

(i) p(y)=y2y+1$p(y) = y^{2}- y + 1$

Answer: p(y)=y2y+1p(0)=(0)2(0)+1=1p(1)=(1)2(1)+1=1p(2)=(2)2(2)+1=3$p(y) = y2 – y + 1\\ ∴ p(0) = (0)^{2}- (0) + 1 = 1\\ p(1) = (1)^{2} – (1) + 1 = 1\\ p(2) = (2)^{2} – (2) + 1 = 3$

(ii) p(a)=2+a+2a2a3$p(a) = 2 + a + 2a^{2}- a^{3}$

Answer: p(a)=2+a+2a2a3p(0)=2+0+2(0)2(0)3=2p(1)=2+1+2(1)2(1)3=2+1+21=4p(2)=2+2+2(2)2(2)3=2+2+88=4$p(a) = 2 +a + 2a^{2}-a^{3}\\ ∴ p(0) = 2 + 0 + 2 (0)^{2} – (0)^{3} = 2\\ p(1) = 2 + 1+ 2(1)^{2} – (1)^{3} = 2 + 1 + 2 – 1 = 4\\ p(2) = 2 + 2 + 2(2)^{2} – (2)^{3} = 2 + 2 + 8 – 8 = 4$

(iii) p(x)=x3$p(x) = x^{3}$

Answer: p(x)=x3p(0)=(0)3=0p(1)=(1)3=1p(2)=(2)3=8$p(x) = x3\\ ∴ p(0) = (0)^{3}= 0\\ p(1) = (1)^{3} = 1\\ p(2) = (2)^{3} = 8$

(iv) p(a)=(a1)(a+1)$p(a) = (a- 1) (a+ 1)$

Answer: p(a)=(a1)(a+1)p(0)=(01)(0+1)=(1)(1)=1p(1)=(11)(1+1)=0(2)=0p(2)=(21)(2+1)=1(3)=3$p(a) = (a – 1) (a + 1)\\ ∴ p(0) = (0 – 1) (0 + 1) = (- 1) (1) = – 1\\ p(1) = (1 – 1) (1 + 1) = 0 (2) = 0\\ p(2) = (2 – 1 ) (2 + 1) = 1(3) = 3$

Q.3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x)=3x+1,x=13$p(x) = 3x + 1, x = -\frac{1}{3}$

Answer: For, x=13$x = -\frac{1}{3}$

p(x)=3x+1p(13)=3(13)+1=1+1=0$p(x) = 3x + 1\\ ∴ p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1=-1+1=0$

13$∴ -\frac{1}{3}$ is a zero of p(x).

(ii) p(x)=5xπ,x=45$p(x) = 5x – \pi , x = \frac{4}{5}$

Answer: For, x=45$x = \frac{4}{5}$

p(x)=5xπp(45)=5(45)π=4π$p(x) = 5x – \pi \\ ∴ p(\frac{4}{5}) = 5(\frac{4}{5}) – \pi=4-\pi$

45$∴ \frac{4}{5}$ is not a zero of p(x).

(iii) p(x)=x21,x=1,1$p(x) = x^{2}- 1, x = 1, -1$

Answer: For, x=1,1$x = 1,-1$

p(x)=x21p(1)=x11=11=0p(1)=x11=11=0$p(x) = x^{2}- 1\\ ∴ p(1) = x^{1}- 1=1-1=0\\ p(-1) = x^{-1}- 1=1-1=0$

1,1$∴ 1,-1$ are zeros of p(x).

(iv) p(x)=(x+1)(x2),x=1,2$p(x) = (x + 1) (x – 2), x = -1, 2$

Answer: For, x=1,2$x = -1,2$

p(x)=(x+1)(x2)p(1)=(1+1)(12)=((0)(3))=0p(2)=(2+1)(22)=(3)(0)=0$p(x) = (x + 1) (x – 2)\\ ∴ p(-1) = (-1 + 1) (-1 – 2)\\ =((0)(-3))\\ =0\\ \\ p(2) = (2 + 1) (2 – 2)\\ =(3)(0)=0$

1,2$∴ -1,2$ are zeros of p(x).

(v) p(x)=x2,x=0$p(x) = x^{2}, x = 0$

Answer: For, x=0$x = 0$

p(x)=02=0$p(x) = 0^{2}=0$

0$∴ 0$ is a  zero of p(x).

(vi) p(x)=lx+m,x=ml$p(x)=lx+m,x=-\frac{m}{l}$

Answer: For, x=ml$x=-\frac{m}{l}$

p(x)=lx+mp(ml)=l(ml)+m=m+m=0$p(x)=lx+m\\ ∴ p(-\frac{m}{l} )=l(-\frac{m}{l} )+m=-m+m=0$

ml$∴-\frac{m}{l}$ is a zero of p(x).

(vii) p(x)=3x21,x=13,23$p(x)=3x^{2}-1,x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$

Answer: For, x=13,23$x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$
p(x)=3x21p(13)=3(13)21=3(13)1=11=0p(23)=3(23)21=3(43)1=41=30$p(x)=3x^{2}-1\\ ∴ p(-\frac{1}{\sqrt{3}})=3(-\frac{1}{\sqrt{3}})^{2}-1\\ =3(\frac{1}{3})-1=1-1=0\\ \\ ∴ p(\frac{2}{\sqrt{3}})=3(\frac{2}{\sqrt{3}})^{2}-1\\ =3(\frac{4}{3})-1=4-1=3\neq 0$

13$∴ -\frac{1}{\sqrt{3}}$ is a zero of p(x) but 23$\frac{2}{\sqrt{3}}$ is not a zero of p(x).

(viii) p(x)=2x+1,x=12$p(x) = 2x + 1, x = \frac{1}{2}$
Answer: For, x=12$x=\frac{1}{2}$

p(x)=2x+1p(12)=2(12)+1=1+1=20$p(x) = 2x + 1\\ ∴ p(\frac{1}{2} ) = 2(\frac{1}{2}) + 1=1+1=2\neq 0$

12$∴\frac{1}{2}  $ is not a  zero of p(x).

Q.4. Find the zero of the polynomial in each of the following cases:

(i)p(x) = x + 5

Answer: p(x)=x+5x+5=0x=5$p(x) = x + 5 \\ \Rightarrow x + 5=0 \Rightarrow x=-5$

$∴$-5 is a zero polynomial of the polinomila p(x).

(ii) p(x) = x – 5

Answer: p(x)=x5x5=0x=5$p(x) = x -5 \\ \Rightarrow x -5=0 \Rightarrow x=5$

$∴$5 is a zero polynomial of the polinomila p(x).

(iii) p(x) = 2x + 5

Answer: p(x)=2x+52x+5=02x=5x=52$p(x) =2 x + 5 \\ \Rightarrow 2x + 5=0 \Rightarrow 2x=-5\\ \Rightarrow x=\frac{-5}{2}$

$∴$x=52$x=\frac{-5}{2}$ is a zero polynomial of the polynomial p(x).

(iv)p(x) = 3x – 2

Answer: p(x)=3x23x2=03x=2x=23$p(x) =3 x – 2 \\ \Rightarrow 3x -2=0 \Rightarrow 3x=2\\ \Rightarrow x=\frac{2}{3}$

$∴$x=23$x=\frac{2}{3}$ is a zero polynomial of the polynomial p(x).

(v)p(x) = 3x

Answer: p(x)=3x3x=0x=0$p(x) =3 x \\ \Rightarrow 3x =0 \Rightarrow x=0$

$∴$0 is a zero polynomial of the polynomial p(x).