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

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

Find : $\int \frac{d x}{x ^{2} - 6 x + 13}$

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Answer:

\frac{1}{2} tan^{- 1} (\frac{x - 3}{2}) + C

$x^{2} - 6 x + 13 = (x - 3)^{2} + 2^{2}$
$\int \frac{d x}{( x - 3 ) ^{2} + 2 ^{2}} = \frac{1}{2} tan^{- 1} (\frac{x - 3}{2}) + C$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q22 marksVery Short AnswerDifferential Equations

Find the general solution of the differential equation : $e^{d y / d x} = x^{2}$ .

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Answer:

y = 2 x lo g x - 2 x + C

Taking log: $\frac{d y}{d x} = lo g x^{2} = 2 lo g x$ (for $x > 0$ )
$y = 2 \int lo g x d x = 2 (x lo g x - x) + C$
So $y = 2 x lo g x - 2 x + C$

Q32 marksVery Short AnswerVector Algebra

Write the projection of the vector $(b + c)$ on the vector $a$ , where $a = 2 i^- 2 j^+ k^$ , $b = i^+ 2 j^- 2 k^$ and $c = 2 i^- j^+ 4 k^$ .

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Answer: 2

$b + c = 3 i^+ j^+ 2 k^$
$a \cdot (b + c) = 6 - 2 + 2 = 6$
$∣ a ∣ = 4 + 4 + 1 = 3$
Projection $= \frac{6}{3} = 2$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q42 marksVery Short AnswerThree Dimensional GeometryNot in current syllabus

If the distance of the point (1, 1, 1) from the plane $x - y + z + λ = 0$ is $\frac{5}{3}$ , find the value(s) of $λ$ .

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Answer:

λ = 4

λ = - 6

Distance $= \frac{∣1 - 1 + 1 + λ ∣}{1 + 1 + 1} = \frac{∣1 + λ ∣}{3}$
$\frac{∣1 + λ ∣}{3} = \frac{5}{3} \Rightarrow ∣1 + λ ∣ = 5$
$1 + λ = \pm 5 \Rightarrow λ = 4$ or $λ = - 6$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q52 marksVery Short AnswerProbabilityNot in current syllabus

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of spade cards.

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

\frac{9}{16}

\frac{6}{16}

\frac{3}{8}

\frac{1}{16}

Let X = number of spades; X = 0, 1, 2
P(spade) $= \frac{13}{52} = \frac{1}{4}$ , P(not spade) $= \frac{3}{4}$ , draws independent (with replacement)
$P (X = 0) = \frac{3}{4} \cdot \frac{3}{4} = \frac{9}{16}$
$P (X = 1) = 2 \cdot \frac{1}{4} \cdot \frac{3}{4} = \frac{6}{16} = \frac{3}{8}$
$P (X = 2) = \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16}$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q62 marksVery Short AnswerProbability

A pair of dice is thrown and the sum of the numbers appearing on the dice is observed to be 7. Find the probability that the number 5 has appeared on atleast one die.

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Answer:

\frac{1}{3}

F: sum is 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, 6 outcomes
E: 5 on at least one die; $E \cap F$ = {(2,5), (5,2)}, 2 outcomes
$P (E ∣ F) = \frac{n ( E \cap F )}{n ( F )} = \frac{2}{6} = \frac{1}{3}$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q6 (OR) (OR)2 marksVery Short AnswerProbability

The probability that A hits the target is $\frac{1}{3}$ and the probability that B hits it, is $\frac{2}{5}$ . If both try to hit the target independently, find the probability that the target is hit.

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Answer:

\frac{3}{5}

P(A misses) $= \frac{2}{3}$ , P(B misses) $= \frac{3}{5}$
P(both miss) $= \frac{2}{3} \cdot \frac{3}{5} = \frac{2}{5}$ (independent)
P(target is hit) $= 1 - \frac{2}{5} = \frac{3}{5}$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q73 marksShort AnswerIntegrals

Evaluate : $\int_{0}^{2 π} \frac{d x}{1 + e ^{s i n x}}$

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Answer:

π

Let $I = \int_{0}^{2 π} \frac{d x}{1 + e ^{s i n x}}$
Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ and $sin (2 π - x) = - sin x$ : $I = \int_{0}^{2 π} \frac{d x}{1 + e ^{- s i n x}} = \int_{0}^{2 π} \frac{e ^{s i n x}}{e ^{s i n x} + 1} d x$
Adding: $2 I = \int_{0}^{2 π} 1 d x = 2 π$
$I = π$

Q83 marksShort AnswerDifferential Equations

Find the particular solution of the differential equation $x \frac{d y}{d x} - y = x^{2} \cdot e^{x}$ , given $y (1) = 0$ .

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Answer:

y = x e^{x} - e x

Divide by x: $\frac{d y}{d x} - \frac{1}{x} y = x e^{x}$ , linear with $P = - \frac{1}{x}$ , $Q = x e^{x}$
I.F. $= e^{- \int \frac{1}{x} d x} = e^{- l o g x} = \frac{1}{x}$
$y \cdot \frac{1}{x} = \int x e^{x} \cdot \frac{1}{x} d x = e^{x} + C$
$y = x e^{x} + C x$
$y (1) = 0$ : $0 = e + C \Rightarrow C = - e$
$y = x e^{x} - e x$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q8 (OR) (OR)3 marksShort AnswerDifferential Equations

Find the general solution of the differential equation $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ .

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Answer:

lo g (\frac{y}{x}) = C x

$\frac{d y}{d x} = \frac{y}{x} (lo g \frac{y}{x} + 1)$ , a homogeneous equation
Put $y = v x$ : $v + x \frac{d v}{d x} = v (lo g v + 1)$
$x \frac{d v}{d x} = v lo g v$
$\int \frac{d v}{v l o g v} = \int \frac{d x}{x}$
$lo g ∣ lo g v ∣ = lo g ∣ x ∣ + lo g ∣ C ∣$
$lo g v = C x$ , i.e. $lo g (\frac{y}{x}) = C x$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q93 marksShort AnswerVector Algebra

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

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Answer: Unit vector

\frac{1}{7} (3 i^- 6 j^+ 2 k^)

(or

\frac{1}{69} (i^- 2 j^+ 8 k^)

along the other diagonal); area

= 115

sq units

Let $a = 2 i^- 4 j^+ 5 k^$ , $b = i^- 2 j^- 3 k^$
Diagonal $a + b = 3 i^- 6 j^+ 2 k^$ , magnitude $9 + 36 + 4 = 7$
Unit vector $= \frac{1}{7} (3 i^- 6 j^+ 2 k^)$
(Other diagonal $a - b = i^- 2 j^+ 8 k^$ gives $\frac{1}{69} (i^- 2 j^+ 8 k^)$ )
$a \times b = i^(12 + 10) - j^(- 6 - 5) + k^(- 4 + 4) = 22 i^+ 11 j^$
Area $= ∣ a \times b ∣ = 484 + 121 = 605 = 115$ sq units

Also asked in: 2022 65/3/3

Q9 (OR) (OR)3 marksShort AnswerVector Algebra

If $a = 2 i^+ 2 j^+ 3 k^$ , $b = - i^+ 2 j^+ k^$ and $c = 3 i^+ j^$ are such that the vector $(a + λ b)$ is perpendicular to vector $c$ , then find the value of $λ$ .

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Answer:

λ = 8

$a + λ b = (2 - λ) i^+ (2 + 2 λ) j^+ (3 + λ) k^$
Perpendicular to $c$ : $(a + λ b) \cdot c = 0$
$3 (2 - λ) + 1 (2 + 2 λ) + 0 = 0$
$8 - λ = 0 \Rightarrow λ = 8$

Also asked in: 2022 65/3/3

Q103 marksShort AnswerThree Dimensional GeometryNot in current syllabus

Show that the lines :
$\frac{1 - x}{2} = \frac{y - 3}{4} = \frac{z}{- 1}$ and $\frac{x - 4}{3} = \frac{2 y - 2}{- 4} = z - 1$ are coplanar.

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Answer: Proved.

Line 1: $\frac{x - 1}{- 2} = \frac{y - 3}{4} = \frac{z - 0}{- 1}$ , through $(1, 3, 0)$ , d.r. $- 2, 4, - 1$
Line 2: $\frac{x - 4}{3} = \frac{y - 1}{- 2} = \frac{z - 1}{1}$ , through $(4, 1, 1)$ , d.r. $3, - 2, 1$
Condition: $4 - 1 - 2 3 1 - 3 4 - 2 1 - 0 - 1 1 = 3 - 2 3 - 2 4 - 2 1 - 1 1$
The first and third rows are equal, so the determinant is 0
Hence the lines are coplanar. (In fact $(1, 3, 0)$ lies on line 2, so they intersect there.)

Also asked in: 2022 65/3/2

Q114 marksLong AnswerApplication of Integrals

Find the area of the region bounded by curve $4 x^{2} = y$ and the line $y = 8 x + 12$ , using integration.

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Answer:

\frac{128}{3}

sq units

Intersection: $4 x^{2} = 8 x + 12 \Rightarrow x^{2} - 2 x - 3 = 0 \Rightarrow x = - 1, 3$
On $[- 1, 3]$ the line lies above the parabola
Area $= \int_{- 1}^{3} (8 x + 12 - 4 x^{2}) d x = [4 x^{2} + 12 x - \frac{4 x ^{3}}{3}]_{- 1}^{3}$
$= (36 + 36 - 36) - (4 - 12 + \frac{4}{3}) = 36 + \frac{20}{3} = \frac{128}{3}$ sq units

Q124 marksLong AnswerIntegrals

Find : $\int \frac{x ^{2}}{( x ^{2} + 1 ) ( 3 x ^{2} + 4 )} d x$

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Answer:

- tan^{- 1} x + \frac{2}{3} tan^{- 1} (\frac{3 x}{2}) + C

Put $x^{2} = t$ : $\frac{t}{( t + 1 ) ( 3 t + 4 )} = \frac{A}{t + 1} + \frac{B}{3 t + 4}$
$t = A (3 t + 4) + B (t + 1)$ ; $t = - 1$ : $A = - 1$ ; $t = - \frac{4}{3}$ : $B = 4$
So integrand $= - \frac{1}{x ^{2} + 1} + \frac{4}{3 x ^{2} + 4}$
$\int \frac{4}{3 x ^{2} + 4} d x = \frac{4}{3} \int \frac{d x}{x ^{2} + ( \frac{2}{3} ) ^{2}} = \frac{4}{3} \cdot \frac{3}{2} tan^{- 1} \frac{3 x}{2} = \frac{2}{3} tan^{- 1} \frac{3 x}{2}$
Answer: $- tan^{- 1} x + \frac{2}{3} tan^{- 1} (\frac{3 x}{2}) + C$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q12 (OR) (OR)4 marksLong AnswerIntegrals

Evaluate : $\int_{- 2}^{1} 5 - 4 x - x^{2} d x$

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Answer:

\frac{9 π}{4}

$5 - 4 x - x^{2} = 9 - (x + 2)^{2}$
$\int 9 - (x + 2)^{2} d x = \frac{x + 2}{2} 5 - 4 x - x^{2} + \frac{9}{2} sin^{- 1} \frac{x + 2}{3}$
At $x = 1$ : $\frac{3}{2} \cdot 0 + \frac{9}{2} \cdot \frac{π}{2} = \frac{9 π}{4}$
At $x = - 2$ : $0 + 0 = 0$
Value $= \frac{9 π}{4}$

Also asked in: 2022 65/3/2, 2022 65/3/3

Q134 marksLong AnswerThree Dimensional GeometryNot in current syllabus

Find the distance of the point $(1, - 2, 9)$ from the point of intersection of the line $r = 4 i^+ 2 j^+ 7 k^+ λ (3 i^+ 4 j^+ 2 k^)$ and the plane $r \cdot (i^- j^+ k^) = 10$ .

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Answer: 10 units

General point of the line: $(4 + 3 λ, 2 + 4 λ, 7 + 2 λ)$
On the plane $x - y + z = 10$ : $(4 + 3 λ) - (2 + 4 λ) + (7 + 2 λ) = 10$
$9 + λ = 10 \Rightarrow λ = 1$ , point $(7, 6, 9)$
Distance from $(1, - 2, 9)$ $= 6^{2} + 8^{2} + 0^{2} = 10$ units

Also asked in: 2022 65/3/2, 2022 65/3/3

Q144 marksCase StudyProbability

A shopkeeper sells three types of flower seeds A1, A2, A3. They are sold in the form of a mixture, where the proportions of these seeds are 4 : 4 : 2, respectively. The germination rates of the three types of seeds are 45%, 60% and 35% respectively.
Based on the above information :
(a) Calculate the probability that a randomly chosen seed will germinate; (2)
(b) Calculate the probability that the seed is of type A2, given that a randomly chosen seed germinates. (2)

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Answer: (a) 0.49 (b)

\frac{24}{49}

$P (A_{1}) = 0.4$ , $P (A_{2}) = 0.4$ , $P (A_{3}) = 0.2$ ; G = seed germinates
$P (G ∣ A_{1}) = 0.45$ , $P (G ∣ A_{2}) = 0.60$ , $P (G ∣ A_{3}) = 0.35$
(a) $P (G) = 0.4 \times 0.45 + 0.4 \times 0.60 + 0.2 \times 0.35 = 0.18 + 0.24 + 0.07 = 0.49$
(b) By Bayes' theorem, $P (A_{2} ∣ G) = \frac{0.24}{0.49} = \frac{24}{49}$

Also asked in: 2022 65/3/2, 2022 65/3/3

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