See below CBSE Class 12 Mathematics Sample Paper Set B 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.
Section – A
1. If A is a square matrix satisfying A2 = I, then what is the inverse of A?
Answer. A–1 = A
2. Show that the points A(3, –5, 1), B(–1, 0, 8) and C(7, –10, –6) are collinear.
Answer.
3. Equation of the plane passing through (2, 3, –1) and is perpendicular to the vector 3î − 4ĵ + 7k̂
Answer. Equation of plane passing through (2, 3, –1) is
a(x – 2) + b(y – 3) + c(z + 1) = 0 …(i)
(i) is perpendicular to the vector 3î − 4ĵ + 7k̂
∴ 3(x – 2) –4(y – 3) + 7(z + 1) = 0
⇒ 3x – 6 – 4y + 12 + 7z + 7 = 0
⇒ 3x – 4y + 7z + 13 = 0
Answer.
(∵ sin x is an odd function
∴ sin3x cos2x and sin x cos3x are odd functions)
Section – B
5. Prove that the function f(x) = x3 – 6x2 + 15x – 18 is strictly increasing on R.
Answer.
Given f(x) = x3 – 6x2 + 15x – 18, Df = R.
Differentiating w.r.t. x, we get
f ′(x) = 3x2 – 6⋅2x + 15⋅1 – 0 = 3(x2 – 4x + 5)
= 3[(x – 2)2 + 1] ≥ 3 (Q(x – 2)2 ≥ 0 for all x ∈ R)
⇒ f ′(x) > 0 for all x ∈ R
⇒ f(x) is strictly increasing function for all x ∈ R.
6. Find the order of differential equation formed by the family of curves represented as xy = Aex + Be–x + x2
Answer. xy = Aex + Be–x + x2
Since there are two unknowns A and B, therefore the given equation has to be differentiated two times.
Hence the order of the given differential equation is 2.
7. Evaluate the definite integral :
Answer.
8. If a̅,, b̅ and c̅ are three vectors such that a̅, ⋅ b̅ = a̅, ⋅ c̅, then show that a̅, = 0¯ or b̅ = c̅ or a̅, is perpendicular to b̅ – c̅
Answer.
9. Give an example to show that the relation R in the set of natural numbers, defined by R = {(x, y), x, y ∈ N, x ≤ y2} is not transitive.
Answer. Given R = {(x, y), x, y ∈ N, x ≤ y2}
Let x = 14, y = 4, z = 2
(x, y) ∈ R as 14 ≤ (4)2, true.
(y, z) ∈ R as 4 ≤ (2)2, true.
Also, (x, z) ∉ R as 14 ≤ (2)2 ⇒ 14 ≤ 4, false.
Hence, the relation is not transitive.
10. Write the vector equation of the following line :
Answer.
Answer.
12. Form the differential equation corresponding to the function y = c(x – c)2.
Answer.
Consider y = c(x – c)2 …(i)
Differentiating (i) w.r.t. x, we get
y′ = 2c(x – c) …(ii)
Dividing (i) by (ii), we get
Answer.
14. Differentiate the function with respect to x :
Answer.
15. Find the probability of drawing
(i) a diamond card (ii) an ace card in each of the two consecutive draws from a well shuffled pack of cards, if the card drawn is not replaced after the first draw.
Answer. Total number of cards = 52
(i) Number of diamond cards = 13
16. Prove that :
Answer.
17. Find the interval in which the value of the determinant of the matrix A lies.
Answer.
∴ |A| = 1[1 + sin2θ] –sinθ [–sinθ + sinθ]
+1 [sin2θ + 1]
= 1 + sin2θ + sin2θ + 1 = 2 + 2sin2θ
We know, –1 ≤ sinθ ≤ 1
⇒ 0 ≤ sin2θ ≤ 1 ⇒ 0 ≤ 2sin2θ ≤ 2
⇒ 2 ≤ 2 + 2sin2θ ≤ 4
Hence, 2 ≤ |A| ≤ 4
Answer.
19. Find the angle between the line
r̅ = (î – ĵ + k̂) + λ ( 2î – ĵ + 3k̂ and the plane
r̅ · (2î + ĵ – k̂) = 4 Find whether the line is parallel to the plane or not.
Answer. Equation of line is r̅ = (î − ĵ + k̂)+ l(2î − ĵ + 3k̂)
Equation of plane is r̅ ⋅(2î + ĵ − k̂) = 4.
On comparing with the general equation of a line r̅ = a̅ + λb̅ and the general equation of plane r̅ ⋅n̅ = d, we have b̅ = 2i − j + 3k and n̅ = 2î + ĵ − k̂.
The angle ϕ between the line and plane is
∴ ϕ = 0
Hence, line is parallel to the plane.
20. Let A = {1, 2, 3, …, 9} and R be the relation in A × A defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation.
OR
Let ‘*’ be a binary operation on the set
Find the identity element and the inverse element of each element of the set for the operation ‘*’.
Answer. Given A = {1, 2, 3, 4,…,9} ⊂ N, the set of natural numbers.
To show : R is an equivalence relation.
(i) Reflexivity : Let (a, b) be an arbitrary
element of A × A. Then, we have
(a, b) ∈ A × A ⇒ a, b ∈ A
⇒ a + b = b + a
(by commutativity of addition on A ⊂ N)
⇒ (a, b) R (a, b)
Thus, (a, b) R (a, b) for all (a, b) ∈ A × A
So, R is reflexive.
(ii) Symmetry: Let (a, b), (c, d) ∈ A × A such that
(a, b) R (c, d) ⇒ a + d = b + c
⇒ b + c = a + d
⇒ c + b = d + a
(by commutativity of addition on A ⊂ N)
⇒ (c, d) R (a, b).
Thus, (a, b) R (c, d) ⇒ (c, d) R (a, b) for all
(a, b), (c, d) ∈ A × A.
So, R is symmetric.
(iii) Transitivity: Let (a, b), (c, d), (e, f) ∈ A × A
such that (a, b) R (c, d) and (c, d) R (e, f)
Now, (a, b) R (c, d) ⇒ a + d = b + c …(1)
and (c, d) R (e, f) ⇒ c + f = d + e …(2)
Adding (1) and (2), we get
(a + d) + (c + f) = (b + c) + (d + e)
⇒ a + f = b + e ⇒ (a, b) R (e, f)
Thus, (a, b) R (c, d) and (c, d) R (e, f)
⇒ (a, b) R (e, f).
So, R is transitive.
∴ R is an equivalence relation.
OR
The operation table for * is given as
From the table, we note that
a * 0 = 0 * a = a ∈ {0, 1, 2, 3, 4, 5}
Hence, 0 is the identity element for the operation *
Now, from the table inverse of 0, 1, 2, 3, 4 and 5 are 0, 5, 4, 3, 2 and 1 respectively
Answer.
22. Determine the values of a, b and c for which the function
Answer. Here, f(0) = c;
L.H.L. at x = 0
f(3) = 27 + 9b + 3a + 5 = 3a + 9b + 32
Now, f(1) = f(3)
⇒ a + b + 6 = 3a + 9b + 32
⇒ 2a + 8b + 26 = 0
⇒ a + 4b + 13 = 0 …(i)
Now, f ′(x) = 3x2 + 2bx + a
23. Solve the differential equation :
Answer.
Section – D
24. A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the
Answer. Let AB (= l) be the hypotenuse of a triangle AOB and let P be a point on the hypotenuse AB such
that PL ⊥ OA and PM ⊥ OB and PL = a, PM = b.
Let ∠OAB = θ, then ∠MPB = ∠OAB = θ
25. Three cards are drawn successively with replacement from a well-shuffled deck of 52 cards. A random variable X denotes the number of hearts in the three cards drawn. Find the mean and variance of X.
OR
Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time, a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability
(i) the patient followed a course of meditation and yoga.
(ii) the patient followed the prescription of certain drug.
Answer. Let E = Event of drawing a heart
OR
Let A, E1 and E2 respectively denotes the event that a person has a heart attack, the selected person followed the course of yoga and meditation and the person adopted the drug prescription.
P(A/E1) = 0.40 × 0.70 = 0.28
P(A/E2) = 0.40 × 0.75 = 0.30
Probability that the patient suffering heart attack followed course of meditation and yoga is
26. A factory owner wants to purchase two types of machines A and B for his factory. The machine A requires an area of 1000 m2 and 12 skilled men for running it and its daily output is 50 units, whereas the machine B requires 1200 m2 area and 8 skilled men, and its daily output is 40 units. An area of 7600 m2 and 72 skilled men be available to operate the machines.
(i) How many machines of each type should be bought to maximize the daily output?
(ii) Write two advantages of keeping skilled men in a firm.
Answer. (i) If x machines of type A and y machines of type B are bought, then our problem is to maximize Z = 50x + 40y subject to constraints x ≥ 0, y ≥ 0
1000x + 1200y ≤ 7600 ⇒ 5x + 6y ≤ 38
12 + 8y ≤ 72 ⇒ 3x + 2y ≤ 18
Feasible region is shown shaded in the figure
i.e., Daily output should be maximized. When4 machines of type A and 3 machines of type B are bought.
(ii) (a) High productivity
(b) Time management
(c) Efficiency
equation of the plane containing these lines.
OR
Find the shortest distance between the lines whose vector equations are
r̅, = (1− t) î + (t − 2) ĵ + (3 − 2t)k̂, and
r̅, = (s +1) î + (2s −1) ĵ − (2s +1)k̂.
Answer. We know that the lines
⇒ x(8 – 9) – (y – 2)(4 – 6) + (z + 3)(3 – 4) = 0
⇒ –x + 2 (y – 2) – (z + 3) = 0
⇒ x – 2y + z + 7 = 0
28. Let f : N → R be a function defined as
f(x) = 4x2 + 12x + 15.
Show that f : N → S is invertible, where S is the range of f. Also, find inverse of f.
Answer. f(x) = 4x2 + 12x + 15
Let f(x1) = f(x2)
⇒ 4x1
2 + 12x1 + 15 = 4x2
2 + 12x2 + 15
⇒ 4x1 (x1 + 3) = 4x2 (x2 + 3)
⇒ x1 + 3 = x2 + 3
⇒ x1 = x2
Hence, f is one-one
Also codomain is equal to the range.
So, f is onto.
Since, f is both one-one and onto.
Hence, f : N → S is invertible.
Let y = f(x)
⇒ 4x2 + 12x + 15 = y
⇒ 4x2 + 12x + 9 – 9 + 15 = y
⇒ (2x + 3)2 + 6 = y ⇒ (2x + 3)2 = y – 6
Prove that the curves y = x2 and x = y2 divide the square bounded by x = 0, y = 0, x = 1 and y = 1 into three parts which are equal in area.
Answer.
We have x = y2 and y = x2
⇒ y = y4 ⇒ y = 0 or 1
When y = 0 ⇒ x = 0 and when y = 1 ⇒ x = 1
So, the points of intersection are 0(0, 0) and A (1, 1)
Let A1, A2, A3 be the area denoted in the figure.
We need to prove A1 = A2 = A3
Hence, y2 = x and x2 = y divide the area of square bounded by x = 0, y = 0, x = 1 and y = 1 into three equal parts.