Exploring Algebraic Identities Class 9 MCQ Maths Chapter 4

Practicing Class 9 Maths MCQ and Ganita Manjari Class 9 Maths Chapter 4 Exploring Algebraic Identities MCQ Questions Online Test with Answers daily helps in time management.

MCQ on Exploring Algebraic Identities Class 9

Exploring Algebraic Identities MCQ Class 9

Class 9 Maths Exploring Algebraic Identities MCQ

Question 1.
Value of x, (x + 2)² = 49
(a) 9
(b) 7
(c) 2
(d) 5
Answer:
(d) 5

Explanation:
Given, (x + 2)² =49
⇒ x² + 4 + 4x = 49
⇒ x² + 4x = 45
⇒ x² + 4x – 45 = 0
⇒ x² + 9x – 5x – 45 = 0
⇒ x(x + 9) – 5(x + 9) = 0
⇒ (x + 9)(x – 5) = 0
⇒ x = -9, 5

Alternate Method
(x + 2)² = 49
⇒ x+2 =+7
⇒ x =±7 – 2
⇒ x = -9, 5

Question 2.
Value of x,(x +2)2 – 4 = (3x – 5) + x²
(a) 5
(b) -5/7
(c) – 5
(d) 5/7
Answer:
(c) – 5

Explanation:
(x + 2)² – 4 =(3x – 5) ± x²
⇒ x² + 4x = 3x – 5 + x²
⇒ 4x – 3x = -5
⇒ x = -5

Question 3.
If x² + \(\frac{1}{x^2}\) = 4, then value of x – \(\frac{1}{x}\) is
(a) 2
(b) 1
(c) 0
(d) 1/2
Answer:
(a) 2

Question 4.
If x – \(\frac{1}{x}\) = 2, then value of x² + \(\frac{1}{x^2}\) is
(a) 16
(b) 20
(c) 18
(d) 14
Answer:
(c) 18

Explanation:
(x – \(\frac{1}{x}\))² = 4²
⇒ x² + \(\frac{1}{x^2}\) = 16 + 2 = 18

Question 5.
Simplify (x + y)³ – (x – y)³
(a) 3(y³ + 3x²y)
(b) 2(y³ + 3x²y)
(c) (y³+ 2x²y)
(d) 2y³ + 3x²y
Answer:
(b) 2(y³ + 3x²y)

Question 6.
Value of x,(x – 7)(x – 9) = x² – 1
(a) 2
(b) 6
(c) -4
(d) 4
Answer:
(d) 4

Explanation:
(x – 7)(x – 9) = x² – 1
⇒ x² – 9x – 7x + 63 = x² – 1
⇒ – 16x + 63 = – 1
⇒ x = 4

Question 7.
Which one is the common factor of 9a²b⁴ + 12a³b³.
(a) 3a³b³
(b) 3ab³
(c) 3a²b³
(d) 3a²b²
Answer:
(c) 3a²b³

Explanation:
9a²b⁴ + 12a³b³.
= 3a²b³(3b + 4a)
3a²b³

Question 8.
By using suitable formula, value of 102² ….98² is
(a) 800
(b) 200
(c) 1800
(d) 400
Answer:
(a) 800

Explanation:
(102 + 98)(102 – 98) = 200 × 4 = 800

Question 9.
Without multip1ication 48 × 52 is equal to
(a) 2486
(b) 2496
(c) 2497
(d) 2476
Answer:
(b) 2496

Explanation:
(50 – 2)(50 + 2) = (50)² – 2²
= 2500 – 4
= 2496

Question 10.
Factorise 4×2² – 12ax – y² – z² – 2yz + 9a².
(a) (2x + y + z – 3a)(2x – y – z – 3a)
(b) (2x + y + z – 3a)(2x + y – z – 3a)
(c) (2x – y + z + 3a)(2x – z + y – 3a)
(d) (2x + y – z + 3a)(2x + y – z – 3a)
Answer:
(a) (2x + y + z – 3a)(2x – y – z – 3a)

Explanation:
(4x² -12ax + 9a²) – (y² +2yz + z²) (2x – 3o)² – (y + z)²
(2x + y + z – 3a)(2x – y – z – 3a)

Question 11.
If \(\frac{x}{y}+\frac{y}{x}\) = 1, then x³ + y³ is
(a) 1
(b) 2y³
(c) -2y³
(d) 0
Answer:
(d) 0

Explanation:
Given, \(\frac{x}{y}+\frac{y}{x}\) = 1
⇒ x² + y² = xy …(i)
We know that x³ + y³ = (x + y)(x² + y² – xy)
= (x + y)(xy – xy) [from Eq. (i)]
= 0

Question 12.
Simplify
\(\frac{0.87 \times 0.87 \times 0.87+0.13 \times 0.13 \times 0.13}{0.87 \times 0.87-0.87 \times 0.13+0.13 \times 0.13}\)
(a) 0.87
(b) 1
(c) 0.74
(d) 0
Answer:
(b) 1

Explanation:
\(\frac{1}{2}\)
= 0.87 + 0.13
[∵ (x³ + y³) = (x + y)(x² + y² – xy)
= 1

Exploring Algebraic Identities Class 9 Assertion and Reason Questions

Direction (Q. Nos. 1-4) In the questions given below, there are two statements marked as Assertion (A) and Reason (R). Read the statements and choose the correct option.

Question 1.
Assertion (A): If x² + 4x +a = (x + 2)², then the value of a is 4.
Reason (R): (a+b)² = a² + 2ab + b²
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
Given,
x² + 4x + a = (x + 2)
⇒ x² + 4x + a = x² + 2 × x × 2 + 2²
[∵ (a + b)² = a² + 2ab + b²]
⇒ x² + 4x + a = x² + 4x + 4
On comparing both sides, we get
a = 4
So, Assertion is true.
Clearly, Reason is true.
Hence, both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 2.
Assertion (A): If 2x – 5y = 10 and xy = 15 then the value of 8x³ – 125y³ is 5300.
Reason (R): (a – b)³ = a³ – b³ – 3ab(a – b)
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(d) A is false but R is true.

Explanation:
We have, 2x – 5y = 10 …(i)
and xy = 15 …(ii)
On cubing both sides of Eq. (i), we get
(2x – 5y)³ = 10³
⇒ (2x)³ – (5y)³ – 3 (2x) (5y) (2x – 5y) = 1000
[∵ (a – b)³ = a³ – b³ – 3ab(a – b)]
⇒ 8x³ – 125y³ – 30xy (2x – 5y) = 1000
⇒ 8x³ -125y³ – 30 (15)(10) = 1000
[from Eqs. (i) and (ii)]
⇒ 8x³ – 125y³ -4500 = 1000
⇒ 8x³ – 125y³ =1000 + 4500 = 5500
So, Assertion is false.
Clearly, Reason is true.

Question 3.
Assertion (A): The value of (—14)³ + (9)³ + (5)³ is -1890.
Reason (R) If a + b + c = 0 then a³ + b³ + c³ = 3abc.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
We have, (-14)³ + (9)³ + (5)³.
We know that if a + b + c = 0, a³ + b³ + c³ = 3abc
Here, (-14) + 9 + 5 = 0
(-14)³ + (9)³ + (5)³ = 3(-14) (9) (5) = -1890
So, Assertion is true.
Clearly, Reason is true.
Hence, both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Question 4.
Assertion (A): The factorisation of ax⁵ – \(\frac{27 a x^2}{y^3}\) is ax² \(\left(x-\frac{3}{y}\right)\left(x^2+\frac{3 x}{y}+\frac{9}{y^2}\right)\)
Reason(R): a³ – b³ = (a + b)(a² + ab + b²)
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(c) A is true but R is false.

Explanation:
We have,