WBBSE Class 9 Maths Algebra Chapter 5 Factorisation Multiple Choice Questions

WBBSE Class 9 Maths Algebra Chapter 5 Factorisation Multiple Choice Questions

Example 1. If a² – b² = 11 × 9 and a & b are positive integers (a > b) then

1. a = 11, b = 9
2. a = 33, b = 3
3. a = 10, b = 1
4. a = 100, b = 1

Solution: The correct answer is 3. a = 100, b = 1

⇒ a² – b² = 11 x 9, (a + b) (a – b) = 11 × 9

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⇒ If a + b = 11, a – b = 9 then a + b + (a – b) = 11 + 9, 2a = 20

⇒ or, a = 10

∴ b = 11 – a = 11 – 10 = 1

a = 10, b = 1

Example 2. If \(\frac{a}{b}+\frac{b}{a}\) = 1 then one of the value of a³ + b³ is

1. 1
2. a
3. b
4. 0

Solution: The correct answer is 4. 0

\(\frac{a^2+b^2}{a b}\), a² + b² – ab = 0 then, a³ + b³ = (a + b) (a² + b² – ab) = (a + b) x 0 = 0

The value of a³ + b³ is  (a + b) (a² + b² – ab) = (a + b) x 0 = 0

Example 3. The value of 25² – 75² + 50² + 3 x 25 x 75 x 50 is

1. 150
2. 0
3. 25
4. 50

Solution: The correct answer is 2. 0

⇒ Here a + b + c = 25 + (-75) + 50 = 0

∴ a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca) = 0

The value of 25² – 75² + 50² + 3 x 25 x 75 x 50 is 0.

Example 4. If a + b + c = 0 then value of \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\) is

1. 0
2. 1
3. -1
4. 3

Solution: The correct answer is 4. 3

\(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}=\frac{a^3+b^3+c^3}{a b c}=\frac{3 a b c}{a b c}\) as a + b + c = 0 = 3

Value of \(\frac{a^2}{b c}+\frac{b^2}{c a}+\frac{c^2}{a b}\) is 3.

Example 5. If x² – px + 12 = (x – 3)(x -a) is an identity then the value of a and p are respectively

1. a = 4, p = 7
2. a = 7, p = 4
3. a = 4, p = -7
4. a = -4, p = 7

Solution: The correct answer is 1. a = 4, p = 7

⇒ x² – px + 12 = x² – (3 + a) x + 3a

⇒ By comparing both sides as co-efficients of x and constant terms, we get

⇒ 3a = 12, a = 4

∴ p = 3 + a = 3 + 4 = 7

The value of a and p are respectively 4 and 7.

Example 6. If a^x.a^y.a^z then x³ + y³ + z³

1. 0
2. -1
3. 3
4. 3xyz

Solution: The correct answer is 4. 3xyz

⇒ a^x.a^y.a^z = a^x+y+z= a⁰, x + y + z = 0

∴ x³+ y³+ z³ = 3xyz

Example 7. \(\frac{(x-y)^3+(y-z)^3+(z-x)^3}{(x-y)(y-z)(z-x)}=\)

1. 3xyz
2. 3
3. xyz
4. 0

Solution: The correct answer is 2. 3

⇒ Let x – y = a, y – z = b, z – x = c

∴ x – y + y – z + z – x = 0,

∴ a + b + c = 0

∴ (x – y)³ + (y – z)³ + (z – x)³ = 3 (x − y)(y – z)(z – x)

⇒ or, \(\frac{3(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}\) = 3

Example 8. If a³ + b³ + c³ – 3abc = 0 and a + b + c ≠ 0 then

1. a = 2b + c
2. b = 2c + a
3. a = b = c
4. c = 2a + b

Solution: The correct answer is 3. a = b = c

⇒ a³ + b³ + c³ – 3abc = 0

⇒ or, \(\frac{1}{2}\)(a+b+c) {(a – b)² + (b – c)² + (c – a)²} = 0

⇒ but, a + b + c ≠ 0

∴ (a – b)² + (b – c)² + (c – a)² ≠ 0

∴ a – b = 0 = b – c = c – a

∴ a = b = c.

Example 9. If x² – px + 8 = (x – 2) (x – 4) be an identity then, p =?

1. 0
2. 2
3. 4
4. 6

Solution: The correct answer is 4. 6

⇒ x² – px + 8 = x² – (4 + 2)x + 8

⇒ By comparing, coefficients of x & constant terms, p = 4 + 2 = 6

p= 6.

Example 10. The number of factors of a⁶ – b⁶ is

1. 1
2. 2
3. 3
4. 4

Solution: The correct answer is 4. 4

⇒ a⁶ – b⁶ = (a³)² – (b³)² = (a³ + b³)(a³ – b³)

= (a + b) (a² – ab + b²) (a – b) (a² + ab + b²)

The number of factors of a⁶ – b⁶ is 4