Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Set 3 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 12 Maths Set 3 with Solutions
Time Allowed: 3 hours
Maximum Marks: 70
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 MCQ’s and 02 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 source based/case based/passage based/integrated units of assessment (4 marks each) with sub parts.
Section – A
(Multiple Choice Questions) Each question carries 1 mark
Question 1.
If A is a matrix of order 3 × 4, then each row of A has:
(a) 3 elements
(b) 12 elements
(c) 7 elements
(d) 4 elements
Answer:
(d) 4 elements
Explanation:
Since 3 × 4 signifies that there are 3 rows and 4 columns. Then each row has 4 elements.
Question 2.
The area of the triangle formed by 3 collinear points is :
(a) one
(b) two
(c) zero
(d) four
Answer:
(c) zero
Explanation:
The area of triangle formed by three collinear points is zero.
Question 3.
Which of the following function is decreasing on \(\left(0,\frac{\pi}{2}\right)\)?
(a) sin 2x
(b) tan x
(c) cos x
(d) cos 3x
Answer:
(c) cos x
Explanation:
In the interval\(\left(0,\frac{\pi}{2}\right)\), f(x) = cos x
⇒ f(x) = – sin x
which gives f'(x) < 0 in \(\left(0,\frac{\pi}{2}\right)\) Hence,f(x) = cos x in decreasing in \(\left(0,\frac{\pi}{2}\right)\) Question 4. If f(x) = \(\begin{cases}\lambda\left(x^2-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}\) then which one of the following is correct :
(a) f(x) is continuous at x = 0 for any value of 1
(b) f(x)is discontinuous at x = 0 for any value of 1
(c) None of the above
(d) f(x) is discontinuous at x = 0 for any value of 1
Answer:
(b) f(x)is discontinuous at x = 0 for any value of 1
Explanation:
\(\lim _{x \rightarrow 0^{-}} f(x)=0 \text { and } \lim _{x \rightarrow 0^{+}} f(x)=1\)
Question 5.
Let f : R → R be defined as
(a) 9
(b) 14
(c) 5
(d) None of these
Answer:
(a) 9
Explanation:
f(-1) = 3(-1) = -3
f(2) = (2)2 = 4
f(4) = 2 × 4 = 8
⇒ f(-1) + f(2) + f(3) = – 3 + 4 + 8 = 9.
Question 6.
The degree of the differential equation
\(\left(1+\frac{d y}{d x}\right)^3=\left(\frac{d^2 y}{d x^2}\right)^2\) is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2
Explanation:
We have,
\(\left(1+\frac{d y}{d x}\right)^3=\left(\frac{d^2 y}{d x^2}\right)^2\) is:
So, order = 2
and degree = 2
Question 7.
An optimisation problem may involve finding:
(a) maximum profit
(b) minimum cost
(c) minimum use of resources
(d) all of the above
Answer:
(d) all of the above
Explanation:
An optimisation problem may involve finding maximum profit, minimum cost or minimum use of resources etc.
Question 8.
Find the interval in which function, f(x) = sin x + cos x, 0 ≤ x ≤ 2π is decreasing :
(a) \(\left(\frac{\pi}{4},\frac{5 \pi}{4}\right)\)
(b) \(\left(\frac{-\pi}{4},\frac{5 \pi}{4}\right)\)
(c) \(\left(\frac{\pi}{4},\frac{-5 \pi}{4}\right)\)
(d) \(\left(\frac{-\pi}{4},\frac{\pi}{4}\right)\)
Answer:
(a) \(\left(\frac{\pi}{4},\frac{5 \pi}{4}\right)\)
Explanation:
f(x) = sin x + cos x
f'(x) = cos x – sin x
f'(x) = 0 gives
sin x = cos x,
which gives that x = \(\frac{\pi}{4},\frac{5 \pi}{4}\) as 0 ≤ x ≤ 2π
Question 9.
\(\int \frac{d x}{\cos 2 x+\sin ^2 x}\)
(a) sin x + C
(b) cos x + C
(c) tan x + C
(d) 0
Answer:
(c) tan x + C
Explanation:
Question 10.
If \(\left[\begin{array}{cc}1 & 2 \\-2 & -b\end{array}\right]+\left[\begin{array}{ll}a & 4 \\3 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\1 & 0\end{array}\right]\) then a2 + b2 is equation to :
(a) 20
(b) 22
(c) 12
(d) 10
Answer:
(a) 20
Explanation:
We have,
Question 11.
The linear inequalities or equations or restrictions on the variables of a linear programming problem are called:
(a) linear relation
(b) constraints
(c) functions
(d) objective functions
Answer:
(b) constraints
Explanation:
The restrictions on the variable of a linear programming problem are called constraints.
Question 12.
If M12 = – 40, M12 = -10 and M13 = 35 of the determinant ∆ = \(\left|\begin{array}{rrr}1 & 3 & -2 \\4 & -5 & 6 \\3 & 5 & 2\end{array}\right|\), then the value of
∆ is :
(a) -80
(b) 60
(c) 70
(d) 100
Answer:
(a) -80
Explanation:
∆ = a11A11 + a12A12 + a13A13
= a11M11 – a12M12 + a13M13
= 1 .(-40) – 3(-10) + (-2) (35)
= – 40 + 30 – 70 = -80
Question 13.
If ∆ = \(\left|\begin{array}{lll}a & h & g \\h & b & f \\g & f & c\end{array}\right|\), then the
cofactor A21 is :
(a) – (hc + fg)
(b) fg – hc
(c) fg + hc
(d) hc – fg
Answer:
(b) fg – hc
Explanation:
A21 = (-1)2+1 M21 = -M21 = –\(\left|\begin{array}{ll}h & g \\f & c\end{array}\right|\)
= -(hg – fg) = fg – hc.
Question 14.
If p(A∩B) = \(\frac{7}{10}\) and p(B) = \(\frac{7}{10}\), then p(A|B) is equal to :
(a) \(\frac{14}{17}\)
(b) \(\frac{17}{20}\)
(c) \(\frac{7}{8}\)
(d) \(\frac{1}{8}\)
Answer:
(a) \(\frac{14}{17}\)
Explanation:
Question 15.
The degree of differential equation \(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3\) + 6y5 = 0
(a) 1
(b) 2
(c) 3
(d) 5
Answer:
(a) 1
Explanation:
\(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3+6 y^5\) = 0
We know that, the degeer of a differential equation is exponent highest of order derivative.
∴ Degree = 1.
Question 16.
If f(x) = 2x and g(x) = \(\frac{x^2}{2}\) + 1, then which of the following can be discontinuous function?
(a) f(x) + g(x)
(b) f(x) – g(x)
(c) f(x).g(x)
(d) \(\frac{g(x)}{f(x)}\)
Answer:
(d) \(\frac{g(x)}{f(x)}\)
Explanation:
We know that, if f and g are constraints function, then
(a) f + g is continuous
(b) f – g is continuous
(c) fg is continuous
(d) \(\frac{f}{g}\) is continuous at these points, where g(x) ≠ 0.
Here, \(\frac{g(x)}{f(x)}=\frac{\frac{x^2}{2}+1}{2 x}=\frac{x^2+2}{4 x}\)
which is discontinuous at x = 0.
Question 17.
Direction ratios of the line represented by the equation x = ay + b, z = cy + d are :
(a) (a, 1, c)
(b) (a, b-d, C)
(c) (a, 1, a)
(d) (b, ac, d)
Answer:
(a) (a, 1, c)
Explanation:
Given, x = ay + b, Z = cy + d
Hence direction ratios are (a, 1, c).
Question 18.
If A = \(\left[\begin{array}{rr}2 & 3 \\-4 & -6\end{array}\right]\)?
(a) A(adj A) ≠ |A|I
(b) A(adj A) ≠ (Adj A)A
(c) A(adj A) = (adj A)A = |A| I = \(\left[\begin{array}{ll}0 & 0 \\0 & 0\end{array}\right]\)
(d) None of the above
Answer:
(c) A(adj A) = (adj A)A = |A| I = \(\left[\begin{array}{ll}0 & 0 \\0 & 0\end{array}\right]\)
Explanation:
We know, if A is any square matrix of order n, then A(adj A) = (adj A) • A = |A|.I.
Question 19.
Assertion (A): A function y =f{x) is defined by x2 – cos-1 y = π, then domain of f(x) is R.
Reason (R): cos-1 y ∈ [0,π].
Answer:
A is false but R is true.
Explanation:
x2 – cos-1y = π
⇒ cos-1y = (x2 – π)
∵ 0 ≤ cos-1 y ≤ π
⇒ 0 ≤ x2 – π ≤ π
⇒ π ≤ x2 ≤ 2π
∴ x ∈ [- √2π – √π] ∪ [√π, √2π]
Hence domain of f(x) is not real number.
Question 20.
Assertion (A): Alf \(\vec{a}\) and \(\vec{b}\) are reciprocal vectors, then \(\vec{a}\) . \(\vec{b}\) = 1.
Reason (R): If \(\vec{a}\) and \(\vec{b}\) are reciprocal, then \(\vec{a}\) = λ \(\vec{b}\), λ ∈ R+ and |\(\vec{a}\)| |\(\vec{b}\)| = 1.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
If \(\vec{a}\) and \(\vec{b}\) are reciprocal, then
\(\vec{a}\) = λ \(\vec{b}\), λ ∈ R+ and |\(\vec{a}\)| |\(\vec{a}\)| = 1.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
If \(\vec{a}\) and \(\vec{b}\) are reciprocal, then
Section – B
[This section comprises of very short answer type-questions (VSA) of 2 marks each]
Question 21.
Using principal value, find the value of cos -1 \(\left(\cos \frac{13 \pi}{6}\right)\)
What is the principal value of sec1(-2)?
Answer:
As principal value of cos-1θ is [0, π]
OR
We know that principal value of sec-1 θ is [0, π] – \(\left\{-\frac{\pi}{2}\right\}\)
Question 22.
A balloon, which always remains spherical has a variable diameter \(\frac{3}{2}\) (2x + 3). The rate of change of value with respect to x is …… .
Answer:
\(\frac{27}{8}\)π (2x + 3)2
Let V be the volume of the balloon.
Radius of spherical balloon = \(\frac{3}{4}\)(2x +3)
Question 23.
Equation of line is \(\frac{4-x}{2}=\frac{y+3}{2}=\frac{z+2}{1}\) Find the direction cosines of a line parallel to above line.
OR
The equations of a line are 5x-3 = 15y + 7 = 3- lOz. Write the direction cosines of the line.
Answer:
Given equation of line can be written as
\(\frac{x-4}{-2}=\frac{y+3}{2}=\frac{z+2}{1}\)
∴ Direction cosines of line parallel to above line are given by
Hence, required direction cosines of a line parallel to the given line is \(\left(-\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\)
OR
Given line is 5x – 3 = 15y + 7 = 3- lOz
Rewritting the equation in standard form :
Question 24.
Write integrating factor of the following differential equations :
Answer:
Given differential equation is,
\(\frac{d y}{d x}+\frac{1}{1+x^2}\)y = sin x
Integrating factor of above equation is given by
Question 25.
The length, x of a rectangle is decreasing at the rate 5 cm/minute and the width, y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rate of change of the area of the rectangle.
Answer:
We have,
Hence, the rate of change of the area of the rectangle is 2 cm2/min.
Section – C
[This section comprises of short answer type questions (SA) of 3 marks each]
Question 26.
Find \(\int \frac{x+2}{\sqrt{(x-2)(x-3)}} d x\)
Answer:
Question 27.
Find ∫tan32x sec 2x dx =
Answer:
Question 28.
\(\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x\)
OR
Find : \(\int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x\)
Answer:
We have,
Multiple both sides by (t + 1) (t + 3)
1 = A(t + 3) + B(t + 1)
put t = -1 in equation (i)
then 1 = A(-1 + 3)
⇒ 1 = 2A
⇒ A = \(\frac{1}{2}\)
It t = -3
Put the value in equation (i)
then 1 = B(-3 + 1)
1 = -2B
B = –\(\frac{1}{2}\)
OR
Question 29.
Find the particular solution of the differential equation (1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = l.
OR
Find the particular solution of the following differential equation :
\(\frac{d y}{d x}\) = 1 + x2 + y2 + x2y2
Given that y = 1, when x = 0.
Answer:
The given differential equation is
(1 – y2) (1 + log x)dx + 2xy dy = 0
\(\frac{(1+\log x)}{x} d x\) = \(\frac{-2 y}{\left(1-y^2\right)} dy\)
On increasing both sides, we have
\(\int\frac{1+\log x}{x} dx\) = \(\int\frac{-2 y}{\left(1-y^2\right)} dy\)
In first integral,
put 1 + log x = t
⇒ \(\frac{1}{x}\)dx = dt
Also in second integral
put 1 – y2 = u
⇒ -2y dy = du
∴\(\int t \cdot d t=\int \frac{1}{u} du\)
⇒ \(\frac{t^2}{2}\)– log |u| = C
or \(\frac{1}{2}\)(1 + log x)2 – log|1 – y2| = C
It is given that y = 0 when x = 1
So, \(\frac{1}{2}\)(1 + log 1)2 – log|1 – 02| = C
⇒ C = \(\frac{1}{2}\)
∴ \(\frac{(1+\log x)^2}{2}\) – log|1 – y2| = \(\frac{1}{2}\)
or (1 + log x)2 – 2 log|1 – y2| = 1
It is the required particular solution
OR
Given the differential equation is
is the required particular solution of given equation.
Question 30.
A box has 2 apples out of which one is red and other is green. What is the probability that both of them are bad, if it is known that (i) the red apple is bad, (ii) and green apple is bad.
OR
An electronic assembly consists of two sub-systems, say A and B. From previous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0 . 2
P(B fails alone) = 0.15
P(A and B fail) = 0.15
Evaluate the following probabilities:
(i) P(A) fails given that B has failed,
(ii) P(A) fails alone.
Answer:
Let Bi and Gi denote the apples to be bad and good respectively where i ∈ {R, G}. R stands for red apple and G for green apple.
Then sample space is S = {GRGG, BRGG, BRBG, GRBG)
Let us consider the events
A = both apples are bad
B = red apple is bad
C = green apple is bad
Then, A = (BRBG), B = {BRGG, BRBG) and C = {BRBG, GRBG}
and A ∩ B – {BRBG}, A ∩ C – {BRBG}
(i) Required probability = P (A/B)
\(\frac{P(A \cap B)}{P(B)}=\frac{1 / 4}{2 / 4}=\frac{1}{2}\)
(ii) Required probability = P (A/C)
\(\frac{P(A \cap C)}{P(C)}=\frac{1 / 4}{2 / 4}=\frac{1}{2}\)
OR
Let the event in which A fails and B fails be denoted by EA and EB respectively.
Then, P(EA) = 0.2
P(EA and EB) = 0.15
P(B fails alone) = P(EB) – P(EA and EB)
∴ 0.15 = P(EB)- 0 . 15
∴ P(EB) = 0.15 + 0.15 = 0.3.
(i) p(EA/EB) = \(\frac{P\left(E_A \cap E_B\right)}{P\left(E_B\right)}\)
= \(\frac{0 \cdot 15}{0 \cdot 3}\) = 0.5
(ii) P(A falis alone) = P(EA) – P(EA and EB
= 0.2 – 0.15
= 0.05.
Question 31.
Maximize Z = x + 2y
subject to the constraints
x + 2y ≥ 100, 2x – y ≤ 0
2x + y ≤ 200, x,y ≥ 0
Solve the above LPP graphically.
Answer:
Given, maximize Z = x + 2y
Subject to the constraints
x + 2y ≥ 100 2x – y ≤ 0
2x + y ≤ 200 x,y ≥ 0
Converting the inequations into equations we obtain the lines x + 2y = 100, 2x – y = 0, 2x + y = 200. Then,
x + 2y = 100
x | 0 | 100 |
y | 50 | 0 |
2x – y = 0
x | 10 | 20 |
y | 20 | 40 |
2x + y = 200
x | 0 | 100 |
y | 200 | 0 |
Plotting these points on the graph, we get the shaded feasible region, i.e., ADECA.
Corner points | Z = x + 2y |
A(0, 50) | (0) + 2(50) = 100 |
D(20, 40) | 20 + 2(40) = 100 |
E(50, 100) | 50 + 2(100) = 250 |
C(0, 200) | 0 + 2(200) = 400 maximum |
Cleariy, the maximum value of Z is 400 at (0,200)
Section – D
[This section comprises of long answer type questions (LA) of 5 marks each]
Question 32.
If A = \(\left[\begin{array}{cc}0 & -\tan \alpha / 2 \\\tan \alpha / 2 & 0\end{array}\right]\)and is a 2 × 2 units matrix, then prove that I + A = (I – A) \(\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\\sin \alpha & \cos \alpha\end{array}\right]\)
Answer:
L.H.S = I + A
Question 33.
Find the length and foot of perpendicular drawn from the point (2, -1, 5) to line
\(\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}\)
OR
By computing the shortest distance, determine whether the following lines intersects or not
\(\frac{x-1}{2}=\frac{y+1}{3}\) = z and \(\frac{x+1}{5}=\frac{y-2}{1}=\frac{z-2}{0}\)
Answer:
Let AB be the line whose equation is
\(\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}\)
Let Q be the foot of perpendicular and PQ be the length of perpendicular. Any point Q on the given line is given by
\(\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}\) = λ(say)
⇒ x – 10λ + 11,y = -4λ -2,z = -llλ – 8
∴ Coordinates of Q are (10λ + 11, – 4λ – 2, – 11λ – 8)
Now, diameter ratios of line PQ are (10λ + 9, -4λ – 1, – 11λ – 8 -5)
∴ Direction ratios of the PQ = (10λ + 9, – 4λ – 1, – 11λ – 8 – 5)
∵ PQ ⊥ AB
∴ a1a2 + b1b2 + c1c2 = 0
Where a1 = 10λ + 9, b1 = -4λ – 1, c1 = -11λ – 13
and a2 = 10, b2 = -4, c1 = -11
∴ We get
10(10λ + 9) -4(-4λ – 1)-11(-11λ – 13) = 0
⇒ 100λ + 90 + 16λ + 4 + 121λ + 143 = 0
⇒ 237λ + 237 = 0
⇒ λ = -1
Putting λ = -1 in coordinates of point Q, we get
Foot of perpendicular = Q(-10 + 11, 4 – 2,11 – 8)
= Q(1,2,3)
Also, length of perpendicular PQ is given by
PQ = \(\sqrt{(2-1)^2+(-1-2)^2+(5-3)^2}\)
= \(\sqrt{1+9+4}\) = √14 units
OR
The given equations of the two lines are
Since d ≠ 0
Hence the given line do not intersect.
Question 34.
Find the area enclosed by the curve y = -x2 and the straight line x + y + 2 = 0. Sol. We have, y = x2 and x + y + 2 = 0
Answer:
⇒ – x – 2 = – x2
⇒ x2 – x – 2 = 0
⇒ x2 + x – 2x – 2 = 0
⇒ x(x +1) – 2(x +1) = 0
⇒ (x – 2)(x + l) = 0
⇒ x = 2,-l
Area of shaded region,
Question 35.
Find the derivative of tan-1 \(\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)\)
with respect to x.
OR
Find the values of a and b such that the following function f(x) is a continuous function.
Answer:
OR
Given, the function is continuous. For the continuity at x = 2 and x = 10.
Continuity at x = 2
Since, f(x) is continuous at x = 10
L.H.L = R.H.L = f(0)
10a + b = 21
Solving (i) and (ii),we get
a = 2 and b= 1
Section – E
[This section comprises of 3 case-study/passage-based questions of 4 marks each with sub-parts. The First two case study questions have three sub-parts (i), (ii), (iii) of marks 1,1, 2 respectively. The third case study question has two sub-parts of 2 marks each.]
Question 36.
Sanjay, Ajay and Vijay are three best friends. After completing their MBA from IIM Lucknow. They apply for the job in the same company for the post of manager in finance department. Chances of selection of Sanjay, Ajay and Vijay are in the ratio of 1 : 2 : 4. The probabilities that Sanjay can increase the profits of the company by his efforts is 0.8 whereas probabilities for the same task of Ajay and Vijay are 0.5 and 0.3 respectively.
From the above information attempt the following questions:
(i) If E1, E2 and E3 are the events of selecting Sanjay, Ajay and Vijay, then find P(E1), P(E2) and P(E3).
(ii) A be event of increase the profit then find if it is due to Ajay.
(iii) Find P(E1/A).
OR
(iii) If E1, E2 and E3 be events of not increase the profits, then find the probability that it is due to the appointment of Vijay?
Answer:
(i) E1 be event of selecting Sanjay
(ii) E2 be event of selecting Ajay
(iii) E3 be event of selecting Vijay.
OR
(iii) P be event of not increase in profits.
Question 37.
A retailer purchases two types of tea priced at ₹120 and ₹160 per kg. He wants to sell the mixtures of these two types of tea at a price of ?150 per kg. If he bought total of 200 kg of tea.
Give the answer of the followng questions :
(i) If f(x) is function for the total purchase price of green tea + black tea, where x denotes the quantity (in kg) for Green tea, formulate the function f(x).
(ii) How much is the quantity of green tea to be purchased to earn the same revenue?
(iii) Find the value of f(400) – f(200)
OR
(iii) If g(x) = 2x + 5 then find g(10):
Answer:
Green tea = x
Black tea = 200 – x
f(x) = 120x + (200 – x) 160
= 120 x + 3200 – 160 x
f(x) = ₹32000 – 40 x
(ii) Total revenue 150 × 200 = 30,000
∴ 32,000 – 40x = 30,000
⇒ 2000 = 40x
⇒ x = 50
(iii) f(400) = 32000 – 40 × 400
= 32000 – 16000
= ₹16000
f(200) = 3200 – 8000
= ₹24000
f(400) – f(200) = 16000 – 24000 = -8000
Or
(iii) g(x) = 2x + 5
Now, g(10) = 2 × 10 + 5 = 25
Question 38.
Raja was flying a kite from a point A in the direction of \(\vec{a}=5 \hat{i}+\hat{j}+4 \hat{k}\) His friend Sohan was watching him from the point B which is horizontal to A with the position vector \(\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}\)The point B is just below the Kite.
Answer the following question :
(i) Find the projection of \(\vec{a}\) on \(\vec{b}\)?
(ii) Find a unit vector in the direction of \(\vec{a}\) and \(\vec{b}\)?
Answer: