RBSE 12th Maths Model Paper Set 2 with Answers in English

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RBSE Class 12 Maths Board Model Paper Set 2 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) If the binary operation * on the set of integers Z, is defined by a * b = a + 3b², then find the value of 8 * 3 is : (1)
(a) 11
(b) 24
(c) 17
(d) 35
Answer:
(d) 35

(ii) If sin^-1 x = y, then : (1)
(a) 0 ≤ y ≤ π
(b) \(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)
(c) 0 < y < π
(d) \(-\frac{\pi}{2}\) < y < \(\frac{\pi}{2}\)
Answer:
(b) \(-\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)

(iii) If \(\left[\begin{array}{cc}
x+y & 4 \\
-5 & 3 y
\end{array}\right]=\left[\begin{array}{rr}
3 & 4 \\
-5 & 6
\end{array}\right]\), then the values of x and y are: (1)
(a) x = 1, y = 2
(b) x – 2, y = 3
(c) x = 2, y = 1
(d) x = 3, y = 2
Answer:
(a) x = 1, y = 2

(iv) If A = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right]\), then det (adj A) equals : (1)
(a) a²⁷
(b) a⁹
(c) a⁶
(d) a²
Answer:
(c) a⁶

(v) If y = cos(√3x), then find \(\frac{dy}{dx}\) is: (1)
(a) \(\frac{\sin \sqrt{3 x}}{2 \sqrt{3 x}}\)
(b) \(\frac{-3 \sin \sqrt{3 x}}{2 \sqrt{3 x}}\)
(c) \(\frac{-3 \sin \sqrt{x}}{2}\)
(d) \(\frac{\sin \sqrt{3 x}}{2 x}\)
Answer:
(b) \(\frac{-3 \sin \sqrt{3 x}}{2 \sqrt{3 x}}\)

(vi) ∫\(\frac{(x-1)(x-\log x)^{3}}{x}\)dx is equal to: (1)
(a) \(\frac{(1-\log x)^{4}}{4}\) + C
(b) \(\frac{(3 x-\log x)^{4}}{3}\) + C
(c) \(\frac{(x-\log x)^{3}}{4}\) + C
(d) \(\frac{(x-\log x)^{4}}{4}\) + C
Answer:
(d) \(\frac{(x-\log x)^{4}}{4}\) + C

(vii) The sum of the order and degree of the differential equation \(\frac{d}{d x}\left\{\left(\frac{d y}{d x}\right)^{3}\right\}\) = 0 is: (1)
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

(viii) If λ(3î + 2ĵ – 6k̂) is a unit vector, then the value of λ is : (1)
(a) ± 7
(b) ± \(\sqrt{43}\)
(c) ± \(\frac{1}{\sqrt{43}}\)
(d) ± \(\frac{1}{7}\)
Answer:
(c) ± \(\frac{1}{\sqrt{43}}\)

(ix) Three dice are thrown simultaneously. The probability of obtaining a total score of 5 is: (1)
(a) \(\frac{5}{216}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{1}{36}\)
(d) \(\frac{1}{49}\)
Answer:
(d) \(\frac{1}{49}\)

(x) If A is square matrix such that A² = A, then(I + A)³ – 7A is equal to: (1)
(a) A
(b) I – A
(c) I
(d) 3A
Answer:
(c) I

(xi) Find \(\frac{d^{2} y}{d x^{2}}\) if y = x³ + tan x. (1)
(a) 3x² + sec² x
(b) 6x + 2 sec² x tanx
(c) 6x + sec² x
(d) sec² x tan x
Answer:
(b) 6x + 2 sec² x tanx

(xii) If \(\vec{a} \cdot \vec{b}=-|\vec{a}||\vec{b}|\) then the angle between a and b is : (1)
(a) 180°
(b) 60°
(c) 45°
(d) 90°
Answer:
(a) 180°

Question 2.
Fill in the blanks :
(i) A relation in a set A is called ____________ relation, if each element of A is related to itself. (1)
Answer:
reflexive

(ii) If tan^-1x + tan^-1y = \(\frac{\pi}{4}\) xy < 1, then the value of x + y + xy is ____________ (1)
Answer:
1

(iii) If [2 1 3]\(\left[\begin{array}{ccc}
-1 & 0 & -1 \\
-1 & 1 & 0 \\
0 & 1 & 1
\end{array}\right]\left[\begin{array}{c}
1 \\
0 \\
-1
\end{array}\right]\) = A, then the order of matrix A is ____________ (1)
Answer:
|x|

(iv) The number of points of discontinuity of f defined by f(x) = |x| – |x + 1| is ____________ (1)
Answer:
zero

(v) The value of ∫^π/2 ₀ cos 2x dx is ____________. (1)
Answer:
0

(vi) The magnitude of \(\left|\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}\right|\) is ____________. (1)
Answer:
1

Question 3.
Very Short Answer Type Questions
(i) Find gof and fog, if f(x) = 8x³ and g(x) = x^1/3. (1)
Answer:
Given, f(x) = 8x³ and g(x) = x^1/3
then gofix) = g(f(x)) = g(8x³)
= (8x³)^1/3 = [(2x)³]^1/3
= 2x

Now ,fog{x) =f(g(x)) = f(x^1/3)
= 8(x^1/3)³ = 8x

(ii) Find the domain of sec^-1(3x – 1). (1)
Answer:
We know that the range of sec x is (- ∞, – 1) ∪ (1, ∞)
3x -1 < -1 and 3x – 1 ≥ 1
⇒ 3x < 0 and 3x ≥ 2
⇒ x < 0 and x ≥ \(\frac{2}{3}\)
∴ x ∈ (- ∞, 0) x ∈ (\(\frac{2}{3}\), ∞)
∴ x ∈ (- ∞, 0) ∪ (\(\frac{2}{3}\), ∞)
Thus, domain is (- ∞, 0) ∪ (\(\frac{2}{3}\), ∞)

(iii) Construct a 2 × 2 matrix, A = [a_ij], whose elements are given by: a_ij = \(\frac{(i+j)^{2}}{2}\) (1)
Answer:
Given a_ij = \(\frac{(i+j)^{2}}{2}\)

(iv) Let A be a square matrix of order 3 × 3. Write the value of |2A|, where | A | =4. (1)
Answer:
Given (A) = 4
|2A| = 2³. | A |
(Using | kA | = kⁿ. | A |)
= 2³ × 4 = 8 × 4 = 32 (here k = 2, n = 3)

(v) If y – x|x|, find \(\frac{dy}{dx}\) for x < 0; (1)
Answer:
We have, y = x|x|
When x < 0, then | x | = – x
y = x(- x) = – x²
Thus, \(\frac{dy}{dx}\) = – 2x

(vi) Evaluate: ∫\(\frac{\sin ^{6} x}{\cos ^{8} x}\)dx (1)
Answer:

(vii) Find the differential equation representing the family of curves y = ae^2x + 5, where a is an arbitrary constant. (1)
Answer:
Given y = ae^2x + 5 ……..(i)

(viii) Find |\(\vec{a}\)| and |\(\vec{b}\)|, if \(|\vec{a}|=2|\vec{b}|\) and \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 12. (1)
Answer:

(ix) If A and B are independent events and P(A) = 0.2, P(B) – 0.5, then find P(A ∪ B). (1)
Answer:
Given,
P(A) = 0.2 and P(B) = 0.5

Now, P(A u B)
= P(A) + P(B) – P(A)P(B)
= 0.2 + 0.5 – 0.2 × 0.5
= 0.70 – 0.10 = 0.60

(x) Find the value of λ, so that the points (1, -5), (4, – 5) and (λ, 7) are collinear. (1)
Answer:
Given, the points (1, – 5), (-4, 5) and (λ, 7) are collinear, then
\(\left|\begin{array}{ccc}
1 & -5 & 1 \\
-4 & 5 & 1 \\
\lambda & 7 & 1
\end{array}\right|\) = 0

On applying R₂ → R₂ – R₁ and R₂ → R₃ – R₁
⇒ \(\left|\begin{array}{ccc}
1 & -5 & 1 \\
-5 & 10 & 0 \\
\lambda-1 & 12 & 0
\end{array}\right|\) = 0
Expanding along C₃, we get
1[- 60 – 10(λ -1)] = 0
⇒ -6o – 10λ + 10 = o ⇒ – 10λ = 5o
Thus, λ = – 5

(xi) Verify that the given functions (explicit or implict) is a solution of the corresponding differential equation : y = cos x + C : y’ + sin x = 0. (1)
Answer:
The given function is :
y = cos x + C …(1)

Differentiating equation (1) w.r.t ‘x’, we get
y’ = – sin x
or y’ + sin x = 0
Hence, the given function y = cos x + C is the solution of the differential equation y’ + sin x = 0.
Hence Proved.

(xii) Find the projection of the vector \(\vec{a}\) = 2î + 3ĵ + 2k̂ on the vector \(\vec{b}\) = î + 2ĵ + k̂ . (1)
Answer:
The projection of vector \(\vec{a}\) on the vector \(\vec{b}\) is

Section – B
Short Answer Type Questions

Question 4.
Check if the relation R on the set A = {1, 2, 3, 4, 5, 6} defined as R = {(x, y) : y is divisible by x} is (i) symmetric (ii) transitive. (2)
Answer:
Given, A = {1, 2, 3, 4, 5, 6}
R = {(x, y) :y is divisible by x}
(i) Symmetric: Now, (2,4) ∈ R [as 4 is divisible by 2]
But (4, 2)jfR [as 2 is not divisible by 4]
∴ R is not symmetric.

(ii) Transitive ; Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y-
∴ z is divisible by x.
⇒ (x, y) ∈ R
∴ R is transitive
Thus, R is reflexive but not symmetric.

Question 5.
Find the matrix A such that \(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right]\)A = \(\left[\begin{array}{rr}
-1 & -8 \\
1 & -2 \\
9 & 22
\end{array}\right]\) (2)
Answer:

On equating the corresponding elements, we get
⇒ 2a – c = — 1 …(i)
2b -d = – 8 …(ii)
a = 1 …(iii)
b = -2 …(iv)
-3a + 4c = 9 …(v)
– 3b+ M = 22 …(vi)

On solving these equations, we get a = 1,b = -2, c = 3, d = 4
Thus, A = \(\left[\begin{array}{cc}
1 & -2 \\
3^{-} & 4
\end{array}\right]\)

Question 6.
Solve the following system of linear equations by matrix method : (2)
2x – y = – 2 and 3x + 4y = 3
Answer:
Given system of equations is:
2x – y = -2
3x + 4y = 3

Writing the given system of equations in matrix form,
AX = B…….(i)

= 2 x 4 – 3 x (-1) = 8 + 3 = 11
∴ | A | = 11 ≠ 0
Since, matrix A is non-singular so A^-1 exists, and system is consistent.
If A_ij is cofactor of a., in A
A₁₁ = (-1)¹⁺¹ 4 = 4
A₁₂ = (- 1)¹⁺² = 3 = – 3
A₂₁ = (-1)²⁺¹ (-1)
= (-1)(-1) = 1
A₂₂ = (-1)²⁺² 2 = 2

Matrix formed by the cofactor of |A| is:

Question 7.
If y = \(\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \ldots \infty}}}\), then find \(\frac{dy}{dx}\). (2)
Answer:
Given y = \(\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \ldots \infty}}}\)
⇒ y = \(\sqrt{\log x+y}\)

Squaring both sides, we get
y² = logx + y

Differentiating both sides by x, we get

Question 8.
Evaluate : ∫ sin x. log cos x dx. (2)
Answer:
Let
I = ∫ sin x. log cos x dx

Put cos x = t ⇒ – sin x dx = dt
I = – ∫ log tdt = -∫ (log t).1 dt
= -[log t∫1 dt – ∫{\(\frac{d}{dx}\)(log t)∫1 dt}dt]
= – t log t + ∫1.dt = – t log t + t + C
= – cos x log cos x + cos x + c

Question 9.
From well shuffled pack of 52 cards, 4 cards are drawn one by one and drawn cards are replaced. Find the probability that : (2)
(i) all cards are of heart (ii) 3 cards are of heart
Answer:
Probability to get card of heart
p = \(\frac{13}{52}=\frac{1}{4}\)

∴ Probability not to get card of heart
q = 1 – p = 1 – \(\frac{1}{4}=\frac{3}{4}\)

If random variable X denotes num¬ber of cards of heart, then by formula P(X = r) = ⁿC. p^r q^n-r where n = 4
(i) P(X = 4) = ⁴C₄p⁴q^4-4
= 1 × \(\left(\frac{1}{4}\right)^{4} \times\left(\frac{3}{4}\right)^{0}=\left(\frac{1}{4}\right)^{4}\)

(ii) P(X = 3) = ⁴C₃p³q^4-3
= 4 × \(\left(\frac{1}{4}\right)^{3} \times\left(\frac{3}{4}\right)^{1}=12\left(\frac{1}{4}\right)^{4}\)

Question 10.
If A = \(\left[\begin{array}{ccc}
0 & 6 & 7 \\
-6 & 0 & 8 \\
7 & -8 & 0
\end{array}\right]\) B = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 2 \\
1 & 2 & 0
\end{array}\right]\) and C = \(\left[\begin{array}{c}
2 \\
-2 \\
3
\end{array}\right]\), then calculate AC, BC and (A + B) Also, verify that (A + B)C = AC + BC (2)
Answer:

Question 11.
If sin y = x cos (a + y), then show that \(\frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\cos a}\). Also, show that \(\frac{d y}{d x}\) = cos a,when x = 0.
Answer:
Given, sin y = x cos (a + y) ……….(i)

Question 12.
By using properties of determination, prove that: \(\left|\begin{array}{lll}
a+b & b+c & c+a \\
b+c & c+a & a+b \\
c+a & a+b & b+c
\end{array}\right|=2\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|\) (2)
Answer:

Question 13.
Evaluate : ∫¹ _-1\(\frac{d x}{x^{2}+2 x+5}\) (2)
Answer:

Question 14.
Find the particular solution of the differential equation (1 – y²)(1 + log | x |) dx + 2xy dy = 0 given that ij = 0, when x = 1. (2)
Answer:
Given differential equation is :
(1 – y²)(1 + log | x |) dx + 2xy dy = 0
⇒ \(\frac{(1+\log |x|)}{x}\)dx + \(\frac{2 y}{1-y^{2}}\)dy = 0

On integrating, we get

Question 15.
Find the magnitude of two vectors \(\vec{a}\) and \(\vec{b}\) , having the same magnitude and such that the angle between them is 60° and their scalar product is \(\frac{1}{2}\). (2)
Answer:

Question 16.
P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are they likely to agree in stating the same fact ? Do you think, when they agree, mean both are speaking truth ? (2)
Answer:
Let P_T: Event that P speaks truth and Q_T: Event that Q speaks truth.
Given, P(P_T) = \(\frac{70}{100}\) ,then
P(P̄_T) = 1 – \(\frac{70}{100}=\frac{30}{100}\)
and P(Q_T) = \(\frac{80}{100}\), then
P̄(Q_T) = 1 – \(\frac{80}{100}=\frac{20}{100}\)
P(A and B are agree to each other)

Percentage of P(A and B are agree to each other)
= \(\frac{62}{100}\) × 100 = 62%
No, agree does not mean that they are speaking truth.

Section – C
Long Answer Type Questions

Question 17.
Prove that: \(\frac{9 \pi}{8}-\frac{9}{4}\) sin^-1 = \(\left(\frac{1}{3}\right)=\frac{9}{4}\)sin^-1\(\left(\frac{2 \sqrt{2}}{3}\right)\) (3)
Or
If (tan^-1x)² + (cot^-1x)² = \(\frac{5 \pi^{2}}{8}\), then find x. (3)
Answer:

Question 18.
If x\(\sqrt{1+y}\) + y\(\sqrt{1+x}\) = 0, then prove that: (1 + x)²\(\frac{d y}{d x}\) + 1 = 0. (3)
Or
If y = sin^-1\(\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right)\), then show that: \(\frac{d y}{d x}=\frac{-1}{2 \sqrt{1-x^{2}}}\) (3)
Answer:
Given, x\(\sqrt{1+y}\) + y\(\sqrt{1+x}\) = 0
⇒ x\(\sqrt{1+y}\) = – y\(\sqrt{1+x}\)
⇒ x²\(\sqrt{1+y}\) = – y²\(\sqrt{1+x}\)
⇒ x² + x²y = y² + xy²
⇒x² – y² = xy² – x²y
⇒ (x – y) (x + y) = – xy(x – y)
⇒ x + y = -xy ⇒ x = -xy – y
⇒ x = -y(x + 1) ⇒ y = \(\frac{-x}{1+x}\)

Differentiating both sides w.r.t.x, we get

Question 19.
Evaluate : ∫\(\frac{x^{2}}{(x \sin x+\cos x)^{2}}\) dx. (3)
Or
Evaluate : ∫\(\frac{\sqrt{1-\sin x}}{1+\cos x}\)e^-x/2 dx. (3)
Answer:
Let I = ∫\(\frac{x^{2}}{(x \sin x+\cos x)^{2}}\) dx
Multiplying N^r and D^r by cos x, we get

Putting x sin x + cos x = t
⇒ (x cos x + sin x – sin x)dx = dt
⇒ xcos xdx=dt
Let I₁ = ∫\(\frac{x \cos x}{(x \sin x+\cos x)^{2}}\) dx
= ∫\(\frac{d t}{t^{2}}=\frac{-1}{t}=\frac{-1}{x \sin x+\cos x}\)

[Putting t = x sin x + cos x]
Now, integrating Eq. (i) by parts, we get

Question 20.
Let \(\vec{a}\) = î + ĵ + k̂,\(\vec{b}\) = 4î – 2ĵ + 3k̂ and \(\vec{c}\) = î – 2ĵ + k̂.Find a vector of magnitude 6 units, which is parallel to the vector 2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\). (3)
Or
Let \(\vec{a}\) = î + 4ĵ + 2k̂, \(\vec{b}\) = 3î -2j + 7k̂ and \(\vec{c}\) =2î – j + 4k̂.Find a vector \(\vec{p}\), which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\vec{p} \cdot \vec{c}\) = 18. (3)
Answer:
Given,
\(\vec{a}\) = î + ĵ + k̂,\(\vec{b}\) = 4î – 2ĵ + 3k̂ and \(\vec{c}\) = î – 2ĵ + k̂
∴ 2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\)
2(î + ĵ + k̂) – (4î – 2ĵ + 3k̂) + 3(î – 2ĵ + k̂)
2î + 2ĵ + 2k̂ – 4î + 2ĵ – 3k̂+ 3î – 6ĵ + 3k̂
= î – 2ĵ + 2k̂
Now, a unit vector in the direction of vector

Hence, vector of magnitude 6 units parallel to the vector
2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\) = 6(\(\frac{1}{3}\)î – \(\frac{2}{3}\)ĵ + \(\frac{2}{3}\)k̂)
= 2î – 4ĵ + 4k̂

Section – D
Essay Type Questions

Question 21.
Evaluate: ∫⁵ ₂[|x – 2| + |x – 3| + |x – 5|]dx. (4)
Or
Prove that : ∫^π/4 ₀(\(\sqrt{\tan x}+\sqrt{\cot x}\)) dx = √2.\(\frac{\pi}{2}\). (4)
Answer:
First, we redefined the integrand of the integral between the given limits (2, 5). After that integrate and simplify it.
For, 2 ≤ x < 5, | x – 2| =(x-2), 2 ≤ x < 3, | x – 3 | = – (x – 3)
3 ≤ x < 5, | x – 3| = (x – 3) and 2 ≤ x < 5, | x – 51 = (5 – x)
Now, I = ∫⁵ ₂[|x – 2| + |x – 3| + |x – 5|]dx

Question 22.
Find the particular solution of the differential equation e^x \(\sqrt{1-y^{2}}\) dx + \(\frac{y}{x}\)dy = o, given that y = 1,when x = 0. (4)
Or
Form the differential equation of the family of circles touching the x-axis at origin. (4)
Answer:
Given, differential equation is
e^x \(\sqrt{1-y^{2}}\) dx + \(\frac{y}{x}\)dy = o
⇒ e^x \(\sqrt{1-y^{2}}\) dx = – \(\frac{y}{x}\)dy

On separating the variables, we get
\(\frac{-y}{\sqrt{1-y^{2}}}\) dy = x e^x dx

On integrating both sides, we get
∫\(\frac{-y}{\sqrt{1-y^{2}}}\) dy = ∫x e^x dx

On putting 1 – y² = t ⇒ -ydy = \(\frac{d t}{2}\) in LHS, we get

using integration by parts]
⇒ \(\sqrt{1-y^{2}}\) = xe^x – ∫e^x dx [put t = 1 – y²]
⇒ \(\sqrt{1-y^{2}}\) = x e^x – e^x + C …(i)

Also, given that y = 1, when x = 0
On putting y = 1 and x = 0 in Eq.(i), we get
\(\sqrt{1-1}\) = 0 – e°+ C
⇒ C = 1 [∵ e° = 1]

On substituting the value of C in Eq. (i), we get
\(\sqrt{1-y^{2}}\) = x e^x – e^x + 1
which is the required particular solution of given differential equation.

Question 23.
An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple-choice question? (4)
Or
Consider the experiment of throwing a dice, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’. (4)
Answer:
The given data can be tabulated as :

Let us denote: E = easy questions, M = multiple choice questions, D = difficult questions, and T = True/False questions
Total number of questions = 1400
Total number of multiple choice questions = 900

Therefore, probability of selecting an easy multiple choice question is
P(E ∩ M) = \(\frac{500}{1400}=\frac{5}{14}\)

Probability of selecting a multiple choice questions, P(M), is
\(\frac{900}{1400}=\frac{9}{14}\)

P(E/M) represents the probability that a randomly selected question will be an easy question, given that it is a multiple choice question.
∴ P(E/M) = \(\frac{P(E \cap M)}{P(M)}=\frac{\frac{5}{14}}{\frac{9}{14}}=\frac{5}{9}\)

Therefore, the required probability is \(\frac{5}{9}\)