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CBSE Class 12 Maths 2022 Question Paper 65/4/1 (Term 2) with Solutions

All 18 questions from the CBSE Class 12 Mathematics board paper, Set 65/4/1 (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 AnswerProbability

A bag contains 3 red and 4 white balls. Three balls are drawn at random, one-by-one without replacement from the bag. If the first ball drawn is red in colour, then find the probability that the remaining two balls drawn are also red in colour.

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Answer:

\frac{1}{15}

After the first red ball is drawn, the bag has 2 red and 4 white balls (6 balls).
P(second red) $= \frac{2}{6}$ ; then 1 red is left among 5 balls, so P(third red) $= \frac{1}{5}$ .
Required probability $= \frac{2}{6} \times \frac{1}{5} = \frac{1}{15}$ .

Q22 marksVery Short AnswerProbabilityNot in current syllabus

A coin is tossed twice. The following table shows the probability distribution of number of tails :
X: 0, 1, 2
P(X): K, 6K, 9K
(a) Find the value of K.
(b) Is the coin tossed biased or unbiased ? Justify your answer.

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Answer: (a)

K = \frac{1}{16}

(b) Biased

(a) Sum of probabilities is 1: $K + 6 K + 9 K = 1$ , so $16 K = 1$ and $K = \frac{1}{16}$ .
(b) So $P (X = 0) = \frac{1}{16}$ , $P (X = 1) = \frac{6}{16}$ , $P (X = 2) = \frac{9}{16}$ .
For an unbiased coin tossed twice, $P (X = 0) = \frac{1}{4}$ , $P (X = 1) = \frac{1}{2}$ , $P (X = 2) = \frac{1}{4}$ .
The given probabilities differ (e.g. $P (X = 2) = \frac{9}{16} \neq = \frac{1}{4}$ ), so the coin is biased (in fact $P (tail) = \frac{3}{4}$ ).

Also asked in: 2022 65/4/2, 2022 65/4/3

Q32 marksVery Short AnswerThree Dimensional GeometryNot in current syllabus

The foot of a perpendicular drawn from the point $(- 2, - 1, - 3)$ on a plane is $(1, - 3, 3)$ . Find the equation of the plane.

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Answer:

3 x - 2 y + 6 z = 27

The normal to the plane is along the perpendicular: $(1 - (- 2), - 3 - (- 1), 3 - (- 3)) = (3, - 2, 6)$ .
The plane passes through the foot $(1, - 3, 3)$ : $3 (x - 1) - 2 (y + 3) + 6 (z - 3) = 0$ .
$3 x - 2 y + 6 z - 27 = 0$ , i.e. $3 x - 2 y + 6 z = 27$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q42 marksVery Short AnswerVector Algebra

If $∣ a \times b ∣^{2} + ∣ a \cdot b ∣^{2} = 400$ and $∣ b ∣ = 5$ , then find the value of $∣ a ∣$ .

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Answer:

∣ a ∣ = 4

$∣ a \times b ∣^{2} + ∣ a \cdot b ∣^{2} = ∣ a ∣^{2} ∣ b ∣^{2} (sin^{2} θ + cos^{2} θ) = ∣ a ∣^{2} ∣ b ∣^{2}$ .
So $25∣ a ∣^{2} = 400$ , giving $∣ a ∣^{2} = 16$ .
$∣ a ∣ = 4$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q4 (OR) (OR)2 marksVery Short AnswerVector Algebra

Find all the possible vectors of magnitude $53$ which are equally inclined to the coordinate axes.

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Answer:

5 i^+ 5 j^+ 5 k^

and

- 5 i^- 5 j^- 5 k^

Equally inclined means direction cosines $l = m = n$ ; with $l^{2} + m^{2} + n^{2} = 1$ , $3 l^{2} = 1$ , so $l = m = n = \pm \frac{1}{3}$ .
Required vector $= 53 (\pm \frac{1}{3}) (i^+ j^+ k^)$ .
So the vectors are $5 i^+ 5 j^+ 5 k^$ and $- 5 i^- 5 j^- 5 k^$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q52 marksVery Short AnswerDifferential Equations

Find the general solution of the differential equation
$sec^{2} x \cdot tan y d x + sec^{2} y \cdot tan x d y = 0$ .

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Answer:

tan x tan y = C

Divide by $tan x tan y$ : $\frac{s e c ^{2} x}{t a n x} d x + \frac{s e c ^{2} y}{t a n y} d y = 0$ .
Integrate: $lo g ∣ tan x ∣ + lo g ∣ tan y ∣ = lo g ∣ C ∣$ .
So $tan x tan y = C$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q62 marksVery Short AnswerIntegrals

Evaluate :
$\int_{0}^{1} x^{2} e^{x} d x$

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Answer:

e - 2

By parts: $\int x^{2} e^{x} d x = x^{2} e^{x} - \int 2 x e^{x} d x = x^{2} e^{x} - 2 (x e^{x} - e^{x})$ .
So $\int x^{2} e^{x} d x = e^{x} (x^{2} - 2 x + 2)$ .
$[e^{x} (x^{2} - 2 x + 2)]_{0}^{1} = e (1) - 1 (2) = e - 2$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q73 marksShort AnswerApplication of Integrals

Find the area of the region ${(x, y) : x^{2} \leq y \leq x + 2}$ , using integration.

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Answer:

\frac{9}{2}

sq. units

The region lies between the parabola $y = x^{2}$ (below) and the line $y = x + 2$ (above).
Intersection: $x^{2} = x + 2 \Rightarrow (x - 2) (x + 1) = 0$ , so $x = - 1, 2$ .
Area $= \int_{- 1}^{2} (x + 2 - x^{2}) d x = [\frac{x ^{2}}{2} + 2 x - \frac{x ^{3}}{3}]_{- 1}^{2}$ .
$= (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = \frac{10}{3} + \frac{7}{6} = \frac{9}{2}$ sq. units.

Q83 marksShort AnswerIntegrals

Find :
$\int \frac{1}{e ^{x} + 1} d x$

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Answer:

- lo g (1 + e^{- x}) + C

(equivalently

x - lo g (e^{x} + 1) + C

)

$\frac{1}{e ^{x} + 1} = \frac{e ^{- x}}{1 + e ^{- x}}$ .
Put $t = 1 + e^{- x}$ , $d t = - e^{- x} d x$ .
$\int \frac{e ^{- x}}{1 + e ^{- x}} d x = - \int \frac{d t}{t} = - lo g ∣ t ∣ + C = - lo g (1 + e^{- x}) + C$ .
This equals $x - lo g (e^{x} + 1) + C$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q8 (OR) (OR)3 marksShort AnswerIntegrals

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

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Answer: 10

On $[1, 4]$ , $∣ x ∣ = x$ . Also $∣3 - x ∣ = 3 - x$ for $1 \leq x \leq 3$ and $x - 3$ for $3 \leq x \leq 4$ .
So the integrand is $3$ on $[1, 3]$ and $2 x - 3$ on $[3, 4]$ .
$\int_{1}^{3} 3 d x = 6$ ; $\int_{3}^{4} (2 x - 3) d x = [x^{2} - 3 x]_{3}^{4} = 4 - 0 = 4$ .
Total $= 6 + 4 = 10$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q93 marksShort AnswerVector Algebra

If $a$ , $b$ and $c$ are mutually perpendicular vectors of equal magnitude, then prove that the vector $(2 a + b + 2 c)$ is equally inclined to both $a$ and $c$ . Also, find the angle between $a$ and $(2 a + b + 2 c)$ .

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Answer: Proved; angle

= cos^{- 1} (\frac{2}{3})

Let $∣ a ∣ = ∣ b ∣ = ∣ c ∣ = k$ and $a \cdot b = b \cdot c = c \cdot a = 0$ . Let $v = 2 a + b + 2 c$ .
$∣ v ∣^{2} = 4 k^{2} + k^{2} + 4 k^{2} = 9 k^{2}$ , so $∣ v ∣ = 3 k$ .
$v \cdot a = 2 k^{2}$ , so $cos α = \frac{2 k ^{2}}{3 k \cdot k} = \frac{2}{3}$ .
$v \cdot c = 2 k^{2}$ , so $cos γ = \frac{2 k ^{2}}{3 k \cdot k} = \frac{2}{3}$ .
Since $cos α = cos γ$ , $v$ is equally inclined to $a$ and $c$ .
Angle between $a$ and $v$ is $cos^{- 1} (\frac{2}{3})$ .

Q103 marksShort AnswerThree Dimensional Geometry

If a line makes $6 0^{\circ}$ and $4 5^{\circ}$ angles with the positive directions of x-axis and z-axis respectively, then find the angle that it makes with the positive direction of y-axis. Hence, write the direction cosines of the line.

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Answer:

6 0^{\circ}

12 0^{\circ}

; direction cosines

\frac{1}{2}, \pm \frac{1}{2}, \frac{1}{2}

Let the angle with the y-axis be $β$ . Then $cos^{2} 6 0^{\circ} + cos^{2} β + cos^{2} 4 5^{\circ} = 1$ .
$\frac{1}{4} + cos^{2} β + \frac{1}{2} = 1$ , so $cos^{2} β = \frac{1}{4}$ and $cos β = \pm \frac{1}{2}$ .
So $β = 6 0^{\circ}$ or $12 0^{\circ}$ .
Direction cosines: $\frac{1}{2}, \pm \frac{1}{2}, \frac{1}{2}$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q10 (OR) (OR)3 marksShort AnswerThree Dimensional Geometry

Check whether the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ are skew or not.

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Answer: Not skew (the lines intersect at

(- 1, - 1, - 1)

)

Line 1: through $A (1, 2, 3)$ , direction $b_{1} = 2 i^+ 3 j^+ 4 k^$ . Line 2: through $B (4, 1, 0)$ , direction $b_{2} = 5 i^+ 2 j^+ k^$ .
The directions are not parallel.
$b_{1} \times b_{2} = (3 - 8) i^- (2 - 20) j^+ (4 - 15) k^= - 5 i^+ 18 j^- 11 k^$ .
$A B = 3 i^- j^- 3 k^$ ; $A B \cdot (b_{1} \times b_{2}) = - 15 - 18 + 33 = 0$ .
So the shortest distance is 0 and the lines intersect (at $(- 1, - 1, - 1)$ , taking parameter $- 1$ on each line). Hence they are not skew.

Also asked in: 2022 65/4/2, 2022 65/4/3

Q114 marksLong AnswerThree Dimensional GeometryNot in current syllabus

Find the equations of the planes passing through the line of intersection of the planes $r \cdot (i^+ 3 j^) = 6$ and $r \cdot (3 i^- j^- 4 k^) = 0$ , which are at a distance of 1 unit from the origin.

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Answer:

2 x + y - 2 z = 3

and

x - 2 y - 2 z + 3 = 0

A plane through the line of intersection: $(x + 3 y - 6) + λ (3 x - y - 4 z) = 0$ , i.e. $(1 + 3 λ) x + (3 - λ) y - 4 λ z - 6 = 0$ .
Distance from origin $= \frac{6}{( 1 + 3 λ ) ^{2} + ( 3 - λ ) ^{2} + 16 λ ^{2}} = 1$ .
$(1 + 3 λ)^{2} + (3 - λ)^{2} + 16 λ^{2} = 36 \Rightarrow 26 λ^{2} + 10 = 36 \Rightarrow λ = \pm 1$ .
$λ = 1$ : $4 x + 2 y - 4 z - 6 = 0$ , i.e. $2 x + y - 2 z = 3$ .
$λ = - 1$ : $- 2 x + 4 y + 4 z - 6 = 0$ , i.e. $x - 2 y - 2 z + 3 = 0$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q124 marksLong AnswerDifferential Equations

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

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Answer:

x y = \frac{π}{4} - tan^{- 1} x

Write as $\frac{d y}{d x} + \frac{1}{x} y = - \frac{1}{x ( 1 + x ^{2} )}$ . Integrating factor $= e^{\int \frac{1}{x} d x} = x$ .
So $\frac{d}{d x} (x y) = - \frac{1}{1 + x ^{2}}$ .
Integrate: $x y = - tan^{- 1} x + C$ .
$y (1) = 0$ : $0 = - \frac{π}{4} + C$ , so $C = \frac{π}{4}$ .
Particular solution: $x y = \frac{π}{4} - tan^{- 1} x$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q12 (OR) (OR)4 marksLong AnswerDifferential Equations

Find the general solution of the differential equation
$x (y^{3} + x^{3}) d y = (2 y^{4} + 5 x^{3} y) d x$ .

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Answer:

y (y^{3} + 4 x^{3}) = C x^{8}

$\frac{d y}{d x} = \frac{2 y ^{4} + 5 x ^{3} y}{x ( y ^{3} + x ^{3} )}$ , homogeneous of degree 4. Put $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
$v + x \frac{d v}{d x} = \frac{2 v ^{4} + 5 v}{v ^{3} + 1}$ , so $x \frac{d v}{d x} = \frac{v ^{4} + 4 v}{v ^{3} + 1}$ .
Separate: $\frac{v ^{3} + 1}{v ( v ^{3} + 4 )} d v = \frac{d x}{x}$ .
Partial fractions: $\frac{v ^{3} + 1}{v ( v ^{3} + 4 )} = \frac{1}{4 v} + \frac{3 v ^{2}}{4 ( v ^{3} + 4 )}$ .
Integrate: $\frac{1}{4} lo g ∣ v ∣ + \frac{1}{4} lo g ∣ v^{3} + 4∣ = lo g ∣ x ∣ + lo g ∣ c ∣$ , so $v (v^{3} + 4) = C x^{4}$ .
Put $v = \frac{y}{x}$ : $\frac{y ( y ^{3} + 4 x ^{3} )}{x ^{4}} = C x^{4}$ , i.e. $y (y^{3} + 4 x^{3}) = C x^{8}$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

Q134 marksLong AnswerIntegrals

Evaluate :
$\int_{0}^{π /2} (2 lo g cos x - lo g sin 2 x) d x$

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Answer:

- \frac{π}{2} lo g 2

(i.e.

\frac{π}{2} lo g \frac{1}{2}

)

$lo g sin 2 x = lo g 2 + lo g sin x + lo g cos x$ .
So the integrand is $lo g cos x - lo g sin x - lo g 2$ .
By $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ , $\int_{0}^{π /2} lo g cos x d x = \int_{0}^{π /2} lo g sin x d x$ , so those two terms cancel.
Integral $= - \int_{0}^{π /2} lo g 2 d x = - \frac{π}{2} lo g 2$ .

Q144 marksCase StudyProbability

In a game of Archery, each ring of the Archery target is valued. The centremost ring is worth 10 points and rest of the rings are allotted points 9 to 1 in sequential order moving outwards.
Archer A is likely to earn 10 points with a probability of 0.8 and Archer B is likely the earn 10 points with a probability of 0.9.
Based on the above information, answer the following questions :
If both of them hit the Archery target, then find the probability that
(a) exactly one of them earns 10 points. (2)
(b) both of them earn 10 points. (2)

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Answer: (a) 0.26 (b) 0.72

Let E: A earns 10 points, F: B earns 10 points, independent, with $P (E) = 0.8$ , $P (F) = 0.9$ .
(a) P(exactly one) $= P (E) P (F ˉ) + P (E ˉ) P (F) = 0.8 \times 0.1 + 0.2 \times 0.9 = 0.08 + 0.18 = 0.26$ .
(b) P(both) $= P (E) P (F) = 0.8 \times 0.9 = 0.72$ .

Also asked in: 2022 65/4/2, 2022 65/4/3

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