Students must start practicing the questions from CBSE Sample Papers for Class 12 Maths with Solutions Set 9 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Maths Set 9 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 80
General Instructions:
- This Question paper contains five sections A, B, C, D, and E. Each section is compulsory. However, there are internal choices in some questions.
- Section A has 18 MCQs and 2 Assertion-Reason-based questions of 1 mark each.
- Section B has 5 Very Short Answer (VSA) type questions of 2 marks each.
- Section C has 6 Short Answer (SA) type questions of 3 marks each.
- Section D has 4 Long Answer (LA) type questions of 5 marks each.
- Section E has 3 sources based/case based/passage based/integrated units of assessment (4 marks each) with subparts.
Section – A (20 Marks)
(Multiple Choice Questions Each question carries 1 mark)
Question 1.
Let f: R → R be defined by f(x) = x2 + 1. The pre-images of 17 and -3 are respectively. [1]
(a) {-4, 3}, {-3, -4}
(b) {-1, 1}, {-3, 4}
(c) {-4, 4}, {-3, 3}
(d) {-4, 4}, not defined
Solution:
(d) {-4, 4}, not defined
Explanation:
For pre-image of 17:
f(x) = x2 + 1 = 17
⇒ x2 = 17 – 1 = 16
⇒ x = ±√16 = ±4
So, x ∈ {-4, 4}
For pre-image of -3:
f(x) = x2 + 1 = -3
⇒ x2 = -3 – 1 = -4
No real value of x is possible.
So, x ∈ φ
Question 2.
Simplify: \(\sin \left(2 \sin ^{-1} \sqrt{\frac{63}{65}}\right)\). [1]
(a) \(\sqrt{\frac{63}{65}}\)
(b) \(\frac{2 \sqrt{126}}{65}\)
(c) \(\sqrt{\frac{123}{65}}\)
(d) Not defined
Solution:
(b) \(\frac{2 \sqrt{126}}{65}\)
Explanation:
Question 3.
If sin-1x – cos-1x = \(\frac{\pi}{6}\), then the value of x is: [1]
(a) \(\frac{1}{2}\)
(b) \(\frac{\sqrt{3}}{2}\)
(c) \(\frac{1}{\sqrt{2}}\)
(d) 1
Solution:
(b) \(\frac{\sqrt{3}}{2}\)
Explanation:
⇒ x = \(\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\)
Question 4.
If A = \(\left[\begin{array}{rrr}
1 & -2 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\), then find A(adj A). [1]
(a) 6I3
(b) 8I3
(c) I3
(d) 2I3
Solution:
(b) 8I3
Explanation:
Here, |A| = 1(8 – 6) + 2(0 + 9) + 2(0 – 6) = 8
Also, we know that
A(adj A) = |A| I3
⇒ A(adj A) = 8I3
Question 5.
If A = \(\left|\begin{array}{rrr}
2 & -1 & 2 \\
2 & x & 4 \\
-1 & 1 & -2
\end{array}\right|\), and |A| = 0, then the value(s) of x is: [1]
(a) -2
(b) 1
(c) -1
(d) 4
Solution:
(a) -2
Explanation:
A = \(\left|\begin{array}{rrr}
2 & -1 & 2 \\
2 & x & 4 \\
-1 & 1 & -2
\end{array}\right|\) = 0, gives
2(-2x – 4) + 1(-4 + 4) + 2(2 + x) = 0
⇒ -4x – 8 + 4 + 2x = 0
⇒ x = -2
Question 6.
If A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), then for any natural number n, det (An) is: [1]
(a) 1
(b) 0
(c) -1
(d) 2
Solution:
(a) 1
Explanation:
An = cos2 nθ + sin2 nθ = 1
Hence, det (An) = 1
Question 7.
If x = sin t and y = 2t, then \(\frac{d^2 y}{d x^2}\) is: [1]
(a) 2 sin t cos2t
(b) 2 cos 2t
(c) 2 sec2t tan t
(d) 2 sin 2t
Solution:
(c) 2 sec2t tan t
Explanation:
Question 8.
\(\int_0^1\left(x^6+7 x^5+6 x^4+5 x^3+4 x^2+3 x+1\right) e^x d x\): [1]
(a) 6e
(b) 7e2 – 1
(c) e – 1
(d) 5e
Solution:
(a) 6e
Explanation:
Question 9.
\(\left.\int_1^4(\mid x-1)+|x-2|+|x-3|\right) d x\) is: [1]
(a) \(\frac{5}{2}\)
(b) 4
(c) \(\frac{3}{5}\)
(d) \(\frac{19}{2}\)
Solution:
(d) \(\frac{19}{2}\)
Explanation:
Question 10.
Find \(\int \frac{1}{x} \log (\log x) d x\). [1]
(a) log(log x) + C
(b) log x |log(log x) – log x + C|
(c) (log x – 1) + C
(d) log(log(log x)) + C
Solution:
(b) log x |log(log x) – log x + C|
Explanation:
Let I = \(\int \frac{1}{x} \log (\log x) d x\)
Let log x = y
Question 11.
Find the integrating factor of the differential equation: [1]
\(\left(1+y^2\right) \frac{d x}{d y}+\left[x-e^{\tan ^{-1} y}\right]=0\)
(a) \(e^{\tan ^{-1}} y\)
(b) 1 + y2
(c) y2
(d) tan-1y
Solution:
(a) \(e^{\tan ^{-1}} y\)
Explanation:
Question 12.
The sum of the order and the degree of the differential equation \(\left(\frac{d y}{d x}\right)^4+3 x \frac{d^2 y}{d x^2}=0\) is: [1]
(a) 3
(b) 2
(c) 5
(d) 6
Solution:
(a) 3
Explanation:
Here, the order of the differential equation is 2 and its degree is 1
∴ sum = 2 + 1 = 3
Question 13.
If |\(\vec{a}\)|= 3, |\(\vec{b}\)| = 5, |\(\vec{c}\)| = 7 and \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is: [1]
(a) \(\frac{\pi}{3}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{2\pi}{3}\)
(d) \(\frac{\pi}{4}\)
Solution:
(a) \(\frac{\pi}{3}\)
Explanation:
Question 14.
If \(\vec{a}\) and \(\vec{b}\) are two unit vectors inclined at an angle \(\frac{\pi}{3}\), then \(|\vec{a}+\vec{b}|\) is: [1]
(a) √5
(b) √2
(c) √3
(d) √6
Solution:
(c) √3
Explanation:
\(|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}||\vec{b}| \cos \theta\)
= 1 + 1 + 2 cos(π/3)
= 3
∴ \(|\vec{a}+\vec{b}|\) = √3
Question 15.
If the vector equation of the line through (3, 5, 4) and (5, 8, 11) is \(\vec{r}=(3 \hat{i}+5 \hat{j}+4 \hat{k})+\lambda(2 \hat{i}+a \hat{j}+b \hat{k})\), then a + b is: [1]
(a) 3
(b) 10
(c) 11
(d) 7
Solution:
(b) 10
Explanation:
The vector equation of the line through (3, 5, 4) and (5, 8, 11) is
\(\vec{r}=3 \hat{i}+5 \hat{j}+4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+7 \hat{k})\)
Comparing this equation of the line with the given equation of the line, we get
a = 3 and b = 7
∴ a + b = 10
Question 16.
If P(A) = \(\frac{4}{5}\) and P(A ∩ B) = \(\frac{7}{10}\), then P(B/A) is: [1]
(a) \(\frac{1}{8}\)
(b) \(\frac{7}{10}\)
(c) \(\frac{4}{5}\)
(d) \(\frac{7}{8}\)
Solution:
(d) \(\frac{7}{8}\)
Explanation:
Question 17.
A line makes an angle of 60° each with a y-axis and z-axis respectively. Then the angle which it makes with the x-axis anticlockwise is: [1]
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Solution:
(b) 45°
Explanation:
⇒ cos α = \(\frac{1}{\sqrt{2}}\)
⇒ α = 45° (anti-clockwise)
Question 18.
The area (in sq. units) bounded by the curve given in the figure y = ex, the x-axis and between x = -1 and x = 2 is: [1]
(a) \(\left(e^2-\frac{1}{e}\right)\)
(b) \(\left(\frac{e-1}{e}\right)\)
(c) \(\left(\frac{1-e^2}{e}\right)\)
(d) \(\left(e^3-\frac{1}{e^2}\right)\)
Solution:
(a) \(\left(e^2-\frac{1}{e}\right)\)
Explanation:
Assertion-Reason Based Questions
In the following questions, a statement of assertion (A) is followed by a statement of the reason (R).
Choose the correct answer out of the following choices.
(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.
Question 19.
Assertion (A): \(\int \frac{d x}{x^2+2 x+3}\) = \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x+1}{\sqrt{2}}\right)+c\)
Reason (R): \(\int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\). [1]
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
Question 20.
Assertion (A): If \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), then angle between |\(\vec{a}\)| and |\(\vec{b}\)| is 90°.
Reason(R): \(\vec{a}+\vec{b}=\vec{b}+\vec{a}\). [1]
Solution:
(b) Both A and R are true but R is not the correct explanation of A.
Explanation:
Section – B (10 Marks)
This section comprises very short answer type-questions (VSA) of 2 marks
Question 21.
Find the value of \(\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)\) \(+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left[\sin \left(-\frac{\pi}{2}\right)\right]\). [2]
Solution:
= \(-\frac{\pi}{12}\)
Question 22.
If \(\left|\begin{array}{lll}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right|\) = 10, then find the value of \(\left|\begin{array}{ccc}
a+g & b+h & c+i \\
g & h & i \\
d+a & e+b & f+c
\end{array}\right|\)
OR
Find the maximum value of \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & 1+\sin \theta & 1 \\
1 & 1 & 1+\cos \theta
\end{array}\right|\). [2]
Solution:
= 1[(1 + sin θ)(1 + cos θ) – 1] – 1[1 + cos θ – 1] + 1[1 – (1 + sin θ)]
= [1 + sin θ + cos θ + sin θ cos θ – 1] – cos θ – sin θ
= sin θ + cos θ + sin θ cos θ – cos θ – sin θ
= sin θ cos θ
The maximum value of sin θ is 1 and that of cos θ is 1
Hence, the maximum value of sin θ cos θ is 1 × 1 = 1
Question 23.
If sin-1x + sin-1y = \(\frac{\pi}{2}\), then find \(\left(\frac{d y}{d x}\right)\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\)
OR
If y = sin 3x cos 8x + sin 7x cos 2x, then find \(\frac{d y}{d x}\) at x = π. [2]
Solution:
Question 24.
Prove that \(\int_0^{\pi / 4} 2 \tan ^3 x d x=1-\log 2\). [2]
Solution:
Question 25.
Solve: x \(\frac{d y}{d x}\) = y(log y – log x + 1). [2]
Solution:
Given differential equation can also be written as
Section – C (18 Marks)
This section comprises short answer type questions (SA) of 3 marks
Question 26.
Let T be the set of all triangles in the Euclidean plane, and let a relation R on T be defined as aRb if a is congruent to b, for all a, b ∈ T. Check if it is an equivalence relation is not. [3]
Solution:
R is reflexive
a R a, for all a ∈ T, as every triangle is congruent to itself.
R is symmetric
Let a R b, where a, b ∈ T
⇒ a is congruent to b
⇒ b is congruent to a
⇒ b R a
R is transitive
Let a R b and b R c, where a, b, c ∈ T
⇒ a is congruent to b and b is congruent to c
⇒ a is congruent to c
⇒ a R c
Hence, R is an equivalence relation.
Question 27.
Find the derivative \(\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]\) with respect to \(\tan ^{-1}\left[\frac{\cos x}{1+\sin x}\right]\). [3]
Solution:
Question 28.
Integrate \(\frac{1}{x^4+1}\) w.r.t. x.
OR
Integrate \(\frac{3 x-2}{(x+3)(x+1)^2}\) w.r.t. x. [3]
Solution:
Let I = \(\int \frac{1}{x^4+1} d x\)
Dividing both numerator and denominator by x2
Question 29.
Using integration, find the area of the region bounded by the lines y = 2x + 1, y = 3x + 1, and x = 4. [3]
Solution:
The region enclosed by the given lines is the shaded region CAB.
Question 30.
Find the particular solution of the following differential equation: x2 dy + (xy + y2) dx = 0, given that y = 1 when x = 1.
OR
Prove that y = a cos (log x) + b sin (log x) is the solution of \(x^2 \frac{d^2 y}{d x^2}+x \frac{d y}{d x}+y=0\). [3]
Solution:
x2 dy + (xy + y2) dx = 0
Rearranging terms
⇒ \(\frac{d y}{d x}=-\frac{\left(x y+y^2\right)}{x^2}\)
Question 31.
The probabilities of A, B, and C hitting a target are \(\frac{1}{3}\), \(\frac{2}{7}\) and \(\frac{3}{8}\) respectively. If all three try to hit the target simultaneously, find the probability that exactly one of them can hit the target. Also, find the probability that the target will be hit.
OR
A random variable X has the following probability distribution:
Determine:
(A) k, (B) P(X < 3), (C) P(X > 6). [3]
Solution:
Section – D (20 Marks)
This section comprises long answer type questions (LA) of 5 marks.
Question 32.
If A = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\), then prove that A3 – 6A2 + 7A + 2I = 0.
OR
Using the matrix method, examine the consistency of the following system of equations:
3x – y – 2z = 2;
2y – z = -1;
3x – 5y = 3. [5]
Solution:
Question 33.
Find the intervals in which the function f defined as f(x) = 1 – 12x – 9x2 – 2x3 is strictly increasing or strictly decreasing. [5]
Solution:
f(x) = 1 – 12x – 9x2 – 2x3 gives
f'(x) = -12 – 18x – 6x2
= -6(2 + 3x + x2)
= -6(x + 2)(x + 1)
gives f'(x) = 0 implies -6(x + 2)(x + 1) = 0,
i.e. x = -1 or -2
f'(x) > 0, when -2 < x -1
f'(x) < 0, when x < -2 or x > -1
Thus, f(x) is strictly increasing in (-2, -1); and strictly decreasing in (-∞, -2) ∪ (-1, ∞).
Question 34.
Solve the following L.P.P. graphically:
Maximise Z = 100x + 300y subject to constraints
x + y ≤ 24; \(\frac{x}{2}\) + y ≤ 16; x ≥ 0, y ≥ 0
OR
Maximise Z = x + y subject to constraints
4x+ 5y ≤ 200; \(\frac{x}{25}+\frac{x}{40}\) ≤ 1; x ≥ 0, y ≥ 0. [5]
Solution:
We have, subject to the constraints
x + y ≤ 24
⇒ \(\frac{x}{2}\) + y ≤ 16
⇒ x + 2y ≤ 32
x ≥ 0, y ≥ 0
Changing inequations to equations
For x + y = 24
For x + 2y = 32
Plotting the equations on a graph paper, we get the bounded feasible region (Shaded region) with corners A(0, 16), B(16, 8), C(24, 0), O(0, 0).
Z = 100x + 300y
ZA = 100 × 0 + 300 × 16 = 4800
ZB = 100 × 16 + 300 × 8 = 4000
ZC = 100 × 24 + 300 × 0 = 2400
Z is maximum at (0, 16) and the maximum value is 4800.
OR
We have, subject to the constraints 4x + 5y ≤ 200
⇒ \(\frac{x}{25}+\frac{y}{40}\) ≤ 1
⇒ 40x + 25y ≤ 1000
x ≥ 0, y ≥ 0
Changing inequations to equations
For 4x + 5y = 200
For 40x + 25y = 1000
Plotting the equations on graph paper, we get the bounded feasible region (Shaded region) with corners A(0, 40), B(25, 0), O(0, 0).
Z = x + y
ZA = 0 + 40 = 40
ZB = 25 + 0 = 25
Z0 = 0 + 0 = 0
Z is maximum at (0, 40) and the maximum value is 40.
Question 35.
Find the equation of a line passing through the point (1, 2, -4) and perpendicular to two lines \(\vec{r}=(8 \hat{i}-19 \hat{j}+10 \hat{k})+\lambda(3 \hat{i}-16 \hat{j}+17 \hat{k})\) and \(\vec{r}=(15 \hat{i}+29 \hat{j}+5 \hat{k})+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})\). [5]
Solution:
Section – E (12 Marks)
This section comprises of 3 case-study/passage-based questions of 4 marks each with two sub-part. First, two case-study questions have three sub-parts (A), (B), (C) of marks 1, 1, 2 respectively. The third case-study question has two subparts of 2 marks.
Question 36.
A laboratory blood test is 99% effective in detecting COVID-19 infection when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e., if a healthy person is tested then with a probability of 0.005, the test will imply he has the COVID-19 infection). Let 0.1% of the population actually have the COVID-19 infection
Based on the above information, answer the following questions:
(A) What is the probability that a person, chosen at random and he/she does not has an infection? [1]
(B) What is the probability that a person has the infection and the test result is positive? [1]
(C) What is the probability that a person has the infection given that the test result is positive?
OR
Find the probability that a person, chosen at random, has the infection and also find the probability that a person does not has the infection and the test result is positive. [2]
Solution:
Let A: A person has the disease;
B: A person does not have the disease;
C: The test is positive
P(A) = 0.1% = 0.001;
P(B) = 99.9% = 0.999;
P(C/A) = 99% = 0.99;
P(C/B) = 0.005
(A) Population does not have infection = 1 – Population has an infection
= 1 – 0.001
= 0.999
(B) The probability that a person has the infection and the test result is positive is 99% = 0.99.
OR
Population actually having the COVID-19 infection = 1% = 0.001
The probability that a person does not has the infection and the test result is positive, is 0.005.
Question 37.
Our Prime minister gave a speech on 15th August 2020 at Red fort. In his speech, he announced to start of a new project named “one sun, one world, one grid”. This program aims to achieve a uniform electricity grid in the whole world which will be powered by solar energy, thereby making India a global leader in solar power.
Scientists from all over the world identified in their preliminary survey 3 grid station points whose vertices are A(1, 1, 1); B(1, 2, 3); C(2, 3, 1).
There are some questions related to the position of the coordinates.
Based on the above information, answer the following questions:
(A) Find vector \(\overrightarrow{\mathrm{AB}}\). [1]
(B) Find vector \(\overrightarrow{\mathrm{AC}}\). [1]
(C) Find \((\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}})\).
OR
What is the area of ∆ABC in sq. units? [2]
Solution:
Question 38.
Three students Piyush, Prateek, and Pankaj are given a rectangular sheet of sides 45 cm and 24 cm. They are asked to work independently and form an open box by cutting the squares of equal length from all four corners as shown and folding up the flaps, they want to check the volume of the box formed.
Based on the above information, answer the following questions:
(A) What should be the value of the x, so that the volume of the box is maximum? [2]
(B) For what value(s) of λ is the function defined by f(x) = \(\left\{\begin{array}{r}
\lambda\left(x^2-2 x\right), \text { if } x \leq 0 \\
4 x+1, \text { if } x>0
\end{array}\right.\) continuous at x = 0? [2]
Solution:
(A) Length = (45 – 2x) cm
Breadth = (24 – 2x) cm
Height = (x) cm
l = (45 – 2x) cm, b = (24 – 2x) cm, h = x cm
∴ Volume = lbh
= (45 – 2x) (24 – 2x) × x cm3
Here, V = (45 – 2x) (24 – 2x)x
\(\frac{d V}{d x}\) = (45 – 2x) (24 – 2x) + (45 – 2x) × (-2) + (-2) (24 – 2x) (x)
= 2[(45 – 2x) (12 – x) – x(45 – 2x) – x(24 – 2x)]
= 2[540 – 24x – 45x + 2x2 – 45x + 2x2 – 24x + 2x2]
= 2[6x2 – 138x + 540]
= 12 [x2 – 23x + 90]
= 12 (x – 18) (x – 5)
x = 18, x = 5
Hence, it is not continuous at x = 0 for any value of λ.