OP Malhotra Class 9 Maths Solutions Chapter 3 Expansions Chapter Test

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S Chand Class 9 ICSE Maths Solutions Chapter 3 Expansions Chapter Test

Question 1.
The coefficient of x in the product (2 – 3x) (5 – 2x) is
(a) 19
(b) – 19
(c) 15
(d) 6
Solution:
(b) – 19
(2 – 3x) (5 – 2x)
Hence, term x has – 4x – 15x = – 19x
∴ Co-efficient of x = – 19

Question 2.
If 3x⁴ + kx² – 8 = (3x² – 2) (x² + 4) for all x, then the value of k is:
(a) – 2
(b) 12
(c) 10
(d) – 8
Solution:
(c) 10
3x⁴ + kx² – 8 = (3x² – 2) (x² + 4)
⇒ 3x⁴ + kx² – 8 = 3x⁴ + 12x² – 2x² – 8
⇒ 3x⁴ + kx² – 8 = 3x⁴ + 1ox² – 8
Comparing, we get
k = 10

Question 3.
The coefficient of x² in (3x + x³) (x + \(\frac { 1 }{ x }\)) is
(a) 3
(b) 1
(c) 4
(d) 2
Solution:
(c) 4
(3x + x³) (x + \(\frac { 1 }{ x }\))
= 3x² + 3 + x⁴ + x² = 4x² + x⁴ + 3
Co-efficient of x² = 4

Question 4.
If a = 3 + b, prove that a³ – b³ – 9ab = 27.
Solution:
a = 3 + b
a – b = 3
Cubing both sides,
(a – b)³ = (3)³
⇒ a³ – b³ – 3ab(a – b) = 27
⇒ a³ – b³ – 3ab x 3 = 27
⇒ a³ – b³ – 9ab = 27
Hence proved.

Question 5.
Simplify:
\(\frac{\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3}{(a-b)^3+(b-c)^3+(c-a)}\)
Solution:
\(\frac{\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3}{(a-b)^3+(b-c)^3+(c-a)}\)
= \(\frac{3\left(a^2-b^2\right)\left(b^2-c^2\right)\left(c^2-a^2\right)}{3(a-b)(b-c)(c-a)}\)
= \(\frac{(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{(a-b)(b-c)(c-a)}\)
= (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
Solution:
(d) 2
a + \(\frac{1}{(a+2)}\) = 0 ⇒ a² + 2a + 1 = 0
⇒ (a + 1) = 0 ⇒ a = – 1
Putting the value a in (a + 2)³ + \(\frac{1}{(a+2)^3}\)
= (- 1 + 2)³+ \(\frac{1}{(-1+2)^3}\) = 1³ + \(\frac{1}{(1)^3}\)
= 1 + 1 = 2

Question 7.
If a + \(\frac { 1 }{ a }\) + 2 = 0, then the value of a \(\left(a^{37}-\frac{1}{a^{100}}\right)\) is
(a) 0
(b) – 2
(c) 1
(d) 2
Solution:
(b) – 2
a + \(\frac { 1 }{ a }\) + 2 = 0 ⇒ a² + 1 +2a = 0
⇒ (a + 1)² = 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
Solution:
(d) 1
∵ (a – 1)² + (b + 2)² + (c + 1)² = 0
∴ (a – 1) = 0, (b + 2) = 0 and (c + 1) = 0
⇒ a = 1, b = – 2, c = – 1
Now, 2a – 3b + 7c = 2 x 1 – 3 x (- 2) + 7 x (- 1)
= 2 + 6 – 7 = 8 – 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
Solution:
(d) 25
ax + by = 3
bx – ay = 4 and x² + y² = 1
Squaring and adding, we get
(ax + by)² + (bx – ay)² = 3² + 4²
a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy
= 9 + 16
⇒ (a² + b²)x² + (a² + b²)y² = 25
(a² + b²) (x² + y²) = 25
⇒ (a² + b²) x 1 = 25 (x² + y² = 1)
∴ a² + b² = 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
Solution:
(b) 18
p + q = 10, pq = 5
Squaring, p + q = 10
p² + q² x 2pq = 100
⇒ p² + q² + 2 x 5 = 100
⇒ p² + q² + 10 = 100
p² + q² = 100 – 10 = 90
Now \(\frac{p}{q}+\frac{q}{p}=\frac{p^2+q^2}{p q}\)
= \(\frac { 90 }{ 5 }\) = 18