SChand CLASS 9 Chapter 3 Expansions TEST

 TEST

Question 1

The coefficient of x in the product (2 – 3x) (5 – 2x) is

(a) 19

(b) – 19

(c) 15

(d) 6

Sol :

=(2-3x)(5-2x)

term x has -4x-15x=-19x

Question 2

If $3x^4$ + kx² – 8 = (3x² – 2) (x² + 4) for all x, then the value of k is:

(a) – 2

(b) 12

(c) 10

(d) – 8

Sol :

$3 x^{4}+k x^{2}-8=\left(3 x^{2}-2\right)\left(x^{2}+4\right)$

$3 x^{4}+k x^{2}-8=3 x^{4}+12 x^{2}-2 x^{2}-8$

$3 x^{4}+k x^{2}-8=3 x^{4}+10 x^{2}-8$

k=10

Question 3

The coefficient of x² in (3x + x³)$\left(x + \frac{1}{x}\right)$ is

(a) 3

(b) 1

(c) 4

(d) 2

Sol :

$\left(3x+x^{3}\right)\left(x+\frac{1}{x}\right)$

$3 x^{2}+3+x^{4}+x^{2}=4 x^{2}+x^{4}+3$

$x^{2}=4$

Question 4

If a = 3 + b, prove that a³ – b³ – 9ab = 27.

Sol :

a=3+b

a-b=3

cubing both sides

$(a-b)^{3}=(3)^{3}$

$a^{3}-b^{3}-3 a b(a-b)=(9)^{3}$

$a^{3}-b^{3}-3 a b \times 3=27$

$a^{3}-b^{3}-9 a b=27$

Question 5

Simplify:

$\frac{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3}{(a-b)^3+(b-c)^3+(c-a)}$

Sol :

$\frac{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}$

$\frac{3\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right)}{3(a-b)(b-c)(c-a)}$

$\frac{(a-b)(a+b)(b-c)(b+ c)(c- a)(c+a)}{(a-b)(b-c)(c-a)}$

(a+b)(b+c)(c+a)

Question 6

If $a+\frac{1}{(a+2)}=0$ , then the value of $(a+2)^3+\frac{1}{(a+2)^3}$ is

(a) 6

(b) 4

(c) 3

(d) 2

Sol :

$a+\frac{1}{(a+2)}=0$

$\frac{a \times(a+2)+1}{a+2)} \Rightarrow 0$

$a^{2}+2 a+1=0$

(a+1)=0

a=-1

Putting value of a in $(a+2)^{3}+\frac{1}{(a+2)^{2}}$

$(-1+2)^{3}+\frac{1}{(-1+2)^{3}}$

$(1)^{3}+\frac{1}{(1)^{3}}$

=1+1=2

Question 7

If $a+\frac{1}{a}+2=0$  , then the value of a $\left(a^{37}-\frac{1}{a^{100}}\right)$

a) 0

(b) – 2

(c) 1

(d) 2

Sol :

$a+\frac{1}{a}+2 \Rightarrow 0$

$a^{2}+1+2a \Rightarrow 0$

$(a-1)^{2}=0$

$(a-1) \Rightarrow 0$

a=1

Putting the value of a in $a^{37}-\frac{1}{a^{100}}$

$(-1)^{37}-\frac{1}{(-1)^{100}}$ 

${-1-\frac{1}{1}}$

-1-1=-2

Question 8

If (a – 1)² + (b + 2)² + (c + 1)² = 0 then the value of 2a – 3b + 7c is

(a) 12

(b) 3

(c) – 11

(d) 1

Sol :

$(a-1)^{2}+(b+2)^{2}+(c+1)^{2} \Rightarrow 0$

$(a-1)^{2} \Rightarrow 0$

a=1

$(b+2)^{2}=0$

b+2=0

b=-2

$(c+1)^{2} \Rightarrow 0$

c+1=0

c=-1

Now putting the value of a,b,c in 2a-3b+7c

=2×1-3×-2+7×-1

=2+6-7=1

Question 9

If ax + by = 3, bx – ay = 4 and x² + y² = 1, then the value of a² + b² is

(a) – 1

(b) – 25

(c) 1

(d) 25

Sol :

ax+by=3...(i)

bx-ay=4....(ii)

$x^{2}+y^{2} \Rightarrow 1$...(iii)

Squaring equation (i) and (ii) and adding

$(a x+b y)^{2}+(b x-a y)^{2} \Rightarrow 3^{2}+4^{2}$

$a^{2} x^{2}+b^{2} y^{2}+2 a b x y+b^{2} x^{2}+a^{2} y^{2}-2 a b x y \Rightarrow 9+16$

$\left(a^{2}+b^{2}\right) x^{2}+\left(a^{2}+b^{2}\right) y^{2} \Rightarrow 25$

$\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right) \Rightarrow 25$

$\left(a^{2}+b^{2}\right) \times 1 \Rightarrow 25$ from equation (3)

$a^{2}+b^{2} \Rightarrow 25$

Question 10

If p + q = 10 and pq = 5, then the numerical value of $\frac{p}{q}+\frac{q}{p}$ will be:

(a) 22

(b) 18

(c) 16

(d) 20

Sol :

p+q⇒10 , p.q⇒5

p+q=10

Squaring both sides

$(p+q)^{2} \Rightarrow 10^{2}$

$p^{2}+q^{2}+2 p q \Rightarrow 100$

$p^{2}+q^{2}+2 \times 5 \Rightarrow 100$

$p^{2}+q^{2} \Rightarrow 100-10$

$p^{2}+q^{2} \Rightarrow 90$

Now $\frac{p}{q}+\frac{q}{p} \Rightarrow \frac{p^{2}+q^{2}}{p q}$

$\frac{90}{50}$

=1.8