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# Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

## Importance of Class 12 Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2 with Solutions

Class 12 Maths Sample Paper for Term 2 CBSE Board Exam with Solution is created hereby Future Study Point for helping class 12 maths students in their preparation of the term 2 exams CBSE Board. Students are required to practice this sample paper with proper understanding because this is the solution of the question paper published by the CBSE Board for 2021- 22.By studying Class 12 Maths Sample Paper for Term 2 CBSE Board Exam with Solution, the students can plan the study of maths for better preparations. All the questions of the Class 12 Maths Sample Paper for Term 2 CBSE Board Exam with Solution are solved by a maths expert of CBSE by a step-by-step method.

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## The points on Class 12 Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

1. This question paper contains three sections – A, B and C. Each part is compulsory.

2. Section – A has 6 short answer type (SA1) questions of 2 marks each.

3. Section – B has 4 short answer type (SA2) questions of 3 marks each.

4. Section – C has 4 long answer-type questions (LA) of 4 marks each.

5. There is an internal choice in some of the questions.

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6.Q14 is a case-based problem having 2 subparts of 2 marks each

Marking Scheme Time duration 2 hour

## Section A

Q1.Find

$\fn_cm \int \frac{logx.dx}{\left ( 1+logx \right )^{2}}$

Ans. The given function is

$\fn_cm \int \frac{logx.dx}{\left ( 1+logx \right )^{2}}$

Adding and subtracting 1 in the numerator

$\fn_cm \int \frac{logx.dx+1-1}{\left ( 1+logx \right )^{2}}dx$

$\fn_cm =\int \frac{1+logx}{\left ( 1+logx \right )^{2}}dx-\int \frac{1}{\left ( 1+logx \right )^{2}}dx$

$\fn_cm =\int \frac{1}{1+logx}dx-\int \frac{1}{\left ( 1+logx \right )^{2}}dx$……(i)

$\fn_cm \int \frac{1}{1+logx}dx=\int \frac{1}{1+logx}.1dx=\frac{1}{1+logx}\int 1.dx-\left ( \int \frac{d}{dx}.\frac{1}{1+logx}\int 1.dx \right )dx$

$\fn_cm = \frac{1}{1+logx}\times x-\int \left ( \frac{-1}{\left ( 1+logx \right )^{2}}\times \frac{1}{x}\times x\right )dx$

$\fn_cm = \frac{x}{1+logx}+\int \frac{dx}{\left ( 1+logx \right )^{2}}$

Putting it in equation (i)

$\fn_cm \therefore I= \frac{x}{1+logx}+\int \frac{dx}{\left ( 1+logx \right )^{2}}-\int \frac{dx}{\left ( 1+logx \right )^{2}}$

$\fn_cm I= \frac{x}{1+logx}+C$

OR

Find

$\fn_cm \int \frac{sin2x}{\sqrt{9-cos^{4}x}}dx$

t =cos2x

dt/dx = 2cosx.(-sin x) =-2sinx.cosx =-sin 2x

dt = -sin 2x.dx

-dt = sin 2x.dx

Substituting sin 2x.dx by -dt and t =cos2x

$\fn_cm -\int \frac{dt}{\sqrt{3^{2}-t^{2}}}$

$\fn_cm \because \int \frac{dx}{\sqrt{a^{2}-x^{2}}}=sin^{-1}\frac{x}{a}$

$\fn_cm \therefore \int \frac{dt}{\sqrt{3^{2}-t^{2}}}=sin^{-1}\frac{t}{3}+C$

Substituting  t =cos2x

$\fn_cm \int \frac{dt}{\sqrt{3^{2}-t^{2}}}=sin^{-1}\frac{cos^{2}x}{3}+C$

Q2.Write the sum of the order and the degree of the following differential equation:

$\fn_cm \frac{d}{dx}\left ( \frac{dy}{dx} \right )=5$

Ans. The given differential equation is

$\fn_cm \frac{d}{dx}\left ( \frac{dy}{dx} \right )=5$

$\fn_cm \frac{d_{2}x}{dx^{2}}=5$

y” = 5

Order of derivative is 2

Degree of y”  is 1

Hence the sum of the order and the degree = 2 +1 =3

Q3.If 𝑎̂ and 𝑏̂ are unit vectors, then prove that

$\fn_cm \left | \hat{a} +\hat{b}\right |=2cos\frac{\theta }{2}$

Where θ is the angle between them.

Ans. It is given that 𝑎̂ and 𝑏̂ are unit vectors

$\fn_cm \therefore \left | \hat{a}\right |=1,\left | \hat{b}\right |=1$

$\fn_cm \left | \overrightarrow{a} \right |^{2}=\overrightarrow{a}.\overrightarrow{a}$

$\fn_cm \left |\hat{a}+\hat{b} \right |^{2}=(\hat{a}+\hat{b}).(\hat{a}+\hat{b})$

$\fn_cm \left |\hat{a}+\hat{b} \right |^{2}=\hat{a}.\hat{a}+\hat{a}.\hat{b}+\hat{b}.\hat{a}+\hat{b}.\hat{b}$

$\fn_cm \left |\hat{a}+\hat{b} \right |^{2}=\hat{a}.\hat{a}+2\hat{a}.\hat{b}+\hat{b}.\hat{b}\left ( \because \overrightarrow{a}.\overrightarrow{b}= \overrightarrow{b}.\overrightarrow{a}\right )$

$\fn_cm \left |\hat{a}+\hat{b} \right |^{2}=\left | a \right |^{2}+2\left | a \right |.\left | b \right |cos\theta +\left | b \right |^{2}\left ( \because \overrightarrow{a}.\overrightarrow{b} =\left | \overrightarrow{a} \right |.\overrightarrow{b}cos\theta \right )$

$\fn_cm \left |\hat{a}+\hat{b} \right |^{2}=1^{2}+2.1.1cos\theta +1^{2}$

$\fn_cm \left |\hat{a}+\hat{b} \right |^{2}=2+2cos\theta$

$\fn_cm \left |\hat{a}+\hat{b} \right |^{2}=2(1+cos\theta )$

$\fn_cm \left |\hat{a}+\hat{b} \right |^{2}=2.2cos^{2}\frac{\theta }{2}$

$\fn_cm \left |\hat{a}+\hat{b} \right |^{2}=4cos^{2}\frac{\theta }{2}$

$\fn_cm \left |\hat{a}+\hat{b} \right |=2cos\frac{\theta }{2}$

Q4.Find the direction cosines of the following line:

$\fn_cm \frac{3-x}{-1}=\frac{2y-1}{2}=\frac{z}{4}$

Ans. The given line is

$\fn_cm \frac{3-x}{-1}=\frac{2y-1}{2}=\frac{z}{4}$

$\fn_cm \frac{-(x-3)}{-1}=\frac{2(y-1/2)}{2}=\frac{z}{4}$

$\fn_cm \frac{(x-3)}{1}=\frac{(y-1/2)}{1}=\frac{z}{4}$

The direction ratios are 1,1 and 4

The direction cosines will be

$\fn_cm \frac{1}{\sqrt{1^{2}+1^{2}+4^{2}}},\frac{1}{\sqrt{1^{2}+1^{2}+4^{2}}},\frac{4}{\sqrt{1^{2}+1^{2}+4^{2}}}$

$\fn_cm \frac{1}{\sqrt{18}},\frac{1}{\sqrt{18}},\frac{4}{\sqrt{18}}$

$\fn_cm \frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}},\frac{4}{3\sqrt{2}}$

Q5.A bag contains 1 red and 3 white balls. Find the probability distribution of the number of red balls if 2 balls are drawn at random from the bag one-by-one without replacement

Ans. Total balls are given (1+3=4)

Let the number of red balls drawn be x

The cases if 2 balls are drawn at random from the bag one-by-one without replacement

First case: The number of red balls drawn is 0 i.e x =0

P(x =0) = Probability that the first ball drawn is white×  Probability that the second ball drawn is white

P(x =0) =( 3/4)×(2/3)=1/2

Second case: The number of red balls drawn is 1 i.e x =1

P(x =1) = Probability that the first ball drawn is red×  Probability that the second ball drawn is white+Probability that the first ball drawn is white×  Probability that the second ball drawn is red

P(x =1) =( 1/4)×(3/3)+( 3/4)×(1/3)

P(x=1) = 1/4+ 1/4 = 1/2

Therefore probability distribution of the number of red balls is

Q6.Two cards are drawn at random from a pack of 52 cards one by one without replacement. What is the probability of getting the first card red and the second card, Jack?

Ans. Probability of getting the first card red and the second card, Jack

= Probability that the first card drawn is red but not jack×Probability that second card drawn is jack if first is not red +Probability that the first card drawn is red jack×Probability that the second card drawn is red if first is red jack

=(24/52)×(4/51) + (2/52)×(3/51)

= (96/(52×51) + (6/(52×51)

=(96+6)/(52×51) =

=102/(52×51) =2/52 =1/26

## Section B

Q7. Find

$\fn_cm \int \frac{x+1}{\left ( x^{2} +1\right )x}.dx$

Ans. Let

$\fn_cm \frac{x+1}{\left ( x^{2} +1\right )x}=\frac{Ax+B}{x^{2}+1}+\frac{C}{x}$

$\fn_cm \frac{x+1}{\left ( x^{2} +1\right )x}=\frac{(Ax+B)x+C\left ( 1+x^{2} \right )}{x\left ( x^{2} +1\right )}$

x + 1 = Ax² +Bx +C + Cx²

x + 1 = x²(A +C)+Bx +C…..(i)

A,B and C are constants

Putting x =0,in equation (i)

C =1

Putting x =1 and C =1 in equation (i)

A +1-B+1 =2

A+B = 0

A = -B……..(ii)

Putting x =-1 and C =1,A=-B in equation (i)

-1 + 1 = (-1)²(A +1)-A(-1) +1

A +1 +A +1 =0

2A +2 =0

A = -2/2 =-1

B =1

Substituting the values A =-1,B =1 and C =1

$\fn_cm \frac{x+1}{\left ( x^{2} +1\right )x}= \frac{(-x+1)}{x^{2}+1}+ \frac{1}{x}$

Integrating it both sides

$\fn_cm \int \frac{x+1}{\left ( x^{2} +1\right )x}.dx= \int \frac{(-x+1)}{x^{2}+1}.dx+ \int \frac{1}{x}.dx$

$\fn_cm \int\frac{-x+1}{\left ( x^{2}+1 \right )x}.dx =\int \frac{-x}{x^{2}+1}.dx+\int \frac{1}{x^{2}+1}.dx+\int \frac{1}{x}.dx$

$\fn_cm \int \frac{-x}{x^{2}+1}.dx$

Let t = x² +1

dt/dx =2x

dt = 2x.dx

x.dx = dt/2

$\fn_cm \therefore \int \frac{-x}{x^{2}+1}.dx=-\frac{1}{2}\int \frac{dt}{t}=-\frac{1}{2}log\left | t \right |+C_{1}$

$\fn_cm \int \frac{-x}{x^{2}+1}.dx=-\frac{1}{2}log\left | x^{2} +1\right |+C_{1}$

Therefore

$\fn_cm \int\frac{-x+1}{\left ( x^{2}+1 \right )x}.dx =-\frac{1}{2}log\left | x^{2}+1 \right |+C_{1}+tan^{-1}x+C_{2}+log\left | x \right |+C_{3}$

$\fn_cm \int\frac{-x+1}{\left ( x^{2}+1 \right )x}.dx=-\frac{1}{2}log\left | x^{2}+1 \right |+tan^{-1}x+log\left | x \right |+C$

Q8.Find the general solution of the following differential equation:

$\fn_cm x.\frac{dy}{dx}=y-xsin\frac{y}{x}$

Ans. The given equation is

$\fn_cm x.\frac{dy}{dx}=y-xsin\frac{y}{x}$

$\fn_cm \frac{dy}{dx}=\frac{y}{x}-sin\frac{y}{x}$

Which is a homogeneous differential equation

Let v =y/x

y = vx

Differentiating both sides

dy/dx = x dv/dx +v

Subtituting the value v=y/x  and dy/dx

x dv/dx +v =v  -sin v

x dv/dx = -sinv

dv/sinv = -dx/x

cosec v.dv = -dx/x

Integrating both sides

∫ cosec v.dv = -∫dx/x

$\fn_cm log\left | cosecv-cotv \right |=-logx+logk$

$\fn_cm log\left | cosec\frac{y}{x}-cot\frac{y}{x} \right |=log\left | \frac{k}{x}\right |$

Taking antilog both sides

$\fn_cm \left | cosec\frac{y}{x}-cot\frac{y}{x} \right |=\left | \frac{k}{x}\right |$

Therefore the required solution is

$\fn_cm cosec\frac{y}{x}-cot\frac{y}{x} = \frac{k}{x}$

OR

Find the particular solution of the following differential equation, given that y = 0 when x = 𝜋/ 4 :

dy/dx +y cotx = 2/(1 +sin x)

Ans. The given differential equation is

dy/dx +y cotx = 2/(1 +sin x)

Comparing the standard form of the linear differential equation

dy/dx +Py = Q

Where P = cot x,Q = 2/(1 +sin x)

Evaluating its Integrating Factor(IF)

IF =  e∫Pdx

IF = e∫cotx dx

∫cotx  = log(sinx)

IF = elog(sinx)

IF = sinx( as elogx = x)

Solution is given by

y ×IF = ∫(Q×IF) dx +C

Where C is a constant

y sinx =∫(Q×sinx) dx +C

$\fn_cm ysinx=2x-2\int \frac{1-sinx}{1-sin^{2}x}+C$

$\fn_cm ysinx=2x-2\int \frac{1-sinx}{cos^{2}x}+C$

$\fn_cm ysinx=2x-2\int \frac{1}{cos^{2}x}+2\int \frac{sinx}{cos^{2}x}+C$

$\fn_cm ysinx=2x-2\int sec^{2}x+2\int \frac{sinx}{cos^{2}x}+C$

$\fn_cm ysinx=2x-2\int sec^{2}x+2\int tanx.secx+C$

ysinx =2x -2tanx +2secx +C

It is given that y = 0 when x = 𝜋/ 4

0×sinx =2x -2tan(𝜋/4) +2sec(π/4) +C

2π/4- 2 ×1+ 2×√2 +C =0

C = 2 -2√2 – π/2=2(1 -√2) – π/2

Hence

ysinx =2x -2tanx +2secx +2(1 -√2) – π/2

Q9. If

$\fn_cm \overrightarrow{a}\neq \overrightarrow{0},\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c},\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{c}$

Then show that

$\fn_cm \overrightarrow{b}=\overrightarrow{c}$

Ans. It is given that

$\fn_cm \overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}\overrightarrow{c}$

$\fn_cm \overrightarrow{a}\left ( \overrightarrow{b}-\overrightarrow{c} \right )=0$

$\fn_cm If\: \overrightarrow{a}\: =0$

$\fn_cm \overrightarrow{a}and \overrightarrow{b-c}$  are perpendicular to each other

It is also given that

$\fn_cm \overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}$

$\fn_cm \overrightarrow{a}\times( \overrightarrow{b}-\overrightarrow{c})=0$

$\fn_cm If\: \overrightarrow{a}\: =0$

$\fn_cm \overrightarrow{a}\: and\: ( \overrightarrow{b}-\overrightarrow{c})$  are parallel to each other

$\fn_cm \overrightarrow{a}\: and\: ( \overrightarrow{b}-\overrightarrow{c})$  can’t be both parallel and perpendicular

Therefore

$\fn_cm \overrightarrow{a}\: \neq 0$

$\fn_cm \therefore \overrightarrow{b}\:-\overrightarrow{c} = 0$

$\fn_cm \overrightarrow{b}\:=\overrightarrow{c}$

Q10.Find the shortest distance between the following lines:

$\fn_cm \overrightarrow{r}=\left ( \hat{i} +\hat{j}-\hat{k}\right )+s\left (2 \hat{i} +\hat{j}+\hat{k} \right )$

$\fn_cm \overrightarrow{r}=\left ( \hat{i} +\hat{j}+2\hat{k}\right )+t\left (4 \hat{i} +2\hat{j}+2\hat{k} \right )$

Ans. The given lines are

$\fn_cm \overrightarrow{r}=\left ( \hat{i} +\hat{j}-\hat{k}\right )+s\left (2 \hat{i} +\hat{j}+\hat{k} \right )$

$\fn_cm \overrightarrow{r}=\left ( \hat{i} +\hat{j}+2\hat{k}\right )+t\left (4 \hat{i} +2\hat{j}+2\hat{k} \right )$

Rewritting the given lines

$\fn_cm \overrightarrow{r}=\left ( \hat{i} +\hat{j}-\hat{k}\right )+s\left (2 \hat{i} +\hat{j}+\hat{k} \right )$

$\fn_cm \overrightarrow{r}=\left ( \hat{i} +\hat{j}+2\hat{k}\right )+2t\left (2 \hat{i} +\hat{j}+\hat{k} \right )$

Since parallel vectors are same i.e    $\fn_cm 2 \hat{i} +\hat{j}+\hat{k}$

The vectors equations of two parallel lines are given as

$\fn_cm \overrightarrow{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b},\overrightarrow{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b}$

The distance(d) between these parallel vector is given as

$\fn_cm d=\left | \frac{\overrightarrow{b}\times \left ( \overrightarrow{a_{2}}- \overrightarrow{a_{1}}\right )}{\left | \overrightarrow{b} \right |} \right |$

Comparing the given vectors

$\fn_cm \overrightarrow{a_{1}}=\hat{i}+\hat{j}-\hat{k},\overrightarrow{a_{2}}=\hat{i}+\hat{j}+2\hat{k}$

$\fn_cm \overrightarrow{b}=2\hat{i}+\hat{j}+\hat{k}$

Now,we have

$\fn_cm \overrightarrow{a_{2}}-\overrightarrow{a_{1}}=\hat{i}+\hat{j}+2\hat{k}-\hat{i}-\hat{j}+\hat{k}=3\hat{k}$

$\fn_cm \left | \overrightarrow{b} \right |=\sqrt{2^{2}+1^{2}+1^{2}}=\sqrt{6}$

$\fn_cm \overrightarrow{b}\times \left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right )=\begin{vmatrix} i & j &k \\ 2& 1 &1 \\ 0& 0& 3 \end{vmatrix}$

= i(3×1-1×0) -j(2×3-1×0) +k(2×0-1×0)

$\fn_cm =3\hat{i}-6\hat{j}$

$\fn_cm \left | \overrightarrow{b}\times \left ( \overrightarrow{a_{2}}-\overrightarrow{a_{1}} \right ) \right |=\sqrt{3^{2}+6^{2}}=\sqrt{45}=3\sqrt{5}$

Therefore the shortest distance between the given lines is

$\fn_cm d=\left | \frac{3\sqrt{5}}{\sqrt{6}} \right |=\frac{3\sqrt{5}}{\sqrt{6}}\: unit$

Q11.Evaluate

$\fn_cm \int_{-1}^{2}\left | x^{3}-3x^{2}+2x \right |dx$

Ans. Factorizing

x³ -3x² +2x

x[x² -3x +2]

x[x² – x-2x +2]

x[x(x -1) -2(x -1)]

x(x-1)(x -2)

∴ The critical point of the function are

x =0, x =1,x =2

$\fn_cm \therefore \left | x^{3} -3x^{2}+2x\right |=\left\{\begin{matrix} -x, &-\left ( x-1 \right ), & -\left ( x-2 \right ), & if\: -1\leq x< 0\\ x,& -\left ( x-1 \right ), & -\left ( x-2 \right ), & if\: 0\leq x< 1\\ x,& x-1 & -\left ( x-2 \right ), & if\: 1\leq x< 2 \end{matrix}\right.$

Now,we have

$\fn_cm \int_{1}^{2}\left | x^{3}-3x^{2}+2x \right |dx$

$\fn_cm =-\int_{-1}^{0}( x^{3}-3x^{2}+2x )dx+\int_{0}^{1}( x^{3}-3x^{2}+2x )dx-\int_{1}^{2}( x^{3}-3x^{2}+2x )dx$

$\fn_cm =-\left [ \frac{x^{4}}{4}-3\frac{x^{3}}{3}+2\frac{x^{2}}{2} \right ]_{-1}^{0}+\left [ \frac{x^{4}}{4}-3\frac{x^{3}}{3}+2\frac{x^{2}}{2} \right ]_{0}^{1}-\left [ \frac{x^{4}}{4}-3\frac{x^{3}}{3}+2\frac{x^{2}}{2} \right ]_{1}^{2}$

$\fn_cm =-\left [ \frac{x^{4}}{4}-x^{3}+x^{2} \right ]_{-1}^{0}+\left [ \frac{x^{4}}{4}-x^{3}+x^{2} \right ]_{0}^{1}-\left [ \frac{x^{4}}{4}-x^{3}+x^{2} \right ]_{1}^{2}$

$\fn_cm =-\left (- \frac{9}{4} \right )+\left ( \frac{1}{4} \right )-\left (- \frac{1}{4} \right )$

$\fn_cm =\frac{9}{4}+\frac{1}{4}+\frac{1}{4}=\frac{11}{4}$

Q12.Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y 2 = x and the x-axis.

Ans. Drawing the graph of line x + y = 2, the parabola y² = x and the x-axis.

Solving x + y = 2 and y² = x

Putt y = 2-x in the equation  y² = x

(2 -x)² = x

4 +x² – 4x -x =0

x² -5x +4 =0

x² -4x -x+4 =0

x(x -4) -1(x -4) =0

(x -4)(x -1) =0

x = 4,1

Since x=1 < 2,  so y =1,therefore the point of intersection of x +y =2 and y²=x is (1,1)

The area enclosed by the curve, given line, and x-axis is AOB

Area of AOC = Area of AOB + Area of BAC

y =±√x , considering y =√x ,since we have to find the area in the first quadrant

And from the equation of the line ,we have y = 2 -x

Area of AOC =Integration of √x within the limits 0 and 1 + Integration of 2-x within the limits 1 and 2

$\fn_cm areaAOC=\int_{0}^{1}\sqrt{x}dx+\int_{1}^{2}\left ( 2-x \right )dx$

$\fn_cm areaAOC=\left [ \frac{x^{3/2}}{3/2} \right ]_{0}^{1}+2\left [ x \right ]_{1}^{2}-\left [ \frac{x^{2}}{2} \right ]_{1}^{2}$

Area of AOC = (1/3/2) +2(2-1) -[2²/2 -1²/2]

Area of AOC = 2/3 +2 -(2 -1/2)

Area of AOC = 2/3 +2 -3/2 =(4+12-9/6)= 7/6

Hence area of AOC is 7/6 square unit

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