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

All 18 questions from the CBSE Class 12 Mathematics board paper, Set 65/2/3 (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

Write the cartesian equation of the line PQ passing through points $P (2, 2, 1)$ and $Q (5, 1, - 2)$ . Hence, find the y-coordinate of the point on the line PQ whose z-coordinate is $- 2$ .

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Answer:

\frac{x - 2}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{- 3}

; y-coordinate

= 1

Direction ratios of PQ: $(5 - 2, 1 - 2, - 2 - 1) = (3, - 1, - 3)$ .
Equation: $\frac{x - 2}{3} = \frac{y - 2}{- 1} = \frac{z - 1}{- 3}$ .
Put $z = - 2$ : $\frac{- 2 - 1}{- 3} = 1$ , so $\frac{y - 2}{- 1} = 1$ , giving $y = 1$ .

Also asked in: 2022 65/2/1, 2022 65/2/2

Q22 marksVery Short AnswerProbability

Probabilities of A and B solving a specific problem are $\frac{2}{3}$ and $\frac{3}{5}$ , respectively. If both of them try independently to solve the problem, then find the probability that the problem is solved.

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Answer:

\frac{13}{15}

P(A fails) $= \frac{1}{3}$ , P(B fails) $= \frac{2}{5}$ .
Since they work independently, P(both fail) $= \frac{1}{3} \times \frac{2}{5} = \frac{2}{15}$ .
P(problem solved) $= 1 - \frac{2}{15} = \frac{13}{15}$ .

Also asked in: 2022 65/2/1, 2022 65/2/2

Q32 marksVery Short AnswerProbability

A pair of dice is thrown. It is given that the sum of numbers appearing on both dice is an even number. Find the probability that the number appearing on at least one die is 3.

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Answer:

\frac{5}{18}

Let E: sum is even, F: at least one die shows 3. Sum is even when both numbers are odd or both even: $9 + 9 = 18$ outcomes, so $P (E) = \frac{18}{36}$ .
$E \cap F = {(1, 3), (3, 1), (3, 3), (3, 5), (5, 3)}$ , so $P (E \cap F) = \frac{5}{36}$ .
$P (F ∣ E) = \frac{P ( E \cap F )}{P ( E )} = \frac{5}{18}$ .

Also asked in: 2022 65/2/1, 2022 65/2/2

Q42 marksVery Short AnswerVector Algebra

$a$ and $b$ are two unit vectors such that $∣2 a + 3 b ∣ = ∣3 a - 2 b ∣$ . Find the angle between $a$ and $b$ .

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Answer:

\frac{π}{2}

Square both sides: $∣2 a + 3 b ∣^{2} = ∣3 a - 2 b ∣^{2}$ .
$4 + 9 + 12 a \cdot b = 9 + 4 - 12 a \cdot b$ (since $∣ a ∣ = ∣ b ∣ = 1$ ).
$24 a \cdot b = 0$ , so $a \cdot b = 0$ .
Hence the angle between $a$ and $b$ is $\frac{π}{2}$ .

Also asked in: 2022 65/2/1, 2022 65/2/2

Q52 marksVery Short AnswerDifferential Equations

Find the product of the order and the degree of the differential equation $[\frac{d}{d x} (x y^{2})] \cdot \frac{d y}{d x} + y = 0$ .

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Answer: 2

$\frac{d}{d x} (x y^{2}) = y^{2} + 2 x y \frac{d y}{d x}$ .
So the equation is $(y^{2} + 2 x y \frac{d y}{d x}) \frac{d y}{d x} + y = 0$ , i.e. $2 x y (\frac{d y}{d x})^{2} + y^{2} \frac{d y}{d x} + y = 0$ .
Highest derivative is $\frac{d y}{d x}$ , so order $= 1$ ; its highest power is 2, so degree $= 2$ .
Product $= 1 \times 2 = 2$ .

Also asked in: 2022 65/2/1

Q62 marksVery Short AnswerIntegrals

Find :
$\int \frac{c o t x}{s i n x c o s x} d x$

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Answer:

- 2 cot x + C

$sin x cos x = sin^{2} x cot x$ , so the integrand is $\frac{cosec ^{2} x}{c o t x}$ .
Put $t = cot x$ , $d t = - cosec^{2} x d x$ : integral $= - \int t^{- 1/2} d t = - 2 t + C$ .
$= - 2 cot x + C$ .

Q6 (OR) (OR)2 marksVery Short AnswerIntegrals

Find :
$\int \frac{1}{x ( x ^{2} + 4 )} d x$

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Answer:

\frac{1}{4} lo g ∣ x ∣ - \frac{1}{8} lo g (x^{2} + 4) + C

Write $\frac{1}{x ( x ^{2} + 4 )} = \frac{A}{x} + \frac{B x + C}{x ^{2} + 4}$ ; then $1 = A (x^{2} + 4) + (B x + C) x$ .
$x = 0$ gives $A = \frac{1}{4}$ ; comparing $x^{2}$ and $x$ terms: $B = - \frac{1}{4}$ , $C = 0$ .
$\int (\frac{1}{4 x} - \frac{x}{4 ( x ^{2} + 4 )}) d x = \frac{1}{4} lo g ∣ x ∣ - \frac{1}{8} lo g (x^{2} + 4) + C$ .

Q73 marksShort AnswerIntegrals

Evaluate :
$\int_{0}^{1} tan^{- 1} x d x$

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Answer:

\frac{π}{4} - \frac{1}{2} lo g 2

By parts: $\int_{0}^{1} tan^{- 1} x d x = [x tan^{- 1} x]_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x ^{2}} d x$ .
$= \frac{π}{4} - \frac{1}{2} [lo g (1 + x^{2})]_{0}^{1}$ .
$= \frac{π}{4} - \frac{1}{2} lo g 2$ .

Also asked in: 2022 65/2/1, 2022 65/2/2

Q7 (OR) (OR)3 marksShort AnswerIntegrals

Find :
$\int \frac{2 x}{x ^{2} + 3 x + 2} d x$

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Answer:

4 lo g ∣ x + 2∣ - 2 lo g ∣ x + 1∣ + C

$x^{2} + 3 x + 2 = (x + 1) (x + 2)$ . Write $\frac{2 x}{( x + 1 ) ( x + 2 )} = \frac{A}{x + 1} + \frac{B}{x + 2}$ .
$2 x = A (x + 2) + B (x + 1)$ . Put $x = - 1$ : $A = - 2$ . Put $x = - 2$ : $B = 4$ .
$\int (\frac{- 2}{x + 1} + \frac{4}{x + 2}) d x = - 2 lo g ∣ x + 1∣ + 4 lo g ∣ x + 2∣ + C$ .

Also asked in: 2022 65/2/1, 2022 65/2/2

Q83 marksShort AnswerThree Dimensional GeometryNot in current syllabus

Find the equation of the plane passing through points $(2, 1, 0)$ , $(3, - 2, - 2)$ and $(1, 1, - 7)$ . Also, obtain its distance from the origin.

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Answer:

7 x + 3 y - z = 17

; distance

= \frac{17}{59}

units

Vectors in the plane: $(3, - 2, - 2) - (2, 1, 0) = (1, - 3, - 2)$ and $(1, 1, - 7) - (2, 1, 0) = (- 1, 0, - 7)$ .
Normal $= (1, - 3, - 2) \times (- 1, 0, - 7) = (21, 9, - 3)$ , i.e. $(7, 3, - 1)$ .
Plane: $7 (x - 2) + 3 (y - 1) - (z - 0) = 0$ , i.e. $7 x + 3 y - z = 17$ .
Distance from origin $= \frac{∣0 - 17∣}{49 + 9 + 1} = \frac{17}{59}$ units.

Also asked in: 2022 65/2/1, 2022 65/2/2

Q8 (OR) (OR)3 marksShort AnswerThree Dimensional Geometry

Find the distance between the lines $x = \frac{y - 1}{2} = \frac{z - 2}{3}$ and $x + 1 = \frac{y + 2}{2} = \frac{z - 1}{3}$ .

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Answer:

\frac{3 21}{7}

units (i.e.

\frac{27}{7}

)

Both lines have direction $b = i^+ 2 j^+ 3 k^$ , so they are parallel.
Points: $a_{1} = j^+ 2 k^$ on the first, $a_{2} = - i^- 2 j^+ k^$ on the second; $a_{2} - a_{1} = - i^- 3 j^- k^$ .
$b \times (a_{2} - a_{1}) = i^1 - 1 j^2 - 3 k^3 - 1 = 7 i^- 2 j^- k^$ , magnitude $54$ .
Distance $= \frac{∣ b \times ( a _{2} - a _{1} ) ∣}{∣ b ∣} = \frac{54}{14} = \frac{27}{7} = \frac{3 21}{7}$ units.

Also asked in: 2022 65/2/1, 2022 65/2/2

Q93 marksShort AnswerDifferential Equations

Solve the following differential equation :
$(1 + e^{y / x}) d y + e^{y / x} (1 - \frac{y}{x}) d x = 0$

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Answer:

y + x e^{y / x} = C

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

Q103 marksShort AnswerVector Algebra

If $a$ and $b$ are two vectors such that $a = i^- j^+ k^$ and $b = 2 i^- j^- 3 k^$ , then find the vector $c$ , given that $a \times c = b$ and $a \cdot c = 4$ .

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Answer:

c = - 3 j^+ k^

Let $c = x i^+ y j^+ z k^$ . $a \times c = i^1 x j^- 1 y k^1 z = (- z - y) i^+ (x - z) j^+ (x + y) k^$ .
Equating with $b$ : $- y - z = 2$ , $x - z = - 1$ , $x + y = - 3$ .
$a \cdot c = 4$ : $x - y + z = 4$ .
From $z = x + 1$ and $y = - 3 - x$ : $x + 3 + x + x + 1 = 4$ , so $x = 0$ , $y = - 3$ , $z = 1$ .
$c = - 3 j^+ k^$ (check: $a \times c = 2 i^- j^- 3 k^$ , $a \cdot c = 4$ ).

Q114 marksLong AnswerIntegrals

Evaluate :
$\int_{0}^{1} x (1 - x)^{n} d x$

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Answer:

\frac{1}{( n + 1 ) ( n + 2 )}

Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : $\int_{0}^{1} x (1 - x)^{n} d x = \int_{0}^{1} (1 - x) x^{n} d x$ .
$= \int_{0}^{1} (x^{n} - x^{n + 1}) d x = [\frac{x ^{n + 1}}{n + 1} - \frac{x ^{n + 2}}{n + 2}]_{0}^{1}$ .
$= \frac{1}{n + 1} - \frac{1}{n + 2} = \frac{1}{( n + 1 ) ( n + 2 )}$ .

Also asked in: 2022 65/2/1

Q124 marksLong AnswerApplication of Integrals

Using integration, find the area of the smaller region enclosed by the curve $4 x^{2} + 4 y^{2} = 9$ and the line $2 x + 2 y = 3$ .

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Answer:

\frac{9}{16} (π - 2)

sq. units

Curve: $x^{2} + y^{2} = \frac{9}{4}$ , a circle of radius $\frac{3}{2}$ . Line: $x + y = \frac{3}{2}$ , meeting the circle at $(\frac{3}{2}, 0)$ and $(0, \frac{3}{2})$ .
Smaller region lies in the first quadrant between the arc and the line.
Area $= \int_{0}^{3/2} \frac{9}{4} - x^{2} d x - \int_{0}^{3/2} (\frac{3}{2} - x) d x$ .
$\int_{0}^{3/2} \frac{9}{4} - x^{2} d x = [\frac{x}{2} \frac{9}{4} - x^{2} + \frac{9}{8} sin^{- 1} \frac{2 x}{3}]_{0}^{3/2} = \frac{9}{8} \cdot \frac{π}{2} = \frac{9 π}{16}$ .
$\int_{0}^{3/2} (\frac{3}{2} - x) d x = [\frac{3 x}{2} - \frac{x ^{2}}{2}]_{0}^{3/2} = \frac{9}{4} - \frac{9}{8} = \frac{9}{8}$ .
Area $= \frac{9 π}{16} - \frac{9}{8} = \frac{9}{16} (π - 2)$ sq. units.

Also asked in: 2022 65/2/1, 2022 65/2/2

Q12 (OR) (OR)4 marksLong AnswerApplication of Integrals

If the area of the region bounded by the curve $y^{2} = 4 a x$ and the line $x = 4 a$ is $\frac{256}{3}$ sq. units, then using integration, find the value of a, where $a > 0$ .

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Answer:

a = 2

The region is symmetric about the x-axis, from $x = 0$ to $x = 4 a$ .
Area $= 2 \int_{0}^{4 a} 2 a x d x = 4 a [\frac{2}{3} x^{3/2}]_{0}^{4 a}$ .
$= 4 a \cdot \frac{2}{3} \cdot 8 a^{3/2} = \frac{64 a ^{2}}{3}$ .
$\frac{64 a ^{2}}{3} = \frac{256}{3}$ gives $a^{2} = 4$ , and since $a > 0$ , $a = 2$ .

Also asked in: 2022 65/2/1, 2022 65/2/2

Q134 marksLong AnswerThree Dimensional Geometry

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the zx-plane.

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Answer:

(\frac{17}{3}, 0, \frac{23}{3})

Direction ratios: $(3 - 5, 4 - 1, 1 - 6) = (- 2, 3, - 5)$ . Line: $\frac{x - 5}{- 2} = \frac{y - 1}{3} = \frac{z - 6}{- 5} = λ$ .
General point: $(5 - 2 λ, 1 + 3 λ, 6 - 5 λ)$ .
On the zx-plane $y = 0$ : $1 + 3 λ = 0$ , so $λ = - \frac{1}{3}$ .
Point: $(5 + \frac{2}{3}, 0, 6 + \frac{5}{3}) = (\frac{17}{3}, 0, \frac{23}{3})$ .

Q144 marksCase StudyProbabilityNot in current syllabus

At the start of a cricket match, a coin is tossed and the team winning the toss has the opportunity to choose to bat or bowl. Such a coin is unbiased with equal probabilities of getting head and tail.
Based on the above information, answer the following questions :
(a) If such a coin is tossed 2 times, then find the probability distribution of number of tails. (2)
(b) Find the probability of getting at least one head in three tosses of such a coin. (2)

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Answer: (a) X = number of tails: X = 0, 1, 2 with P(X) =

\frac{1}{4}, \frac{1}{2}, \frac{1}{4}

(b)

\frac{7}{8}

(a) Two tosses give HH, HT, TH, TT, each with probability $\frac{1}{4}$ .
Let X be the number of tails. $P (X = 0) = \frac{1}{4}$ , $P (X = 1) = \frac{2}{4} = \frac{1}{2}$ , $P (X = 2) = \frac{1}{4}$ .
(b) P(no head in 3 tosses) $= P (T T T) = (\frac{1}{2})^{3} = \frac{1}{8}$ .
P(at least one head) $= 1 - \frac{1}{8} = \frac{7}{8}$ .

Also asked in: 2022 65/2/1, 2022 65/2/2

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