CBSE Maths › Class 12 PYQs › 2022

CBSE Class 12 Maths 2022 Question Paper 65/5/3 (Term 2) with Solutions

All 18 questions from the CBSE Class 12 Mathematics board paper, Set 65/5/3 (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

The Cartesian equation of a line AB is :
$\frac{2 x - 1}{12} = \frac{y + 2}{2} = \frac{z - 3}{3}$
Find the direction cosines of a line parallel to line AB.

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Answer:

\frac{6}{7}, \frac{2}{7}, \frac{3}{7}

(or

- \frac{6}{7}, - \frac{2}{7}, - \frac{3}{7}

)

Rewrite: $\frac{x - \frac{1}{2}}{6} = \frac{y + 2}{2} = \frac{z - 3}{3}$ , so direction ratios are 6, 2, 3.
$6^{2} + 2^{2} + 3^{2} = 49 = 7$ .
Direction cosines of a parallel line: $\pm (\frac{6}{7}, \frac{2}{7}, \frac{3}{7})$ .

Q22 marksVery Short AnswerIntegrals

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

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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 65/5/2

Q32 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.

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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 65/5/2

Q3 (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.

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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 65/5/2

Q42 marksVery Short AnswerDifferential Equations

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

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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 65/5/2

Q52 marksVery Short AnswerVector Algebra

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

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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 65/5/2

Q62 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.

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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 65/5/2

Q73 marksShort AnswerVector Algebra

If $a$ and $b$ are two vectors such that $∣ a + b ∣ = ∣ b ∣$ , then prove that $(a + 2 b)$ is perpendicular to $a$ .

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Answer: Proved.

$∣ a + b ∣^{2} = ∣ b ∣^{2} \Rightarrow ∣ a ∣^{2} + 2 a \cdot b + ∣ b ∣^{2} = ∣ b ∣^{2}$ .
So $a \cdot a + 2 a \cdot b = 0$ , i.e. $a \cdot (a + 2 b) = 0$ .
Hence $(a + 2 b)$ is perpendicular to $a$ .

Also asked in: 2022 65/5/1

Q7 (OR) (OR)3 marksShort AnswerVector Algebra

If $a$ and $b$ are unit vectors and $θ$ is the angle between them, then prove that
$sin \frac{θ}{2} = \frac{1}{2} ∣ a - b ∣$ .

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Answer: Proved.

$∣ a - b ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} - 2 a \cdot b = 1 + 1 - 2 cos θ$ .
$= 2 (1 - cos θ) = 4 sin^{2} \frac{θ}{2}$ .
Since $0 \leq θ \leq π$ , $sin \frac{θ}{2} \geq 0$ , so $∣ a - b ∣ = 2 sin \frac{θ}{2}$ .
Hence $sin \frac{θ}{2} = \frac{1}{2} ∣ a - b ∣$ .

Also asked in: 2022 65/5/1

Q83 marksShort AnswerIntegrals

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

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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 65/5/2

Q8 (OR) (OR)3 marksShort AnswerIntegrals

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

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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 65/5/2

Q93 marksShort AnswerIntegrals

Evaluate : $\int_{1}^{3} \frac{x}{x + 4 - x} d x$

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Answer: 1

Using $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ with $a + b = 4$ : $I = \int_{1}^{3} \frac{4 - x}{4 - x + x} d x$ .
Adding the two forms: $2 I = \int_{1}^{3} 1 d x = 2$ .
So $I = 1$ .

Q103 marksShort AnswerThree Dimensional GeometryNot in current syllabus

Find the distance of the point $(2, 3, 4)$ measured along the line $\frac{x - 4}{3} = \frac{y + 5}{6} = \frac{z + 1}{2}$ from the plane $3 x + 2 y + 2 z + 5 = 0$ .

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Answer: 7 units

Line through $(2, 3, 4)$ parallel to the given line: $x = 2 + 3 t$ , $y = 3 + 6 t$ , $z = 4 + 2 t$ .
Substitute in the plane: $3 (2 + 3 t) + 2 (3 + 6 t) + 2 (4 + 2 t) + 5 = 0 \Rightarrow 25 + 25 t = 0$ , so $t = - 1$ .
Point on the plane: $(- 1, - 3, 2)$ .
Distance $= (2 + 1)^{2} + (3 + 3)^{2} + (4 - 2)^{2} = 9 + 36 + 4 = 7$ units.

Q114 marksLong AnswerApplication of Integrals

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

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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 65/5/2

Q124 marksLong AnswerDifferential Equations

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

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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 65/5/2

Q12 (OR) (OR)4 marksLong AnswerDifferential Equations

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

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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 65/5/2

Q134 marksLong AnswerProbability

There are two boxes, namely box-I and box-II. Box-I contains 3 red and 6 black balls. Box-II contains 5 red and 5 black balls. One of the two boxes, is selected at random and a ball is drawn at random. The ball drawn is found to be red. Find the probability that this red ball comes out from box-II.

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Answer:

\frac{3}{5}

Let $E_{1}$ , $E_{2}$ be choosing box-I, box-II: $P (E_{1}) = P (E_{2}) = \frac{1}{2}$ . Let R = red ball drawn.
$P (R ∣ E_{1}) = \frac{3}{9} = \frac{1}{3}$ , $P (R ∣ E_{2}) = \frac{5}{10} = \frac{1}{2}$ .
$P (E_{2} ∣ R) = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{5}{12}} = \frac{3}{5}$ .

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)

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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 65/5/2

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