See below CBSE Class 12 Mathematics Sample Paper Set D with solutions. We have provided CBSE Sample Papers for Class 12 Mathematics as per the latest paper pattern issued by CBSE for the current academic year. All sample papers provided by our Class 12 Mathematics teachers are with answers. You can see the sample paper given below and use them for more practice for Class 12 Mathematics examination.
PART – A
Section – I
1. Check whether the function f : R → R defined as f(x) = x3 is one-one or not.
Answer: Let f(x1) = f(x2) for some x1, x2 ∈ R.
⇒ (x1)3 = (x2)3 ⇒ x13 – x23 = 0
⇒ (x1 – x2)(x12 + x22 + x1x2) = 0
OR
How many reflexive relations are possible in a set A whose n(A) = 3 ?
Answer: Number of reflexive relations on a set having n
elements = 2n(n – 1)
So, required number of reflexive relations = 23(3 – 1) = 26
2. A relation R in S = {1, 2, 3} is defined as R = {(1, 1), (1, 2), (2, 2), (3, 3)}. Which element(s) of relation R be removed to make R an equivalence relation?
Answer: We have, R = {(1, 1), (2, 2), (3, 3), (1, 2)}
which is reflexive and transitive.
For R to be symmetric (1, 2) should be removed from R.
3. A relation R in the set of real numbers R defined as R = {(a, b) : a = b} is a function or not. Justify
Answer: Since √a is not defined for a ∈ (–∞, 0)
∴ a = b is not a function.
OR
An equivalence relation R in A divides it into equivalence classes A1, A2, A3. What is the value of A1 ∪ A2 ∪ A3 and A1 ∩ A2 ∩ A3.
Answer: As we know that, union of all equivalence classes of a set is the set itself.
∴ A1 ∪ A2 ∪ A3 = A
Also, A1 ∩ A2 ∩ A3 = φ
[∴ Equivalence classes are either equal or disjoint]
4. If A and B are matrices of order 3 × n and m × 5 respectively, then find the order of matrix 5A – 3B, given that it is defined.
Answer: For addition or subtraction of two matrices to be defined, the two matrices should be of same order.
∴ 3 × n = m × 5 ⇒ m = 3 and n = 5
So, order of matrix (5A – 3B) is 3 × 5.
5. Find the value of A2, where A is a 2 × 2 matrix whose elements are given by
Answer:
OR
Given that A is a square matrix of order 3 × 3 and |A| = –4. Find |adj A|.
Answer: We know, |adjA| = |A|n – 1, where n × n is the order of non-singular matrix A.
∴ |adj A| = (–4)3 – 1 = 16
6. Let A = [aij] be a square matrix of order 3 × 3 and |A|= –7. Find the value of a11 A21+ a12A22+ a13 A23 where Aij is the cofactor of element aij.
Answer: We know that, if elements of a row are multiplied
with cofactors of any other row, then their sum is 0.
∴ a11A21 + a12A22 + a13A23 = 0.
7. Find ∫ex (1-cosx + cosec2 x) dx.
Answer:
OR
Evaluate :
Answer: ∴ f(x) = x2sinx is an odd function.
8. Find the area bounded by y = x2, the x- axis and the lines x = –1 and x =1.
Answer: Required area,
9. How many arbitrary constants are there in the particular solution of the differential equation dy/dx = -4xy2 ; y(0) = 1 ?
Answer: There is no arbitrary constant in a particular solution of differential equation.
OR
For what value of n is the following a homogeneous differential equation: dy/dx = x3 – yn/x2y + xy2
Answer: For n = 3, the given differential equation becomes homogeneous.
10. Find a unit vector in the direction opposite to -3/4 ĵ .
Answer: Let a̅ a be the unit vector in the direction opposite
11. Find the area of the triangle whose two sides are represented by the vectors 2ĵ and -3ĵ .
Answer: Area of the triangle
12. Find the angle between the unit vectors
Answer:
13. Find the direction cosines of the normal to YZ plane.
Answer: Since, normal to the YZ plane is a line parallel to X-axis.
∴ Direction cosines of normal to YZ plane are 〈1, 0, 0〉.
14. Find the coordinates of the point where the line x+3/3 = y-1/-1 = z-5/-5 cuts the XY plane.
Answer: The given line is
Now, as given line cuts XY-plane.
So, z-coordinate of point of intersection = 0.
⇒ x+3/3 = y-1/-1 = ⇒ x = 0, y = 0
∴ Required point is (0, 0, 0).
15. The probabilities of A and B solving a problem independently are 1/3 and 1/4 respectively. If both of them try to solve the problem independently, what is the probability that the problem is solved?
Answer: Required probability
= 1 – P(problem is not solved)
= 1−(A̅).P(B̅) = 1 – 2/3 x 3/4 = 1/
16. The probability that it will rain on any particular day is 50%. Find the probability that it rains only on first 4 days of the week.
Answer: Let A be the event that it will rain on any particular day.
Then, P(A) = 50% = 1/2 ⇒ P(A̅) = 1-1/2 = 1/2
Now, required probability
Section – II
Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark.
17. An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200 m as shown below:
Based on the above information answer the following:
(i) If x and y represents the length and breadth of the rectangular region, then the relation between the variables is
(a) x + π y = 100
(b) 2x + π y = 200
(c) π x + y = 50
(d) x + y = 100
Answer
B
(ii) The area of the rectangular region A expressed as a function of x is
(a) 2/π(100x-x2)
(b) 1/π(100x-x2)
(c) x/π(100-x)
(d) πy2(100x-x2)
Answer
A
(iii) The maximum value of area A is
(a) π/3200 m2
(b) 3200/π m2
(c) 5000/π m2
(d) 1000/π m2
Answer
C
(iv) The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen the value of x should be
(a) 0 m
(b) 30 m
(c) 50 m
(d) 80 m
Answer
A
(v) The extra area generated if the area of the whole floor is maximized is
(a) 3000/π m2
(b) 5000/π m2
(c) 7000/π m2
(d) No change, Both areas are equal
Answer
D
18. In an office three employees Vinay, Sonia and Iqbal process incoming copies of a certain form. Vinay process 50% of the forms, Sonia processes 20% and Iqbal the remaining 30% of the forms. Vinay has an error rate of 0.06, Sonia has an error rate of 0.04 and Iqbal has an error rate of 0.03.
Based on the above information answer the following:
(i) The conditional probability that an error is committed in processing given that Sonia processed the form is
(a) 0.0210
(b) 0.04
(c) 0.47
(d) 0.06
Answer
B
(ii) The probability that Sonia processed the form and committed an error is
(a) 0.005
(b) 0.006
(c) 0.008
(d) 0.68
Answer
C
(iii) The total probability of committing an error in processing the form is
(a) 0
(b) 0.047
(c) 0.234
(d) 1
Answer
B
(iv) The manager of the company wants to do a quality check. During inspection he selects a form at random from the days output of processed forms. If the form selected at random has an error, the probability that the form is NOT processed by Vinay is
(a) 1
(b) 30/47
(c) 20/47
(d) 17/47
Answer
D
(v) Let A be the event of committing an error in processing the form and let E1, E2 and E3 be the events that Vinay, Sonia and Iqbal processed the form. The value of
(a) 0
(b) 0.03
(c) 0.06
(d) 1
Answer
D
PART – B
Section – III
19. Express tan-1(cos x/1-sin x) , -3π/2 < x < π/2 in the simplest form.
Answer: We have,
20. If A is a square matrix of order 3 such that A2 = 2A, then find the value of |A|.
Answer: We have, A2 = 2A
⇒ |AA| = |2A|
⇒ |A||A| = 8|A| (Q |AB| = |A||B| and |kA| = kn|A|,
where n is order of square matrix A)
⇒ |A| (|A| – 8) = 0 ⇒ |A| = 0 or 8
OR
show that A2 – 5A + 7I = O. Hence find A–1.
Answer:
21. Find the value(s) of k so that the following function is continuous at x = 0.
Answer:
22. Find the equation of the normal to the curve y = x+1/x , x > 0 perpendicular to the line 3x − 4y = 7.
Answer: The given curve is
y = x + 1/x ⇒ dy/dx = 1 – 1/x2
Since, normal is perpendicular to 3x – 4y = 7, so tangent is parallel to it.
23. Find
Answer:
OR
Evaluate :
Answer:
24. Find the area of the region bounded by the parabola y2 = 8x and the line x = 2.
Answer: Required area
25. Solve the following differential equation: dy/dx = x3 cosec y, given that y(0) = 0
Answer: The given differential equation is
26. Find the area of the parallelogram whose one side and a diagonal are represented by coinitial vectors î – ĵ + k̂and 4î + 5k̂respectively.
Answer: Let ABCD is a parallelogram such that
27. Find the vector equation of the plane that passes through the point (1, 0, 0) and contains the line r̅ = λĵ .
Answer: Let the normal vector to the plane be n̅ .
28. A refrigerator box contains 2 milk chocolates and 4 dark chocolates. Two chocolates are drawn at random.
Find the probability distribution of the number of milk chocolates. What is the most likely outcome?
Answer: Let X denotes the number of milk chocolates drawn. Then probability distribution table is
Most likely outcome is getting one chocolate of each type.
OR
Given that E and F are events such that P(E) = 0.8, P(F) = 0.7, P(E ∩ F) = 0.6. Find P(E̅ | F̅).
Answer:
Section – IV
29. Check whether the relation R in the set Z of integers defined as R = {(a, b) : a + b is “divisible by 2”} is reflexive, symmetric or transitive. Write the equivalence class containing 0, i.e., [0].
Answer: (i) Reflexive :
Since, a + a = 2a which is even.
∴ (a, a) ∈ R ∀ a ∈ Z
Hence R is reflexive.
(ii) Symmetric:
If (a, b) ∈ R, then a + b = 2l ⇒ b + a = 2l
⇒ (b, a) ∈ R. Hence R is symmetric.
(iii) Transitive :
If (a, b) ∈ R and (b, c) ∈ R
then a + b = 2l …. (i) and b + c = 2m … (ii)
Adding (i) and (ii), we get
a + 2b + c = 2(l + m) ⇒ a + c = 2(l + m – b)
⇒ a + c = 2k, where k = l + m – b ⇒ (a, c) ∈ R
Hence R is transitive.
Equivalence class containing 0 i.e.,
[0] = {…, –4, –2, 0, 2, 4, …}
30. If y = exsin2 x + (six)x , find dy/dx .
Answer: We have, y = exsin2x + (sinx)x ⇒ y = u + v,
where u = exsin2x and v = (sinx)x
⇒ dy/dx = exsin2 x [x.(2sin x cos x ) + sin2 x]
= exsin2x [x(sin2x) + sin2x] … (ii)
Also, v = (sinx)x
⇒ logv = xlog(sinx)
Differentiating both sides with respect to x, we get
31. Prove that the greatest integer function defined by f(x) = [x], 0 < x < 2 is not differentiable at x = 1.
Answer: We have, f(x) = [x], 0 < x < 2.
OR
If x = a secθ, y = b tanθ, find d2y/dx2 at θ = π/6 .
Answer: We have, y = b tanθ ⇒ dy/dθ = bsec2 θ
32. Find the intervals in which the function f given by f(x) = tanx – 4x, x ∈ (0,π/2) is
(a) strictly increasing (b) strictly decreasing
Answer: We have, f(x) = tanx – 4x
⇒ f ′(x) = sec2x – 4
(a) For f(x) to be strictly increasing, f ′(x) > 0
⇒ sec2x – 4 = 0 ⇒ sec2x > 4
33. Find
Answer:
34. Find the area of the region bounded by the curves x2 + y2 = 4, y = √3x and x-axis in the first quadrant.
Answer: Solving y = √3x and x2 + y2 = 4, we get
x2 + 3x2 = 4 ⇒ x2 = 1 ⇒ x = 1
OR
Find the area of the ellipse x2 + 9y2 = 36 using integration.
Answer: Given equation of ellipse is
35. Find the general solution of the following differential equation:
xdy – (y + 2×2)dx = 0
Answer: The given differential equation can be written as
Section – V
36. If A=
find A–1. Hence solve the system of equations;
x – 2y = 10
2x – y – z = 8
–2y + z = 7
Answer:
OR
Evaluate the product AB, where A =
Hence solve the system of linear equations
x – y = 3
2x + 3y + 4z = 17
y + 2z = 7
Answer: We have,
37. Find the shortest distance between the lines r̅ = 3î + 2ĵ – 4k̂+ λ (î + 2ĵ +2k̂) and r̅ = 5î – 2ĵ + μ (3î + 2ĵ + 6k̂) .
If the lines intersect, then find their point of intersection.
Answer:
OR
Find the foot of the perpendicular drawn from the point (–1, 3, –6) to the plane 2x + y – 2z +5=0. Also find the equation and length of the perpendicular.
Answer: Let P be the given point and Q be the foot of the perpendicular.
Then, equation of PQ will be,
38. Solve the following linear programming problem (L.P.P) graphically.
Maximize Z = 3x + y
subject to constraints ;
x + 2y ≥ 100
2x – y ≤ 0
2x + y ≤ 200
x, y ≥ 0
Answer: Maximize Z = 3x + y
Subject to x + 2y ≥ 100
2x – y ≤ 0
2x + y ≤ 200
x ≥ 0, y ≥ 0
Converting the given inequations into equations, we get
x + 2y = 100 … (i)
2x – y = 0 … (ii)
2x + y = 200 … (iii)
Now, draw the graphs of (i), (ii) and (iii).
The feasible region is shaded region and corner points are A(0, 50), B(0, 200), C(50, 100) and D(20, 40).
The values of Z at corner points are shown in the following table:
OR
The corner points of the feasible region determined by the system of linear constraints are as shown below:
Answer each of the following:
(i) Let Z =3x – 4y be the objective function. Find the maximum and minimum value of Z and also the corresponding points at which the maximum and minimum value occurs.
(ii) Let Z = px + qy ,where p, q > 0 be the objective function. Find the condition on p and q so that the maximum value of Z occurs at B(4,10) and C(6, 8). Also mention the number of optimal solutions in this case.
Answer: (i)
Thus, maximum value of Z is 12 at E(4, 0).
and minimum value of Z is –32 at A(0, 8).
(ii) Since maximum value of Z occurs at B(4, 10) and C(6, 8).
∴ 4p + 10q = 6p + 8q ⇒ 2q = 2p ⇒ p = q
Number of optional solutions are infinite.
[∴ Every point on the line segment BC joining the two corner points B and C also give the same maximum value]
