Students can use CBSE Previous Year Question Papers Class 12 Maths with Solutions and CBSE Class 12 Maths Question Paper 2018 Comptt (Delhi & Outside Delhi) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Maths Question Paper 2018 Comptt (Delhi & Outside Delhi) with Solutions
Time Allowed: 3 hours
Maximum Marks: 100
General Instructions:
- All questions are compulsory.
- The Question Paper consists of 29 questions divided into four Sections A, B, C and D. Section-A comprises of 4 questions of one mark each, Section-B comprises of 8 questions of two marks each, Section-C comprises of 11 questions of four marks each and Section-D comprises of 6 questions of six marks each.
- All questions in Section-A are to be answered in one word, one sentence or as per the exact requirement of the question.
- There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 3 questions of six marks each.
- Use of calculators is not permitted. You may ask for logarithmic tables, if required.
-
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Section-A
Questions number 1 to 4 carry one mark each.
Question 1.
Find the value of tan
-1
\(\sqrt{3}\) – sec
-1
(-2).
Solution:
Given tan
-1
\((\sqrt{3})\) – sec
-1
(-2)
= \(\frac{\pi}{3}\) – [π – sec
-1
(2)]
….[∵ sec
-1
(-x) = π – sec-1x, |x| ≥ 1]
= \(\frac{\pi}{3}\) – π + \(\frac{\pi}{3}\) = –\(\frac{\pi}{3}\)
Question 2.
If A = \(\left(\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & x \\
-2 & 2 & -1
\end{array}\right)\) is a matrix satisfying AA’ = 9I, find x.
Solution:
Question 3.
Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card Jack?
Solution:
The Required Probability = P(The first is a red jack and the second is a jack card) or (The first is a red non-jack card and the second is a jack card)
= \(\frac{2}{52} \times \frac{3}{51}\) + \(\frac{24}{52} \times \frac{4}{51}\) = \(\frac{1}{26}\)
Question 4.
The position vectors of point A and B are \(\vec{a}\) and \(\vec{b}\) respectively.
P divides AB in the ratio 3 : 1 and Q is mid-point of AP. Find the position vector of Q.
Solution:
Coordinates of point P
= \(\frac{3(\vec{a})+1(\vec{b})}{3+1}\) = \(\frac{3 \vec{a}+\vec{b}}{4}\)
Section – B
Questions number 5 to 12 carry 2 marks each.
Question 5.
Prove that: 3 cos
-1
x = cos
-1
(4x
3
– 3x), x ∈ [\(\frac{1}{2}\), 1].
Solution:
RH.S. = cos
-1
(4x
3
– 3x)
= cos
-1
(4 cos
3
θ – 3 cos θ) [Let x = cosθ ∴ cos
-1
x = θ]
= cos
-1
(cos3θ) = 3θ
= 3 cos
-1
x = L.H.S. (Hence proved)
Question 6.
If A = \(\left[\begin{array}{rr}
2 & 3 \\
5 & -2
\end{array}\right]\) be such that A
-1
= kA, then find the value of k.
Solution:
⇒ 2k = \(\frac{2}{19}\)
………… [Equating the corresponding elements
∴ k = \(\frac{1}{19}\)
Question 7.
Differentiate tan
-1
\(\left[\frac{\cos x-\sin x}{\cos x+\sin x}\right]\) with respect to x.
Solution:
Let y = tan
-1
\(\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\)
Dividing Numerator and Denominator by cos x, we have
y = tan
-1
\(\left(\frac{1-\tan x}{1+\tan x}\right)\)
y = tan
-1
\(\left(\tan \left(\frac{\pi}{4}-x\right)\right)\)
y = \(\frac{\pi}{4}\) – x
Differentiating both sides w.r.t. x, we have \(\frac{d y}{d x}\) = -1
\(\frac{d y}{d x}\) = -1
Question 8.
The total revenue received from the sale of x units of a product is given by R(x) = 3x
2
+ 36x + in rupees. Find the marginal revenue when x = 5, where by marginal revenue we mean the rate of change of total revenue with respect to the number of items sold at an instant.
Solution:
R(x) = 3x
2
+ 36x + 5
Differentiating both sides w.r.t. x, we have
Marginal Revenue, R'(x) = 6x + 36
When x = 5, R'(5) = 6(5) + 36
= 30 + 36 = ₹ 66
∴ Marginal Revenue is ₹ 66.
Question 9.
Find: \(\int \frac{3-5 \sin x}{\cos ^2 x}\)dx
Solution:
Question 10.
Solve the differential equation cos\(\left(\frac{d y}{d x}\right)\) = a, (a ∈ IR).
Solution:
cos \(\left(\frac{d y}{d x}\right)\) = a
\(\frac{d y}{d x}\) = cos
-1
a
dy = cos
-1
a∫dy ……… [Integrating both sides]
y = x . cos
-1
a + C
y – C = x cos
-1
a
\(\frac{y-C}{x}\) = cos
-1
a
cos\(\left(\frac{y-C}{x}\right)\) = a is the solution of the differential equation.
Question 11.
If \(\vec{a}+\vec{b}+\vec{c}\) = \(\overrightarrow{0}\) and \(|\vec{a}|\) = 5, \(|\vec{b}|\) = 6 and \(|\vec{c}|\) = 9, then find the angle between a and b.
Solution:
Question 12.
Evaluate: P(A ∪ B), if 2P(A) = P(B) = \(\frac{5}{13}\) and P(A | B) = \(\frac{2}{5}\).
Solution:
Section-C
Questions number 13 to 23 carry 4 marks each.
Question 13.
Evaluate: \(\int_{-1}^2\left|x^3-3 x^2+2 x\right|\) dx
Solution:
The given definite integral
Question 14.
If sin y = x cos(a + y), then show that \(\frac{d y}{d x}\) = \(\frac{\cos ^2(a+y)}{\cos a}\)
Also, show that \(\frac{d y}{d x}\) = cos a, when x = 0.
Solution:
Question 15.
If x = a sec
3
θ and y = a tan
3
θ, find \(\frac{d^2 y}{d x^2}\) at θ = \(\frac{\pi}{3}\).
Answer:
Or
If y = \(e^{\tan ^{-1} x}\), prove that (1 + x
2
)\( + (2x – 1)[latex]\frac{d y}{d x}\) = 0.
Solution:
Question 16.
Find the intervals in which the function f(x) = -2x
3
– 9x
2
– 12x + 1 is
(i) Strictly increasing,
(ii) Strictly decreasing.
Solution:
f(x) = -2x
3
– 9x
2
– 12x + 1
Differentiating both sides w.r.t. x, we have
f’(x) = -6x
2
– 18x – 12
= -6(x
2
+ 3x + 2)
= -6(x
2
+ 2t + x + 2)
= -6[x(x + 2) + 1(x + 2)]
= -6(x + 2) (x + 1)
When f’(x) = 0, x = -2, -1
(i) f is strictly increasing in (-2, -1).
(ii) f is strictly decreasing in (-∞, -2) ∪ (-1, ∞).
Question 17.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 metres. Find the dimensions of the window to admit maximum light through the whole opening.
Solution:
Let length of rectangular window = 2x m
Let breadth of rectangular window = y m
∴ Radius of rectangular window = x m
Perimeter of window = 10 m
∴ y + 2x + y + πx = 10
…..[∵ Perimeter of semi-circle = πr
2y = 10 – 2x – πx
y = \(\frac{10-2 x-\pi x}{2}\) ……….. (i)
y = \(\left(5-x-\frac{\pi}{2} x\right)\)
Area of window = Area of rectangle + Area of semi-circle
∴ Area is maximum at x = \(\frac{10}{4+\pi}\) m
∴ Maximum light will enter the window when radius = \(\frac{10}{4+\pi}\) m.
Length of rectangle window
= 2x = \(\frac{20}{4+\pi}\) m,
Breadth of rectangle window, y
= 5 – \(\frac{10}{4+\pi}\) – \(\frac{\pi}{2}\left(\frac{10}{4+\pi}\right)\)
= \(\frac{20+5 \pi-10-5 \pi}{4+\pi}\) = \(\frac{10}{4+\pi}\) m
Question 18.
Find: \(\int \frac{4}{(x-2)\left(x^2+4\right)}\) dx.
Solution:
Question 19.
Solve the differential equation (x
2
– y
2
) dx + 2xy dy = 0.
Solution:
Or
Find the particular solution of the differential equation (1 + x
2
)\(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^2}\), given that y = 0 when x = 1.
Solution:
Question 20.
Find the value of λ so that the lines \(\frac{1-x}{3}\) = \(\frac{7 y-14}{2 \lambda}\) = \(\frac{5 z-10}{11}\) and \(\frac{7-7 x}{3 \lambda}\) = \(\frac{y-5}{1}\) = \(\frac{6-z}{5}\) are perpendicular to each other.
Solution:
Question 21.
Find the shortest distance between the lines \(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\) and \([\frac{x-2}{3}/latex] = [latex]\frac{y-4}{4}\) = \(\frac{z-5}{5}\).
Answer:
Question 22.
Two groups are competing for the positions of the Board of Directors of a corporation. The probabilities that the first and second group will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Solution:
Let E
1
, E
2
, A be the following events.
E
1
: 1
st
group will win
E
2
: 2
nd
group will win
A : Introducing a new product
P(E
1
) = 0.6, P(A | E
1
) = 0.7
P(E
2
) = 0.4, P(A | E
2
) = 0.3
Using Bayes’ theorem,
Question 23.
From a lot of 20 bulbs which include 5 defectives, a sample of 3 bulbs is drawn by random, one by one with replacement. Find the probability distribution of the number of defective bulbs. Also, find the mean of the distribution.
Solution:
Section-D
Questions number 24 to 29 carry 6 marks each.
Question 24.
Show that the relation R on the set Z of all integers defined by (x, y) ∈ R ⇔ (x – y) is divisible by 3 is an equivalence relation.
Solution:
R = {(x, y): x, y ∈ Z and (x – y) is divisible by 3}
Reflexive: R is reflexive, for x ∈ Z
x – x = 0, which is divisible by 3
(x, x) ∈ R ∴ R is reflexive.
Symmetric:
R is symmetric, (x, y) ∈ R where x, y ∈ Z
x – y is divisible by 3.
Let (x – y) = 3 m, where in m ∈ Z
-(y – x) = 3 m
(y – x) = -3 m
(y – x) is divisible by 3.
(y – x) ∈ R ∴ R is symmetric.
Transitive:
R is transitive, (x, y) ∈ R where x, y ∈ Z
x – y is divisible by 3.
x – y = 3 in, where m ∈ Z
For (y, z) ∈ Z, where y, Z ∈ Z
y – z is divisible by 3.
y – z = 3 n, where n ∈ Z …….(ii)
Adding (i) and (ii),
(x – y) + (y – z) = 3m + 3n
x – z = 3(m+n)
⇒ (x – z) is divisible by 3.
(x, y) ∈ R, (y, z) ∈ R
∴ (x, z) ∈ R where x, y, z ∈ Z
R is transitive.
Since R is reflexive, symmetric and transitive.
∴ R is an equivalence relation.
Question 25.
Given A = \(\left[\begin{array}{lll}
5 & 0 & 4 \\
2 & 3 & 2 \\
1 & 2 & 1
\end{array}\right]\), B
-1
= \(\left[\begin{array}{lll}
1 & 3 & 3 \\
1 & 4 & 3 \\
1 & 3 & 4
\end{array}\right]\), compute (AB)
-1
.
Solution:
Question 26.
Using integration, find the area of the region: {(x, y) : 0 ≤ 2y ≤ x
2
, 0 ≤ y ≤ x, 0 ≤ x ≤ 3}
Solution:
Question 27.
Evaluate: \(\int_0^{\pi / 2} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x}\) dx.
Solution:
Question 28.
Find the angle between the following pair of lines : \(\frac{-x+2}{-2}\) = \(\frac{y-1}{7}\) = \(\frac{z+3}{-3}\) and \(\frac{x+2}{-1}\) = \(\frac{2 y-8}{4}\) = \(\frac{z-5}{4}\) and check whether the lines are parallel or perpendicular.
Solution:
Question 29.
A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3 g of silver and 1 g of gold while that of B requires 1 g of silver and 2 g of gold. The company can use at most 9 g of silver and 8 g of gold. If each unit of type A brings a profit of ₹40 and that of type B ₹50, find the number of units of each type that the company should produce to maximize the profit. Formulate and solve graphically the LPP and find the maximum profit.
Solution:
Maximum value of Z is ₹ 230 at B(2, 3) i.e., the company should produce 2 units of Type A and 3 units of Type B goods to maximize the Profit.
