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CBSE Class 12 Maths 2025 Question Paper 65/5/2 with Solutions

All 45 questions from the CBSE Class 12 Mathematics board paper, Set 65/5/2 (2025), with answers and step-by-step solutions. Total 80 marks. Tap “Show answer & solution” under any question.

Set 65/1/1Set 65/1/2Set 65/1/3Set 65/2/1Set 65/2/2Set 65/2/3Set 65/4/1Set 65/4/2Set 65/4/3Set 65/5/1Set 65/5/2Set 65/5/3Set 65/6/1Set 65/6/2Set 65/6/3Set 65/7/1Set 65/7/2Set 65/7/3

If
is continuous at x = 0, then the value of a is :

  1. (A)1
  2. (B)–1
  3. (C)
  4. (D)0
Show answer & solution
Answer: (C)
  1. For continuity at :
  2. So .
Also asked in: 2025 65/5/1, 2025 65/5/3

The principal value of is :

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (D)
  1. The principal value branch of is and .
Also asked in: 2025 65/5/1, 2025 65/5/3
Q31 markMCQMatrices

If A and B are two square matrices of the same order, then (A + B) (A – B) is equal to :

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (A)
  1. Matrix multiplication is not commutative, so cannot be cancelled.
Q41 markMCQDeterminants

If is a diagonal matrix such that , and , then is :

  1. (A)0
  2. (B)–10
  3. (C)10
  4. (D)1
Show answer & solution
Answer: (B) –10
  1. The determinant of a diagonal matrix is the product of its diagonal entries.
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Q51 markMCQMatrices

If , then is :

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (B)
  1. , where is the identity matrix of order 3.
  2. So .
  3. Hence every diagonal entry of is 125 and the other entries are 0, which is option (B).
Also asked in: 2025 65/5/1, 2025 65/5/3
Q61 markMCQDeterminants

If , then the value of x is :

  1. (A)3
  2. (B)7
  3. (C)
  4. (D)
Show answer & solution
Answer: (C)
  1. LHS , RHS .
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Q71 markMCQProbability

If and , then is :

  1. (A)0.3
  2. (B)1
  3. (C)1.3
  4. (D)0.7
Show answer & solution
Answer: (D) 0.7
Also asked in: 2025 65/5/1, 2025 65/5/3
Q81 markMCQMatrices

If a matrix A is both symmetric and skew-symmetric, then A is a :

  1. (A)diagonal matrix
  2. (B)zero matrix
  3. (C)non-singular matrix
  4. (D)scalar matrix
Show answer & solution
Answer: (B) zero matrix
  1. Symmetric: ; skew-symmetric: .
  2. So , the zero matrix.

The slope of the curve is maximum at :

  1. (A)(1, –10)
  2. (B)(1, 10)
  3. (C)(10, 1)
  4. (D)(–10, 1)
Show answer & solution
Answer: (A) (1, –10)
  1. Slope
  2. , and , so the slope is maximum at .
  3. At : . Point .
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The area of the region enclosed between the curve , x-axis, x = – 2 and x = 2 is :

  1. (A)
  2. (B)
  3. (C)0
  4. (D)8
Show answer & solution
Answer: (B)
  1. for and for ; the curve is symmetric about the origin.
  2. Area
  3. sq units
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Q111 markMCQIntegrals

is equal to :

  1. (A)
  2. (B)
  3. (C)
  4. (D)
Show answer & solution
Answer: (B)
Q121 markMCQIntegrals

If , then the value of 'a' is :

  1. (A)
  2. (B)
  3. (C)
  4. (D)4
Show answer & solution
Answer: (B)

If is the greatest integer function, then the correct statement is :

  1. (A)f is continuous but not differentiable at x = 2.
  2. (B)f is neither continuous nor differentiable at x = 2.
  3. (C)f is continuous as well as differentiable at x = 2.
  4. (D)f is not continuous but differentiable at x = 2.
Show answer & solution
Answer: (B) f is neither continuous nor differentiable at x = 2.
  1. LHL at : ; RHL: .
  2. LHL ≠ RHL, so is not continuous at .
  3. A function that is not continuous at a point cannot be differentiable there.
Also asked in: 2025 65/5/1, 2025 65/5/3

The integrating factor of the differential equation
is :

  1. (A)
  2. (B)e
  3. (C)
  4. (D)
Show answer & solution
Answer: (D)
  1. , linear with
  2. I.F.
Q151 markMCQVector Algebra

Let be a position vector whose tip is the point (2, – 3). If , where coordinates of A are (– 4, 5), then the coordinates of B are :

  1. (A)(– 2, – 2)
  2. (B)(2, – 2)
  3. (C)(– 2, 2)
  4. (D)(2, 2)
Show answer & solution
Answer: (C) (– 2, 2)
  1. So B is .
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Q161 markMCQVector Algebra

The respective values of and , if given and , are :

  1. (A)48 and 16
  2. (B)3 and 1
  3. (C)24 and 8
  4. (D)6 and 2
Show answer & solution
Answer: (C) 24 and 8
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Q171 markMCQLinear Programming

For a Linear Programming Problem (LPP), the given objective function Z = 3x + 2y is subject to constraints :



The correct feasible region is :

Diagram for CBSE 2025 Class 12 Maths question 17
  1. (A)ABC
  2. (B)AOEC
  3. (C)CED
  4. (D)Open unbounded region BCD
Show answer & solution
Answer: (B) AOEC
  1. Both constraints are of the type , so the feasible region lies on the origin side of both lines and in the first quadrant.
  2. On the origin side of and of , with , the region is bounded by O, E(5, 0), C(4, 3) and A(0, 5).
  3. So the feasible region is AOEC.
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The sum of the order and degree of the differential equation
is :

  1. (A)2
  2. (B)
  3. (C)3
  4. (D)4
Show answer & solution
Answer: (C) 3
  1. Highest order derivative is , so order = 2.
  2. The equation is a polynomial in derivatives and appears with power 1, so degree = 1.
  3. Sum
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Q191 markAssertion–ReasonLinear Programming

Assertion (A) : The shaded portion of the graph represents the feasible region for the given Linear Programming Problem (LPP).
Min Z = 50x + 70y
subject to constraints
, ,
Z = 50x + 70y has a minimum value = 380 at B(2, 4).
Reason (R) : The region representing 50x + 70y < 380 does not have any point common with the feasible region.

Diagram for CBSE 2025 Class 12 Maths question 19
  1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. (C)Assertion (A) is true, but Reason (R) is false.
  4. (D)Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (D) Assertion (A) is false, but Reason (R) is true.
  1. The feasible region for , , is the unbounded region in the first quadrant lying on or above A(0, 8), B(2, 4), C(10, 0) (away from the origin).
  2. The hatched portion in the graph also covers points below these lines and points outside the first quadrant, so it is not the feasible region: Assertion is false.
  3. Corner values: Z(A) = 560, Z(B) = 100 + 280 = 380, Z(C) = 500, smallest is 380 at B.
  4. The line (i.e. ) has slope , between the slopes and of the two boundary lines, so it meets the feasible region only at B and the half-plane has no common point with it: Reason is true.
Also asked in: 2025 65/5/1, 2025 65/5/3
Q201 markAssertion–ReasonRelations and Functions

Assertion (A) : Let . If be defined as , then f is not an onto function.
Reason (R) : If , then .

  1. (A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. (B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. (C)Assertion (A) is true, but Reason (R) is false.
  4. (D)Assertion (A) is false, but Reason (R) is true.
Show answer & solution
Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  1. Range of on is , so is not onto: Assertion is true.
  2. For , gives , which is not a real number, so it is not in A: Reason is true.
  3. The element of the codomain has no pre-image, which is exactly why is not onto, so R explains A.
Also asked in: 2025 65/5/1, 2025 65/5/3
Q212 marksVery Short AnswerInverse Trigonometric Functions

Find the domain of .

Show answer & solution
Answer:
  1. The domain of is , so or .
  2. ;
  3. Domain
Q222 marksVery Short AnswerApplication of Derivatives

The radius of a cylinder is decreasing at a rate of 2 cm/s and the altitude is increasing at the rate of 3 cm/s. Find the rate of change of volume of this cylinder when its radius is 4 cm and altitude is 6 cm.

Show answer & solution
Answer: cm³/s, i.e. the volume is decreasing at cm³/s
  1. With , , , :
  2. cm³/s
  3. So the volume is decreasing at the rate of cm³/s.
Q232 marksVery Short AnswerVector Algebra

Find a vector of magnitude 5 which is perpendicular to both the vectors and .

Show answer & solution
Answer:
  1. Let , .
  2. Required vector
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OR
Q23 (OR) (OR)2 marksVery Short AnswerVector Algebra

Let , and be three vectors such that and , . Show that .

Show answer & solution
Answer: Proved.
  1. From : , so or .
  2. From : , so or .
  3. A non-zero vector cannot be both perpendicular and parallel to the non-zero vector .
  4. Hence , i.e. .
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Q242 marksVery Short AnswerVector Algebra

A man needs to hang two lanterns on a straight wire whose end points have coordinates A (4, 1, – 2) and B (6, 2, – 3). Find the coordinates of the points where he hangs the lanterns such that these points trisect the wire AB.

Show answer & solution
Answer: and
  1. The points of trisection divide AB internally in the ratios 1 : 2 and 2 : 1.
  2. P (1 : 2)
  3. Q (2 : 1)
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Q252 marksVery Short AnswerContinuity and Differentiability

Differentiate with respect to x.

Show answer & solution
Answer:
  1. Let .
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OR
Q25 (OR) (OR)2 marksVery Short AnswerContinuity and Differentiability

If , prove that .

Show answer & solution
Answer: Proved.
  1. Hence .
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Q263 marksShort AnswerApplication of Derivatives

Show that is an increasing function in .

Show answer & solution
Answer: Proved.
  1. The denominator is always positive.
  2. For , , so ; only at the end point .
  3. Hence is increasing in .
Q273 marksShort AnswerProbability

The probability that a student buys a colouring book is 0.7 and that she buys a box of colours is 0.2. The probability that she buys a colouring book, given that she buys a box of colours, is 0.3. Find the probability that the student :
(i) Buys both the colouring book and the box of colours.
(ii) Buys a box of colours given that she buys the colouring book.

Show answer & solution
Answer: (i) 0.06 (ii)
  1. Let C: buys a colouring book, B: buys a box of colours. , , .
  2. (i)
  3. (ii)
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OR
Q27 (OR) (OR)3 marksShort AnswerProbabilityNot in current syllabus

A person has a fruit box that contains 6 apples and 4 oranges. He picks out a fruit three times, one after the other, after replacing the previous one in the box. Find :
(i) The probability distribution of the number of oranges he draws.
(ii) The expectation of the random variable (number of oranges).

Show answer & solution
Answer: (i) X = 0, 1, 2, 3 with P(X) = (ii)
  1. Let X = number of oranges drawn. Draws are with replacement, so each draw gives an orange with probability , .
  2. ,
  3. ,
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Q283 marksShort AnswerDifferential Equations

Find the particular solution of the differential equation
; given that y = 0, when x = 1.

Show answer & solution
Answer:
  1. , a homogeneous equation.
  2. Put :
  3. Integrating: , i.e.
  4. At , :
  5. Particular solution:
Q293 marksShort AnswerIntegrals

Find :

Show answer & solution
Answer:
  1. Put , :
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OR
Q29 (OR) (OR)3 marksShort AnswerIntegrals

Evaluate :

Show answer & solution
Answer: 7
  1. On , :
  2. Required value
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Q303 marksShort AnswerLinear Programming

In the Linear Programming Problem (LPP), find the point/points giving maximum value for Z = 5x + 10y
subject to constraints



Show answer & solution
Answer: Maximum Z = 600 at (120, 0) and (60, 30), and at every point on the segment joining them.
  1. Corner points of the (bounded) feasible region:
  2. and give (40, 20); and give (60, 30); (60, 0) and (120, 0) on the x-axis.
  3. Z at (60, 0) = 300; at (120, 0) = 600; at (60, 30) = 300 + 300 = 600; at (40, 20) = 200 + 200 = 400.
  4. Maximum Z = 600 occurs at both (120, 0) and (60, 30).
  5. Hence every point on the line segment joining (120, 0) and (60, 30) gives the maximum value 600.
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Q313 marksShort AnswerVector Algebra

If such that , , , then find the angle between and .

Show answer & solution
Answer: (i.e. )
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OR
Q31 (OR) (OR)3 marksShort AnswerVector Algebra

If and are unit vectors inclined with each other at an angle , then prove that .

Show answer & solution
Answer: Proved.
  1. Since , , so .
  2. Hence .
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Q325 marksLong AnswerApplication of Integrals

Draw a rough sketch of the curve . Using integration, find the area of the region bounded by the curve , x = 4 and x-axis, in the first quadrant.

Show answer & solution
Answer: sq units
  1. is the upper half of the parabola (starting at the origin, rising to the right); at , .
  2. The region lies between the curve, the x-axis and the line .
  3. Area
  4. sq units
Q335 marksLong AnswerDeterminants

An amount of ₹ 10,000 is put into three investments at the rate of 10%, 12% and 15% per annum. The combined annual income of all three investments is ₹ 1,310, however the combined annual income of the first and the second investments is ₹ 190 short of the income from the third. Use matrix method and find the investment amount in each at the beginning of the year.

Show answer & solution
Answer: ₹ 2,000 at 10%, ₹ 3,000 at 12% and ₹ 5,000 at 15%
  1. Let the amounts be ₹ x, ₹ y, ₹ z at 10%, 12%, 15%.
  2. ; ;
  3. ,
  4. Investments: ₹ 2,000, ₹ 3,000 and ₹ 5,000.
Q345 marksLong AnswerThree Dimensional Geometry

Find the foot of the perpendicular drawn from the point (1, 1, 4) on the line .

Show answer & solution
Answer:
  1. The line is : through with direction ratios 5, 2, 3.
  2. Any point on it: .
  3. With :
  4. line:
  5. Foot of perpendicular
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OR
Q34 (OR) (OR)5 marksLong AnswerThree Dimensional Geometry

Find the point on the line at a distance of units from the point (–1, –1, 2).

Show answer & solution
Answer: (1, –1, 4) or
  1. Any point on the line: .
  2. Distance from :
  3. or
  4. : ; :
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For a positive constant 'a', differentiate with respect to , where t is a non-zero real number.

Show answer & solution
Answer:
  1. Let and .
  2. (for )
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OR
Q35 (OR) (OR)5 marksLong AnswerContinuity and Differentiability

Find if , where a and b are constants.

Show answer & solution
Answer:
  1. Let , , ; then .
  2. Adding:
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Q364 marksCase StudyProbability

A gardener wanted to plant vegetables in his garden. Hence he bought 10 seeds of brinjal plant, 12 seeds of cabbage plant and 8 seeds of radish plant. The shopkeeper assured him of germination probabilities of brinjal, cabbage and radish to be 25%, 35% and 40% respectively. But before he could plant the seeds, they got mixed up in the bag and he had to sow them randomly.
Based upon the above information, answer the following questions :
(i) Calculate the probability of a randomly chosen seed to germinate. (2)
(ii) What is the probability that it is a cabbage seed, given that the chosen seed germinates ? (2)

Show answer & solution
Answer: (i) 0.33 (ii)
  1. Let : the seed is brinjal, cabbage, radish; G: the seed germinates.
  2. , , ; , ,
  3. (i)
  4. (ii)
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Q374 marksCase StudyApplication of Derivatives

A carpenter needs to make a wooden cuboidal box, closed from all sides, which has a square base and fixed volume. Since he is short of the paint required to paint the box on completion, he wants the surface area to be minimum.
On the basis of the above information, answer the following questions :
(i) Taking length = breadth = x m and height = y m, express the surface area (S) of the box in terms of x and its volume (V), which is constant. (1)
(ii) Find . (1)
(iii) (a) Find a relation between x and y such that the surface area (S) is minimum. (2)
OR
(iii) (b) If surface area (S) is constant, the volume (V) , x being the edge of base. Show that volume (V) is maximum for . (2)

Show answer & solution
Answer: (i) (ii) (iii)(a) (the box is a cube) OR (iii)(b) Proved.
  1. (i) ;
  2. (ii)
  3. (iii)(a) ; , so S is minimum when .
  4. (iii)(b)
  5. at , so V is maximum there.
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Q384 marksCase StudyRelations and Functions

Let A be the set of 30 students of class XII in a school. Let , N is a set of natural numbers such that function f(x) = Roll Number of student x.
On the basis of the given information, answer the following :
(i) Is f a bijective function ? (1)
(ii) Give reasons to support your answer to (i). (1)
(iii) (a) Let R be a relation defined by the teacher to plan the seating arrangement of students in pairs, where
R = {(x, y) : x, y are Roll Numbers of students such that y = 3x}.
List the elements of R. Is the relation R reflexive, symmetric and transitive ? Justify your answer. (2)
OR
(iii) (b) Let R be a relation defined by
R = {(x, y) : x, y are Roll Numbers of students such that }.
List the elements of R. Is R a function ? Justify your answer. (2)

Show answer & solution
Answer: (i) No (ii) f is one-one (different students have different roll numbers) but not onto, since only 30 natural numbers are images (iii)(a) R = {(1, 3), (2, 6), (3, 9), (4, 12), (5, 15), (6, 18), (7, 21), (8, 24), (9, 27), (10, 30)}; R is neither reflexive, nor symmetric, nor transitive OR (iii)(b) R = {(1, 1), (2, 8), (3, 27)}; R is not a function on the set of roll numbers, since roll numbers 4 to 30 have no image
  1. (i) and (ii) Two different students cannot have the same roll number, so f is one-one. The codomain N is infinite but the range has only 30 elements, so f is not onto. Hence f is not bijective.
  2. (iii)(a) Taking roll numbers 1 to 30: R = {(1, 3), (2, 6), (3, 9), (4, 12), (5, 15), (6, 18), (7, 21), (8, 24), (9, 27), (10, 30)}.
  3. Not reflexive: (1, 1) ∉ R. Not symmetric: (1, 3) ∈ R but (3, 1) ∉ R. Not transitive: (1, 3), (3, 9) ∈ R but (1, 9) ∉ R.
  4. (iii)(b) only for : R = {(1, 1), (2, 8), (3, 27)}.
  5. Each of 1, 2, 3 has exactly one image, but roll numbers 4 to 30 have no image, so R is not a function from the set of roll numbers to itself (it is a function only on its domain {1, 2, 3}).
Also asked in: 2025 65/5/1, 2025 65/5/3
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