Students must start practicing the questions from RBSE 12th Maths Model Papers Set 3 with Answers in English Medium provided here.

## RBSE Class 12 Maths Board Model Paper Set 3 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) Number of binary operations on the set {a, b} are:

(a) 10

(b) 16

(c) 20

(d) 8

Answer:

(b) 16

(ii) sin (tan^{-1} x), ∀ | x | < 1 is equal to :

(a) \(\frac{x}{\sqrt{1-x^{2}}}\)

(b) \(\frac{1}{\sqrt{1-x^{2}}}\)

(c) \(\frac{1}{\sqrt{1+x^{2}}}\)

(d) \(\frac{x}{\sqrt{1+x^{2}}}\)

Answer:

(d) \(\frac{x}{\sqrt{1+x^{2}}}\)

(iii) If A, B are symmetric matrices of same order, then AB – BA is a :

(a) skew symmetric matrix

(b) symmetric matrix

(c) zero matrix

(d) identity matrix

Answer:

(a) skew symmetric matrix

(iv) If \(\left|\begin{array}{lll}

2 & 3 & 2 \\

x & x & x \\

4 & 9 & 1

\end{array}\right|\) + 3 = 0, then the value of x is:

(a) 3

(b) 0

(c) – 1

(d) 1

Answer:

(c) – 1

(v) If

is continuous at x = \(\frac{\pi}{3}\), then value of m is : (1)

(a) 3

(b) 6

(c) – 3

(d) – 6

Answer:

(a) 3

(vi) \(\int \frac{d x}{e^{x}+e^{x}}\) x is equal to

(a) tan^{-1} (e^{x}) + C

(b) tan^{-1} (e^{-x}) + C

(c) log (e^{x} – e^{-x}) + C

(d) log (e^{x} + e^{-x}) + C

Answer:

(a) tan^{-1} (e^{x}) + C

(vii) The order and degree (if defined) of the differential equation

\(\frac{d^{2} y}{d x^{2}}\) + x\(\left(\frac{d y}{d x}\right)^{2}\) = 2x^{2} log\(\left(\frac{d^{2} y}{d x^{2}}\right)\) are:

(a) 2, 1

(b) 1, 2

(c) 0, 2

(d) 2, not defined

Answer:

(d) 2, not defined

(viii) If θ is the angle between two vectors \(\vec{a}\) and \(\vec{b}\), then \(\vec{a} \cdot \vec{b}\) ≥ 0 only when :

(a) 0 < θ < \(\frac{\pi}{2}\)

(b) 0 ≤ θ ≤ \(\frac{\pi}{2}\)

(c) 0 < θ < π

(d) 0 ≤ θ ≤ π

Answer:

(b) 0 ≤ θ ≤ \(\frac{\pi}{2}\)

(ix) If A and B are two events such that P (A) ≠ 0 and P (B/A) = 1, then:

(a) A ⊂ B

(b) B ⊂ A

(c) B = Φ

(d) A = Φ

Answer:

(a) A ⊂ B

(x) If the matrix A is both symmetric and skew-symmetric, then

(a) A is a diagonal matrix

(b) A is a zero matrix

(c) A is a square matrix

(d) None of these

Answer:

(b) A is a zero matrix

(xi) The derivative of f given by f(x) = tan^{-1} x assuming it exists is :

(a) \(\frac{1}{1-x^{2}}\)

(b) \(\frac{1}{\sqrt{1-x^{2}}}\)

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

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

Answer:

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

(xii) Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors and θ is the angle between them. Then \(\vec{a}+\vec{b}\) is a unit vector if

(a) θ = \(\frac{\pi}{4}\)

(b) θ = \(\frac{\pi}{3}\)

(c) θ = \(\frac{\pi}{2}\)

(d) θ = \(\frac{2 \pi}{3}\)

Answer:

(d) θ = \(\frac{2 \pi}{3}\)

Question 2.

Fill in the blanks:

(i) If f: R → b R be given by f(x) = (3 – x^{3})^{1/3}, then fof (x) = _____________ .

Answer:

x

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

Answer:

\(\frac{1}{5}\)

(iii) If A + B = \(\left[\begin{array}{ll}

1 & 0 \\

1 & 1

\end{array}\right]\) and A – 2B = \(\left[\begin{array}{cc}

-1 & 0 \\

1 & -1

\end{array}\right]\) then A = _____________ .

Answer:

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

\frac{1}{3} & \frac{1}{3} \\

\frac{2}{3} & \frac{1}{3}

\end{array}\right]\)

(iv) If f is discontinuous at c, then c is called a point of _____________ of f.

Answer:

discontinuity

(v) \(\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\) dx is equal to _____________ .

Answer:

\(\frac{x^{2}}{2}\) + log_{e} x + 2x + C

(vi) If \(\vec{a}\) is a non-zero vector, then \((\vec{a} \cdot \hat{i}) \hat{i}+(\vec{a} \cdot \hat{j}) \hat{j}+(\vec{a} \cdot \hat{k}) \hat{k}\) equals _____________ .

Answer:

0

Question 3.

Very Short Answer Type Questions

(i) If f: R → R is defined by/(x) = 3x + 2, then find f[f(x)]. (1)

Answer:

Given, f(x) = 3x + 2

∴ f[f(x)] = f(3x + 2)

= 3(3x + 2) + 2

= 9x + 6 + 2 = 9x + 8

(ii) Find the principal value of cot [sin^{-1} {cos (tan^{-1}1)}] (1)

Answer:

Given, cot [sin^{-1} {cos (tan^{-1})}]

cot \(\left[\sin ^{-1}\left\{\cos \frac{\pi}{4}\right\}\right]\)

= cot \(\left[\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right]\)

= cot \(\frac{\pi}{4}\) = 1

(iii) Write the element a_{23} of a 3 × 3 matrix A = [a_{ij}], whose elements a_{ij} are given by a_{ij} = \(\frac{|i-j|}{2}\). (1)

Answer:

Given, A = [a_{ij}]_{3 × 3},

where, a_{ij} = \(\frac{|i-j|}{2}\)

Now, a_{23} = \(\frac{|2-3|}{2}=\frac{|-1|}{2}=\frac{1}{2}\)

(iv) What positive value of x makes following pair of determinants equal? (1)

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

2 x & 3 \\

5 & x

\end{array}\right|,\left|\begin{array}{cc}

16 & 3 \\

5 & 2

\end{array}\right|\)

Answer:

Let \(\left|\begin{array}{cc}

2 x & 3 \\

5 & x

\end{array}\right|=\left|\begin{array}{cc}

16 & 3 \\

5 & 2

\end{array}\right|\)

On expanding, we get

2x^{2} – 15 = 32 – 15

⇒ 2x^{2} – 15 = 17

⇒ 2x^{2} = 32

⇒ x^{2} = 16

⇒ x = ±4

Thus, for x = ± 4, given pair of determinants is equal.

(v) Differentiate cos √x with respect to x. (1)

Answer:

Let y = cos √x

⇒ \(\frac{d y}{d x}\) = – sin √x × \(\frac{d}{d x}\) √x

⇒ – sin √x × \(\frac{1}{2 \sqrt{x}}\) = – \(\frac{\sin \sqrt{x}}{2 \sqrt{x}}\)

(vi) Evaluate : \(\int \sqrt{1-\sin 2 x}\) dx, ∀\(\frac{\pi}{4}\) < x < \(\frac{\pi}{2}\) (1)

Answer:

(vii) Write the order and degree, if defined of the following differential equation : (1)

y”’ y^{2} + e^{y} = 0

Answer:

Order = 3, Degree = not defined

(viii) Find the magnitude of each of the two vectors \(\vec{a}\) and \(\vec{b}\) , having the same magnitude such that the angle between them is 60° and their scalar product is \(\frac{9}{2}\). (1)

Answer:

We know that

(ix) If A and B are independent events and P(A) = 0.3, P(B) = 0.4, then find P(A ∪ B). (1)

Answer:

Given, P(A) = 0.3, P(B) = 0.4, P(A ∪ B) = ?

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

For independent event,

P(A ∩ B) = P(A).P(B) = (0.3) (0.4) = 0.12

Thus, P(A ∪ B) = 0.3 + 0.4 – 0.12 = 0.58

(x) Evaluate the following determinant: (1)

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

a+i b & c+i d \\

-c+i d & a-i b

\end{array}\right|\)

Answer:

Le A = \(\left|\begin{array}{cc}

a+i b & c+i d \\

-c+i d & a-i b

\end{array}\right|\)

= (a + ib) (a – ib) – (- c + id) (c + id)

= (a^{2} – i^{2}b^{2}) – (- c^{2} + i^{2}d^{2})

= [a^{2} – (- 1)b^{2}] – [- c^{2} + (- 1)d^{2}]

= (a^{2} + b^{2}) – (- c^{2} – d^{2})

= a^{2} + b^{2} + c^{2} + d^{2}

(xi) Find the general solution of \(\frac{d y}{d x}-\frac{y}{x}\) = 0. (1)

Answer:

The given differential equation is:

⇒ log y = log x + log C = log Cx

y = Cx, which is the required solution.

(xii) Write the number of vectors of unit length perpendicular to the vectors

\(\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{b}=\hat{j}+\hat{k}\) (1)

Answer:

Unit vectors perpendicular to \(\vec{a}\) and \(\vec{b}\) are ±\(\left(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\right)\)

So, there are two unit vectors perpendicular to the given vectors.

Section – B

Short Answer Type Questions

Question 4.

If binary operation * defined on set of real numbers R such that a * b = \(\frac{a b}{2}\) + 1, where a, b ∈ R then test the associativity of binary operation *. (2)

Answer:

If a, b, c ∈ R, then

∴ (a*b) * c ≠ a* (b * c)

Thus, binary operation * does not follow associative law.

Question 5.

If \(\left[\begin{array}{cc}

x-y & z \\

2 x-y & w

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

-1 & 4 \\

0 & 5

\end{array}\right]\), find x, y, z and w.

Answer:

Given, matrix equation is:

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

x-y & z \\

2 x-y & w

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

-1 & 4 \\

0 & 5

\end{array}\right]\)

On equating the corresponding

elements, we get

x – y = – 1 …. (i)

2x – y = 0 ….. (ii)

z = 4 …..(iii)

w = 5 …… (iv)

Subtracting equation (ii), from (i), we get

Substituting the value of x in Eq.

(ii), we get

y = 2

Thus, x = 1, y = 2, z = 4 and w = 5

Question 6.

Write the value of Δ = \(\left|\begin{array}{ccc}

x+y & y+z & z+x \\

z & x & y \\

-3 & -3 & -3

\end{array}\right|\) (2)

Answer:

Question 7.

If the function f defined as

is continuous at x = 3, find the value of k. (2)

Answer:

Question 8.

Evaluate: ∫\(\frac{\cos x d x}{\sqrt{\sin ^{2} x-2 \sin x+5}}\)

Answer:

Question 9.

If a fair coin is tossed 10 times. Find the probability of appearing exactly four tails. (2)

Answer:

Let number of tails in 10 trials is X.

∴ In X distribution, n = 10 and p = \(\frac{1}{2}\).

Question 10.

If A is a square matrix such that A^{2} = I, then find the simplified value of (A – I)^{3} + (A + I)^{3} – 7A. (2)

Answer:

We have, A^{2} = I

∴ A^{3} = A^{2}.A = IA = A

We know that

(A + B)^{3} = A^{3} + 3A^{2}B + 3AB^{2} + B^{2}

(A – B)^{3} = A^{3} – 3A^{2}B + 3AB^{2} – B^{3}

Provided that AB = BA

Since AI = IA = A

∴ (A + I)^{3} = A^{3} + 3A^{2}I + 3AI^{2} + I

⇒ (A – I)^{3} = A^{3} – 3A^{2}I + 3AI^{2} – I

(A + I)^{3} + (A – I)^{3} = 2(A^{3} + 3A)

⇒ (A + I)^{3} × (A – I)^{3} = 2(A + 3A)

⇒ (A + I)^{3} + (A – I)^{3} = 8A

Thus, (A – I)^{3} + (A + I)^{3} – 7A

= 8A – 7A = A

Question 11.

Differentiate the following w.r.t. x : sin^{-1}\(\left\{\sqrt{\frac{1+\cos x}{2}}\right\}\)

Answer:

Question 12.

Given A = \(\left[\begin{array}{cc}

2 & -3 \\

-4 & 7

\end{array}\right]\), compute A^{-1}

Answer:

We have,

= 14 – 12 = 2 ≠ 0

So, A is a non-singular matrix and its inverse exists.

Cofactors of | A | are :

a_{11} = (- 1)^{1+1}(7) = 7

a_{12} = (- 1)^{1+2}(- 4) = + 4

a_{21} = (- 1)^{2+1}(-3) = 3

a_{22} = (- 1)^{2+2}(2) = 2

Matrix formed by the cofactors of |A| is:

Question 13.

Prove that:

Answer:

Question 14.

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

cos \(\left(\frac{d y}{d x}\right)\) = a (a ∈ R); y = 1 when x = 0 (2)

Answer:

Given differential equation is

cos \(\left(\frac{d y}{d x}\right)\) = a

or dy = cos^{-1} a dx …(1)

Integrating both sides of equation

(1), we get

∫ dy = cos^{-1} a ∫dx

or y = x cos^{-1} a + C …(2)

When x = 0, then y = 1,

∴ 1 = 0 × cos^{-1} a + C

∴ C = 1

Putting the value of C in equation

(2), we get

y = x cos^{-1} a + 1

or y – 1 = x cos^{-1} a

or \(\frac{y-1}{x}\) = cos^{-1} a

Thus a = cos\(\left(\frac{y-1}{x}\right)\)

This is the required solution.

Question 15.

Find λ and µ if (2î + 6ĵ + 27k̂) × (î + λĵ + µk̂) = 0.

Answer:

Given:

Question 16.

A fair coin and an unbiased die are tossed. Let A be the event head appears on the coin and B be the event 3 on the die. Check whether A and B are independent events or not. (2)

Answer:

If a fair coin and an unbiased die are tossed, then the sample space S is given by,

S = {(H,1), (H,2),(H,3),(H,4),(H,5),(H,6) (T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}

Let A : Head appears on the coin

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

⇒ P(A) = \(\frac{6}{12}=\frac{1}{2}\)

B: 3 on die = {(H, 3), (T, 3)}

P(B) = \(\frac{2}{12}=\frac{1}{6}\)

∴ A ∩ B = {(H, 3)}

P(A ∩ B) = \(\frac{1}{12}\)

P(A).P(B) = \(\frac{1}{2} \times \frac{1}{6}\) = P(A ∩ B)

Therefore, A and B are independent events.

Section – C

Long Answer Type Questions

Question 17.

Prove: tan^{-1}\(\frac{1}{4}\) + tan^{-1}\(\frac{2}{9}\) = \(\frac{1}{2}\) cos^{-1}\(\frac{3}{5}\) = \(\frac{1}{4}\) sin^{-1}\(\frac{4}{5}\) (3)

Or

Prove: tan^{-1}\(\left(\frac{1}{7}\right)\) + 2 tan^{-1}\(\left(\frac{1}{3}\right)\) = \(\frac{\pi}{4}\) (3)

Answer:

Question 18.

Differentiate (x^{2} – 5x + 8) (x^{3} + 7x + 9) by using product rule. (3)

Or

Verify Rolle’s Theorem for function f(x) = x^{2} – 5x + 6 in interval [2, 3] (3)

Answer:

Let y = (x^{2} – 5x + 8) (x^{3} + 7x + 9)

⇒ \(\frac{d y}{d x}\) =(x^{2} – 5x + 8). {\(\frac{d y}{d x}\)(x^{3} + 7x + 9) + (x^{3} + 7x + 9) . {\(\frac{d y}{d x}\)(x^{2} – 5x + 8)}

= (x^{2} – 5x + 8) (3x^{2} + 7) + (x^{3} + 7x + 9) (2x – 5)

= x^{2}(3x^{2} + 7) – 5x(3x^{2} + 7) + 8(3x^{2} + 7) + x^{3}(2x – 5) + 7x(2x – 5) + 9(2x – 5)

= 3x^{4} + 7x^{2} – 15x^{3} – 35x + 24ax^{2} + 56 + 2x^{4} – 5x^{3} + 14x^{2} – 35x + 18x – 45

= 5x^{4} – 20x^{3} + 45x^{2} – 52x + 11

Question 19.

Evaluate: ∫\(\frac{\cos x}{(1+\sin x)(2+\sin x)}\) dx

Or

Evaluate: ∫\(\frac{4}{(x-2)\left(x^{2}+4\right)}\) dx

Answer:

1 = 2A + tA + B + Bt

1 = (2A + B) + t(A + B)

On comparing the coefficients of t and constant term on both sides, we get 2A + B = 1 and A + B = 0

Now, on solving these two equations, we get

⇒ A = 1 and B = -1

Question 20.

Find the unit vector perpendicular to the plane ABC where the position vectors of A, B and C are 2î – ĵ + k̂, î + ĵ + 2k̂ and 2 î + 3k̂ respectively. (3)

Or

Show that the points A(2î – ĵ + k̂), B(î – 3ĵ – 5k̂), C(3î – 4ĵ – 4k̂) are the vertices of a right-angled triangle.

Answer:

Let O be the origin of reference.

Section – D

Essay Type Questions

Question 21.

Evaluate: (4)

Or

Evaluate: (4)

Answer:

Question 22.

Show that the differential equation 2ye^{x/y} dx + (y – 2xe^{x/y})dy = 0 is homogeneous and find its particular solution, given that, x = 0 when y = 1. (4)

Or

Solve the differential equation (xdy – y dx)y sin\(\left(\frac{y}{x}\right)\) = (ydx + x dy)x cos\(\left(\frac{y}{x}\right)\) (4)

Answer:

The given differential equation is:

2ye^{x/y} dx + (y – 2x e^{x/y})dy = 0

⇒ \(\frac{d x}{d y}\) = \(\frac{2 x e^{x / y}-y}{2 y e^{x / y}}\) ……. (1)

Let F(x, y) = \(\frac{2 x e^{x / y}-y}{2 y e^{x / y}}\), then

Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation.

To solve it, we make the substitution x = vy ……. (2)

Differentiating equation (2) with respect to y, we get

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

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

⇒ 2e^{v} = – log|y| = C ……. (3)

Replacing v by \(\frac{x}{y}\), we get

2e^{x/y} + log|y| = C …(3)

Substituting x = 0 and y = 1 in equation (3), we get

2e° + log |1| = C ⇒ C = 2

Substituting the value of C in equation (3), we get

2e^{x/y} + log | y | = 2

which is the particular solution of the given differential equation.

Question 23.

Two numbers are selected at random (without replacement) from the first five positive integers. Let X denotes the larger of the two numbers obtained. Find the ‘mean and variance of X. (4)

Or

A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y. (4)

Answer:

Total number of possibles outcomes

= ^{5}p_{2} = \(\frac{5 !}{3 !}\) = 5 × 4 = 20

Here, X denotes the larger of two numbers obtained.

∴ X can take values 2, 3,4 and 5.

Now, P(X = 2) = P (getting (1, 2) or (2,1))

= \(\frac{2}{20}\) = \(\frac{1}{10}\)

P(X = 3) = P (getting (1,3) or (3,1) or (2, 3) or (3, 2)) = \(\frac{4}{20}\) = \(\frac{2}{10}\)

P(X = 4) = P(getting (1, 4) or (4,1) or (2,4) or (4, 2) or (3, 4) or (4, 3)) = \(\frac{6}{20}\) = \(\frac{3}{10}\)

and P(X = 5) = P (getting (1, 5) or (5,1) or (2, 5) or (5,2) or (3, 5) or (5, 3) or (4, 5) or (5, 4))

= \(\frac{8}{20}\) = \(\frac{4}{10}\)

Thus, the probability distribution of X is

Now, mean of X = E(X) = ΣX.P (X)

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