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# Solutions of class 12  maths question paper 2021 preboard exam CBSE

Solutions of class 12 CBSE maths question paper 2021 preboard exam are created for helping the students of 12 class in CBSE board exam. The solutions of the question paper will give a complete idea about the type of questions and the ways of solving them. All solutions of class 12 CBSE maths question paper 2021 preboard exams are solved by an expert of maths subject. We hope these solutions will definitely help to boost the preparation of maths question papers in the board exam of CBSE.

This question paper contains two parts-A and B. Each part is compulsory. Part A carries 24 marks and part B has descriptive-type questions of 56 marks.

Part A

Section I -16 questions(1 marks each)

Section II -2 case studies. Each case study comprises of 5 case-based MCQ’s.questions (1 marks each)

Part B

Section III-10 questions of 2 marks each

Section IV-07 questions of 3 marks each

Section V-03 questions of 5 marks each

## Part -A

### Section-I

Q1. Determine whether relation R in the set N of natural number defined as:

R = {(x,y) : y = x + 5 and x < 4} is symmetric.

Ans. The relation given us in the the set N of natural number

R = {(x,y) : y = x + 5 and x < 4}

x < 4 implies that x= 1,2,3

For x = 1,y = 1 +5 =6,for x = 2,y = 2+ 5 =7, for x =3,y = 3 +5=8

Therefore

R = {(1,6),(2,7),(3,8)}

Since (1,6) ∈ R but (6,1) ∉ R,so R is not symmetric

Q2.Let A = R – {3}, B = R – {1},consider the function f : A →B defined by

f(x) = (x -2)/(x -3). Check whether function is one-one or not.

Ans. We are given: A = R – {3}

B = R – {1}

F: A→B defied as

$F(x)\, =\, \frac{x-2}{x-3}$

$F(x_{1})\, =\, \frac{x_{1}-2}{x_{1}-3}$

(x1 ∉ 3)

$F(x_{2})\, =\, \frac{x_{2}-2}{x_{2}-3}$

If f(x1) = f(x2) then

$\frac{x_{1}-2}{x_{1}-3}=\frac{x_{2}-2}{x_{2}-3}$

x1= x2

Tis implies that x1= x2 therefore is on-on

Q3.Let Set A = {1,2,3} define the relation R in the set A as follows:R ={(1,1),(2,2),(3,3),(1,3)} which order to be added to R make it smallest equivalence relation.

Ans. A relation is said to be an equivalence relation when it is transitive,reflexive and symmetric

For the  Set A = {1,2,3}, R={(1,1),(2,2),(3,3),(1,3)}

For the Set A the relation is reflexive a ∈A, (a,a) ∈ R ,so minimum order  added should be (1,1),(2,2),(3,3)

For the Set A the relation is symmetric , a ∈A,b∈A, (a,b) ∈ R and (b,a) ∈R ,so minimum order added should be {(1,3),(3,1}

For the Set A the relation is transitive , (a,b) ∈A,(b,c) ∈A then (a,c) ∈ R ,the order must be  {(1,3),(3,3)}∈R,so (1,3) ∈ R

Therefore order to be added to R make it smallest equivalence relation is (3,1)

OR

Show that function f from z →z defined as f(x) = x + 2 is bijective.

Ans. For every value of x ∈z, f(x) = x +2: f(1) = 1+2 =3,f(2) =2+2=4 ,f(1) ≠f(2) for 1 ≠ 2, for every x ∈z,there is distintive image of z in z,so given function is one to one relation

For every value of x ∈ z,there is atleast one image of f in z: f(1) =2, 2∈z,f(o) =2 it indicates that for f(1) = f(2) ,0≠1,so function is onto relation

Since the function is one to one as well as onto, therefore it is bijective

Q4. Find the number of all possible matrices of order 3 ×3 with each entry 0 or 1.

Ans. The number of all possible entries in the matrix of order 3 ×3 =9

The number of choices to fill every entry is,2 = 0,9

So, the number of total possible   matrices are =29 =512

$\boldsymbol{Q5.If\: \:A= \begin{bmatrix}1 \: & -5\\ 2 & 4 \end{bmatrix}\: B=\begin{bmatrix}0 & 8\\ 8 & -6 \end{bmatrix}.}$

Ans. We are given A, B and C matrix, such that 2A + 3B – 2C ia a null matrix where

$A=\begin{bmatrix} 1 & -5\\ 2& 4 \end{bmatrix}$ $B=\begin{bmatrix} 0 & 8\\ 8& -6 \end{bmatrix}$

$null \, matrix\, is=\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}$

According to question

$2\begin{bmatrix} 1& -5\\ 2& 4 \end{bmatrix}+3\begin{bmatrix} 0 & 8\\ 8 & -6 \end{bmatrix}-2c=\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}$

$\begin{bmatrix} 2& -10\\ 4& 8 \end{bmatrix}+\begin{bmatrix} 0 & 24\\ 24 & -18 \end{bmatrix}-2c=\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}$

$\begin{bmatrix} 2& -14\\ 28& -10 \end{bmatrix}-2c=\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}$

$2c=\begin{bmatrix} 2& -14\\ 28& -10 \end{bmatrix}-\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}$

$2c=\begin{bmatrix} 2& -14\\ 28& -10 \end{bmatrix}$

$c=\frac{1}{2}\begin{bmatrix} 2& -14\\ 28& -10 \end{bmatrix}$

$c=\begin{bmatrix} 1& -7\\ 14& -5 \end{bmatrix}$

OR

Q7.Find ∫ ex sec x(1+tanx)dx

Ans.∫ ex secx(1+tanx)dx

= ∫ ex (secx + secx.tanx)dx

It is of the form

∫ ex[f(x) + f′(x)]dx = ex f(x) + C

where f(x) = secx

f′(x) = secx.tanx

so, our equations becomes

∫ ex (secx + secx.tanx)dx = ex secx + C

OR

$\boldsymbol{If\: \int_{0}^{a}\frac{dx}{1+4x^{2}}=\frac{\Pi }{8},find \: the \: value\: of\: a.}$

Ans. We are given

$\fn_cm \int_{0}^{a}\frac{dx}{1+4x^{2}}=\frac{\pi }{8}$

$\fn_cm \int_{0}^{a}\frac{dx}{4(\frac{1}{4}+x^{2})}=\frac{\pi }{8}$

$\fn_cm \frac{1}{4}\int_{0}^{a}\frac{dx}{(\frac{1}{4}+x^{2})}=\frac{\pi }{8}$

Using the integral formula

$\boldsymbol{\int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}tan^{-1}\frac{x}{a}+c}$

$\fn_cm =\frac{1}{4}\left [ \frac{1}{1/2} tan^{-1}\frac{x}{1/2} \right ]^{a}_{0}$

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

$\fn_cm \frac{1}{2}tan^{-1}2a =\frac{\pi }{8}$

$\fn_cm tan^{-1}2a =\frac{\pi }{4}$

$\fn_cm 2a =tan\frac{\pi }{4}$

2a = 1

a = 1/2

### Solutions of class 12  maths question paper 2021 preboard exam CBSE

Q8. Find the area of region bounded by curve y = cos x between x = 0 and x = π.

Ans. The graph of the curve  y = cos x  between x = 0 and x = π is as shown below

The graph y = cosx is a periodic function,so the area from x =0 to x =π/2 = area from π/2 to π is same.

Area from x= 0 to π = 2× area of graph from 0 to π/2

Area from x= 0 to π = 2×Integration of cos x within the limit 0 to π/2

$\fn_cm =2\int_{0}^{\pi /2}cosx\: dx$

$\fn_cm =2\left [ sinx \right ]^{\pi /2}_{0}$

= 2(sinπ/2 – sin0)

= 2

Therefore area of region bounded by curve is 2 sq.unit

Q9. Find the degree of differential equation.

$\boldsymbol{\left [ \frac{d^{2}y}{dx^{2}} \right ]^{2}+\left [ \frac{dy}{dx} \right ]^{2}=x\: sin\: x\left [ \frac{dy}{dx} \right ]}$

Ans. Highest order of derivative is (d²x/dx²) and its degree is 2

Hence the degree of differential equation is 2

OR

Find the integrating factor of differential equation

$\boldsymbol{x\frac{dy}{dx}-y=2x^{2}}$

Ans. The differential equation  given to us is

$\boldsymbol{x\frac{dy}{dx}-y=2x^{2}}$

The integration factor (I) of linear equation dy/dx +py = Q is given as following

$\fn_cm I =e^{\int P\: dx}$

Rearranging the  given equation in the standard form

dy/dx -y/x = 2x²

On comparing the standard equation and given equation ,we have p = -1/x

$\fn_cm I =e^{\int -\frac{dx}{x}}$

$\fn_cm I=e^{-logx}$

$\fn_cm =e^{log(1/x)}$

I = 1/x ( Since elogx = x)

Pintrest future study poin

Q10. Find the direction ratio of vector  $\boldsymbol{\vec{a}=\hat{i}+\hat{j}-2\hat{k}}$  and hence calculate its direction cosines.

Ans. The given vector is $\fn_cm \vec{a}$ = i +j -2k, its direction ratios are a =1,b = 1 and c = -2

The magnitude of $\fn_cm \vec{a}$ = i +j -2k is = $\fn_cm \left | a \right |$ = √[1²+1²+(-2)²] =√(1 +1 +4) =√6

The direction cosines of vector $\fn_cm \vec{a}$ are

$\fn_cm \left ( \frac{a}{l{a} l},\frac{b}{l\vec{bl}},\frac{c}{l\vec{c}l} \right )$

$\fn_cm =\left ( \frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}} \right )$

Q11. Find the area of the parallelogram whose adjacent sides are determined by vectors$\boldsymbol{\vec{a}=\hat{i}-\hat{j}+3\hat{k}\: and\:\vec{b}=2\hat{i}-7\hat{j}+\hat{k}. }$

Ans. Area of the parallelogram whose adjacent sides are (i-j +3k) and (2i -7j +k) is given by = $\fn_cm l\vec{a}\times \vec{b}l$

$\fn_cm \vec{a}\times \vec{b}=\begin{vmatrix}i & j &k \\ 1 & -1 &3 \\ 2 & -7 & 1 \end{vmatrix}$

= [-1 – (-21) ]i – [1-6]j + [-7-(-2)]k

= 20i + 5j -5k

$\fn_cm l\vec{a}\times \vec{b}l=\sqrt{20^{2}+5^{2}+\left (-5 \right )^{2}}$

$\fn_cm =\sqrt{400+25+25}=\sqrt{450}=15\sqrt{2}$

Area of the parallelogram is = 15√2 square unit.

Q12.If $\boldsymbol{\hat{a}}$ is a unit vector and    $\boldsymbol{\left ( \vec{x}-\vec{a} \right )\left ( \vec{x} +\vec{a}\right )=8.Find\: \left |\vec{x} \right |.}$

Ans. We are given $\boldsymbol{\hat{a}}$ is a unit vector such that

$\fn_cm \left ( \vec{x} -\vec{a}\right )\left ( \vec{x} +\vec{a}\right )=8$

$\fn_cm \begin{vmatrix}x \end{vmatrix}^{2}-\begin{vmatrix}a \end{vmatrix}^{2}=8$

$\fn_cm \begin{vmatrix}x \end{vmatrix}^{2}-1=8$

$\fn_cm \begin{vmatrix}x \end{vmatrix}^{2}=8+1 =9$

$\fn_cm \begin{vmatrix}x \end{vmatrix}^{2} =9$

$\fn_cm \begin{vmatrix}x \end{vmatrix} =3$

Q13.Find the equation of the line which is parallel to    $\boldsymbol{2\hat{i}+\hat{j}+3\hat{k}}$  and passes through the point (5,2,-4).

Ans. Equation of the line passing throgh (5,2,-4) is

$\fn_cm \vec{a}=5\hat{i}+2\hat{j}-4\hat{k}$

The given line is

$\fn_cm \vec{b}=2\hat{i}+\hat{j}+3\hat{k}$

Equation of a line passing through a point and parallel to another vector $\fn_cm \vec{b}$ is given as

$\fn_cm \vec{r}=\vec{a}+\lambda \vec{b}$, where $\fn_cm \vec{r}$ is the position vector of the required line and λ is a scaler

$\fn_cm \vec{r}=\left ( 5\hat{i} +2\hat{j}-4\hat{k}\right )+\lambda \left ( 2\vec{i}+j+3\vec{k} \right )$

### Solutions of class 12  maths question paper 2021 preboard exam CBSE

Q14. Find the distance between two planes:2x +3y +4z=4 and 4x +6y +8z =12

Ans. The distance between two planes Ax +By +Cz = d1 and Ax +By +Cz = d2

is given by

$\fn_cm \left | \frac{d_{1}-d_{2}}{\sqrt{A^{2}+B^{2}+C^{2}}} \right |$

The given planes are 2x +3y +4z=4 and 4x +6y +8z =12,rearranging the given second plains =2(2x+3y +4z)=12⇒2x +3y +4z=6

On comparing the equations of given plains with the equations standard plains, we have

A =2, B =3, C = 4, d1= 4, d2=6, Let D is the distance between two plains

$\fn_cm D=\left | \frac{4-6}{\sqrt{2^{2}+3^{2}+4^{2}}} \right |$

$\fn_cm D=\frac{2}{\sqrt{29}}$

Q15. Evaluate P(A∪B) if 2P(A) = P(B) = 5/13 and P(A/B) = 2/5.

Ans. Given, 2P(A) = P(B) = 5/13

So, P(A) = 1/2 × 5/13 = 5/26

And P(B) = 5/13

Also, P(A|B) = 2/5

We know, P(A|B)= P(A $\fn_cm \cap$ B)/P(B)

$\fn_cm \frac{2}{5}=\frac{P(A\cap B)}{\frac{5}{13}}$

$\fn_cm P(A\cap B)=\frac{2}{5}\times \frac{5}{13}$

$\fn_cm P(A\cap B)= \frac{2}{13}$

Also, P(A $\fn_cm \cap$ B) = P(A) + P(B) – P(A $\fn_cm \cap$ B)

$=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}$

$=\frac{5}{26}+\frac{3}{13}$

$=\frac{5}{26}+\frac{6}{26}$

$=\frac{11}{26}$

$\therefore P(A\cup B)=\frac{11}{26}$

Q16. Event A and B are such that P(A) =1/2, P(B) =7/12, and P(not A nor B)=1/4. State whether A and B are independent?

Ans. It is given that P(A) = 1/2, P(B) = 7/12, and P(not A nor B)=1/4

P(A′ $\cup$ B′) = 1/4

$P\begin{pmatrix} (A\cap B){}' \end{pmatrix}=\frac{1}{4}$

$[A'\cup{}B'=(A\cap B){}']$

$1-P(A\cap B)=\frac{1}{4}$

$P(A\cap B)=\frac{3}{4}...(1)$

However, P(A) – P(B) = 1/2 × 7/12 = 7/24…(2)

Here, 2/4 ≠ 7/24

∴ P(A $\cap$ B) ≠ P(A).P(B)

Therefore, A and B are not independent events.

### SECTION-II

Both case study-based questions are compulsory. Attempt any 4 sub-parts of each question 17 and 18. Each question carries 1 mark.

Q17.In a box manufacturing company during practical ability test of B.Tech for the post of an engineer,instructor of company instructed a participant to construct an open box with square base with a given quantity of cardboard of area C².

(i) If length of side of square base be x units and height is y units .Then area of material used is:

(a) x² + 4xy              (b) x + 4x²y

(c) x + 4xy²             (d) y² +4xy

Ans. Since the box is open,therefore the  area of cardboard required

2× y(x +x) + x×x

x² +4xy

(ii) The relation between x and y is :

(a) x = (c² -x²)/4y    (b) y =(c²-x²)/4x

(c) x = (x²-c²)/4y     (d) y =(x²-c²)/4x

Ans. Area of the cardboard = area of the box =c²

Area of the open box = x² +4xy

x² +4xy = c²⇒y = (c²-x²)/4x

(iii) Volume of box expressed as function of x is :

(a) 1/4(c² x -x³)       (b) 1/4(x³- c²x)

(c) 1/4(x²- cx)         (d) 1/4(cx – x²)

Ans. Volume of the box,V = l×b×h = x × x×y= x²y

As it is calculated above in (i) y = (c²-x²)/4x

V = x²(c²-x²)/4x = x(c²-x²)/4= 1/4(c² x -x³)

(iv) At what value of ‘x’ volume is maximum ?

(a) x = c/3               (b) x = c²/√3

(c) x = c/√3            (d) x =√3 c

Ans.

For getting maximum value of volume ,the slope of the V-x curve should be 0.

V = 1/4(c²x – x³)

dV/dx = 1/4 d/dx(c²x – x³) = 1/4(c² – 3x²)

1/4(c² – 3x²) = 0

c² – 3x² = 0

x = c/√3

(v) Maximum value of V is :

(a) c²/6√3            (b) c/3√6

(c) c²/6√3           (d) c/6√3

Ans. As we have determined in (iv) the value of x for which volume is maximum

So putting the value of x = c/√3 in volume equation

V = 1/4(c²x – x³)

V(max.) = 1/4[c² ×c/√3 – (c/√3)³]

= 1/4[c³/√3 -c³/3√3]

= c³/4√3[1-1/3] = 2c³/12√3 = c³/6√3

### Solutions of class 12  maths question paper 2021 preboard exam CBSE

Q18. A car manufacturing factory has two plants X and Y.Plant X manufactures 70 % of cars and plant Y manufactures 30 %. 80% of cars at plant X and 90 % of cars at plant Y are rated of standard quality. A car is chosen at random and is found to be of standard quality.

Let A = Event car of standard quality

B1  =Event car is manufactured in X

B2 = Event car is manufactured in Y

Based on the information given above answer the following (any four)

(i) Value of P(B1 ) will be:

(a) 70 %                 (b) 7/10

(c) 70/10               (d) 3/10

Ans.B1  =Event car is manufactured in X

P(B1) →Probability of the car that is manufactured in X plant

P(B1) = 70 %

(ii) Value of P(B2 ) will be:

(a) 10/3                (b) 30%

(c) 3/10               (d) 7/10

B=Event car is manufactured in Y

P(B2) →Probability of the car that is manufactured in Y plant

P(B2) = 30 %

(iii) Probability that standard quality car is manufactured in plant X will be:

(a) 9/10               (b) 8/10

(c) 7/10               (d) 3/10

Ans. The probability that a standard quality car is manufactured in plant X

It means if the selected car is  manufactured in plant X then it is of standard quality , its probability is given by = P(E/B1) =80% =  8/10

(iv) Probability that standard quality car is manufactured in plant Y will be:

(a) 9/10               (b) 8/10

(c) 7/10               (d) 3/10

Ans. The probability that a standard quality car is manufactured in the plant Y

It means if the selected car is  manufactured in plant Y then it is of standard quality, its probability is given by = P(E/B2) =90% =  9/10

(v) If a car is selected at random is of standard quality. What is the probability standard quality car has manufactured from plant X.

(a) 83/56              (b) 56/100

(c) 56/83               (d) 27/100

Ans. The probability that a standard quality car is manufactured in plant X

It means if the selected car is of standard quality then it is manufactured in plant X ,its probability is given by = P(B1/E)

Applying Bays theorem

$\fn_cm P\left ( B_{1}/E \right )=\frac{P\left ( B_{1} \right ).P\left ( E/B_{1} \right )}{P\left ( B_{1} \right ).P\left ( E/B_{1} \right )+P\left ( B_{2} \right ).P\left ( E/B_{2} \right )}$

Where P(B1) = 70 %  =0.70, P(B2) = 30 %=0.30

P(E/B1) = 80% =0.80, P(E/B2) = 90%=0.90

$\fn_cm P\left ( B_{1}/E \right )=\frac{0.70\times 0.80}{0.70\times0.80+0.30\times 0.90}$

$\fn_cm =\frac{0.56}{0.56+0.27}$

$\fn_cm =\frac{0.56}{0.83}=\frac{56}{83}$

## Part -B

### SECTION-III

Q19. Write     $Cot^{-1}\begin{pmatrix} \frac{1}{\sqrt{x^{2}-1}} \end{pmatrix}|x|>1$      in the simplest form.

Ans. We are given the expression

$Cot^{-1}\begin{pmatrix} \frac{1}{\sqrt{x^{2}-1}} \end{pmatrix}|x|>1$

Let x = secθ, θ = Sec-1x

$\fn_cm cot^{-1}\left [ \frac{1}{sec^{2}x-1} \right ]$

$\fn_cm cot^{-1}\left [ \frac{1}{\sqrt{sec^{2}\Theta -1}} \right ]$

$\fn_cm =cot^{-1}\left [ \frac{1}{tan\Theta } \right ]$

$\fn_cm =cot^{-1}\left ( cot\Theta \right )$

$\fn_cm \Theta =sec^{-1}x$

Q20.Find the value of x, y, and z if the matrix

$\boldsymbol{A=\begin{bmatrix}0 & 2y & z\\ x & y &-z \\ x & -y & z \end{bmatrix}}$

satisfying equation AA´= I

Ans. The matrix(A) given to us

$\boldsymbol{A=\begin{bmatrix}0 & 2y & z\\ x & y &-z \\ x & -y & z \end{bmatrix}}$

A’ → Transpose of the matrix A

$\fn_cm A'=\begin{bmatrix}0 & x & x\\2y &y &-y \\ z & -z & z \end{bmatrix}$

I→ Unit matrix of the order 3× 3

$\fn_cm I=\begin{bmatrix}1 &0 & 0\\ 0 & 1 & 0\\ 0 &0 & 1 \end{bmatrix}$

According to question

AA’ = I

$\fn_cm \begin{bmatrix}0 & 2y & z\\x & y & -z\\ x & -y & z \end{bmatrix}\begin{bmatrix}0 & x & x\\ 2y &y &-y \\z & -z & z \end{bmatrix}=\begin{bmatrix}1 & 0 &0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$

$\begin{bmatrix}0+x^{2}+x^{2} & 0+xy-xy & 0-xz+xz\\o+xy-xy & 4y^{2} +y^{2}+y^{2}& 2yz-yz-yz\\ 0-zx+zx & 2yz-zy-zy & z^{2}+z^{2}+z^{2} \end{bmatrix}$$\fn_cm =\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$

$\fn_cm \begin{bmatrix}2x^{2} & 0 & 0\\ 0 & 6y^{2} & 0\\0 & 0& 3z^{2} \end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$

Since matrices are equal so corresponding elements will be equal

2x² =1⇒ x =±1/√2, 6y² = 1⇒y =±1/√6,3z²=1⇒z=±1/√3

OR

Find the non-zero values of x satisfying, Matrix equation:

$\boldsymbol{x\begin{bmatrix}2x & 2\\ 3 & x \end{bmatrix}+2\begin{bmatrix}8 & 5x\\ 4 & 4x \end{bmatrix}=2\begin{bmatrix}x^{2}+8 & 24\\10 & 6x \end{bmatrix}}$

Ans.The matrix equation is given to us

$\boldsymbol{x\begin{bmatrix}2x & 2\\ 3 & x \end{bmatrix}+2\begin{bmatrix}8 & 5x\\ 4 & 4x \end{bmatrix}=2\begin{bmatrix}x^{2}+8 & 24\\10 & 6x \end{bmatrix}}$

$\fn_cm \Rightarrow \begin{bmatrix}2x^{2} & 2x\\ 3x & x^{2} \end{bmatrix}+\begin{bmatrix}16 & 10x\\8 & 8x \end{bmatrix}=\begin{bmatrix}2x^{2}+16 & 48\\ 20 & 12x \end{bmatrix}$

$\fn_cm \Rightarrow \begin{bmatrix}2x^{2}+16 & 2x+10x\\ 3x+8 & x^{2}+8x \end{bmatrix}=\begin{bmatrix}2x^{2}+16 & 48\\ 20 & 12x \end{bmatrix}$

On comparing the corresponding elements

3x +8 = 20 ⇒ x = 4, this value of x also satisfy other equations 2x+12x = 48,x² +8x = 12x

Q21.Find the relationship between a and b. So that function ‘f’ defined.

$\boldsymbol{f(x)=\left\{\begin{matrix}ax+1 & if & x\leq 3\\ bx+3 & if & x> 3 \end{matrix}\right.}$

is continuous at x = 3

Ans. As we know a function f(x) is continuous at x= a, when its left-hand limit , right-hand limit at x =a is equal to the value of the function at x =a.

Left-hand limit of the function( the function  defined  for x<3 is f(x) = ax +1)

$\fn_cm \lim_{x\rightarrow 3^{-}}ax+1=3a+1$

Right-hand limit of the function( the function  defined  for x>3 is f(x) = bx +3)

$\fn_cm \lim_{x\rightarrow 3^{+}}bx+3=3b+3$

Since it is given to us that function is continuous at x =3

∴ LHL = RHL

3a + 1 = 3b +3

3a – 3b =2

Therefore required relation is 3a – 3b =2

Q22.Find the equations of all lines having slope zero which are tangent to the curve      $y=\frac{1}{x^{2}-2x+3}$

Ans. We are given the curve

$\fn_cm y=\frac{1}{x^{2}-2x+3}$

$\fn_cm y=\left ( x^{2}-2x+3 \right )^{-1}$

$\fn_cm \frac{dy}{dx}=-\left ( x^{2}-2x+3 \right )^{-2} \frac{d}{dx}\left (x^{2}-2x+3 \right )$

$\fn_cm =-\frac{2x-2}{(x^{2}-2x+3)^{2}}$

We are given,the slope of the curve is 0, so dy/dx = 0

$\fn_cm -\frac{2x-2}{(x^{2}-2x+3)^{2}}=0$

x = 1

Value of y when x = 1

$\fn_cm y=\frac{1}{x^{2}-2x+3}$

$\fn_cm y=\frac{1}{1-2+3}=\frac{1}{2}$

The slope is passing through the coordinates (1,1/2)

The equations of the lines through the point (x1,y1) is given as

(y-y1) = m(x-x1)

(y-1/2) = 0(x -1)

y-1/2 = 0

Therefore the equation of the tangent is y-1/2 = 0

### Solutions of class 12  maths question paper 2021 preboard exam CBSE

Q23.Integrate: ∫ cos³ x elog sinx dx

Ans. The integral function given to us is

∫ cos³ x elog sinx dx

I = ∫ cos³ x sinx dx   [elog x = x]

Let t = sinx

dt/dx = cos x

$\fn_cm dx=\frac{dt}{cosx}$

Putting the value of dt and t in the function

$\fn_cm I=\int cos^{3}x\: .t.dt/cosx$

$\fn_cm I=\int cos^{2}x\: .t.dt$

$\fn_cm =\int (1-sin^{2}x)t.dt$

Putting the value sinx =t

$\fn_cm I=\int (1-t^{2})\: .t.dt$

$\fn_cm =\int t.dt-\int t^{3}dt$

$\fn_cm =\frac{t^{2}}{2}-\frac{t^{4}}{4}+c$

Substituting the value of t= sin x

$\fn_cm =\frac{sin^{2}x}{2}-\frac{sin^{4}x}{4}+c$

Therefore the integration of the given function is

$\fn_cm I=\frac{sin^{2}x}{2}-\frac{sin^{4}x}{4}+c$

OR

$\boldsymbol{Find \: the \: value \: of\: \int_{-1}^{1}x^{17}cos^{4}\: x\: dx}$

Ans. The given function is

f(x)=x17.cos4 x

Putting the value x = -x

f(-x) =(-x)17.cos4 (-x)

f(-x) = -x17.cos4 x = -f(x)

∴ f(x) is an odd function

Using the property

$\fn_cm \int_{-a}^{a}f\left ( x \right )dx=0,if\: f\left ( -x \right )=-f\left ( x \right )$

$\fn_cm \therefore \int_{-1}^{1}x^{17}cos^{4}x.dx=0$

Q24. Find the area of the region bounded by line x = 2 and parabola y² = 8x.

Ans. The equations of parabola y² = 8x and of the line  is x = 2

Drawing the graph of the parabola and the line

The area bounded by the parabola and the line is = 2× area of AOC

Area of AOC = Integration of the function y with respect to x within the limits x =0 and x =2.

$\fn_cm Area\: of\: AOC=\int_{0}^{2}y.dx$

y² = 8x

y  =±2√2√x

Since y is in the first quadrant

∴ y =  2√2√x

$\fn_cm Area\: of\: AOC=\int_{0}^{2}2\sqrt{2}\sqrt{x}.dx$

$\fn_cm =2\sqrt{2}\int_{0}^{2}\sqrt{x}.dx$

$\fn_cm =2\sqrt{2}\int_{0}^{2}x^{1/2}.dx$

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

$\fn_cm =2\sqrt{2}\left [ \frac{x^{3/2}}{3/2}\right ]^{2}_{0}$

$\fn_cm =\frac{4\sqrt{2}}{3}\left [ x^{3/2} \right ]^{2}_{0}$

$\fn_cm =\frac{4\sqrt{2}}{3}\left [ 2^{3/2} -0\right ]$

$\fn_cm =\frac{2^{2}\times 2^{1/2}\times 2^{3/2}}{3}=\frac{2^{4}}{3}=\frac{16}{3}$

The area bounded by the parabola and the line is (AOBC)

$\fn_cm =2\times \frac{16}{3}=\frac{32}{3}\: sq.unit$

Q25.For given differential equations find the particular solutions satisfying giving condition:

$\boldsymbol{\frac{dy}{dx}=y\: tanx\:: y=1\: when\: x=0}$

Ans. The differential equation given is

$\fn_cm \frac{dy}{dx}=y\: tanx$

$\fn_cm dy =y\: tanx \: dx$

$\fn_cm \int \frac{dy}{y}=\int tan\: x.dx$

$\fn_cm \because \int tanx.dx=log\left | secx \right |$

$\fn_cm log\left | y \right |=log\left | secx \right |+logc$

$\fn_cm log\left | y \right |=log\: \left | c\: secx \right |\left ( \because loga+logb=log\: ab \right )$

y = c sec x

Putting values of y =1, x = 0

1 =c sec 0⇒ c = 1`

Putting value of c = 1,we get

y = 1 ×sec x

y = sec x

Therefore required particular solution of the given differential equation is y= sec x.

Q26. If  $\boldsymbol{\vec{a}=2\hat{i}+2\hat{j}+3\hat{k},\vec{b}=-\hat{i}+2\hat{j}+\hat{k},\vec{c}=3\hat{i}+\hat{j}}$  such that

$\boldsymbol{\vec{a}+\lambda \vec{b}}$  is perpendicular to  $\boldsymbol{\vec{c}.}$ Find the value of λ.

Ans. We are given that

$\fn_cm \vec{a}=2\hat{i}+2\hat{j}+3\hat{k,\vec{b}}=-\hat{i}+2\hat{j}+\hat{k}\: and\: \vec{c}=3\hat{i}+\hat{j}$

$\fn_cm \vec{a}+\lambda \vec{b}$  is perpendicular to $\fn_cm \vec{c}$

$\fn_cm \vec{a}+\lambda \vec{b}=2\hat{i}+2\hat{j}+3\hat{k}+\lambda \left ( -\hat{i}+2\hat{j}+\hat{k} \right )$

$\fn_cm =\left ( 2-\lambda \right )\hat{i}+\left ( 2+2\lambda \right )\hat{j}+\left ( 3+\lambda \right )\hat{k}$

Since  $\fn_cm \vec{a}+\lambda \vec{b}$  is perpendicular to $\fn_cm \vec{c}$

$\fn_cm \therefore \left ( \vec{a} +\lambda \vec{b}\right ).\vec{c}=0$

$\fn_cm \left [ \left ( 2-\lambda \right )\hat{i}+\left ( 2+2\lambda \right )\hat{j} +\left ( 3+\lambda \right )\hat{k}\right ].\left ( 3\hat{i} +\hat{j}\right )=0$

(2-λ)3 + (2 + 2λ) = 0

6- 3λ + 2 + 2λ = 0

8 – λ = 0

λ = 8

Therefore the required value of λ is = 8

### Solutions of class 12  maths question paper 2021 preboard exam CBSE

Q27. Show that line joining the origin to the point (2,1,1) is perpendicular to the line determined by points (3,5,-1) and (4,3,-1).

Ans. Let the origin is O(0,0,0) and the point A(2,1,1) , the points of another line B(3,5,-1) and C(4,3,-1)

A line passing through (x1, y1, z1) and (x2, y2, z2) has the direction ratios

(x2 – x1), (y2 – y1), (z2 – z1)

Direction ratios of OA =(2-0),(1-0),(1-0)= 2,1,1

Direction ratios of BC =(4-3),(3-5),(-1+1)= 1,-2,0

Two lines having direction ratios a1, b1, c1 and a2, b2, c2 are perpendicular to each other if

a1 a2 + b1 b2 + c1 c2 = 0

a1 = 2, b1 = 1, c1 = 1, a2 = 1, b2 = -2, c2 = 0

2×1+1×-2+1×0 =2-2+0 = 0

Therefore given two lines OA and BC are perpendicular to each other

Q28. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Ans.We are given that an urn contains ,10-black balls,5-white balls

Total balls in the urn are = 10+5=15 and number of black balls=10

The probability of first ball drawn is black

= (number of black ball)/(Total ball) = 10/15= 2/3

Since two balls are drawn from the urn one after the other without replacement

∴ The probability of second ball drawn is black if first ball drawn is black = (number of remaining black ball)/(Total remaining ball)= 9/14

According to multiplication rule of probability = (2/3)× (9/14) =3/7

The probability that both drawn balls are black is 3/7

OR

If P(A) = 2/5, P(B) = 3/10, P(A∩B) = 1/5. Find the value of P(A’/B’)

Ans. We have to find out P(A’/B’) =?

As we know, P(A’/B’) = P(A’∩B’)/P(B’)

Since P(A’∩B’) = P(A∪B)

P(A’/B’) = P(A∪B)/P(B’)…….(i)

P(B’) = 1- P(B) = 1 – 3/10 = 7/10

Let’s find out P(A∪B),for this applying the following formula

P(A∪B) = P(A) + P(B) – P(∩B)

From the question,we have P(A) = 2/5, P(B) = 3/10, P(A∩B) = 1/5

P(A∪B) = 2/5 + 3/10 – 1/5 = 5/10 =1/2

Putting the value of P(B’) and P(A∪B) in (i)

P(A’/B’) = (1/2)/(7/10)= 10/14 = 5/7

## Part -B

### SECTION-IV

Q29. Let A = {1,2,3……9} and R be the relation in A × A defined by (a,b)R(c,d) if a +d = b+c for (a,b),(c,d) in A ×A. Prove that R is equivalence relation.Also obtain equivalence class [2,5].

Ans. We are given that  A = {1,2,3……9}and R be the relation in A × A defined by (a,b)R(c,d) if a +d = b+c for (a,b),(c,d) in A ×A

R in A × A

(a,b)R(c,d) if   (a,b),(c,d) ∈ A ×A

a +d = b+c

Considering (a,b) R (a,b)

a +b = b+a which is true for every value of (a,b) ∈ R(A ×A)

Hence R is reflexive

(a,b)R(c,d) if   (a,b),(c,d) ∈ A ×A

a +d = b+c

Checking it for (c,d)R(a,b)

c + b = d +a, which is true for every value of (c,d) ,(a,b)∈ R(A ×A)

(c,d) R(a,b) is true

Hence R is symmetric

Let (a,b)R(c,d)and(c,d)R(e,f)
a+d = b+c and c+f = d+e
⇒ a+d = b+c and d+e = c+f
⇒ (a+d) – (d+e) = (b+c) – (c+f)
⇒ a-e = b-f
⇒ a+f = b+e
(a,b)R(e,f)
So,R is transitive
Hence, R is an equivalence relation

Equivalence class [2,5] in A = {1,2,3,4,5,6,7,8,9}

a +d = b+c ⇒ a -b = c-d

2 – 5 = -3⇒ a-b =c-d =-3 for every (a,b),(c,d) ∈ A [2,5]

∴ Forming the ordered pair (c,d) satisfying the condition d= c+3

(2,5) R (c,d)

(2,5) R (1,4),(2,5),(3,6),(4,7),(5,8),(6,9)

It is symmetric also,so the following relations also true

(1,4)R(2,5),(3,6)R(2,5),(4,7)R(2,5),(5,8)R(2,5),(6,9)R(2,5)

Therefore there 11 equivalent class of [2,5]

Q30. Differentiate the function:

$\boldsymbol{\sqrt{\frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )}}}$

w.r.t. ‘x’

Ans.

$\fn_cm Let\: y=\sqrt{\frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )}}$

$\fn_cm y=\left [ \frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )} \right ]^{1/2}$

Taking logarithms of both sides

$\fn_cm log\: y=log\left [ \frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )} \right ]^{1/2}$

$\fn_cm log\: y=\frac{1}{2}log\left [ \frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )} \right ]$

$\fn_cm log\: y=\frac{1}{2}\left [ log\left ( x-1 \right ) \left ( x-2 \right )-log\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )\right ]$

$\fn_cm log\: y=log\left ( x-1 \right )+log\left ( x-2 \right )-log\left ( x-3 \right )-log\left ( x-4 \right )-log\left ( x-5 \right )$

Now, differentiating both sides

$\fn_cm \frac{d}{dx}\left ( log\: y \right )=\frac{d}{dx}\left [ log\left ( x-1 \right )+log\left ( x-2 \right )-log\left ( x-3 \right )-log\left ( x-4 \right )-log\left ( x-5 \right ) \right ]$

$\fn_cm \frac{1}{y}.\frac{dy}{dx}=\frac{1}{2}\left ( \frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5} \right )$

$\fn_cm \frac{dy}{dx}=\frac{y}{2}\left ( \frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5} \right )$

$\fn_cm \frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{\left ( x-1 \right )\left ( x-2 \right )}{\left ( x-3 \right )\left ( x-4 \right )\left ( x-5 \right )}}\left ( \frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4} -\frac{1}{x-5}\right )$

Q31. Find dy/dx, if x2/3 + y2/3 = a 2/3

Ans. $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$

Differentiating w.r.t x

$\frac{d(x^{\frac{2}{3}})}{dx}+\frac{d(y^{\frac{2}{3}})}{dx}=\frac{d(a^{\frac{2}{3}})}{dx}$

$\frac{2}{3}x^{\frac{2}{3}-1}+\frac{d(y^{\frac{2}{3}})}{dx}\times \frac{dy}{dy}=0$

$\frac{2}{3}x^{\frac{-1}{3}}+\frac{d(y^{\frac{2}{3}})}{dy}\times \frac{dy}{dx}=0$

$\frac{2}{3}x^{\frac{-1}{3}}+\frac{2}{3}y^{\frac{2}{3}}\times \frac{dy}{dx}=0$

$\frac{2}{3}\, \frac{1}{x^{\frac{1}{3}}}+\frac{2}{3}y^{\frac{-1}{3}}\times \frac{dy}{dx}=0$

$\frac{2}{3}\, \frac{1}{x^{\frac{1}{3}}}+\frac{2}{3}\, \frac{1}{y^{\frac{1}{3}}}\times \frac{dy}{dx}=0$

$\frac{2}{3}\, \frac{1}{y^{\frac{1}{3}}}\times \frac{dy}{dx}=-\frac{2}{3}\, \frac{1}{x^{\frac{1}{3}}}$

$\frac{1}{y^{\frac{1}{3}}}\times \frac{dy}{dx}=\frac{-1}{x^{\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$

$\frac{dy}{dx}=-\left ( \frac{y}{x} \right )^{\frac{1}{3}}$

OR

If y = sin-1x show that:

(1 – x²) d²y/dx² – x(dy/dx) = 0

Ans. We are given y = sin-1x

Differentiating both sides

$\fn_cm \frac{dy}{dx}=\frac{d}{dx}\left ( sin^{-1}x \right )$

$\fn_cm \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}$

$\fn_cm \sqrt{1-x^{2}}\: .\frac{dy}{dx}=1$

Squaring both sides

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

Differentiating both sides

$\fn_cm \frac{d}{dx}\left [ \left ( 1-x^{2} \right )\left ( \frac{dy}{dx} \right )^{2} \right ]$

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

$\fn_cm -2x.\left ( \frac{dy}{dx} \right )^{2}+\left ( 1-x^{2} \right ).2\frac{dy}{dx}.\frac{d_{2}y}{dx^{2}}=0$

$\fn_cm \frac{d_{2}y}{dx^{2}}\left ( 1-x^{2} \right )-x\frac{dy}{dx}=0$

Hence Proved

### Solutions of class 12  maths question paper 2021 preboard exam CBSE

Q32. Prove that y = (4sinθ/2+cosθ)-θ is an increasing function of θ in [0, π/2]

Ans. The given function is

$\fn_cm y=\frac{4sin\Theta }{2+cos\: \Theta }-\Theta$

Differentiating both sides w.r.t θ

$\fn_cm \frac{dy}{d\Theta }=\frac{d}{d\Theta }\left ( \frac{4sin\: \Theta }{2+cos\: \Theta }-\Theta \right )$

$\fn_cm =\frac{\left ( 2+cos\: \Theta \right ).4cos\: \Theta -4sin\Theta \left ( 0-sin\Theta \right )}{\left ( 2+cos\Theta \right )^{2}}-1$

$\fn_cm =\frac{8cos\: \Theta +4cos^{2}\Theta +4sin^{2}\Theta -\left ( 2+cos\Theta \right )^{2}}{\left ( 2+cos\Theta \right )^{2}}$

$\fn_cm =\frac{8cos\: \Theta +4-4sin^{2}\Theta +4sin^{2}\Theta -4-cos^{2}\Theta -4cos\: \Theta }{\left ( 2+cos\: \Theta \right )^{2}}$

$\fn_cm \frac{dy}{d\Theta }=\frac{4cos\: \Theta -cos^{2}\Theta }{\left ( 2+cos\: \Theta \right )^{2}}$

$\fn_cm \frac{dy}{d\Theta }=\frac{cos\: \Theta \left ( 4-cos\: \Theta \right )}{\left ( 2+cos\Theta \right )^{2}}$

Observing right hand side cosθ > 0 for θ ∈[0,π/2],(4-cosθ)>0 since -1≤cosθ≤1

(2+cosθ)² is always + ve.

$\fn_cm \therefore \frac{dy}{d\Theta }= +ve$

Hence y = (4sinθ/2+cosθ)-θ is an increasing function of θ in [0, π/2]

$\fn_cm Q33.Find\: \int_{0}^{\pi /2}sin^{3}x\: dx$

The integral function is given to us

$\fn_cm \int_{0}^{\pi /2}sin^{3}x.dx$

$\fn_cm =\int_{0}^{\pi /2}sin^{2}x.sinx.dx$

$\fn_cm =\int_{0}^{\pi /2}(1-cos^{2}x).sinx\: dx$

Let  u = cos x

$\fn_cm \frac{du}{dx}=sin\: x$

du = sin x.dx

Substituting the value u = cos x and du =sin x.dx

$\fn_cm =\int_{0}^{\pi /2}\left ( 1-u^{2} \right )du$

$\fn_cm =\int_{0}^{\pi /2}du-\int_{0}^{\pi /2}u^{2}du$

$\fn_cm =\left | u-\frac{u^{3}}{3} \right |^{\pi /2}_{0}$

Substituting back the value u = cos x

$\fn_cm =\left |cos\: x-\frac{cos^{3}x}{3} \right |^{\pi /2}_{0}$

$\fn_cm = cos\pi /2-1/3(cos^{3}\pi /2)-cos0+1/3(cos^{3}0)$

= 1/3 -1 = -2/3

OR

Evaluate:

$\boldsymbol{\int \frac{e^{logx^{3}}}{x^{4}+1}dx}$

Ans. Evaluating the given integral function

$\fn_cm \int \frac{e^{logx^{3}}}{x^{4}+1}dx$

$\fn_cm \int \frac{x^{3}}{x^{4}-1}dx\: \: \left [ \because e^{loga} =a\right ]$

t = x4 + 1

dt/dx = 4x³

dt = 4x³ dx ⇒ dx = dt/4x³

Substituting x4 + 1 = t and value of dx in the function

$\fn_cm =\int \frac{x^{3}dt}{4x^{3}t}$

$\fn_cm =\frac{1}{4}\int \frac{dt}{t}$

$\fn_cm =\frac{1}{4}log\left | t \right |+C$

$\fn_cm =\frac{1}{4}log\left | x^{4}+1 \right |+C$

Since x4 + 1 is always positive

$\boldsymbol{\int \frac{e^{logx^{3}}}{x^{4}+1}dx}$

$\fn_cm \therefore \int \frac{e^{logx^{3}}}{x^{4}+1}=\frac{1}{4}(x^{4}+1)+C$

Q34. Find the area of the region bounded by y² = 9x. x = 2, x = 4 and x-axis in the first quadrant.

Ans.  The graph of the curve  y² = 9x and the lines x = 2, x = 4  is shown below.

We have to find out the area bounded by x -axis, y² = 9x, y = 2 and y = 4 which is ABCD

The area of ABCD = Integration of  y w.r.t x within the limits x =2 and x =4

$\fn_cm arABCD=\int_{2}^{4}y.dx$

y² = 9x ⇒ y = 3√x

$\fn_cm arABCD=\int_{2}^{4}3\sqrt{x}.dx$

$\fn_cm =3\int_{2}^{4}x^{1/2}.dx$

$\fn_cm =3\left [\frac{x^{3/2}}{3/2} \right ]^{4}_{2}$

$\fn_cm =2\left [ 8- 2\sqrt{2}\right ]$

= 16 -4√2

Hence area bounded by x -axis, y² = 9x, y = 2 and y = 4 is (16 -4√2) sq.unit

Q35. Find the general solution of the following differential equation

Cos²x (dy/dx) + y = tan x, 0 ≤ x ≤ π/2

Ans. The given equation is

$\fn_cm cos^{2}x.\frac{dy}{dx}+y=tanx$

Writing the given equation in the form of the general equation  dy/dx +Py = Q

$\fn_cm \frac{dy}{dx}+\frac{y}{cos^{2}x}=\frac{tanx}{cos^{2}x}$

$\fn_cm \frac{dy}{dx}+y.sec^{2}x=tanx.sec^{2}x$

Where P = sec²x and Q = tanx.sec²x

We know the general solution of the equation dy/dx +Py = Q

y ×(I.F) = ∫(Q × I.F) dx  +C

It is known to us that the integrating factor(I.F) is calculated as follows

I.F = e∫Pdx=e∫sec²dx =etanx

Putting the value of Q = tanx.sec²x and of I.F in the general solution

y etanx =∫(tanx.sec²x) etanx dx  +C

Let t = tanx ⇒ dt/dx = sec²x ⇒ dt = dx.sec²x

Substituting tanx = t and dx.sec²x = dt

yet=∫t.dt .et+ C

yet= t.et– et+ C

y = t -1 + C/et

Putting the value of t = tanx

y = tanx  + C’ is the required solution where -1+ C/e= C’

## Part -B

### SECTION-V

Q36. Use product:

$\boldsymbol{\boldsymbol{\begin{bmatrix} 3 & -2& 3\\ 2& 1& -1\\ 4& -3 & 2\boldsymbol{}\end{bmatrix}}\begin{bmatrix} -1 & -5 & -1\\ -8& -6 & 9\\ -10& 1& 7 \end{bmatrix}}$

to solve the system of equations:

3x – 2y + 3z = 8, 2x +y – z = 1, 4x – 3y + 2z = 4

OR

Solve the following system of equations by matrix method:

3x + 2y – 2z = 3

x + 2y + 3z = 6

2x – y + z = 2

Ans. Given system of equations is:

3x + 2y – 2z = 3

x + 2y + 3z = 6

and 2x – y + z = 2

or AX = B

$i.e.\begin{bmatrix} 3 & 2 &-2 \\ 1& 2& 3\\ 2& -1& i \end{bmatrix}\begin{bmatrix} x\\ y\\ z\end{bmatrix}=\begin{bmatrix} 3\\ 6\\ 2\end{bmatrix}$

X = A-1 B

∴ For A-1, Cofactors are

A11 = 5, A12= 5, A13 = -5,

A21 = 0, A22= 7, A23 = 7,

A31 = 10, A32= -11, A33 = 4

$\therefore\, adj A =\begin{bmatrix} 5 & 5 &-5 \\ 0 & 7& 7\\ 10 & -11& 4 \end{bmatrix}^{r}=\begin{bmatrix} 5 & 0& 10\\ 5& 7& -11\\ -5 & 7&4 \end{bmatrix}$

|A| = 3(5) + 2(5) + (-2)(-5) = 35

$\therefore\, A^{-1} =\frac{adj\, A}{|a|}=\frac{1}{35} \begin{bmatrix} 5 & 0& 10\\ 5& 7& -11\\ -5 & 7&4 \end{bmatrix}$

$Now \, X = A^{-1} B$

$\small \begin{bmatrix} x\\ y\\ z\end{bmatrix}=\frac{1}{35}\begin{bmatrix} 5 & 0 & 10\\ 5& 7&-11 \\ -5& 7& 4 \end{bmatrix}\begin{bmatrix} 3\\ 6\\ 2\end{bmatrix}=\frac{1}{35}\begin{bmatrix} & 15+20 & \\ & 15+42-22 & \\ & -15+42+8& \end{bmatrix}=\frac{1}{35}\begin{bmatrix} 35\\ 35\\ 35\end{bmatrix}=\begin{bmatrix} 1\\ 1\\ 1\end{bmatrix}$

∴ x = 1, y =1 and z = 1

Q37. Find the shortest distance between lines whose vector equations are:

$\boldsymbol{\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (\hat{i}-3\hat{j}+2\hat{k})}$

$\boldsymbol{\vec{r}=\hat{4i}+\hat{5j}+6\hat{k}+\mu(2\hat{i}+3\hat{j}+\hat{k})}$

Ans. The shortest distance between two lines $\fn_cm \vec{r}=\vec{a_{1}}+\lambda \vec{b_{1}}\: and\: \vec{r}=\vec{a_{2}}+\mu \vec{b_{2}}$  is

$\fn_cm =\left | \frac{\vec{(b_{1}}\times\vec{b_{2}} )\left ( \vec{a_{2}}- \vec{a_{1}}\right )}{\left | \vec{b_{1}} \times \vec{b_{2}}\right |} \right |$

We are given here the vector equations

$\boldsymbol{\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (\hat{i}-3\hat{j}+2\hat{k})}$

$\boldsymbol{\vec{r}=\hat{4i}+\hat{5j}+6\hat{k}+\mu(2\hat{i}+3\hat{j}+\hat{k})}$

On comparing these equations with standard vector equations $\fn_cm \vec{r}=\vec{a_{1}}+\lambda \vec{b_{1}}\: and\: \vec{r}=\vec{a_{2}}+\mu \vec{b_{2}}$

We have

$\fn_cm \vec{a_{1}}=\hat{i}+2\hat{j}+3\hat{k}\:, \vec{b_{1}}=\hat{i}-3\hat{j}+2\hat{k}$

$\fn_cm \vec{a_{2}}=4\hat{i}+5\hat{j}+6\hat{k}\:, \vec{b_{2}}=2\hat{i}+3\hat{j}+\hat{k}$

$\fn_cm \vec{a_{2}}-\vec{a_{1}}=(4\hat{i}+5\hat{j}+6\hat{k})- (\hat{i}+2\hat{j}+3\hat{k})$

$\fn_cm =\left ( 4-1 \right )\hat{i}+\left ( 5-2 \right )\hat{j}+\left ( 6-3 \right )\hat{k}=3\hat{i}+3\hat{j}+3\hat{k}$

$\fn_cm \vec{b_{1}}\times \vec{b_{2}}=\begin{vmatrix}\hat{i} &\hat{j} &\hat{k} \\ 1& -3 & 2\\ 2& 3 & 1 \end{vmatrix}$

$\fn_cm = i\left ( -3-6 \right )-\hat{j}\left ( 1-4 \right )+\hat{k}\left ( 3+6 \right )=-9\hat{i}+3\hat{j}+9\hat{k}$

$\fn_cm \left | \vec{b_{1}} \times \vec{b_{2}}\right |=\sqrt{\left ( -9 \right )^{2}+3^{2}+9^{2}}=\sqrt{171}=3\sqrt{19}$

$\fn_cm \left ( \vec{b_{1}} \times \vec{b_{2}}\right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )$

$\fn_cm =\left ( -9\hat{i} +3\hat{j}+9\hat{k}\right ).\left ( 3\hat{i} +3\hat{j}+3\hat{k}\right )=-27+9+27=9$

Therefore the shortest distance between both of the lines is

$\fn_cm \left | \frac{\left ( \vec{b_{1}} \times \vec{b_{2}}\right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |=\left | \frac{9}{3\sqrt{19}} \right |=\frac{3}{\sqrt{19}}$

OR

Find the equation of the plane which contains the line of intersection of the planes $\boldsymbol{\vec{r}.(\hat{i}+\hat{2j}+3\hat{k})-4=0}$ and $\boldsymbol{\vec{r}.(\hat{2i}+\hat{j}-\hat{k})+5=0}$ which is perpendicular to the plane $\boldsymbol{\vec{r}.(\hat{5i}+\hat{3j}-\hat{6k})+8=0}.$

Q38.Minimize z = 13x – 15y

Such that x + y ≤ 7

2x – 3y + 6 ≥ 0, x ≥0, y ≥0

Ans. We have to minimize  z=13x – 15y such that x + y ≤ 7,2x – 3y + 6 ≥ 0, x ≥0, y ≥0

Solutions of the equation x + y = 7 , for x = 0,y = 7 and for y=0,x = 7,ploting the line between (0,7) and (7,0)

Solutions of the equation 2x – 3y + 6 = 0, for x = 0, y= 2, for y=0, x= -3,plotting the lines between (0,2) and (-3,0)

The solutions of the equation x = 0,is y axis and of y= 0 is x axis

Locating the co-ordinates (0,7),(7,0),(0,2),(-3,0) and (0,0)

The corner points of the graph of inequalities are (0,0),(3,4),(0,2),(7,0)

Evaluating z=13x – 15y on every points (0,0),(3,4),(0,2),(7,0)

On (0,0), z = 0, on (3,4) , z = 13×3 -15×4 =39-60 =-21, on (0,2), z =-30,on (7,0), z =91

The minimum value of z is -30

OR

The corner points of feasible region of LPP are (0,8), (4,10), (6,8), (6,5), (0,0) and (5,0). Let the objective function be z = 3x – 4y. Answer the following:

(a) Find the point at which the minimum value of z occurs.

(b) Find the point at which the maximum value of z occurs.

(c) Find the sum of the minimum value of z and the maximum value of z.

(d) Let z = px + qy, where p, q > 0

Ans.Find the relation between p and q, so that minimum of z occurs at (0,8) and (4,10).

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

### NCERT Solutions for class 10 maths

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### NCERT Solutions for class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT solutions for class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

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

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution