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Class 12 Mathematics
CBQ Practice

Competency Based Questions · 3 chapters · 6 CBQ sets

Question types:Case StudySource BasedAssertion–Reason
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Ch 1

Relations and Functions

2 sets
CBQ 1Case StudyModelling Real-Life Relationships4 marks

Read the passage

A school has students and their roll numbers. The relation R defined on the set of students by 'student a is related to student b if they belong to the same class' is an equivalence relation. The teacher of mathematics wants to use this context to explain types of relations. She defines various relations on the set A = {1, 2, 3, 4, 5} and asks students to classify them. She also discusses functions, explaining that a function is a special type of relation where each input has exactly one output. She uses the function f: R → R defined by f(x) = 2x + 3 as an example, showing it is both one-one (injective) and onto (surjective), making it a bijective function.
1

A relation R on a set A is said to be an equivalence relation if it is:

1M
(A)Only reflexive
(B)Reflexive and symmetric only
(C)Reflexive, symmetric, and transitive
(D)Symmetric and transitive only
2

For the function f(x) = 2x + 3, f is one-one because:

1M
(A)f(x) = 0 has a solution
(B)f(x₁) = f(x₂) implies x₁ = x₂ for all x₁, x₂ in the domain
(C)Range of f equals codomain
(D)f has a horizontal asymptote
3

The relation 'student a is related to student b if they belong to the same class' is:

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(A)Reflexive only
(B)Symmetric only
(C)An equivalence relation
(D)Neither reflexive nor symmetric
4

Define a bijective function and verify that f: R → R, f(x) = 2x + 3 is bijective.

1M
CBQ 2Assertion–Reason1 mark
A
Assertion

The function f: R → R defined by f(x) = x² is neither one-one nor onto.

R
Reason

f(1) = f(−1) = 1 (so not one-one), and negative real numbers have no preimage under f (so not onto for codomain R).

(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
Ch 5

Continuity and Differentiability

2 sets
CBQ 1Case StudySpeed of a Particle4 marks

Read the passage

The position of a particle moving along a straight line is given by the function s(t) = t³ − 6t² + 9t + 2, where s is in metres and t is in seconds. A physicist uses calculus concepts of continuity and differentiability to study the motion. Since s(t) is a polynomial, it is continuous and differentiable everywhere. The velocity of the particle is v(t) = s'(t) (first derivative) and acceleration is a(t) = v'(t) = s''(t) (second derivative). The physicist also uses the chain rule and product rule while studying composite motion. He needs to find when the particle is at rest and when its acceleration is zero.
1

The velocity function v(t) = s'(t) for s(t) = t³ − 6t² + 9t + 2 is:

1M
(A)3t² − 12t + 9
(B)3t² + 12t − 9
(C)t³ − 12t + 9
(D)3t² − 6t + 9
2

At what value of t is the particle at rest (v(t) = 0)?

1M
(A)t = 1 and t = 3
(B)t = 2 and t = 4
(C)t = 0 and t = 3
(D)t = 1 and t = 2
3

Every polynomial function is:

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(A)Continuous but not differentiable everywhere
(B)Differentiable but not continuous everywhere
(C)Both continuous and differentiable everywhere
(D)Neither continuous nor differentiable
4

Find the acceleration of the particle at t = 2 seconds.

1M
CBQ 2Assertion–Reason1 mark
A
Assertion

The function f(x) = |x| is continuous at x = 0 but not differentiable at x = 0.

R
Reason

Continuity at a point requires the limit to equal the function value, but differentiability requires the left-hand derivative to equal the right-hand derivative at that point.

(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
Ch 6

Application of Derivatives

2 sets
CBQ 1Case StudyMaximising Area of a Garden4 marks

Read the passage

A farmer has 200 metres of fencing wire. He wants to enclose a rectangular plot using all the wire, with one side along a wall (so only 3 sides need fencing). He wants to find the dimensions that will give the maximum area. The maths teacher uses this as a problem to illustrate application of derivatives — specifically finding maxima using the first and second derivative tests. She also explains that if the area function A(x) has A'(x) = 0 at x = c and A''(c) < 0, then x = c gives the maximum value. She further discusses how rates of change of quantities (like volume of a balloon increasing) are computed using derivatives.
1

If x is the side perpendicular to the wall, the length of the side parallel to the wall is:

1M
(A)200 − x
(B)200 − 2x
(C)100 − x
(D)200 − 3x
2

The area function A(x) for this rectangular plot is:

1M
(A)A(x) = x(200 − 2x)
(B)A(x) = x(200 − x)
(C)A(x) = 2x(100 − x)
(D)A(x) = x² + 200x
3

For the second derivative test, f has a local maximum at x = c if:

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(A)f'(c) = 0 and f''(c) > 0
(B)f'(c) = 0 and f''(c) < 0
(C)f'(c) > 0 and f''(c) = 0
(D)f'(c) < 0 and f''(c) < 0
4

Find the dimensions of the rectangular plot that maximise the area and state the maximum area.

1M
CBQ 2Assertion–Reason1 mark
A
Assertion

A function f(x) can have a local maximum at a point x = c even if f'(c) does not exist.

R
Reason

The first derivative test for local maxima and minima applies only to differentiable functions; at points where the derivative does not exist, no conclusion can be drawn.

(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

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