Explore numerous Class 9 Maths MCQ and Ganita Manjari Class 9 Maths Chapter 4 Exploring Algebraic Identities MCQ Questions Online Test with Answers provided with detailed solutions by looking below.
MCQ on Exploring Algebraic Identities Class 9
Class 9 Maths Exploring Algebraic Identities MCQ
Choose the correct option from the given options:
Question 1.
2x – 3 = x, is a/an:
(a) Linear Equation
(b) Quadratic Equation
(c) Cubic Equation
(d) Identity
Solution:
(a) Linear Equation
Explanation:
It is a linear equation as it is true for only one value of variable which is given by
2x – 3 – x ⇒ 2x – x – 3 ⇒ x – 3.
Question 2.
(2u + 3u)2 = 4u2 + 12uv + 9u2, is a/an:
(a) Linear Equation
(b) Quadratic Equation
(c) Cubic Equation
(d) Identity
Solution:
(d) Identity
Explanation:
It is an Identity as it is true for all values of variable.
To check, we take arbitrary values u = 1 and v = 2,
LHS = (2 × 1 + 3 × 2)2 = (2 + 6)2 = (8)2 = 64
RHS = 4 × (1)2 + 12 × (1) × (2) + 9 × (2)2 = 4 × 1 + 12 × 1 × 2 + 9 × 4
= 4 + 24 + 36
= 64
So, LHS = RHS
Question 3.
The value of 122 + 142 – 2 × 132 is :
(a) 0
(b) 2
(c) -2
(d) 1
Solution:
(b) 2
Explanation:
Since, (n – 1)2 + (n + 1)2 – 2n2 = (n2 – 2n+ 1) + (n2 + 2n + 1) – 2n2
= 2n2 + 2 – 2n2 = 2.
Therefore, 122 + 142 – 2 × 132 = 2.
Question 4.
If (2x + 3)2 = ax2 + bx + c, then the value of a+ b + c is:
(a) (x + 2)(x + 2)
(b) x2 + 22
(c) (x – 2)(x + 2)
(d) (x + 2)(x + 1)
Solution:
(a) (x + 2)(x + 2)
Explanation:
Here, x2 + 4x + 4 = x2 + 2 × x × 2 + 22 = (x + 2)2, (using (a + b)2 = a2 + 2ab + b2)
= (x + 2)(x + 2).
Question 5.
If (2x + 3)2 = ax2 + bx + c, then the value of a + b + c is :
(a) 10
(b) 12
(c) 21
(d) 25
Solution:
(d) 25
Explanation:
Here, (2x + 3)2 = 4x2 + 12x + 9
Comparing it with ax2 + bx + c we get a = 4, b = 12 and c = 9.
So, a + 6 + c = 4 + 12 + 9 = 25.
Exploring Algebraic Identities MCQ Class 9
Question 6.
If (x + 2)(x + 3) = x2 + kx + 6, then the value of k is :
(a) 2
(b) 3
(c) 5
(d) -1
Solution:
(c) 5
Explanation:
Here, (x + 2)(x + 3) = x2 + (2 + 3)x + 2x3, (using (x + a)(x + b) = x2 + (a + b)x + a × b)
= x2 + 5x + 6.
Comparing it with x2 + kx + 6 we get k = 5.
Question 7.
If (x + 4)(x + 6) = x2 + 10x + 6k, then the value of k is :
(a) 2
(b) 4
(c) 5
(d) – 1
Solution:
(c) 5
Explanation:
Here, (x + 4)(x + 6) = x2 + (4 + 6)x + 4×6, (using (x + a)(x + b) = x2 + (a + b)x + a × b)
= x2 + 10x + 24.
Comparing it with x2 + 10x + 24 we get 6k = 24 ⇒ k = \(\frac{24}{6}\) = 4.
Question 8.
If (5x – 2y)2 = ax2 + 106xy + cy2, then the value of 2a + 36 – 4c is :
(a) -28
(b) 60
(c) 28
(d) 26
Solution:
(c) 28
Explanation:
Here, (5x – 2y)2 = 25x2 – 20xy + 4y2
Comparing it with ax2 + 106xy + cy2 we get a
b = \(\frac{-20}{10}\) = -2 and c = 4
So, 2a + 3b – 4c = 2 × 25 + 3 × (-2) – 4 × 4 = 50 – 6 – 16 = 28
Question 9.
What is the side of the square whose area is 4a2 + 32uv + 64c2 square units?
(a) (2u + u)
(b) (2u – 8v)
(c) (2u + 8v)
(d) (u – 8u)
Solution:
(c) (2u + 8v)
Explanation:
The area of the square = 4u2 + 32uv + 64c2 = (2u)2 + 2 × (2a) × (8c) + (8c)2
= (2a + 8c)2 …(i)
Let the side of the square be x units.
So, volume of the square = x2 ……..(ii)
⇒ x2 = (2a + 8c)2 (using (i) and (ii))
⇒ x = 2a + 8c.
Question 10.
9a2 + b2 + 4c2 – 6ab + 12ac – 4bc =
(a) (3a + b + c)2
(b) (3a + b + 2c)2
(c) (a + b + 2c)2
(d) (3a – b + 2c)2
Solution:
(d) (3a – b + 2c)2
Explanation:
Here, 9a2 + b2 + 4c2 – 6ab + 12ac – 4bc
= (3a)2 + (- b)2 + (2c)2 + 2 × (3a) × (- b) + 2 × (3a) × (2c) + 2 × (- b) × (2c)
= {3a + (- b) + 2c}2
= ( 3a – b + 2c)2
Question 11.
What is the side of the cube whose volume is a3 + 6a2b + 12ab2 + 8b3 cubic units?
(a) (2a + b)
(b) (2a – b)
(c) (a + 2b)
(d) (a – 2b)
Solution:
(c) (a + 2b)
Explanation:
Here, volume of the cube = a3 + 6a2b + 12ab2 + 8b3
= a3 + 3 × a2 × (2b) + 3 × a × (2b)2 + (2b)3
= (a + 2b)3 …(i)
Let the side of the cube be x units.
So, volume of the cube = x3 …(ii)
⇒ x3 = (a + 2b)3 (using (i) and (ii))
⇒ x = a + 26.
Question 12.
The value of ( -5)3 + (2)3 + (3)3 is :
(a) 0
(b) 38
(c) -90
(d) 90
Solution:
(c) -90
Explanation:
Let a = – 5, 6 = 2,c = 3 ⇒ a + 6 + c = 0
We know that if a + b + c – 0,then a3 + b3 + c3 = 3abc.
So, (- 5)3 + (2)3 + (3)3 = 3 × (- 5) × (2) × (3)
= – 90.
Question 13.
Simplest form of \(\frac{x^2-7 x+12}{\left(5 x^2+5 x-100\right)}\), 5x2 + 5x – 100 ≠ 0 is :
(a) \(\frac{x-3}{5(x+5)}\)
(b) \(\frac{x-4}{x+5}\)
(c) \(-\frac{x-3}{5(x+5)}\)
(d) \(\frac{x-3}{x+5}\)
Solution:
(a) \(\frac{x-3}{5(x+5)}\)
Explanation:
im-1
= \(\frac{x-3}{5(x+5)}\) which is the required simplest form.
Question 14.
Simplest form of \(\bar{x}\), 4x2 – 1 ≠ 0 is :
(a) \(\frac{2 x-1}{2 x+1}\)
(b) \(\frac{2 x+1}{2 x-1}\)
(c) \(-\frac{2 x-1}{2(x+1)}\)
(d) \(\frac{2 x-3}{2 x+1}\)
Solution:
(b) \(\frac{2 x+1}{2 x-1}\)
Explanation:
im-2
which is the required simplest form.
Question 15.
The value of 4x2 + 9y2 + 12xy – 4, when 2x + 3y = 6 is :
(a) 32
(b) 31
(c) 30
(d) -34
Solution:
(a) 32
Explanation:
Here, 2x + 3y – 6, x2 is :
Therefore, (2x + 3y)2 = 4x2 + 9y2 + 12xy
⇒ (6)2 = 4x2 + 9y2 + 12xy
Now, 4x2 + 9y2 + 12xy – 4 = (6)2 – 4 = 36 – 4 = 32.
Question 16.
The value of x3 + y3 – 12xy + 64, when x + y = – 4 is :
(a) 2
(b) 1
(c) 0
(d) – 4
Solution:
(c) 0
Explanation:
Here, x + y = – 4,
Therefore, (x + y)3 = x3 + y3 + 3xy(x + y)
⇒ (-4)3 = x3 + y3 + 3xy( -4)
⇒ -64 = x3 + y3 – 12xy
⇒ x3 + y3 – 12xy + 64 = 0.
Question 17.
The value of x3 – 8y3 – 36xy – 216, when x = 2y + 6 is :
(a) 0
(b) -1
(c) – 3
(d) – 4
Solution:
(a) 0
Explanation:
Here, x = 2y + 6 ⇒ x – 2y = 6,
Therefore, (x – 2y)3 = x3 – 8, y3 – 3x(2y)(x – 2y)
⇒ (6)3 = x3 – 8,y3 – 3x(2y)(6)
⇒ 216 = x3 – 8y3 – 36xy
⇒ x3 – 8y3 – 36xy – 216 =0.
Question 18.
If x + \(\bar{x}\), x ≠ 0 then the value of x2 + \(\frac{1}{x^2}\) is :
(a) \(\frac{3}{2}\)
(b) \(
(c) [latex]\frac{1}{4}\)
(d) -4
Solution:
(c) \(\frac{1}{4}\)
Explanation:
im-3
Question 19.
If x – \(\frac{1}{x}\) = \(\frac{4}{5}\); x ≠ 0 then the value of x2 + \(\frac{1}{x^2}\) is:
(a) \(\frac{18}{25}\)
(b) \(-\frac{18}{25}\)
(c) \(\frac{66}{25}\)
(d) \(-\frac{64}{25}\)
Solution:
(c) \(\frac{66}{25}\)
Explanation:
im-4
Question 20.
If x – 2 is a factor of px2 + 5x + p, then the value of p is :
(a) -2
(b) -1
(c) 2
(d) 1
Solution:
(a) -2
Explanation:
Here, x – 2 is a factor of px2 + 5x + p
Therefore, p( 2)2 + 5x2 + p = 0
⇒ 4p + 10 + p – 0
⇒ 5p + 10 = 0
⇒ p = – \(\frac{10}{5}\) = – 2.
Question 21.
If x – 1 is a factor of 2x2 + kx – 3, then the value of ‘k’ is:
(a) -2
(b) -1
(c) 2
(d) 1
Solution:
(d) 1
Explanation:
Here, x – 1 is a factor of 2x2 + kx – 3
Therefore, 2(1)2 + k × (1) – 3 = 0
⇒ 2 + k – 3 = 0
⇒ k – 1 = 0
⇒ k = 1.
Question 22.
3x – 2 = 5x, is a/an:
(a) Linear Equation
(b) Quadratic Equation
(c) Cubic Equation
(d) Identity
Solution:
(a) Linear Equation
Question 23.
(\(\frac{1}{2}\)x – 3y)2 = \(\frac{1}{4}\) x2 – 3xy + 9, is a/an:
(a) Linear Equation
(b) Quadratic Equation
(c) Cubic Equation
(d) Identity
Solution:
(b) Quadratic Equation
Question 24.
(\(\frac{2}{5}\)a + \(\frac{1}{3}\)b)2 = \(\frac{4}{25}\)a2 + \(\frac{4}{15}\)ab + \(\frac{1}{2}\)b2 is a/an:
(a) Linear Equation
(b) Quadratic Equation
(c) Cubic Equation
(d) Identity
Solution:
(d) Identity
Question 25.
The value of 342 + 362 – 2 × 352 is :
(a) 0
(b) 2
(c) -2
(d) 1
Solution:
(b) 2
Question 26.
x2 – 6x + 9 =
(a) (x – 3)(x – 3)
(b) x2 + 32
(c) (x – 3)(x + 3)
(d) (x + 3)(x – 1)
Solution:
(a) (x – 3)(x – 3)
Question 27.
If (\(\frac{2}{3}\) x + \(\frac{1}{2}\))2 = : ax2 + bx + c. then the value of a + b + c is :
(a) \(\frac{43}{36}\)
(b) \(\frac{49}{36}\)
(c) \(\frac{36}{49}\)
(d) \(\frac{53}{36}\)
Solution:
(b) \(\frac{49}{36}\)
Question 28.
If (x + 5)(x – 2) = x2 + kx + 6. , then the value of k is :
(a) 2
(b) 3
(c) 5
(d) -3
Solution:
(b) 3
Question 29.
If (x – 7)(x + 3) = x2 – 4x + 3k, then the value of k is :
(a) 7
(b) 0
(c) -7
(d) -1
Solution:
(c) -7
Question 30.
If (3x – 4y)2 = ax2 + 12bxy + cy2, then the value of 2a + 36 – 4c is :
(a) -54
(b) 52
(c) -52
(d) 54
Solution:
(c) -52
Question 31.
What is the side of the square whose area is 4a2 – 12a6 + 962 square units?
(a) (2a + b)
(b) (2a – b)
(c) (2a – 35)
(d) (a – 36)
Solution:
(c) (2a – 35)
Exploring Algebraic Identities Class 9 Assertion and Reason Questions
Direction: A statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from the following options.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Question 1.
Assertion (A): 2x – 3 = 0 is not an identity.
Reason (R): 2x – 3 = 0 is true for only one value of x.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
2x – 3 = 0 is not true for all value of variable.
As for x = 0, LHS = 2 × 0 – 3 = 0 – 3= – 3 ≠ RHS. So, Assertion (A) is true.
2x – 3 = 0 ⇒ 2x = 3 ⇒ x = \(\frac{3}{2}\). So, it is true for only one value of x. So, Reason (R) is true and Reason (R) is the correct explanation for the Assertion (A).
Question 2.
Assertion (A): (u + 2)2 = u2 + 4u + 4 is an identity.
Reason (R): (u + 2)2 = u2 + 4u + 4 is true for u – 0.
Solution:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Explanation:
Here, (u + 2)2 = u2 + 2 × u × 2 + 22
⇒ (u + 2)2 = u2 + 4u + 4, is true for all values of variable u.
So, Assertion (a) is true.
For u = 0, LHS = (0 + 2)2 = (2)2 = 4,RHS = (0)2 + 4(0) + 4 = 4.
Hence, LHS = RHS. So, Reason (R) is true but it does not explain the existence of the Assertion (A).
Question 3.
Assertion (A): 232 + 252 – 2 × 242 = 2
Reason (R): (a + b)2 – (a – b)2 = 4ab.
Solution:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Explanation:
Since, (n – 1)2 + (n + 1)2 – 2n2 = (n2 – 2n + 1) + (n2 + 2n + 1) – 2n2 = 2.
Therefore, (24 – 1)2 + (24 + 1)2 – 2 x (24)2 = 2.
So, Assertion (a) is true.
Also, (a + b)2 – (a – b)2 = (a2 + 2ab + b2) – (a2 – 2ab + b2) = 4 ab.
So, Reason (R) is true.
But Reason (R) is not the correct explanation for the Assertion (A).
Question 4.
Assertion (A): If x + \(\frac{1}{x}\) = 7, then x2 + \(\frac{1}{x^2}\) = 49
Reason (R): (x + y)2 = x2 + 2xy + y2.
Solution:
(d) Assertion (A) is false but Reason (R) is true.
Explanation:
We know that
im-5
– 2 = (7)2 – 2 = 49 – 2 = 47.
So, Assertion (A) is false.
We know that (x + y)2 = x2 + 2xy + y2 is an identity.
So, Reason (R) is true.
Question 5.
Assertion (A): 62 – 42 = 4 × 5 × 1.
Reason (R): (n + 1)2 – (n – 1)2 = 4 × n × 1.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation:
Here, 62 – 42 = (5 + 1)2 – (5 – 1)2 = 4 × 5 × 1.
Therefore, Assertion (A) is true.
Also, (n + 1)2 -(n – 1)2 = (n2 + 2n + 1) – (n2 – 2n + 1) = 4 × n × 1.
So, Reason (R) is true and Reason (R) explains the existence of Assertion (A).
Question 6.
Assertion (A): If x – 3 is a factor of x2 – kx + 9, then k = 2.
Reason (R): For k = 2, x2 – kx + 1 becomes (x – 1)2.
Solution:
(d) Assertion (A) is false but Reason (R) is true.
Explanation:
If x – 3 is a factor of x2 – kx + 9, then (3)2 – k x (3) + 9 = 0
Hence, 9 – 3k + 9 = 0
⇒ – 3k + 18 = 0
⇒ – 3k = – 18
Therefore, Assertion (A) is false.
Also, for k – 2, x2 – kx + 1 can be written as x2 – 2x + 1 = (x – 1)2.
So, Reason (R) is true.
Question 7.
Assertion (A): \(\frac{1}{3}\) x – 7 = 0 is not an identity.
Reason (R): 2x – 3 = 0 is true for only one value of x.
Solution:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Question 8.
Assertion (A): (b + 7)2 = b2 + 14b + 49 is an identity.
Reason (R): (b + 2)2 = b2 + 4b + 4 is true for b = 0.
Solution:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Question 9.
Assertion (A): 34\(\frac{1}{2}\) + 36\(\frac{1}{2}\) – 2 x 35\(\frac{1}{2}\) = 2.
Reason (R): (n + 1)\(\frac{1}{2}\) + (n – 1)\(\frac{1}{2}\) = 2n\(\frac{1}{2}\) + 2.
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 10.
Assertion (A): If x – \(\frac{1}{x}\) = \(\frac{4}{3}\), then x2 + \(\frac{1}{x^2}=\frac{16}{9}\)
Reason (R): (x – y)2 = x2 – 2xy + y2.
Solution:
(d) Assertion (A) is false but Reason (R) is true.
Question 11.
Assertion (A): \(\left(\frac{3}{4}\right)^2-\left(\frac{1}{4}\right)^2\) = 1.
Reason (R): a2 – b2 = (a + b)(a – b).
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Question 12.
Assertion (A): If x – 1 is a factor of x2 – kx + 13, then k = 2.
Reason (R): For k = 2, x2 – kx + 1 becomes (x – 1)2.
Solution:
(d) Assertion (A) is false but Reason (R) is true.
Question 13.
Assertion (A): If x – \(\frac{1}{5}\) is a factor of x2 – \(\frac{2}{5}\) x + \(\frac{1}{25}\)
Reason (R): For x = \(\frac{1}{5}\) the value of x2 – \(\frac{2}{5}\)x + \(\frac{1}{25}\) = 0
Solution:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).