Students must start practicing the questions from RBSE 12th Maths Model Papers Board Model Paper 2022 with Answers in English Medium provided here.

## RBSE Class 12 Maths Board Model Paper 2022 with Answers in English

Time : 2 Hours 45 Min.

Maximum Marks : 80

General Instructions to the Examinees:

- Candidate must write first his/her Roll. No. on the question paper compulsorily.
- All the questions are compulsory.
- Write the answer to each question in the given answer book only.
- For questions having more than one part the answers to those parts are to be written together in continuity.
- Write down the serial number of the question before attempting it.

Section – A

Question 1.

Multiple Choice Questions

(i) Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer : (1)

(a) (2, 4) ∈ R

(b) (3, 8) ∈ R

(c) (6, 8) ∈ R

(d) (8, 7) ∈ R

Answer:

(c) (6, 8) ∈ R

(ii) tan^{-1}√3 – sec^{-1} (-2) is equal to : (1)

(a) π

(b) \(-\frac{\pi}{3}\)

(c) \(\frac{\pi}{3}\)

(d) \(2\frac{\pi}{3}\)

Answer:

(b) \(-\frac{\pi}{3}\)

(iii) A = [a_{ij}]_{m} is a square matrix, if:

(a) m < n (b) m > n

(c) m = n

(d) none of these

Answer:

(c) m = n

(iv) Let A be a square matrix of order 3 × 3, then |kA| is equal to: (1)

(a) k|A|

(b) k^{2}|A|

(c) k^{3}|A|

(d) 3k|A|

Answer:

(c) k^{3}|A|

(v) Find \(\frac{dy}{dx}\) if x – y = π. (1)

(a) 0

(b) 1

(c) -1

(d) 2

Answer:

(b) 1

(vi) ∫(√x + \(\frac{1}{\sqrt{x}}\)) is equal to: (1)

(a) \(\frac{1}{3}\)x^{1/3} + 2x^{1/2} + C

(b) \(\frac{2}{3}\)x^{2/3} + \(\frac{1}{2}\)x^{2} + C

(c) \(\frac{2}{3}\)x^{3/2} + 2x^{1/2} + C

(d) \(\frac{3}{2}\)x^{3/2} + \(\frac{1}{2}\)x^{1/2} + C

Answer:

(c) \(\frac{2}{3}\)x^{3/2} + 2x^{1/2} + C

(vii) The order of the differential equation

2x^{2}\(\frac{d^{2} y}{d x^{2}}\) – 3\(\frac{d y}{d x}\) + y = 0 is: (1)

(a) 2

(b) 1

(c) 0

(d) not defined

Answer:

(a) 2

(viii) Let the vectors \(\vec{a}\) and \(\vec{b}\) be such that |\(\vec{a}\)| = 3 and |\(\vec{b}\)| = \(\frac{\sqrt{2}}{3}\) then \(\vec{a} \times \vec{b}\) is a unit vector, if the angle between \(\vec{a}\) and \(\vec{b}\) is: (1)

(a) π/6

(b) π/4

(c) π/3

(d) π/2

Answer:

(b) π/4

(ix) Two events A and B will be independent, if: (1)

(a) A and B are mutually exclusive

(b) P(A’B’) [1 – P(A)] [1 – P(B)]

(c) P(A) = (P(B)

(d) P(A) + P(B) = 1

Answer:

(b) P(A’B’) [1 – P(A)] [1 – P(B)]

(x) Matrices A and B will be inverse of: (1)

(a) AB = BA

(b) AB = BA = 0

(c) AB = 0, BA = 1

(d) AD = BA = 1

Answer:

(d) AD = BA = 1

(xi) Find \(\frac{d^{2} y}{d x^{2}}\), if y = x^{2} + 3x + 2: (1)

(a) 2x + 3

(b) x + 3

(c) 2

(d) 3

Answer:

(c) 2

(xii) The value of

î.(ĵ × k̂) + ĵ.(î × k̂) + k̂.(î × ĵ) is: (1)

(a) 0

(b) -1

(c) 1

(d) 3

Answer:

(c) 1

Question 2.

Fill in the blanks:

(i) gof(x) = _____________ , If f(x) = | x | and g(x) = | 5x – 2 |, then g(x) = _____________ (1)

Answer:

|5|x |-2|

(ii) The value of cos^{-1} (cos\(\frac{7 \pi}{6}\)) is _____________. (1)

Answer:

\(\frac{5 \pi}{6}\)

(iii) if A = \(\left[\begin{array}{ll}

2 & 4 \\

3 & 2

\end{array}\right]\) and B = \(\left[\begin{array}{cc}

1 & 3 \\

-2 & 5

\end{array}\right]\) then A + B = _____________ (1)

Answer:

\(\left[\begin{array}{ll}

3 & 7 \\

1 & 7

\end{array}\right]\)

(iv) The value of \(\frac{d}{dx}\)[tan(2x + 3)] is _____________ (1)

Answer:

2 sec^{2}(2x + 3)

(v) The value of ∫^{3} _{2}\(\frac{1}{x}\)dx is _____________ (1)

Answer:

log\(\frac{3}{2}\)

(vi) The value of angle between the vectors \(\vec{a}\) = î – 2ĵ + 3k̂ and \(\vec{b}\) = 3î – 2ĵ + k̂ is _____________ (1)

Answer:

cos^{-1}\(\left(\frac{5}{7}\right)\)

Question 3.

Very Short Answer Type Questions

(i) If A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4,), (2, 5), (3, 6)} be a function from A to B. Show that f is one-one. (1)

Answer:

According to the questions,

A = {1, 2, 3}, B = {4, 5, 6, 7}

f: A → B is such that

f = {(1,4), (2, 5), (3, 6)}

Then f(1) = 4, f(2) = 5, f(3) = 6

i.e., Each elements of A has distinct image.

Thus, f is one-one.

(ii) If sin (sin^{-1} (\(\frac{1}{5}\))+ cos^{-1} x) = 1, then find the value of x. (1)

Answer:

Given, sin (sIn^{-1}\(\frac{1}{5}\) + cos^{-1} x) 1

⇒ sin^{-1} (\(\frac{1}{5}\)) + cos^{-1} x = sin^{-1}1

⇒ sin^{-1} (\(\frac{1}{5}\)) + cos^{-1}(x) = \(\frac{\pi}{2}\)

⇒ sin^{-1} (\(\frac{1}{5}\)) = – – cos^{-1} x

⇒ sin^{-1} (\(\frac{1}{5}\)) = sin^{-1}x

(∵ sin^{-1}x + cos^{-1}x = π/2)

Thus, x = \(\frac{1}{5}\)

(iii) Find the values of x, y and z from the following equation: (1)

\(\left[\begin{array}{ll}

4 & 3 \\

x & 5

\end{array}\right]=\left[\begin{array}{ll}

y & z \\

1 & 5

\end{array}\right]\)

Answer:

Given, \(\left[\begin{array}{ll}

4 & 3 \\

x & 5

\end{array}\right]=\left[\begin{array}{ll}

y & z \\

1 & 5

\end{array}\right]\)

On comparing,

4 = y, 3 = z, x = 1

Thus, x = 1,y = 4 ,z = 3

(iv) Using cofactors of elements of second row, evaluate ∆ = \(\left|\begin{array}{lll}

5 & 3 & 8 \\

2 & 0 & 1 \\

1 & 2 & 3

\end{array}\right|\) (1)

Answer:

Given ∆ = \(\left|\begin{array}{lll}

5 & 3 & 8 \\

2 & 0 & 1 \\

1 & 2 & 3

\end{array}\right|\)

Expanding along R2

∆ = -2(3 × 3 – 2 × 8) + 0(5 × 3 – 8 × 1)

= -1 (5 × 2 – 3 × 1)

= -2(9 – 16) + 0 – 1(10 – 3)

= -2(-7) – 1(7) = 14 – 7 = 7

(v) Differentiate e^{sin-1}x with respect x. (1)

Answer:

(vi) Find : ∫\(\frac{\sec ^{2} x d x}{\sqrt{\tan ^{2} x+4}}\) (1)

Answer:

(vii) Find the general solution of the differential equation \(\frac{d y}{d x}=\sqrt{4-y^{2}}\); (-2 < y < 2) (1)

Answer:

The given differential equation is

or y = 2 sin (x + C) …(2)

Equation (2) is the general solution of the given differential equation.

(viii) Show that the vectors 2 î – 3 ĵ + 4k̂ and – 4 î + 6 ĵ – 8k̂ are collinear. (1)

Answer:

Let \(\vec{a}\) = 2 î – 3 ĵ + 4k̂

and \(\vec{b}\) = – 4 î + 6 ĵ – 8k̂ = – 2(2 î – 3ĵ + 4k̂)

∴ \(\vec{b}\) = – 2 \(\vec{a}\) or \(\vec{a}\) = –\(\frac{1}{2}\) \(\vec{b}\)

Vector \(\vec{b}\) can be represented in the form of \(\vec{a}\),

So, vectors \(\vec{a}\) and \(\vec{b}\) have the same direction, therefore they are collinear.

\(\vec{b}\) = λ \(\vec{a}\), Here λ = -2, \(\vec{a}\) = \(\frac{1}{2}\) \(\vec{b}\), then λ = – \(\frac{1}{2}\)

(ix) If P(A) = 3/5 and P(B) = 1/5, find P(A ∩ B) if A and B are independent events. (1)

Answer:

It is given that,

P(A) =3/5 and P(B) = 1/5

A and B are independent events.

Therefore, .

P(A ∩ B) = P(A) – P(B)

= 3/5 . 1/5 = 3/25

(x) If area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4). Find the value of k. (1)

Answer:

If vertices of a traingle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the area of the triangle is given by,

∆ = \(\frac{1}{2}\)\(\left|\begin{array}{lll}

x_{1} & y_{1} & 1 \\

x_{2} & y_{2} & 1 \\

x_{3} & y_{3} & 1

\end{array}\right|\)

Now, by substituting given values in above formula

Expanding along R., we get

⇒ k(- 6 – 4) – 4(2 – 5) + 1 (8 + 30) = ±70

⇒ [- 10k + 12 + 38] = ± 70

⇒ – 10k + 50 = ± 70

Taking positive sign, we get

⇒ – 10 k + 50 = 70

⇒ – 10 k = 20 ⇒ k = – 2

Taking negative sign, we get

⇒ – 10k + 50= – 70

⇒ – 10k = -120 ⇒ k = 12

Thus, k = – 2, 12

(xi) Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constants. (1)

Answer:

The given equation is :

y = a sin (x + b) …(1)

Differentiating equation (1) w.r.t., we get

\(\frac{d y}{d x}\) = a cos(x + b) …(2)

Again, differenting equation (2) w.r.t. ‘X’,

we get

\(\frac{d^{2} y}{d x^{2}}\) = a sin(x + b) …(3)

Eliminating a and h from equations (1), (2) and (3), we get

\(\frac{d^{2} y}{d x^{2}}\) + y = 0 ……….(4)

which is free from the arbitrary constants a and b and hence this the required differential equation.

(xii) Find the area of a parallelgram whose adjacent sides Aare given by the vector \(\vec{a}\) = 3î + ĵ + 4k̂ and \(\vec{b}\) = î – ĵ + k̂. (1)

Answer:

The area of a parallelogram with a and b as its adjacent sides is given by \(|\vec{a} \times \vec{b}|\).

Now, \(\vec{a} \times \vec{b}\) = \(\left|\begin{array}{ccc}

\hat{i} & \hat{j} & \hat{k} \\

3 & 1 & 4 \\

1 & -1 & 1

\end{array}\right|\)

= 5î + ĵ + 4k̂ (on solving)

∴ \(|\vec{a} \times \vec{b}|\) = \(\sqrt{25+1+16}=\sqrt{42}\)

This, the required area is \(\sqrt{42}\) sq.units

Section – B

Short Answer Type Questions

Question 4.

Show that the relation RinR defined as R = {(a, b), is reflexive and transitive but not symmetric. (2)

Answer:

In the set of real numbers relation R is defined as

R = {(a, b) : a ≤ b}

(i) Relation R is reflexive, since for any real number.

a ≤ a ⇒ (a, a) ∈ R

Thus, aRa is true.

(ii) Relation R is transitive, since for real numbers a, b and c,

a ≤ b, b ≤ c ⇒ a ≤ c i.e., (aRb) and (bRc)

⇒ aRc or (a, b) ∈ R, (b, c) ∈ R

⇒ (a, c) ∈ R

(iii) R is not symmetiric, since for any two real numbers a and b,

a ≤ b \(\not \Rightarrow\) b ≤ a

i.e., aRb \(\not \Rightarrow\) bRa

Thus, (a, b) e R \(\not \Rightarrow\) (b, a) e R

As (3, 4) ∈ R ⇒ 3 < 4

but 4 < 3 or (4, 3) ∉ R

Thus, R is reflexive and transitive but not symmetric.

Question 5.

If A = \(\left[\begin{array}{ll}

3 & -2 \\

4 & -2

\end{array}\right]\) and I = \(\left[\begin{array}{ll}

1 & 0 \\

0 & 1

\end{array}\right]\) and A^{2} = kA – 2I, then find k.

Answer:

Question 6.

Solve the following system of linear equations, by matrix method : (2)

5x + 2y = 4

7x + 3y = 5

Answer:

Given system of equations is :

5x + 2y =4

7x + 3 y = 5

Writing the given system of equations in matrix form

AX = B ……….(1)

where A = \(\left[\begin{array}{ll}

5 & 2 \\

7 & 3

\end{array}\right]\)

X = \(\left[\begin{array}{l}

x \\

y

\end{array}\right]\)

B = \(\left[\begin{array}{l}

4 \\

5

\end{array}\right]\)

Determinant of matrix |A| is

|A| = \(\left|\begin{array}{ll}

5 & 2 \\

7 & 3

\end{array}\right|\)

= 5 × 3 – 7 × 2 = 15 – 14 = 1

∴ |A| = 1 ≠ 0

Since, matrix A is non-singular so A^{-1} exists and system is consistent.

If A_{ij} is cofactor of a in A_{ij}, then

A_{11} =(-1)^{1+1}(3) = 3

A_{12} =(-1)^{1+2} = 7 = -7

A_{21} =(- 1)^{2+1} (2) = -2

A_{22} = (1)^{2+2} (5) = 5

Matrix formed by the cofactors of |A| is:

Question 7.

Show that the function defined by g(x) = x – [ x ] is discontinuous at all integral points. Here [ x ] denotes the greatest integer less than or equal to x. (2)

Answer:

Given function, g(x) = x – [x]

Let x = c 0, is any arbitrary real integer, then for continuity or discontinuity at x = c

Thus, function is not continuous at x = c, i.e., discontinuous.

Since, c is an arbitrary integer. Thus, g(x) is discontinuous at all integer.

Question 8.

Find ∫\(\frac{x e^{x}}{(1+x)^{2}}\)dx

Answer:

Question 9.

Find the probability of getting 5 exactly twice in 7 throws of a die. (2)

Answer:

The repeated tossing of a die are Bernoulli trials.

Let X represent the number of times of getting 5 in 7 throws of the die Probability of getting 5 in a single throw of the dice,

p = \(\frac{1}{6}\)

∴ q = 1 – p = 1 – \(\frac{1}{6}=\frac{5}{6}\)

Clearly, X has the probability distribution with n = 7 and p = \(\frac{1}{6}\).

∴ p(X = x) = ^{n}C_{x} q^{n-x}p^{x} = ^{7}C_{x}\(\left(\frac{5}{6}\right)^{7-x}+\left(\frac{1}{6}\right)^{x}\)

P(getting 5 exactly twice) = P(X = 2)

Question 10.

For what values of x :

Answer:

We have

4 + 2x + 2x = 0 ⇒ 4x = -4 Thus, x = -1

Question 11.

Find, \(\frac{dy}{dx}\), if x = a(θ + sinθ) and y = a(1 – cos θ) (2)

Answer:

Question 12.

By using properties of determinants, show that:

\(\left|\begin{array}{lll}

1 & a & a^{2} \\

1 & b & b^{2} \\

1 & c & c^{2}

\end{array}\right|\) = (a – b)(b – c)(c – a)

Answer:

Let ∆ = \(\left|\begin{array}{lll}

1 & a & a^{2} \\

1 & b & b^{2} \\

1 & c & c^{2}

\end{array}\right|\)

Applying R_{2} → R_{2} – R_{1} and R_{3} → R_{3} – R_{1}

Taking (b – a) and (c – a) common from R_{2} and R_{3} respectively

Question 13.

Answer:

Putting, cos x = t ⇒ – sin x dx = at,

⇒ sin x dx = – dt

When x = 0, then t = cos θ = 1,

Question 14.

For the following differential equation, find a particular solution satisfying the given condition :

\(\frac{dy}{dx}\) = y tan x; y = 1 if x = 0 (2)

Answer:

Given differential equations is

\(\frac{dy}{dx}\) = y tan x

or \(\frac{dy}{y}\) = tan x dx ……(1)

On integrating

∫\(\frac{dy}{y}\) = ∫tan x dx

or log y = – log cos x + C …(2)

Putting x = 0 and y = 1 in equation (2), we get

log 1 = – log cos 0 + C

or 0 = – log 1 + C

or 0 = 0 + C ⇒ C = 0

Putting C = 0 in equation (2), we get

log y = – log cos x

or log y = -log\(\frac{1}{secx}\) = log sec x

or y = sec x ………….(3)

Question 15.

If \(\vec{a}\) = 2i + 2j + 3k, \(\vec{b}\) = – i + 2j+k and \(\vec{c}\) = 3 i + j are such that \(\vec{a}\) + λ\(\vec{b}\) is perpendicular to \(\vec{c}\), then find the value of λ. (2)

Answer:

Given, \(\vec{a}\) + λ\(\vec{b}\) is perpendicular to c .

(\(\vec{a}\) + λ\(\vec{b}\)). \(\vec{c}\) = 0

{(2 î + 2 ĵ+3 k̂) + λ(- î+ 2 ĵ + k̂)}.(3 î + ĵ) = 0

or {(2 – λ)î + (2 + 2λ)ĵ + (3 + î)k̂}.(3î + ĵ) = 0

or 3(2 – λ)î .î + (2 – λ)(î .ĵ) + 3(2 + 2λ)( ĵ.î) + (2 + 2λ)ĵ.ĵ + 3(3 + λ)k̂.î + (3 + λ)k̂.ĵ = 0

3 (2 – λ)î.î + (2 + 2λ)ĵ.ĵ = 0

or 3(2 – λ) + (2 + 2λ) = 0

or 6 – 3λ + 2 + 2λ = 0

or 8 – λ = 0 ⇒ λ = 8.

Question 16.

Probability of solving specific problem independently by A and B are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively. If both try to solve the problem independently, find the probability that exactly one of them solves the problem. (2)

Answer:

Probability of solving the problem by

P(A) = \(\frac{1}{2}\)

Probability of solving the problem by B,

P(B) = \(\frac{1}{3}\)

Since the problem is solved independently by A and B,

P(A’) = 1 – P(A) = 1 – \(\frac{1}{2}=\frac{1}{2}\)

P(B’) = 1 – P(B) = 1 – \(\frac{1}{3}=\frac{2}{3}\)

Probability that exactly one of them solves the problem is given by, P(A). P(B’) + P(B). P(A’) = \(\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)

Section – C

Long Answer Type Questions

Question 17.

Prove that :

tan^{-1}√x = \(\frac{1}{2}\)cos^{-1} \(\left(\frac{1-x}{1+x}\right)\), x ∈ [0, 1]. (3)

Answer:

Let x = tan^{2}θ => tan θ = √x

⇒ θ = tan^{-1}1√x

Thus, L.H.S = R.H.S

Or

Solve the following equation :

tan^{-1}\(\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\) = tan^{-1}1 (x), (x > 0). (3)

Answer:

Let x = tan θ

Now, tan^{-1} \(\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\) = tan^{-1}1 x

Question 18.

Find \(\frac{d y}{d x}\) of the function x^{y} + y^{x} = 1. (3)

Answer:

Let u= x^{y} and v = yx

Then u + v = 1 …(1)

In u = xy and v = yx taking logarithm

log u = y log x and log V = x log y

Now, differentiating w.r.t. x, we get

Now, differentiating both sides of (1) w.r.t x, we get

\(\frac{d u}{d x}+\frac{d v}{d x}\) = 0

Putting the value of \(\frac{du}{dx}\) and \(\frac{dv}{dx}\) from equa¬tion (4) in (2) and (3)

Or

Verify Mean Value Theorem if f(x) = x^{2} – 4x – 3, in the interval [a, b] where a = 1 and b = 4. (3)

Answer:

Given, f(x) = x^{2} – 4x – 3 is a polynomial function. Polynomial function is continuous in interval [1, 4] and differentiable in (1, 4).

Now f'(x) = 2x – 4, exists in interval (1,4).

There exist a point c is interval (1,4) such

f'(c) = \(\frac{f(4)-f(1)}{4-1}\) ……(1)

f(4) = 4^{2} – 4 × 4 – 3 = 16 – 16 – 3 = – 3

f(1) = 1^{2} – 4 × 1 – 3 = 1 – 4 – 3 = -6

Putting values of f(4) and f(1) in equation (1)

Hence, mean value theorem verified.

Question 19.

∫\(\frac{x+2}{\sqrt{x^{2}+2 x+3}}\) dx (3)

Answer:

Then, I_{2} = ∫\(\frac{d x}{\sqrt{x^{2}+2 x+3}}\)

= log | t + \(\sqrt{t^{2}+(\sqrt{2})^{2}}\)| + C_{2}

= log|(x + l) + \(\sqrt{(x+1)^{2}+(\sqrt{2})^{2}}\) + C_{2}

= log | (x +1) + \(\sqrt{x^{2}+2 x+3}\)| + C_{2}

Putting the values of I_{1} and I_{2} in (1), we get

Or

Find: ∫\(\frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}\) dx (3)

Answer:

or 1 = A(t + 3) + B(t + 1)

1 = (A + B)t + 3A + B

On comparing,

A + B = 0 and 3A + B = 1

Now, solving equations, we get

A = \(\frac{1}{2}\) and B = –\(\frac{1}{2}\)

Question 20.

Show that the points A, B and C, with position vectors, \(\vec{a}\) = 3î – 4ĵ – 4k̂,\(\vec{b}\) = 2î – ĵ + k̂ and \(\vec{c}\) = î – 3ĵ – 5k̂ respectively form the vertices of a right angled triangle. (3)

Answer:

Let O is origjn, then

Given, \(\overrightarrow{O A}\) = \(\vec{a}\) = 3î – 4ĵ – 4k̂

\(\overrightarrow{O B}\) = \(\vec{b}\) = 2î – ĵ + k̂

\(\overrightarrow{O C}\) = \(\vec{c}\) = î – 3ĵ – 5k̂

Now \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\vec{b}-\vec{a}\)

= (2î – ĵ + k̂) – (3î – 4ĵ – 4k̂)

= 2 î – ĵ + k̂ – 3 î + 4 ĵ + 4 k̂

= (2 – 3) î + (- 1 + 4) ĵ + (1 + 4)k̂

or \(\overrightarrow{A B}\) = – î + 3ĵ + 5k̂

∴ |\(\overrightarrow{A B}\)| = |- î + 3ĵ + 5k̂|

= î – 3ĵ – 5k̂ – (2î – ĵ + k̂)

= î – 3ĵ – 5 k̂- 2î + ĵ – k̂

= – î – 2ĵ – 6k̂

= 3î – 4ĵ – 4k̂-(i – 3ĵ – 5k̂)

= 3î – 4ĵ – 4k̂ – î + 3ĵ + 5k̂

= (3 – 1)î + (-4 + 3)ĵ + (-4 + 5)k̂

= 2î – ĵ + k̂

Hence, ∆ABC is right-angled triangle i.e., points A, B and C form the vertices of a right-angled triangle.

Or

Find a unit vector perpendicular to each of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) where \(\vec{a}\) = î + ĵ + k̂ \(\vec{b}\) = î + 2ĵ + 3k̂. (3)

Answer:

We have,

\(\vec{a}+\vec{b}\) = 2î + 3ĵ + 4k̂ and

\(\vec{a}-\vec{b}\) = – ĵ – 2k̂

A vector which is perpendicular to both \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) is

Section – D

Essay Type Questions

Question 21.

Find : ∫^{π} _{0} log (1 + cosx) dx (4)

Answer:

When f(2a – x) = f(x)

Adding (4) and (5), we get

Putting 2x = t

⇒ 2dx = dt ⇒ dx = \(\frac{1}{2}\)dt

When x = 0, then t = 0

When x = \(\frac{\pi}{2}\), then t = π

From equation (6), we get

2I_{1} = I_{1} – \(\frac{\pi}{2}\)log2(∵I_{2} = I_{1})

∴I = –\(\frac{\pi}{2}\)log2

Putting the value of I_{1} in equation (3), we get

I = 2I_{1} = 2(-\(\frac{\pi}{2}\)log2) ⇒ I = -πlog 2

Or

Find: ∫^{π/4} _{0}\(\frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx (4)

Answer:

Let I = ∫^{π/4} _{0}\(\frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx

Putting sin x – cos x = t

⇒ (cos x + sin x) dx = dt

and (sin x – cos x)^{2} = t^{2}

or sin^{2} x + cos^{2} x – 2 sin x cos x = t^{2}

or 1 – 2 sin x.cos x = t

or 1 – sin 2x = t^{2}

or sin 2x = 1 – t^{2}

When x = 0, then sin 0 – cos 0 = t

or – 1 = t or t = -1

Question 22.

Show that the differential equation x cos\(\left(\frac{y}{x}\right) \frac{d y}{d x}\) = ycos\(\left(\frac{y}{x}\right)\) + x is homogeneous and solve it. (4)

Answer:

The given difierential equation is:

Thus, F(x, y) is a homogenous function of degree zero.

Therefore, the given differential equation is a homogeneous differential equation.

To solve it we make the substitution

y = vx …(2)

Differentiating equation (2). with respect to x, we get

\(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) …………..(3)

Substituting the value of y and \(\frac{d y}{d x}\) in equation (1)

we get,

which is the general solution of the differential equation (1).

Or

Solve the differential equation. (4)

(tan^{-1}y – x)dy = (1 + y^{2})dx

Answer:

The given differential equation is (tan^{-1}y – x)dy = (1 + y^{2})dx

Now (1) is a linear differential equation of the form \(\frac{d y}{d x}\) + P_{1}x = Q_{1}, where P_{1} = \(\frac{1}{1+y^{2}}\) and Q_{1} = \(\frac{\tan ^{-1} y}{1+y^{2}}\)

Therefore, I.F = e^{∫\(\frac{1}{1+y^{2}}\)dy} = e^{tan-1y}

Thus, the solution of the given differential equation is

Substituting tan^{-1} y = t so that \(\left(\frac{1}{1+y^{2}}\right)\)dy = dt, we get

I = ∫te^{t}dt = te^{t} – ∫1.e^{t}dt

= te^{t} – e^{t}

= e^{t}{t – 1)

⇒ I = e^{tan-1y}(tan^{-1}y – 1)

Substituting the value of I in equation (2), we get

x.e^{tan-1}y = e^{tan-1}y (tan^{-1}y – 1) + C

⇒ x(tan^{-1}y – 1) + Ce^{-tan-1}y

which is the general solution of the given differential equation.

Question 23.

A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A ? (4)

Answer:

Let E_{1} E_{2}, and E_{3} be the respective events of the time consumed by machines A, B, and C for the job.

P(E_{1}) = \(\frac{50}{100}=\frac{1}{2}\)

P(E_{2}) = \(\frac{30}{100}=\frac{3}{10}\)

P(E_{3}) = \(\frac{20}{100}=\frac{1}{5}\)

Let X be the event of producing defective items.

The probability that thern defective item was produced by A is given by P\(\left(\frac{E_{1}}{A}\right)\)

By using Bayes’ theorem, we obtain P\(\left(\frac{E_{1}}{X}\right)\)

Or

Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean and variance of the number of kings.

Answer:

Let X denote the number of kings in a draw of two cards. X is a random vairable which can assume the values 0. 1 or 2.

Now, P(X = 0) = P (no king) = \(\frac{{ }^{48} C_{2}}{{ }^{52} C_{2}}\)

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