CBSE Maths › Class 12 PYQs › 2022

CBSE Class 12 Maths 2022 Question Paper 65/5/2 (Term 2) with Solutions

All 18 questions from the CBSE Class 12 Mathematics board paper, Set 65/5/2 (2022, Term 2), with answers and step-by-step solutions. Total 40 marks. Tap “Show answer & solution” under any question.

Set 65/1/1 Set 65/1/2 Set 65/1/3 Set 65/2/1 Set 65/2/2 Set 65/2/3 Set 65/3/1 Set 65/3/2 Set 65/3/3 Set 65/4/1 Set 65/4/2 Set 65/4/3 Set 65/5/1 Set 65/5/2 Set 65/5/3

Q12 marksVery Short AnswerThree Dimensional Geometry

Find the vector equation of a line passing through a point with position vector $2 i^- j^+ k^$ and parallel to the line joining the points $- i^+ 4 j^+ k^$ and $i^+ 2 j^+ 2 k^$ .

Show answer & solution

Answer:

r = (2 i^- j^+ k^) + λ (2 i^- 2 j^+ k^)

Direction of the line joining the points: $(i^+ 2 j^+ 2 k^) - (- i^+ 4 j^+ k^) = 2 i^- 2 j^+ k^$ .
Required line: $r = (2 i^- j^+ k^) + λ (2 i^- 2 j^+ k^)$ .

Q22 marksVery Short AnswerIntegrals

Find : $\int \frac{d x}{4 x - x ^{2}}$

Show answer & solution

Answer:

sin^{- 1} (\frac{x - 2}{2}) + C

$4 x - x^{2} = 4 - (x^{2} - 4 x + 4) = 2^{2} - (x - 2)^{2}$ .
So $\int \frac{d x}{2 ^{2} - ( x - 2 ) ^{2}} = sin^{- 1} (\frac{x - 2}{2}) + C$ .

Also asked in: 2022 65/5/1, 2022 65/5/3

Q32 marksVery Short AnswerDifferential Equations

Find the general solution of the following differential equation :
$\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$

Show answer & solution

Answer:

e^{y} = e^{x} + \frac{x ^{3}}{3} + C

$\frac{d y}{d x} = e^{- y} (e^{x} + x^{2})$ , so $e^{y} d y = (e^{x} + x^{2}) d x$ .
Integrating both sides: $e^{y} = e^{x} + \frac{x ^{3}}{3} + C$ .

Also asked in: 2022 65/5/1, 2022 65/5/3

Q42 marksVery Short AnswerProbability

Events A and B are such that
$P (A) = \frac{1}{2}$ , $P (B) = \frac{7}{12}$ and $P (A ˉ \cup B ˉ) = \frac{1}{4}$
Find whether the events A and B are independent or not.

Show answer & solution

Answer: Not independent

$P (A ˉ \cup B ˉ) = P (A \cap B) = 1 - P (A \cap B)$ , so $P (A \cap B) = 1 - \frac{1}{4} = \frac{3}{4}$ .
$P (A) \cdot P (B) = \frac{1}{2} \cdot \frac{7}{12} = \frac{7}{24}$ .
Since $P (A \cap B) \neq = P (A) \cdot P (B)$ , A and B are not independent.

Also asked in: 2022 65/5/1, 2022 65/5/3

Q4 (OR) (OR)2 marksVery Short AnswerProbability

A box $B_{1}$ contains 1 white ball and 3 red balls. Another box $B_{2}$ contains 2 white balls and 3 red balls. If one ball is drawn at random from each of the boxes $B_{1}$ and $B_{2}$ , then find the probability that the two balls drawn are of the same colour.

Show answer & solution

Answer:

\frac{11}{20}

P(both white) $= \frac{1}{4} \times \frac{2}{5} = \frac{2}{20}$ .
P(both red) $= \frac{3}{4} \times \frac{3}{5} = \frac{9}{20}$ .
P(same colour) $= \frac{2}{20} + \frac{9}{20} = \frac{11}{20}$ .

Also asked in: 2022 65/5/1, 2022 65/5/3

Q52 marksVery Short AnswerProbabilityNot in current syllabus

Let X be a random variable which assumes values $x_{1}$ , $x_{2}$ , $x_{3}$ , $x_{4}$ such that
$2 P (X = x_{1}) = 3 P (X = x_{2}) = P (X = x_{3}) = 5 P (X = x_{4})$ .
Find the probability distribution of X.

Show answer & solution

Answer:

P (X = x_{1}) = \frac{15}{61}

P (X = x_{2}) = \frac{10}{61}

P (X = x_{3}) = \frac{30}{61}

P (X = x_{4}) = \frac{6}{61}

Let each expression equal k. Then $P (x_{1}) = \frac{k}{2}$ , $P (x_{2}) = \frac{k}{3}$ , $P (x_{3}) = k$ , $P (x_{4}) = \frac{k}{5}$ .
Sum of probabilities is 1: $k (\frac{1}{2} + \frac{1}{3} + 1 + \frac{1}{5}) = \frac{61 k}{30} = 1$ , so $k = \frac{30}{61}$ .
Hence X: $x_{1}, x_{2}, x_{3}, x_{4}$ with P(X): $\frac{15}{61}, \frac{10}{61}, \frac{30}{61}, \frac{6}{61}$ .

Also asked in: 2022 65/5/1, 2022 65/5/3

Q62 marksVery Short AnswerVector Algebra

If $a = i^+ j^+ k^$ , $a \cdot b = 1$ and $a \times b = j^- k^$ , then find $∣ b ∣$

Show answer & solution

Answer:

∣ b ∣ = 1

$∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$ .
$∣ a \times b ∣^{2} = 0 + 1 + 1 = 2$ , $(a \cdot b)^{2} = 1$ , $∣ a ∣^{2} = 3$ .
So $2 + 1 = 3∣ b ∣^{2}$ , giving $∣ b ∣ = 1$ .
(Indeed $b = i^$ satisfies both conditions.)

Also asked in: 2022 65/5/1, 2022 65/5/3

Q73 marksShort AnswerIntegrals

Find :
$\int e^{x} \cdot sin 2 x d x$

Show answer & solution

Answer:

\frac{e ^{x}}{5} (sin 2 x - 2 cos 2 x) + C

Let $I = \int e^{x} sin 2 x d x = e^{x} sin 2 x - 2 \int e^{x} cos 2 x d x$ .
$\int e^{x} cos 2 x d x = e^{x} cos 2 x + 2 \int e^{x} sin 2 x d x = e^{x} cos 2 x + 2 I$ .
So $I = e^{x} sin 2 x - 2 e^{x} cos 2 x - 4 I$ , giving $5 I = e^{x} (sin 2 x - 2 cos 2 x)$ .
$I = \frac{e ^{x}}{5} (sin 2 x - 2 cos 2 x) + C$ .

Also asked in: 2022 65/5/1, 2022 65/5/3

Q7 (OR) (OR)3 marksShort AnswerIntegrals

Find :
$\int \frac{2 x}{( x ^{2} + 1 ) ( x ^{2} + 2 )} d x$

Show answer & solution

Answer:

lo g \frac{x ^{2} + 1}{x ^{2} + 2} + C

Put $x^{2} = t$ , $2 x d x = d t$ : $I = \int \frac{d t}{( t + 1 ) ( t + 2 )}$ .
$\frac{1}{( t + 1 ) ( t + 2 )} = \frac{1}{t + 1} - \frac{1}{t + 2}$ .
$I = lo g ∣ t + 1∣ - lo g ∣ t + 2∣ + C = lo g \frac{x ^{2} + 1}{x ^{2} + 2} + C$ .

Also asked in: 2022 65/5/1, 2022 65/5/3

Q83 marksShort AnswerThree Dimensional GeometryNot in current syllabus

Find the equation of the plane passing through the line of intersection of the planes $r \cdot (i^+ j^+ k^) = 10$ and $r \cdot (2 i^+ 3 j^- k^) + 4 = 0$ and passing through the point $(- 2, 3, 1)$ .

Show answer & solution

Answer:

3 x + 4 y = 6

, i.e.

r \cdot (3 i^+ 4 j^) = 6

Planes: $x + y + z - 10 = 0$ and $2 x + 3 y - z + 4 = 0$ .
Required plane: $(x + y + z - 10) + λ (2 x + 3 y - z + 4) = 0$ .
At $(- 2, 3, 1)$ : $(- 8) + λ (8) = 0$ , so $λ = 1$ .
Plane: $3 x + 4 y - 6 = 0$ , i.e. $r \cdot (3 i^+ 4 j^) = 6$ .

Also asked in: 2022 65/5/1

Q93 marksShort AnswerVector Algebra

Let $a = i^+ j^$ , $b = i^- j^$ and $c = i^+ j^+ k^$ . If $n^$ is a unit vector such that $a \cdot n^= 0$ and $b \cdot n^= 0$ , then find $∣ c \cdot n^∣$ .

Show answer & solution

Answer: 1

$n^$ is perpendicular to both $a$ and $b$ , so it is parallel to $a \times b$ .
$a \times b = (i^+ j^) \times (i^- j^) = - k^- k^= - 2 k^$ , so $n^= \pm k^$ .
$∣ c \cdot n^∣ = ∣ \pm 1∣ = 1$ .

Q9 (OR) (OR)3 marksShort AnswerVector Algebra

If $a$ and $b$ are unit vectors inclined at an angle $3 0^{\circ}$ to each other, then find the area of the parallelogram with $(a + 3 b)$ and $(3 a + b)$ as adjacent sides.

Show answer & solution

Answer: 4 sq. units

$(a + 3 b) \times (3 a + b) = a \times b + 9 b \times a = - 8 a \times b$ .
Area $= 8∣ a \times b ∣ = 8∣ a ∣∣ b ∣ sin 3 0^{\circ} = 8 \times 1 \times 1 \times \frac{1}{2} = 4$ sq. units.

Q103 marksShort AnswerIntegrals

Evaluate :
$\int_{0}^{π /2} \frac{1}{1 + ( t a n x ) ^{2/3}} d x$

Show answer & solution

Answer:

\frac{π}{4}

$I = \int_{0}^{π /2} \frac{c o s ^{2/3} x}{c o s ^{2/3} x + s i n ^{2/3} x} d x$ .
Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : $I = \int_{0}^{π /2} \frac{s i n ^{2/3} x}{s i n ^{2/3} x + c o s ^{2/3} x} d x$ .
Adding: $2 I = \int_{0}^{π /2} 1 d x = \frac{π}{2}$ , so $I = \frac{π}{4}$ .

Q114 marksLong AnswerDifferential Equations

Solve the following differential equation :
$(y - sin^{2} x) d x + tan x d y = 0$

Show answer & solution

Answer:

y sin x = \frac{s i n ^{3} x}{3} + C

Divide by $tan x d x$ : $\frac{d y}{d x} + y cot x = \frac{s i n ^{2} x}{t a n x} = sin x cos x$ .
Linear with $P = cot x$ ; I.F. $= e^{\int c o t x d x} = sin x$ .
$y sin x = \int sin^{2} x cos x d x = \frac{s i n ^{3} x}{3} + C$ .
So $y = \frac{s i n ^{2} x}{3} + C cosec x$ .

Also asked in: 2022 65/5/1, 2022 65/5/3

Q11 (OR) (OR)4 marksLong AnswerDifferential Equations

Find the general solution of the differential equation :
$(x^{3} + y^{3}) d y = x^{2} y d x$

Show answer & solution

Answer:

lo g ∣ y ∣ = \frac{x ^{3}}{3 y ^{3}} + C

$\frac{d y}{d x} = \frac{x ^{2} y}{x ^{3} + y ^{3}}$ , which is homogeneous. Put $y = v x$ .
$v + x \frac{d v}{d x} = \frac{v}{1 + v ^{3}}$ , so $x \frac{d v}{d x} = \frac{v}{1 + v ^{3}} - v = - \frac{v ^{4}}{1 + v ^{3}}$ .
$\frac{1 + v ^{3}}{v ^{4}} d v = - \frac{d x}{x}$ , i.e. $(\frac{1}{v ^{4}} + \frac{1}{v}) d v = - \frac{d x}{x}$ .
$- \frac{1}{3 v ^{3}} + lo g ∣ v ∣ = - lo g ∣ x ∣ + C$ , so $lo g ∣ v x ∣ = \frac{1}{3 v ^{3}} + C$ .
With $v = \frac{y}{x}$ : $lo g ∣ y ∣ = \frac{x ^{3}}{3 y ^{3}} + C$ .

Also asked in: 2022 65/5/1, 2022 65/5/3

Q124 marksLong AnswerApplication of Integrals

Find the area bounded by the curves $y = ∣ x - 1∣$ and $y = 1$ , using integration.

Show answer & solution

Answer: 1 sq. unit

$y = ∣ x - 1∣$ meets $y = 1$ where $∣ x - 1∣ = 1$ , i.e. at $x = 0$ and $x = 2$ .
Area $= \int_{0}^{2} (1 - ∣ x - 1∣) d x = \int_{0}^{1} (1 - (1 - x)) d x + \int_{1}^{2} (1 - (x - 1)) d x$ .
$= \int_{0}^{1} x d x + \int_{1}^{2} (2 - x) d x = \frac{1}{2} + [2 x - \frac{x ^{2}}{2}]_{1}^{2} = \frac{1}{2} + \frac{1}{2} = 1$ sq. unit.

Also asked in: 2022 65/5/1, 2022 65/5/3

Q134 marksLong AnswerProbability

In a factory, machine A produces 30% of total output, machine B produces 25% and the machine C produces the remaining output. The defective items produced by machines A, B and C are 1%, 1.2%, 2% respectively. An item is picked at random from a day's output and found to be defective. Find the probability that it was produced by machine B?

Show answer & solution

Answer:

\frac{1}{5}

(= 0.2)

$P (A) = 0.30$ , $P (B) = 0.25$ , $P (C) = 0.45$ .
P(defective | A) $= 0.01$ , P(defective | B) $= 0.012$ , P(defective | C) $= 0.02$ .
P(defective) $= 0.30 (0.01) + 0.25 (0.012) + 0.45 (0.02) = 0.003 + 0.003 + 0.009 = 0.015$ .
P(B | defective) $= \frac{0.003}{0.015} = \frac{1}{5}$ .

Q144 marksCase StudyThree Dimensional Geometry

Two motorcycles A and B are running at the speed more than the allowed speed on the roads represented by the lines $r = λ (i^+ 2 j^- k^)$ and $r = (3 i^+ 3 j^) + μ (2 i^+ j^+ k^)$ respectively.
Based on the above information, answer the following questions :
(a) Find the shortest distance between the given lines. (2)
(b) Find the point at which the motorcycles may collide. (2)

Show answer & solution

Answer: (a) 0 (the lines intersect) (b)

(1, 2, - 1)

(a) $a_{1} = 0$ , $b_{1} = i^+ 2 j^- k^$ ; $a_{2} = 3 i^+ 3 j^$ , $b_{2} = 2 i^+ j^+ k^$ .
$b_{1} \times b_{2} = 3 i^- 3 j^- 3 k^$ , $a_{2} - a_{1} = 3 i^+ 3 j^$ .
$(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 9 - 9 + 0 = 0$ , so shortest distance $= 0$ .
(b) Equate points: $λ = 3 + 2 μ$ , $2 λ = 3 + μ$ , $- λ = μ$ .
From the first and third: $λ = 3 - 2 λ$ , so $λ = 1$ , $μ = - 1$ ; the second gives $2 = 2$ , consistent.
Point of collision: $(1, 2, - 1)$ .

Also asked in: 2022 65/5/1, 2022 65/5/3

← 2022 65/5/1 2022 65/5/3 →

Practise smarter: chapter-wise revision, formula sheets and step-by-step NCERT solutions on MonoMath CBSE →