RBSE 12th Maths Board Model Paper 2022 with Answers in English 

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) k²|A|
(c) k³|A|
(d) 3k|A|
Answer:
(c) k³|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 + \(\frac{1}{2}\)x² + 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
2x²\(\frac{d^{2} y}{d x^{2}}\) – 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 = x² + 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 sec²(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^-11
⇒ 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^-1x
(∵ sin^-1x + cos^-1x = 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-1x 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 (x₁, y₁), (x₂, y₂) and (x₃, y₃), 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 A² = 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
A₁₁ =(-1)¹⁺¹(3) = 3
A₁₂ =(-1)¹⁺² = 7 = -7
A₂₁ =(- 1)²⁺¹ (2) = -2
A₂₂ = (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) = ⁿC_x q^n-xp^x = ⁷C_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 R₂ → R₂ – R₁ and R₃ → R₃ – R₁

Taking (b – a) and (c – a) common from R₂ and R₃ 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 = tan²θ => tan θ = √x
⇒ θ = tan^-11√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^-11 (x), (x > 0). (3)
Answer:
Let x = tan θ
Now, tan^-1 \(\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\) = tan^-11 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) = x² – 4x – 3, in the interval [a, b] where a = 1 and b = 4. (3)
Answer:
Given, f(x) = x² – 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) = 4² – 4 × 4 – 3 = 16 – 16 – 3 = – 3
f(1) = 1² – 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, I₂ = ∫\(\frac{d x}{\sqrt{x^{2}+2 x+3}}\)
= log | t + \(\sqrt{t^{2}+(\sqrt{2})^{2}}\)| + C₂
= log|(x + l) + \(\sqrt{(x+1)^{2}+(\sqrt{2})^{2}}\) + C₂
= log | (x +1) + \(\sqrt{x^{2}+2 x+3}\)| + C₂

Putting the values of I₁ and I₂ 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
2I₁ = I₁ – \(\frac{\pi}{2}\)log2(∵I₂ = I₁)
∴I = –\(\frac{\pi}{2}\)log2

Putting the value of I₁ in equation (3), we get
I = 2I₁ = 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 = t²
or sin² x + cos² x – 2 sin x cos x = t²
or 1 – 2 sin x.cos x = t
or 1 – sin 2x = t²
or sin 2x = 1 – t²
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^-1y – x)dy = (1 + y²)dx
Answer:
The given differential equation is (tan^-1y – x)dy = (1 + y²)dx

Now (1) is a linear differential equation of the form \(\frac{d y}{d x}\) + P₁x = Q₁, where P₁ = \(\frac{1}{1+y^{2}}\) and Q₁ = \(\frac{\tan ^{-1} y}{1+y^{2}}\)

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^tdt = te^t – ∫1.e^tdt
= te^t – e^t
= e^t{t – 1)
⇒ I = e^tan-1y(tan^-1y – 1)

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

x.e^tan-1y = e^tan-1y (tan^-1y – 1) + C
⇒ x(tan^-1y – 1) + Ce^-tan-1y
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 E₁ E₂, and E₃ be the respective events of the time consumed by machines A, B, and C for the job.
P(E₁) = \(\frac{50}{100}=\frac{1}{2}\)
P(E₂) = \(\frac{30}{100}=\frac{3}{10}\)
P(E₃) = \(\frac{20}{100}=\frac{1}{5}\)

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) = \(\frac{{ }^{48} C_{2}}{{ }^{52} C_{2}}\)