Ch- Test of Expansions Class 9 OP Malhotra ICSE Maths Solutions Ch-3.  We Provide Step by Step Solutions / Answer of Expansions OP Malhotra Maths. Visit official Website CISCE  for detail information about ICSE Board Class-9 Mathematics.

Expansions Class 9 OP Malhotra Ch-Test ICSE Maths Solutions Ch-3

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-3	Expansions
Writer	OP Malhotra
Ch-Test	Extra Practice Questions
Edition	2026-2027

Ch-Test on Expansions

Expansions Class 9 OP Malhotra ICSE Maths Solutions Ch-3

Que-1: The coefficient of x in the product (2 – 3x) (5 – 2x) is

(a) 19
(b) – 19
(c) 15
(d) 6

Sol: (b) – 19
(2 – 3x) (5 – 2x)
Hence, term x has – 4x – 15x = – 19x
∴ Co-efficient of x = – 19

Que-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: (c) 10
3x^4 + kx² – 8 = (3x² – 2) (x² + 4)
⇒ 3x^4 + kx² – 8 = 3x^4 + 12x² – 2x² – 8
⇒ 3x^4 + kx² – 8 = 3x^4 + 1ox² – 8
Comparing, we get
k = 10

Que-3: The coefficient of x² in (3x + x³) (x + 1x) is

(a) 3
(b) 1
(c) 4
(d) 2

Sol: (c) 4
(3x + x³) (x + 1x)
= 3x² + 3 + x4 + x² = 4x² + x4 + 3
Co-efficient of x² = 4

Que-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
⇒ a³ – b³ – 3ab(a – b) = 27
⇒ a³ – b³ – 3ab x 3 = 27
⇒ a³ – b³ – 9ab = 27
Hence proved.

Que-5: Simplify : [(a²−b²)³+(b²−c²)³+(c²−a²)³] / [(a−b)³+(b−c)³+(c−a)]

Sol: [(a²−b²)³+(b²−c²)³+(c²−a²)³] / [(a−b)³+(b−c)³+(c−a)]
= [3{(a²−b²)(b²−c²)(c²−a²)}] / [3{(a−b)(b−c)(c−a)}]
= [(a−b)(a+b)(b−c)(b+c)(c−a)(c+a)] / [(a−b)(b−c)(c−a)]
= (a + b) (b + c) (c + a)

Que-6: If a + {1/(a+2)} = 0, then the value of (a + 2)³ + {1/(a+2)³} is

(a) 6
(b) 4
(c) 3
(d) 2

Sol: (d) 2
a + {1/(a+2)} = 0
⇒ a² + 2a + 1 = 0
⇒ (a + 1) = 0 ⇒ a = – 1
Putting the value a in (a + 2)³ + {1/(a+2)³}
= (- 1 + 2)³ + {1/(−1+2)³} = 1³ + 1/(1)³
= 1 + 1 = 2

Que-7: If a + (1/a) + 2 = 0, then the value of a [a^37 − {1/(a^100)}] is

(a) 0
(b) – 2
(c) 1
(d) 2

Sol: (b) – 2
a + (1/a) + 2 = 0
⇒ a² + 1 +2a = 0
⇒ (a + 1)² = 0 ⇒ a = – 1
Putting the value of a in a^37 − {1/(a^100)}
= (-1)^37 – {1/(−1)^100}
= – 1 – (1/1)
= – 1 – 1 = – 2

Que-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: (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

Que-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: (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

Que-10: If p + q = 10 and pq = 5, then the numerical value of (p/q) + (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 (p/q) + (q/p) = (p²+q²)/pq
= 90/5 = 18

— : End of Expansion 9 OP Malhotra Ch-Test ICSE Maths Step by step Solutions :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-9 Maths

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