CBSE Maths Class 12: Assertion and Reason Based Questions with Solutions
Chapter 1:
Relations and Functions
1. Assertion: If a function f : R→R is defined
by f(x) = x3 + 1, then f is both one-one and onto.
Reason: A cubic function of the form f(x) = x3 + c, where c is
a constant, is strictly increasing over all real numbers, so it is one-one.
Also, since for every y ∈ R, there exists an x ∈ R such that f(x) = y, the function is onto.
Solution: (a) We have f(x) = x3 + 1
Let f(x1) = f(x2)
ð x13
+ 1 = x23 + 1
ð x1 = x2 So, f is one-one.
Let f(x) = x3 + 1 = y
ð x3
= y – 1
ð x
= 3√ y – 1
Thus, for every y ∈ R,
there exists an x ∈ R such that f(x) = y. Therefore,
the function is onto.
Both A and R are true and R is the correct explanation of A.
2. Assertion: Let Q be the set of rational numbers. If R is a
relation on Q defined by R = 
, then R is not a transitive
relation.
Reason: Let a = -1, b = 0, c = 2. Then, 1 + ab = 1 > 0 and therefore, a
R b. Again, b R c as 1 + bc = 1
> 0. But a is not related to c as 1 + ac = -1 < 0.
Solution: (a) Both A and R are true and R is
the correct explanation of A. a R b and
b R c, but a is not related to c, implies that the relation R is not
transitive.
3. Assertion: A relation R on set A = {x : x 
W, 0 
10}
defined by R = 
is an
equivalence relation.
Reason: The relation R is an equivalence relation, if
it is reflexive, symmetric and transitive.
Solution: (a) Both A and R are true and R is the
correct explanation of A.
Reason is correct as per the definition of
equivalence relation.
Now, A = {0, 1, 2, 3, …, 10}
For any x A, 
0 (which is a multiple of 2)
Thus, (x, x) R, which means
R is reflexive.
If (x, y) ∈ R, then ∣x − y∣ is a multiple of 2.
But ∣x − y∣ = ∣y − x∣, so (y, x) ∈ R, which means R is symmetric.
If ∣x
− y∣ and ∣y − z∣ are multiple of 2, then ∣x − z∣ is also a multiple of 2. Thus, (x,
z) ∈ R, which means R is transitive.
4. Assertion: The relation R on Z defined by R = 
is an
equivalence relation on Z.
Reason: Let (a, b) R, where a, b Z.
Then a + b is even, which implies that (b, a) R.
Hence, R is symmetric.
Solution: (b) Reason is true because it explains the symmetric
property of R. Let us check if R is reflexive and transitive or not. a + b is
even only if both a and b are even or both are odd. Therefore, (a, a) R, which means
R is reflexive. Again, (a, b) R and (b, c) R implies that
(a, c) R. Therefore, R
is transitive. Thus, R is equivalence relation.
Hence, both A and R are true but R is not
the correct explanation of A.
Chapter-2:
Inverse Trigonometric Functions
1. Assertion: tan-1 (√3) = 
.
Reason: The principal value of tan-1 x lies in the interval [0, π].
Solution: (c) The value of tan-1 (√3) is . So, the assertion is
true.
The principal value of tan-1 x lies in the interval (
, 
). So,
the reason is false.
Hence, the assertion is true, but
reason is false.
2. Assertion: 
= 
.
Reason: The principal value of cos-1(x) lies in the interval [0, π].
Solution: (a) Let 
= y
ð cos
y = - ½ = - cos 
= cos (
)
= cos (
)
Therefore, y = . Thus, assertion is true.
The principal value of cos-1(x) lies in the interval [0, π].
Thus, reason is also true and it is the correct explanation of
assertion.
3. Assertion: The value of cot
= 
.
Reason: The numbers 7, 24 and 25
form the Pythagorean triplets. Using this, we can find the value of all inverse
trigonometric functions.
Solution: (a) Let 
= x
ð
sin x = 24/25
Therefore, cot
x = 7/24 => x = cot-1 (7/24)
Thus, cot = cot
= . Hence, assertion is true.
Reason is also
true, and it is the correct explanation of assertion.
Chapter 3: Matrices
1. Assertion: The matrix A is a square matrix
such that A2 = I, then 2A3 – 7A2 – 2A + 7I is
a zero matrix.
Reason:
Matrix multiplication is commutative.
Solution:
c. Assertion is true because A2 = I implies that A3 = A2.A
= I.A = A. So, the given expression equals 2A – 7I – 2A + 7I = O. But the
statement “matrix multiplication is commutative” is false.
2.
Assertion: Let A and B are two symmetric matrices, then AB is symmetric, if
and only if AB = BA.
Reason:
If A and B are symmetric matrices, then (AB)T = BTAT
= BA.
Solution:
a. Assertion and reason both are correct because if A and B are symmetric, we
have A’ = A and B’ = B. Now, AB is symmetric if and only if (AB)’ = AB. But
(AB)’= B’A’ = BA, since A and B are symmetric. So, if A and B are symmetric,
then AB is symmetric, if and only if AB = BA.
3. Assertion: Let A = ( 
) and B = 
, then the product of the matrices A and B is
not defined.
Reason:
The number of columns in A is not equal to the number of rows in
B.
Solution:
(a) Here, the order of matrix A is 3 × 2 and the order of matrix B is 3 × 3.
The multiplication of two matrices A and B is defined, if the number of columns
in A is equal to the number of rows in B. But this is not so. Therefore, the
product of A and B is not defined.
Chapter 4: Determinants
1.
Assertion: The determinant of matrix A = 
is zero.
Reason: The
determinant of a skew-symmetric matrix of odd order is zero.
Solution:
(a) Matrix A is a skew-symmetric matrix of order 3.
In
a skew-symmetric matrix of odd order, |A| = |-AT| = -|AT|.
But |A| = |A|T.
Hence,
|A| = -|AT| = -|A|. Therefore, |A| = 0.
The
Reasoning is false because AB is symmetric if and only if (AB)’=AB.
2. Assertion: If A = 
, then 
is not equal to 0.
Reason: The
determinant of an upper triangular matrix is product of its diagonal elements.
Solution: (a) Both assertion and reasoning are
true. The reason is a true statement, and it is the correct explanation of the
assertion that |A| = 3 x 4 x 5. Thus, |A| is not equal to zero.
3. Assertion: If A = 
, then |adj (adj A)| = x4y4z4.
Reason: If A is a
square matrix of order n, then |adj (adj A)| = 
Solution: (a) We
know that if A is a matrix of order n, then |adj A| = |
.
Using this repeatedly, we get |adj
(adj A)| = 
Here, |A| = 
Since A is a matrix of order 3, |adj
(adj A)| = =
= x4y4z4
Chapter-5:
Continuity and Differentiability
1. Assertion: If the left-hand
derivative and right-hand derivative of a function at x = a are
equal, then the function is differentiable at x = a.
Reason:
Differentiability requires the existence and equality of the left-hand and
right-hand derivatives at that point.
Solution: (a) Differentiability at a point means the derivative from both sides must
exist and be equal. So, if the left-hand and right-hand derivatives are equal,
the function is differentiable there. Thus, A is true.
Reason is also true and it explains the assertion
correctly.
2.
Assertion: The function f(x) =
is continuous
in R.
Reason: A polynomial function is continuous everywhere.
Solution: (d) Reason is true because a polynomial function is continuous everywhere. f(x)
is continuous at all points in its domain, but f(x) is not
defined at x = 2 and x = 3. So, assertion is false.
3. Assertion:
A function can be differentiable at a point
but not continuous at that point.
Reason: Differentiability implies continuity at that point.
Solution:
(d) Differentiability
implies continuity but the converse is not true. For example, y = |x|
is continuous but not differentiable.
Thus, A
is false, but R is true.
Chapter 6: Application of Derivatives
1. Assertion: The function f(x) = ax – sin x is strictly increasing for a > 1.
Reason: A function f defined on I is strictly increasing, if f’(x) > 0 for every x in I.
Solution: (a) We have, f’(x) = a – cos x.
Since f is increasing, a – cos x > 0 for all x. The
minimum value of a – cos x is when cos x =
1.
This means a – 1 > 0 for all x. This implies a
> 1. So, assertion is true.
Reason is also true and
it is the correct explanation of A.
2. Assertion: The function f(x) = x3 + 6x2 + 15 is increasing in (-
and decreasing in (-4, 0).
Reason: A cubic function is always
increasing in R.
Solution: (c) We have, f(x) = x3 + 6x2 +
15
f’(x) = 3x2 + 12x = 3x(x + 4)
To find critical point f’(x) = 0 => 3x(x + 4) = 0 => x = 0 and x = -4
Let us analyse the sign
of f’(x) in the intervals (-
and (-4, 0).
In interval (-
f’(x) > 0
In interval 
, f’(x) > 0
In interval (-4, 0), f’(x) < 0
Therefore, f is increasing in (- and decreasing in (-4, 0).
Thus, assertion is true
but reason is false.
3. Assertion: The least value of f(x) = x3 - 9x2 + 24x in the
interval [0, 10] is 0.
Reason: The least value is obtained by
substituting the end points and the critical point in f(x) and verifying which is
smaller.
Solution: (a) We have, f(x) = x3 – 9x2 + 24x
=> f’(x) = 3x2 – 18x + 24
To find critical point f’(x) = 0 => 3(x2 – 6x + 8) = 0
i.e., 3(x – 4)(x – 2) =
0. The critical points are x = 2 and x = 4.
Evaluating f(x) at
critical points and end points, we get
f(0) = 0 at x = 0, f(2) =
20 at x = 2, f(4) = 16 at x = 4 and f(10) = 340 at x = 10.
Out of these, the least
value in the interval [0, 10] is 0. Thus, A is true.
Reason is also true and
it explains assertion correctly.
Chapter 7: Integrals
1. Assertion: The value of the integral 
is 4 units.
Reason: 
dx = 
+ 
, where a < c < b.
Solution: (a)
= 
dx + 
dx
= [-x]0-1
+ [x]03
= (0 + 1) + (3 – 0)
= 4
2. Assertion:
(1 + tan x + tan2 x) dx = ex tan
x + C
Reason: 
Solution: (a)
Let I = 
(1 + tan x + tan2 x) dx
= 
) dx
= 
) dx
= 
tan x + C
3. Assertion:
+ C
Reason: The
derivative of 
is 
Solution: (a)
Let y = = 
.
Taking log on both sides, we get
log y = x log 10
Differentiating both sides, we get
log 10
=
4. Assertion:
The value of the integral 
is 
Reason: The value of
the integral 
= 
+ C.
Solution: (c) 
= 
= 
(Since 
= 
+ C)
5.
Assertion: 
= 
.
Reason: 
dx = 0, if f(x) is an even function.
Solution:
(c)
dx =2 
, if
f(x) is an even function. So, the reason is false.
dx = 2 
Let I = 2 --------------(1)
= 2
I
= 2
---------------(2)
Adding (1) and (2), we get
2I = 2 
2I = 2 
This implies 2I =2
Or I =
Chapter-8: Application of Integrals
1. Assertion: The area enclosed between the
curves y = x2 and y = x + 2 is obtained
by integrating (x + 2 − x2) between the points of
intersection.
Reason: The area between two curves y = f(x) and y
= g(x) from x = a to x = b is given
by 
f(x)−g(x)∣dx.
Solution: (a) Points of intersection: Solve x2
= x + 2 ⇒ x2 – x –
2 = 0 ⇒ x = −1.
Between x = −1 to x = 2, y = x + 2 is above y
= x2.
So, the required area = 
(x + 2)− x2]dx.
Thus, both A and R are correct, and R explains why we subtract the
lower curve from the upper one.
2. Assertion: The area bounded by the curve y
= sin x, the x-axis, and the lines x = 0 and x = π
is zero.
Reason: The definite integral
dx
= 2.
Solution: (d) The area bounded by the curve y
= sin x, the x-axis, and the lines x = 0 and x = π
isdx
= [-cos x]0Ï€ =
-[-1 – 1] = 2.
Thus, A is false, but R is true.
3. Assertion: The area enclosed between the
curves y = x and y = x2 from x = 0 to x = 1 is 1/6.
Reason: The area under y = x from 0 to 1 is 
dx
= 1/2, and the area under y = x2 from 0 to 1 is 
dx
= 1/3. So, the area enclosed between the curves y = x and y = x2
from x = 0 to x = 1 is ½ + 1/3 = 5/6
Solution: (c) The area enclosed between the
curves y = x and y = x2 from x = 0 to x = 1 is given
by 
dx
= [x2/2 – x3/3]01 = ½ - 1/3 = 1/6.
Thus, assertion is true, but reason is false.
Chapter 9: Differential Equations
1. Assertion: tan-1 x + tan-1 y
= k is a solution of the differential equation
(1 + x2) dy + (1 + y2)
dx = 0.
Reason: On
differentiating tan-1 x + tan-1 y = k, we get
(1 + x2) dy + (1 + y2) dx = 0.
Solution: (a)
Consider tan-1 x + tan-1 y = k.
Differentiating, we get 
+ 
= 0.
This gives, (1 + x2) dy + (1 + y2)
dx = 0.
2. Assertion: A homogeneous differential
equation of the form 
= f(
) can be solved by making a
substitution.
Reason: If = f(), then substituting x = vy will
reduce it to a form where variable separable method will help to solve for y.
Solution: (a)
Since a homogeneous differential equation of the form = f() can be solved by substituting x
= vy.
Therefore, both A and R are true and R is the correct explanation
of A.
3. Assertion:
The equation of the curve whose slope is given by 
and passing through (1, -1) is given by y = 
.
Reason: Separating the variables help to
solve for y in terms of x.
Solution: (d)
= 
implies log y = 3 log x + log c.
This gives y = c
.
Substituting (1, -1), we get -1 = c.1 or c =
-1
Thus, y = -
4. Assertion: The sum of the order and degree
of the differential equation
+ 2 (
= cos y is 3.
Reason: The degree of
the differential equation is not defined.
Solution: (c)
The order is 2 and degree is 1. The sum of the order and degree is 3.
Chapter 10: Vector Algebra
1.
Assertion (A): If the
angle between the vectors a and b is 
/6 and area of triangle with adjacent sides parallel
to a and b is 9 units, then a.b is 6
units.
Reason
(R): Area of a triangle
with adjacent sides a and b is 
|a × b| sq. units.
Solution: (d) Area of a
triangle with adjacent sides a and b = 
|a ×
b|
Thus, reason is true.
ð
9 = |a ×
b|
ð |a × b| = 18
We have, |a × b| = |a| |b| sin /6
18 = |a| |b| × ½
|a| |b| = 36
Now, a.b = |a| |b| cos
/6
= 36 × 
= 18 units.
Hence, assertion is
false.
2. Assertion (A): If a and b are two vectors such that |a × b|2 = 25, then |b
× a|2 = 25 units.
Reason (R): The cross product of any two vectors is commutative.
Solution: (c) Assertion is true but the
reason is false, because a × b = -b × a.
3.
Assertion (A): If the
difference between two unit vectors is a unit vector, then the magnitude of
their sum is 
Reason
(R): For any two unit
vectors a and b, |a + b|2 + |a – b|2 = 4.
Solution: (d)
We have, |a| = |b|
= 1
|a + b|2 +
|a – b|2 = |a|2 + |b|2 + 2a.b + |a|2
+ |b|2 – 2b.a
= 1 + 1 + 2a.b
+ 1 + 1 – 2b.a = 4
Thus, reason is true.
If a and b are two
unit vectors, then |a – b| = 1
Using identity, |a – b|2
= |a|2 + |b|2 – 2a.b
1 =
1 + 1 – 2a.b
=> a.b = ½
Now, |a + b|2
= |a|2 + |b|2 + 2a.b
= 1 + 1 + 2 × ½ = 3
Hence, |a + b| =
√3
Thus, assertion is
false.
4. Assertion (A): The projection of a = i + j – k on b = i – j + k is 
.
Reason (R): The projection of vector a on vector b is (
).
Solution: (a) The projection of a on b =
() = (i + j – k).(i – j + k)
/ |i – j + k| =
Hence, both A and R are true and R is the correct explanation of A.
Chapter-11: Three Dimensional Geometry
1. Assertion: If 
are the direction angles, then cos 2
cos 2
+ cos 2
= -1.
Reason: If cos 
cos , cos are the direction cosines, then cos2 
cos2 + cos2 = 1.
Solution: (a) We have, cos 2cos 2+ cos 2 =2 cos2 
- 1 + 2 cos2 - 1 + 2 cos2 - 1 = 2(cos2 cos2 + cos2 
= 2 – 3 = -1
Thus, the assertion is true.
If cos cos , cos are the direction cosines, then cos2 cos2 + cos2 = 1.
Hence, the reason is also true,
and it is the correct explanation of assertion.
2. Assertion: The direction ratios of a line perpendicular to the lines 
= 
and 
are proportional to 0, -1 and 2.
Reason: The scalar product of two vectors is a vector perpendicular to both the
vectors.
Solution. (c) Let the direction ratios of the line perpendicular to the
given lines be a, b and c. Then, 2a + 2b + c = 0 and -a + 8b + 4c = 0.
This implies 
(Using cross-product rule)
Therefore, a, b and c are proportional to 0, -1, 2.
Thus, assertion is true, but
reason is false.
3. Assertion: The direction cosines of a vector making equal angles with positive
directions of x,
y and z axes are 
, 
.
Reason: If l, m and n are direction
cosines, then l2 + m2 + n2 =
1.
Solution: (a) If a vector makes equal angles θ with the x, y
and z axes, then all three direction cosines are equal, i.e., l = m = n = x
(say)
If l, m and n are direction cosines, then l2
+ m2 + n2 = 1.
ð x2 + x2
+ x2 = 1
ð 3x2 = 1
ð x2 = 1/3
ð
x =
Hence, both assertion and reason are true and reason is the
correct explanation of assertion.
Chapter-12:
Linear Programming
1. Assertion (A): The optimal value of a
linear objective function always occurs at a corner point (vertex) of the
feasible region.
Reason (R): The feasible region for a linear programming problem is always a
bounded convex polygon.
Solution: (c) The optimal value of a linear objective function occurs
at a corner point of the feasible region. Thus, A is true.
But the feasible region is not always bounded. It can also be
unbounded. So, R is false.
2. Assertion (A): A linear programming problem can have
infinitely many optimal solutions.
Reason (R): If the objective function is parallel to a constraint boundary
segment of the feasible region, then every point on that segment gives the same
optimal value.
Solution: (a) When the objective function is parallel to a boundary
of the feasible region and the entire edge lies within the region, then each
point on that edge yields the same optimal value. This results in infinitely
many optimal solutions, as all those points satisfy the constraints and yield
the same maximum or minimum value.
Thus, both A and R are true, and R is the correct explanation of A.
3. Assertion (A): If a linear programming problem has no
feasible solution, then it has no optimal solution.
Reason (R): The existence of an optimal solution depends on whether the
objective function is linear.
Solution: (c) The assertion is true because if there is no feasible region,
then there is no point that satisfies all constraints. So, no optimal solution
exists.
The reason is false because the linearity of the objective
function does not guarantee an optimal solution. The existence of an optimal
solution depends on the feasibility of the region, not just the form of the
function.
Chapter 13: Probability
1.
Assertion: If the sum of the numbers obtained on
throwing a pair of dice is a prime number, then the probability that the number
obtained on one of them is 4 is 4/15.
Reason: The number of elements in the
sample space, when a pair of dice is thrown, is 36. Out of which, 15 have the
sum of their numbers to be prime.
Solution:
(a) The total number of elements in the sample
space, when a pair of dice is thrown, is 36.
The elements of the sample space whose sum
is a prime number:
(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2
,3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5).
Those with one of the elements as 4 is (1,
4), (3, 4), (4, 1), (4, 3).
Thus, the probability that the number
obtained on one of the dice as 4 = 4/15
Hence, both A and R are true and R is the
correct explanation of A.
2. Assertion: If P(A) = 0.4,
P(B’) = 0.1, and A and B are independent events, then P(A U B) = 0.94.
Reason: If two events are
independent, then P(A Ç B) = P(A) P(B).
Solution: (a) If
two events are independent, then P(A Ç B) = P(A) P(B). Therefore, reason is true.
P(A) = 0.4, P(B) = 1 – P(B’) = 1 – 0.1 = 0.9
We know that P(A U B) = P(A) + P(B) – P(A Ç B)
=
P(A) + P(B) – P(A) P(B) [Since A and B
are independent events.]
= 0.4
+ 0.9 – 0.36 = 1.30 – 0.36 = 0.94
Thus, assertion is also true and reason is the
correct explanation of assertion.
3. Assertion: The
probability distribution of a certain random variable is shown in the following
table.
|
X |
0 |
1 |
2 |
3 |
|
P(X) |
0 |
4k |
2k |
4k |
Then, the value of k is 1/10.
Reason: The probability of a sure event is 1.
Solution:
(a) The sum of all probabilities must be 1, because the probability of
a sure event is always 1. Thus, reason is true.
P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 1
0 + 4k + 2k + 4k = 1
10k = 1
k = 1/10
Thus, both A and R are true and R is the
correct explanation of A.
4.
Assertion: A family
has three children. The probability that the eldest is a boy given that the
youngest one is a girl is ½.
Reason: Conditional probability of an event A given that the
event B has already occurred is P(A/B) = P(A Ç B)/P(A)
Solution:
(c) The sample space = BBB, BBG, BGB, BGG,
GBB, GBG, GGB, GGG
The sample space is now reduced to BBG,
BGG, GBG, GGG because the youngest one is a girl. Out of this, the favourable
outcomes are BBG and BGG.
P(Eldest is B / Youngest is G) = 2/4 =
½
So, assertion is true.
But reason is false as conditional
probability, P(A/B) = P(A Ç B)/P(B).