# CBSE Class 12 Maths Question Paper 2018 Comptt (Delhi & Outside Delhi) with Solutions

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.

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.