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CBSE Class 12 Maths 2022 Question Paper 65/1/3 (Term 2) with Solutions

All 18 questions from the CBSE Class 12 Mathematics board paper, Set 65/1/3 (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 AnswerProbability

Let A and B be two events such that $P (A) = \frac{5}{8}$ , $P (B) = \frac{1}{2}$ and $P (A / B) = \frac{3}{4}$ . Find the value of $P (B / A)$ .

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Answer:

\frac{3}{5}

$P (A \cap B) = P (A / B) \cdot P (B) = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$ .
$P (B / A) = \frac{P ( A \cap B )}{P ( A )} = \frac{3/8}{5/8} = \frac{3}{5}$ .

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

Q22 marksVery Short AnswerProbabilityNot in current syllabus

Two balls are drawn at random from a bag containing 2 red balls and 3 blue balls, without replacement. Let the variable X denotes the number of red balls. Find the probability distribution of X.

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Answer: X = 0, 1, 2 with P(X) =

\frac{3}{10}

\frac{3}{5}

\frac{1}{10}

X can take values 0, 1, 2.
$P (X = 0) = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$ .
$P (X = 1) = \frac{2}{5} \times \frac{3}{4} + \frac{3}{5} \times \frac{2}{4} = \frac{6}{10} = \frac{3}{5}$ .
$P (X = 2) = \frac{2}{5} \times \frac{1}{4} = \frac{1}{10}$ .
Check: $\frac{3}{10} + \frac{6}{10} + \frac{1}{10} = 1$ .

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

Q32 marksVery Short AnswerDifferential Equations

Find the sum of the order and the degree of the differential equation :
$(x + \frac{d y}{d x})^{2} = (\frac{d y}{d x})^{2} + 1$

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Answer: 2

Expand the left side: $x^{2} + 2 x \frac{d y}{d x} + (\frac{d y}{d x})^{2} = (\frac{d y}{d x})^{2} + 1$ .
Cancel $(\frac{d y}{d x})^{2}$ : $x^{2} + 2 x \frac{d y}{d x} = 1$ .
Highest derivative is $\frac{d y}{d x}$ , so order $= 1$ ; it appears to power 1, so degree $= 1$ .
Sum $= 1 + 1 = 2$ .

Also asked in: 2022 65/1/1

Q42 marksVery Short AnswerIntegrals

If $\frac{d}{d x} [F (x)] = \frac{s e c ^{4} x}{cosec ^{4} x}$ and $F (\frac{π}{4}) = \frac{π}{4}$ , then find $F (x)$ .

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Answer:

F (x) = \frac{t a n ^{3} x}{3} - tan x + x + \frac{2}{3}

$\frac{s e c ^{4} x}{cosec ^{4} x} = \frac{s i n ^{4} x}{c o s ^{4} x} = tan^{4} x$ .
$tan^{4} x = tan^{2} x (sec^{2} x - 1) = tan^{2} x sec^{2} x - sec^{2} x + 1$ .
So $F (x) = \frac{t a n ^{3} x}{3} - tan x + x + C$ (using $t = tan x$ for the first term).
$F (\frac{π}{4}) = \frac{1}{3} - 1 + \frac{π}{4} + C = \frac{π}{4}$ gives $C = \frac{2}{3}$ .
Hence $F (x) = \frac{t a n ^{3} x}{3} - tan x + x + \frac{2}{3}$ .

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

Q4 (OR) (OR)2 marksVery Short AnswerIntegrals

Find : $\int \frac{l o g x - 3}{( l o g x ) ^{4}} d x$ .

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Answer:

\frac{x}{( l o g x ) ^{3}} + C

Put $lo g x = t$ , so $x = e^{t}$ and $d x = e^{t} d t$ .
Integral $= \int e^{t} \frac{t - 3}{t ^{4}} d t = \int e^{t} (\frac{1}{t ^{3}} - \frac{3}{t ^{4}}) d t$ .
This is $\int e^{t} (f (t) + f^{'} (t)) d t$ with $f (t) = \frac{1}{t ^{3}}$ , $f^{'} (t) = - \frac{3}{t ^{4}}$ .
So the integral $= \frac{e ^{t}}{t ^{3}} + C = \frac{x}{( l o g x ) ^{3}} + C$ .

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

Q52 marksVery Short AnswerThree Dimensional GeometryNot in current syllabus

Find the values of $λ$ , for which the distance of point $(2, 1, λ)$ from plane $3 x + 5 y + 4 z = 11$ is $22$ units.

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Answer:

λ = \pm 5

Distance $= \frac{∣3 ( 2 ) + 5 ( 1 ) + 4 λ - 11∣}{9 + 25 + 16} = \frac{∣4 λ ∣}{5 2}$ .
Set $\frac{∣4 λ ∣}{5 2} = 22$ : $∣4 λ ∣ = 20$ .
So $λ = 5$ or $λ = - 5$ .

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

Q62 marksVery Short AnswerVector Algebra

If $a = 2 i^+ y j^+ k^$ and $b = i^+ 2 j^+ 3 k^$ are two vectors for which the vector $(a + b)$ is perpendicular to the vector $(a - b)$ , then find all the possible values of y.

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Answer:

y = \pm 3

$(a + b) \cdot (a - b) = 0$ gives $∣ a ∣^{2} = ∣ b ∣^{2}$ .
$4 + y^{2} + 1 = 1 + 4 + 9$ , so $y^{2} = 9$ .
Hence $y = 3$ or $y = - 3$ .

Q73 marksShort AnswerIntegrals

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

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Answer:

2 x - 3 3 x + 6 6 x - 6 lo g (1 + 6 x) + C

Put $x = t^{6}$ , $d x = 6 t^{5} d t$ ; then $x = t^{3}$ , $3 x = t^{2}$ .
Integral $= \int \frac{6 t ^{5}}{t ^{3} + t ^{2}} d t = 6 \int \frac{t ^{3}}{t + 1} d t$ .
$\frac{t ^{3}}{t + 1} = t^{2} - t + 1 - \frac{1}{t + 1}$ .
So integral $= 2 t^{3} - 3 t^{2} + 6 t - 6 lo g ∣ t + 1∣ + C$ .
With $t = x^{1/6}$ : $2 x - 3 3 x + 6 6 x - 6 lo g (1 + 6 x) + C$ .

Also asked in: 2022 65/1/1

Q7 (OR) (OR)3 marksShort AnswerIntegrals

Evaluate : $\int_{0}^{π /2} \frac{c o s x}{( 1 + s i n x ) ( 4 + s i n x )} d x$ .

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Answer:

\frac{1}{3} lo g \frac{8}{5}

Put $sin x = t$ , $cos x d x = d t$ ; limits 0 to 1.
Integral $= \int_{0}^{1} \frac{d t}{( 1 + t ) ( 4 + t )} = \frac{1}{3} \int_{0}^{1} (\frac{1}{1 + t} - \frac{1}{4 + t}) d t$ .
$= \frac{1}{3} [lo g \frac{1 + t}{4 + t}]_{0}^{1} = \frac{1}{3} (lo g \frac{2}{5} - lo g \frac{1}{4})$ .
$= \frac{1}{3} lo g \frac{8}{5}$ .

Also asked in: 2022 65/1/1

Q83 marksShort AnswerThree Dimensional GeometryNot in current syllabus

Find the co-ordinates of point where the line $\frac{x - 3}{- 1} = \frac{y + 4}{1} = \frac{z + 5}{6}$ crosses the plane passing through points $(\frac{7}{2}, 0, 0)$ , (0, 7, 0) and (0, 0, 7).

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Answer:

(1, - 2, 7)

Intercept form of the plane: $\frac{x}{7/2} + \frac{y}{7} + \frac{z}{7} = 1$ , i.e. $2 x + y + z = 7$ .
A general point on the line is $(3 - λ, λ - 4, 6 λ - 5)$ .
Substitute: $2 (3 - λ) + (λ - 4) + (6 λ - 5) = 7$ , so $5 λ - 3 = 7$ , $λ = 2$ .
Point: $(1, - 2, 7)$ .

Q93 marksShort AnswerDifferential Equations

Solve the differential equation : $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}$ .

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Answer:

y^{2} = 2 x^{2} lo g ∣ x ∣ + C x^{2}

The equation is homogeneous. Put $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
$v + x \frac{d v}{d x} = \frac{1 + v ^{2}}{v} = \frac{1}{v} + v$ , so $x \frac{d v}{d x} = \frac{1}{v}$ .
Separate: $v d v = \frac{d x}{x}$ ; integrate: $\frac{v ^{2}}{2} = lo g ∣ x ∣ + C_{1}$ .
Put $v = \frac{y}{x}$ : $y^{2} = 2 x^{2} lo g ∣ x ∣ + C x^{2}$ (where $C = 2 C_{1}$ ).

Q103 marksShort AnswerVector Algebra

If $a$ , $b$ , $c$ and $d$ are four non-zero vectors such that $a \times b = c \times d$ and $a \times c = 4 b \times d$ , then show that $(a - 2 d)$ is parallel to $(2 b - c)$ where $a \neq = 2 d$ , $c \neq = 2 b$ .

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Answer: Proved.

$(a - 2 d) \times (2 b - c) = 2 a \times b - a \times c - 4 d \times b + 2 d \times c$ .
Using $d \times b = - b \times d$ and $d \times c = - c \times d$ , this is $2 (a \times b - c \times d) - (a \times c - 4 b \times d)$ .
By the given conditions both brackets are $0$ , so the cross product is $0$ .
Since $a \neq = 2 d$ and $c \neq = 2 b$ , both vectors are non-zero, so $(a - 2 d) ∥ (2 b - c)$ .

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

Q10 (OR) (OR)3 marksShort AnswerVector Algebra

The two adjacent sides of a parallelogram are represented by $2 i^- 4 j^- 5 k^$ and $2 i^+ 2 j^+ 3 k^$ . Find the unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram also.

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Answer: Unit vectors

\pm \frac{1}{6} (2 i^- j^- k^)

and

\pm \frac{1}{5} (3 j^+ 4 k^)

; area

= 2101

sq units

Let $a = 2 i^- 4 j^- 5 k^$ , $b = 2 i^+ 2 j^+ 3 k^$ .
Diagonals: $d_{1} = a + b = 4 i^- 2 j^- 2 k^$ , $∣ d_{1} ∣ = 26$ ; unit vector $\pm \frac{1}{6} (2 i^- j^- k^)$ .
$d_{2} = b - a = 6 j^+ 8 k^$ , $∣ d_{2} ∣ = 10$ ; unit vector $\pm \frac{1}{5} (3 j^+ 4 k^)$ .
$d_{1} \times d_{2} = i^40 j^- 2 6 k^- 2 8 = - 4 i^- 32 j^+ 24 k^$ .
$∣ d_{1} \times d_{2} ∣ = 16 + 1024 + 576 = 1616 = 4101$ .
Area $= \frac{1}{2} ∣ d_{1} \times d_{2} ∣ = 2101$ sq units.

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

Q114 marksLong AnswerProbability

A card from a pack of 52 playing cards is lost. From the remaining cards, 2 cards are drawn at random without replacement, and are found to be both aces. Find the probability that lost card being an ace.

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Answer:

\frac{1}{25}

Let $E_{1}$ : lost card is an ace, $E_{2}$ : lost card is not an ace; A: both drawn cards are aces.
$P (E_{1}) = \frac{4}{52} = \frac{1}{13}$ , $P (E_{2}) = \frac{12}{13}$ .
$P (A ∣ E_{1}) = \frac{^{3} C _{2}}{^{51} C _{2}} = \frac{3}{1275}$ , $P (A ∣ E_{2}) = \frac{^{4} C _{2}}{^{51} C _{2}} = \frac{6}{1275}$ .
By Bayes' theorem, $P (E_{1} ∣ A) = \frac{\frac{1}{13} \cdot 3}{\frac{1}{13} \cdot 3 + \frac{12}{13} \cdot 6} = \frac{3}{75} = \frac{1}{25}$ .

Also asked in: 2022 65/1/1

Q124 marksLong AnswerIntegrals

Evaluate : $\int_{0}^{π} \frac{x}{1 + s i n x} d x$ .

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Answer:

π

Let $I = \int_{0}^{π} \frac{x}{1 + s i n x} d x$ . Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ : $I = \int_{0}^{π} \frac{π - x}{1 + s i n x} d x$ .
Adding: $2 I = π \int_{0}^{π} \frac{d x}{1 + s i n x}$ .
$\frac{1}{1 + s i n x} = \frac{1 - s i n x}{c o s ^{2} x} = sec^{2} x - sec x tan x$ , with antiderivative $tan x - sec x$ .
$[tan x - sec x]_{0}^{π} = (0 + 1) - (0 - 1) = 2$ (the integrand is continuous on $[0, π]$ , and $tan x - sec x = \frac{s i n x - 1}{c o s x}$ extends continuously at $\frac{π}{2}$ ).
So $2 I = 2 π$ , giving $I = π$ .

Also asked in: 2023 65/5/2, 2022 65/1/1, 2022 65/1/2

Q134 marksLong AnswerApplication of Integrals

Using integration, find the area of the region enclosed by the curve $y = x^{2}$ , the x-axis and the ordinates $x = - 2$ and $x = 1$ .

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Answer: 3 sq units

$y = x^{2} \geq 0$ , so the region lies above the x-axis for $- 2 \leq x \leq 1$ .
Area $= \int_{- 2}^{1} x^{2} d x = [\frac{x ^{3}}{3}]_{- 2}^{1}$ .
$= \frac{1}{3} - (- \frac{8}{3}) = 3$ sq units.

Q13 (OR) (OR)4 marksLong AnswerApplication of Integrals

Using integration, find the area of the region enclosed by line $y = 3 x$ , semi-circle $y = 4 - x^{2}$ and x-axis in first quadrant.

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Answer:

\frac{2 π}{3}

sq units

Intersection: $3 x^{2} = 4 - x^{2}$ gives $x = 1$ (first quadrant), $y = 3$ .
The region is bounded above by the line for $0 \leq x \leq 1$ and by the circle for $1 \leq x \leq 2$ .
Area $= \int_{0}^{1} 3 x d x + \int_{1}^{2} 4 - x^{2} d x$ .
$\int_{0}^{1} 3 x d x = \frac{3}{2}$ .
$\int_{1}^{2} 4 - x^{2} d x = [\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{1}^{2} = π - (\frac{3}{2} + \frac{π}{3})$ .
Area $= \frac{3}{2} + \frac{2 π}{3} - \frac{3}{2} = \frac{2 π}{3}$ sq units.

Q144 marksCase StudyThree Dimensional GeometryNot in current syllabus

Electrical transmission wires which are laid down in winters are stretched tightly to accommodate expansion in summers.
Two such wires lie along the following lines :
$l_{1} : \frac{x + 1}{3} = \frac{y - 3}{- 2} = \frac{z + 2}{- 1}$
$l_{2} : \frac{x}{- 1} = \frac{y - 7}{3} = \frac{z + 7}{- 2}$
Based on the given information, answer the following questions :
(i) Are the lines $l_{1}$ and $l_{2}$ coplanar ? Justify your answer. (2)
(ii) Find the point of intersection of the lines $l_{1}$ and $l_{2}$ . (2)

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Answer: (i) Yes, the lines are coplanar. (ii)

(2, 1, - 3)

(i) $l_{1}$ passes through $(- 1, 3, - 2)$ with direction ratios $3, - 2, - 1$ ; $l_{2}$ passes through $(0, 7, - 7)$ with direction ratios $- 1, 3, - 2$ .
$0 + 1 3 - 1 7 - 3 - 2 3 - 7 + 2 - 1 - 2 = 1 (4 + 3) - 4 (- 6 - 1) - 5 (9 - 2) = 7 + 28 - 35 = 0$ .
So the lines are coplanar (they are not parallel, so they intersect).
(ii) General points: $(3 λ - 1, - 2 λ + 3, - λ - 2)$ on $l_{1}$ and $(- μ, 3 μ + 7, - 2 μ - 7)$ on $l_{2}$ .
Equate x and y: $3 λ - 1 = - μ$ and $- 2 λ + 3 = 3 μ + 7$ ; solving gives $λ = 1$ , $μ = - 2$ .
Check z: $- 1 - 2 = - 3$ and $4 - 7 = - 3$ . Point of intersection is $(2, 1, - 3)$ .

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

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