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 at

(i) a= 0

(ii) a = – 1

(iii) a = 2

Answer:

Let f(x)=5a4a2+3

 

(i) When a=0

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

 

(iii) When a=2

f(a)=5a+4a2+3f(2)=5(2)+4(2)2+3=1016+3=3

 

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

 (i) p(y)=y2y+1

Answer: p(y)=y2y+1p(0)=(0)2(0)+1=1p(1)=(1)2(1)+1=1p(2)=(2)2(2)+1=3

 

 (ii) p(a)=2+a+2a2a3

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

 

(iii) p(x)=x3

Answer: p(x)=x3p(0)=(0)3=0p(1)=(1)3=1p(2)=(2)3=8

 

(iv) p(a)=(a1)(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

 

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

(i) p(x)=3x+1,x=13

Answer: For, x=13

p(x)=3x+1p(13)=3(13)+1=1+1=0

13 is a zero of p(x).

 

 (ii) p(x)=5xπ,x=45

Answer: For, x=45

p(x)=5xπp(45)=5(45)π=4π

45 is not a zero of p(x).

 

 (iii) p(x)=x21,x=1,1

Answer: For, x=1,1

p(x)=x21p(1)=x11=11=0p(1)=x11=11=0

1,1 are zeros of p(x).

 

(iv) p(x)=(x+1)(x2),x=1,2

Answer: For, x=1,2

p(x)=(x+1)(x2)p(1)=(1+1)(12)=((0)(3))=0p(2)=(2+1)(22)=(3)(0)=0

1,2 are zeros of p(x).

 

(v) p(x)=x2,x=0

Answer: For, x=0

p(x)=02=0

0 is a  zero of p(x).

 

(vi) p(x)=lx+m,x=ml

Answer: For, x=ml

p(x)=lx+mp(ml)=l(ml)+m=m+m=0

ml is a zero of p(x).

 

(vii) p(x)=3x21,x=13,23

Answer: For, x=13,23
p(x)=3x21p(13)=3(13)21=3(13)1=11=0p(23)=3(23)21=3(43)1=41=30

13 is a zero of p(x) but 23 is not a zero of p(x).

 

(viii) p(x)=2x+1,x=12
Answer: For, x=12

p(x)=2x+1p(12)=2(12)+1=1+1=20

12 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

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

 

(ii) p(x) = x – 5

Answer: p(x)=x5x5=0x=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

x=52 is a zero polynomial of the polynomial p(x).

 

(iv)p(x) = 3x – 2 

Answer: p(x)=3x23x2=03x=2x=23

x=23 is a zero polynomial of the polynomial p(x).

 

 (v)p(x) = 3x 

Answer: p(x)=3x3x=0x=0

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

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