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