Students must start practicing the questions from RBSE 12th Maths Model Papers Set 1 with Answers in English Medium provided here.

## RBSE Class 12 Maths Board Model Paper Set 1 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) The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is: (1)

(a) reflexive

(b) symmetric

(c) transitive

(d) equivalence

Answer:

(b) symmetric

(ii) The value of tan^{-1} (√3) – cot^{-1} (- √3) is: (1)

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

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

(c) \(\frac{-\pi}{6}\)

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

Answer:

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

(iii) Which of the given values of x and y make the following pair of matrices are equal: (1)

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

3 x+7 & 5 \\

y+1 & 2-3 x

\end{array}\right],\left[\begin{array}{cc}

0 & y-2 \\

8 & 4

\end{array}\right]\)

(a) x = – \(\frac{1}{3}\), y = 7

(b) Not impossible to find

(c) y = 7, x = – \(\frac{2}{3}\)

(d) x = –\(\frac{1}{3}\), y = –\(\frac{2}{3}\)

Answer:

(b) Not impossible to find

(iv) If A = \(\left|\begin{array}{ll}

p & 2 \\

2 & p

\end{array}\right|\) and |A^{3}| = 125, then find the value of p :

(a) + 5

(b) ± 5

(c) ± 3

(d) – 2

Answer:

(c) ± 3

(v) If 2x + 3y = sin y, then \(\frac{d y}{d x}\) is:

(a) \(\frac{9}{\cos y-3}\)

(b) \(\frac{2}{\cos y-3}\)

(c) \(\frac{2}{\cos y+3}\)

(d) \(\frac{3}{\cos y+3}\)

Answer:

(b) \(\frac{2}{\cos y-3}\)

(vi) ∫2x sin (x^{3} + 1)dx is equal to:

(a) – cos (x^{2} + 1) + c

(b) cos (x^{2} – 2) + c

(c) cos (x^{3} – 1) + c

(d) – sin (x^{2} + 1) + c

Answer:

(a) – cos (x^{2} + 1) + c

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

(a) 4

(b) 3

(c) 0

(d) 2

Answer:

(d) 2

(viii) If θ is the angle between any two vectors \(\vec{a}\) and \(\vec{b}\), then \(|\vec{a} \cdot \vec{b}|\) = \(|\vec{a} \times \vec{b}|\) when θ is equal to: (1)

(a) 0

(b) \(\frac{\pi}{4}\)

(c) \(\frac{\pi}{2}\)

(d) π

Answer:

(b) \(\frac{\pi}{4}\)

(ix) A card is picked at random from a pack of 52 playing cards. Given that the picked card is a queen, the probability of this card to be a card of spade is: (1)

(a) \(\frac{1}{3}\)

(b) \(\frac{4}{13}\)

(c) \(\frac{1}{4}\)

(d) \(\frac{1}{2}\)

Answer:

(c) \(\frac{1}{4}\)

(x) If A = \(\left[\begin{array}{rr}

\alpha & \beta \\

\gamma & -\alpha

\end{array}\right]\) is such that A^{2} = I, then :

(a) 1 + α^{2} + βγ = 0

(b) 1 – α^{2} + βγ = 0

(c) 1 – α^{2} – βγ = 0

(d) 1 + α^{2} – βγ = 0

Answer:

(c) 1 – α^{2} – βγ = 0

(xi) If y = log x , then \(\frac{d^{2} y}{d x^{2}}\) is equal to

(a) – \(\frac{1}{x^{2}}\)

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

(c) \(\frac{2}{x}\)

(d) \(\frac{-3}{x^{2}}\)

Answer:

(a) – \(\frac{1}{x^{2}}\)

(xii) The projection of \(\vec{a}\) = 3î – ĵ + 5k̂ on \(\vec{b}\) = 2î + 3ĵ + k is

(a) \(\sqrt{14}\)

(b) \(\frac{8}{\sqrt{14}}\)

(c) \(\frac{8}{\sqrt{39}}\)

(d) \(\frac{8}{\sqrt{35}}\)

Answer:

(b) \(\frac{8}{\sqrt{14}}\)

Question 2.

Fill in the blanks :

(i) If f: R → R is defined by f(x) = x^{2} – 3x + 2, then f(f(x)) is _____________ . (1)

Answer:

x^{4} – 6x^{3} + 10x^{2} – 3x

(ii) The value of sin^{-1}\(\left(-\frac{\sqrt{3}}{2}\right)\) is _____________ . (1)

Answer:

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

(iii) If \(\left[\begin{array}{ll}

2 & 3 \\

5 & 7

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

1 & -3 \\

-2 & 4

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

-4 & 6 \\

-9 & x

\end{array}\right]\), then the value of x _____________. (1)

Answer:

13

(iv) If f(x) = 2|x| + 3|sin x| + 6, then the right hand derivative of f(x) at x = 0 is _____________ (1)

Answer:

5

(v) The value of

is _____________ . (1)

Answer:

0

(vi) The unit vector in the direction of vector \(\vec{a}\) = 2î + 3ĵ + k̂ _____________ (1)

Answer:

\(\frac{2}{\sqrt{14}} \hat{i}+\frac{3}{\sqrt{14}} \hat{j}+\frac{1}{\sqrt{14}} \hat{k}\)

Question 3.

Very Short Answer Type Questions

(i) Let * is a binary operation on the set of all non-zero real numbers, given by a * b = ab/5 for all a, b ∈ R – {0}. Find the value of x, given that 2* (x * 5) = 10. (1)

Answer:

Given, a* b = \(\frac{a b}{5}\) ,∀ a, b ∈ R – {0} ……….. (i)

Also, given, 2* (x* 5) = 10

⇒ 2 * \(\left[\frac{x \times 5}{5}\right]\) = 10 ⇒ 2* x = 10

⇒ \(\frac{2 x}{5}\) = 10 ⇒ x = 25 5 ,

(ii) Write the range of sec^{-1} x, ∀ x ε R :

Answer:

Required range is [0, π] – [π/2]

(iii) If a matrix has 18 elements, what are the possible oders it can have ? What if it has 5 elements? 1

Answer:

When matrix has 18 elements, then possible orders are as follows:

1 × 18, 18 × 1, 2 × 9, 9 × 2, 3 × 6, 6 × 3

If it has 5 elements then orders are 1 × 5 and 5 × 1.

(iv) Using determinants, find the area of the triangle with vertices A (5, 4), B (- 2, 4) and C (2, -6). 1

Answer:

The area of triangle is given by:

Δ = \(\frac{1}{2}\left|\begin{array}{ccc}

5 & 4 & 1 \\

-2 & 4 & 1 \\

2 & -6 & 1

\end{array}\right|\)

= \(\frac{1}{2}\) [5(4 + 6) – 4(- 2 – 2) + 1(12 – 8)]

= \(\frac{1}{2}\) [5 × 10 – 4(- 4) + 1 × 4]

= \(\frac{1}{2}\) [50 + 16 + 4]

= \(\frac{1}{2}\) × 70 = 35 sq.units

(v) If sin^{2} x + cos^{2} y = 1, then find \(\frac{d y}{d x}\).

Answer:

Given, sin^{2} x + cos^{2} y = 1

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

(vi) Evaluate : \(\int \frac{\tan ^{2} x \sec ^{2} x}{1-\tan ^{6} x} d x\) dx

Answer:

Let I = \(\int \frac{\tan ^{2} x \cdot \sec ^{2} x}{1-\tan ^{6} x} d x\)

Putting tan^{3} x = t

= 3 tan^{2}x sec^{2}x dx = dt

tan^{2}x sec^{2}dx = dt

(vii) Form the differential equation representing the family of curves y = mx, where, m is an arbitrary constant. (1)

Answer:

The given equation is:

y = mx …(1)

Differentiating equation (1) w.r.t. ‘x’,

we get

\(\frac{d y}{d x}\) = m

Substituting the value of m in equation (1), we get

y = \(\frac{d y}{d x}\)∙x = x\(\frac{d y}{d x}\) – y = 0

which is free from the parameter m and hence this is the required differential equation.

(viii) Consider two points P and Q with position vectors \(\overrightarrow{O P}\) = \(3 \vec{a}-2 \vec{b}\) and \(\overrightarrow{O Q}\) = \(\vec{a}+\vec{b}\). Find the position vector of a point R which divides the line joining P and Q in the ratio 2 : 1, externally. (1)

Answer:

The position vector of the point R dividing the join of P and Q externally in the ratio 2:1 is

\(\overrightarrow{O R}\) = \(\frac{2(\vec{a}+\vec{b})-(3 \vec{a}-2 \vec{b})}{2-1}\) = \(4 \vec{b}-\vec{a}\)

(ix) Let E and F be events with P(E) = \(\frac{3}{5}\), P(F) = \(\frac{3}{10}\) and P(E ∩ f) = \(\frac{1}{5}\). Are E and F independent ?

Answer:

It is given that

P(E) = \(\frac{3}{5}\), P(F) = \(\frac{3}{10}\)

and P(EF) = P(E ∩ F) = \(\frac{1}{5}\)

P(E)P(F) = \(\frac{3}{5} \cdot \frac{3}{10}\) = \(\frac{9}{50} \neq \frac{1}{5}\)

⇒ P(E) ∙ P(F) ≠ P(EF)

Therefore, E and F are not independent.

(x) In the interval π/2 < x < π, find Hie value of x for which the matrix \(\left[\begin{array}{cc}

2 \sin x & 3 \\

1 & 2 \sin x

\end{array}\right]\) is singular.

Answer:

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

2 \sin x & 3 \\

1 & 2 \sin x

\end{array}\right]\)

∵ A is a singular matrix.

∴ |A| = 0 ⇒ \(\left[\begin{array}{cc}

2 \sin x & 3 \\

1 & 2 \sin x

\end{array}\right]\) = 0

⇒ 4 sin^{2}x – 3 = 0 ⇒ sin^{2}x = \(\frac{3}{4}\)

⇒ sin x = \(\frac{\sqrt{3}}{2}\)

Thus x, = \(\frac{2 \pi}{3}\)

(xi) For the following differential equation, find the general solution:

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

Answer:

Given differential equation is

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

or \(\frac{d y}{1+y^{2}}\) = (1 + x^{2})dx …… (1)

On integrating,

∫\(\frac{d y}{1+y^{2}}\) = ∫((1 + x^{2}) dx

or tan^{-1}y = x + \(\frac{x^{3}}{3}\) + C …….. (2)

This is the required solution.

(xii) Find the value of x for which x(î + ĵ + k̂) is a unit vector.

Answer:

Given, x(î + ĵ + k̂) is a unit vector,

∴ \(|x \hat{i}+x \hat{j}+x \hat{k}|\) = 1

⇒ \(\sqrt{3 x^{2}}\)

⇒ x√3 = + 1

⇒ x = ± \(\frac{1}{\sqrt{3}}\)

Section – B

Short Answer Type Questions

Question 4.

Show that the relation R in the set of integers Z, defined as xRy ⇒ x – y, is divisible by 5, where x, y ∈ Z, is an equivalence relation. (2)

Answer:

Reflexive: ∀ x ∈ Z, x – x = 0 and (x – x)/5 = 0

i.e.r x – x, is divisible by 5.

⇒ xRx (reflexive).

So, R is reflexive.

(ii) Symmetric: Let ∀ x, y ∈ Z, xRy is true, then

xRy ⇒ x – y, is divisible by 5

⇒ y – x, is divisible by 5

[∵(y – x) = – (x – y)]

⇒ yRx (symmetric)

So, R is symmetric.

(iii) Transitive: Let ∀x,y,z ∈ Z, xRy and yRz is true so xRy and yRz ⇒ (x – y) and (y – z) both are divisible by 5.

⇒ (x – y) + (y – z) also divisible by 5.

⇒ x – z, is divisible by 5.

⇒ xRz (transitive)

So, R is transitive.

Thus, given relation is an equivalence relation.

Question 5.

Find tite inverse of matrix A = \(\left[\begin{array}{ccc}

1 & 2 & -2 \\

-1 & 3 & 0 \\

0 & -2 & 1

\end{array}\right]\) by using elementary row transformations. (2)

Answer:

Question 6.

Solve the following system of equation by matrix methods :

x – y + 2z = 7, 2x – y +3z = 12, 3x + 2y -z = 5

Answer:

Writing given equation in matrix form

Here |A| = 1 (1 – 6) + 1 (- 2 – 9)

+ 2(4 + 3) = – 5 – 11 + 14 = – 2 ≠ 0

Now, the cofactors of |A| are:

a_{11} = (- 1)^{1 + 1} (1 – 6) = – 5

a_{12} = (- 1)^{1 + 2} (- 2 – 9) = 11

a_{13} = (- 1)^{1 + 3} (4 + 3) = 7

a_{21} = (- 1)^{2 + 1} (1 – 4) = 3

a_{22} = (- 1)^{2 + 2} (- 1 – 6) = – 7

a_{23} = (- 1)^{2 + 3} (2 + 3) = – 5

a_{31} = (- 1)^{3 + 1} (- 3 + 2) = – 1

a_{32} = (- 1)^{3 + 2} (3 – 4) = 1

a_{33} = (- 1)^{3 + 3} (- 1 + 2) = 1

Matrix formed by the cofactors of |A| is

Thus, x = 2, y = 1, z = 3

Question 7.

Find the values of k so that the function:

Answer:

Question 8.

Evaluate: ∫\(\frac{1}{\sqrt{8+3 x-x^{2}}}\) dx

Answer:

Question 9.

In a college 25% students failed in maths, 15% in chemistry and 10% in maths and chemistry both. A student is randomly chosen: (2)

(i) If he failed in chemistry what will be the probability that he will failed in mathematics.

(ii) If he failed in maths, then what will be probability that he will failed in chemistry?

Answer:

Let failing in mathematics is event E_{1} and failing in chemistry is event E_{2}, then

We have P(E_{1}) = \(\frac{25}{100}\) = 0.25

P(E_{2}) = \(\frac{15}{100}\) = 0.15

and P(E_{1} ∩ E_{2}) = \(\frac{10}{100}\) = 0.10

If one student is randomly selected, then

(i) Probability of failing in mathematics when failed in chemistry

\(P\left(\frac{E_{1}}{E_{2}}\right)\) = \(\frac{P\left(E_{1} \cap E_{2}\right)}{P\left(E_{2}\right)}\) = \(\frac{0.10}{0.15}=\frac{2}{3}\)

(ii) Probability of failing in chemistry, when he failed in mathematics

\(\left(\frac{E_{2}}{E_{1}}\right)\) = \(\frac{P\left(E_{1} \cap E_{2}\right)}{P\left(E_{1}\right)}\) = \(\frac{0.10}{0.25}=\frac{2}{5}\)

Question 10.

Solve the equation for x, y, z and t, if :

Answer:

Comparing corresponding elements, we get

2x + 3 = 9,2z – 3 = 15,

2y = 12, 2t + 6 = 18

⇒ 2x = 9 – 3, 2z = 15 + 3,

y = \(\frac{12}{2}\), 2t = 18 – 6

⇒ 2x = 6, 2z = 18, y = 6, 2t = 12

∴ x = 3, z = 9, y = 6, t = 6

Question 11.

If x = 2 cos θ – cos 2θ and y = 2 sin θ – sin 2θ, then prove that \(\frac{d y}{d x}\) = tan\(\left(\frac{3 \theta}{2}\right)\). (2)

Answer:

Given, x = 2 cos θ – cos 2θ

\(\frac{d x}{d \theta}\) = – 2 sin θ + 2 sin 2θ

and y = 2 sin θ – sin 2θ

⇒ \(\frac{d x}{d \theta}\) = 2 cos θ – 2 cos 2θ

Question 12.

If a, b and c are all non-zero and \(\left|\begin{array}{ccc}

1+a & 1 & 1 \\

1 & 1+b & 1 \\

1 & 1 & 1+c

\end{array}\right|\) = 0 then prove that: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\) = 0. (2)

Answer:

Expanding along R_{1}, we get

a[b(1 + c) + c] – 0 + 1 [bc – 0] = 0

⇒ a(b + bc + c) +1 (bc) = 0

⇒ ab + abc + ca + bc = 0

⇒ ab + bc + ca + abc = 0

Dividing both sides by abc, we get

∴ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\) = 0

Hence proved.

Question 13.

Evaluate:

Answer:

Question 14.

Verify that the given function is a solution of the corresponding differential equation:

y = \(\sqrt{a^{2}-x^{2}}\), x ∈ (- a, a) : x + y \(\frac{d y}{d x}\) = 0, (y ≠ 0). (2)

Answer:

The given function is:

y = \(\sqrt{a^{2}-x^{2}}\) …….. (1)

Differentiating equation (1) w.r.t. ‘x’, we get

Section – C

Long Answer Type Questions

Question 17.

What is the principal valu of cos^{-1}\(\left(\cos \frac{2 \pi}{3}\right)\) + sin^{-1}\(\left(\sin \frac{2 \pi}{3}\right)\)? (3)

Or

Prove the following:

\(\cot ^{-1}\left(\frac{x y+1}{x-y}\right)+\cot ^{-1}\left(\frac{y z+1}{y-z}\right)+\cot ^{-1}\left(\frac{z x+1}{z-x}\right)\) = 0; (o < xy, yz, zx < 1) (3)

Answer:

We know that, the principal value branch of cos^{-1}x is [0, π] and for

which is the required principal value.

Question 18.

Verify Rolle’s theorem for the function f(x) = x^{2} – 5x + 4 on the interval [1, 4]. (3)

Or

Discuss the continuity of the function defined by: (3)

Answer:

Here, f(x) = x^{2} – 5x + 4

(i) f(x) is continuous on [1, 4] being algebraic function.

(ii) f'(x) = 2x – 5 exists for all x ∈ (1,4).

∴ f(x) is derivable on (1, 4).

(iii) f(1) = 1 – 5 + 4 = 0

f(4) = 16 – 20 + 4 = 0

∴ f(1) = f(4)

Rolle’s theorem is satisfied

Now, there exists a number c such that f'(c) = 0 ⇒ 2c – 5 = 0

c = \(\frac{5}{2}\) ∈ (1, 4)

Thus, Rolle’s theorem is verified.

Question 19.

Evaluate: ∫\(\frac{\cos x}{\sqrt{\sin ^{2} x-2 \sin x-3}}\) dx (3)

Or

Evaluate: ∫\(\frac{x+2}{\sqrt{x^{2}+5 x+6}}\) dx (3)

Answer:

Question 20.

The two adjacent sides of a parallelogram are \(2 \hat{i}-4 \hat{j}+5 \hat{k}\) and \(\hat{i}-2 \hat{j}-3 \hat{k}\). Find the unit vector parallel to its diagonal. (3)

Or

Prove that \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})\) = \(|\vec{a}|^{2}+|\vec{b}|^{2}\), if and only if \(\vec{a}, \vec{b}\) are perpendicular, given \(\vec{a} \neq 0, \vec{b} \neq 0\). (3)

Answer:

Let two adjacent sides of the parallelogram are \(\vec{a}\) and \(\vec{b}\).

Section – D

Essay Type Questions

Question 21.

Evaluate:

Or

Evaluate:

Answer:

Question 22.

Find the equation of the curve passing through the point \(\left(0, \frac{\pi}{4}\right)\) whose differential equation is sin x cos y dx + cos x sin y dy = 0 (4)

Or

Show that the following differential equation is hom0geneous and solve it: (4)

\(\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y\) dx = \(\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x\) dy

Answer:

Given differential equation is

sin x cos y dx + cos x sin y dy = 0

or \(\frac{\sin x}{\cos x}\) dx + \(\frac{\sin y}{\cos y}\) dy = 0

or tan x dx + tan y dy = 0

Integrating both sides, we get

∫ tan x dx + ∫ tan y dy = 0

or log sec x + log sec y = log C

or log (sec x sec y) = log C

or sec x.sec y = C …(1)

Curve passes through (0, π/4),

putting x = 0, y = π/4 in equation (1),

sec 0 × sec \(\frac{\pi}{4}\) = C

or 1 × √2 = C

C = √2

Putting the value of C in equation (1), we get

sec x sec y = √2

or \(\frac{1}{\cos y}\) × sec x = √2

or cos y = \(\frac{\sec x}{\sqrt{2}}\)

which is the required equation of the curve

Question 23.

A coin is biased so that the head is three times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails. Hence, find the mean of the number of tails. (4)

Or

There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two Cards are drawn at random without replacement. Let X denotes the sum of the numbers on the two drawn cards. Find the mean and variance of X. (4)

Answer:

Let X be the random variable which denotes the number of tails when a biased coin is tossed twice.

So, X may have values 0,1 or 2. Since, the coin is biased in which head is 3 times as likely to occur as a tail.

∴ P(H) = \(\frac{3}{4}\) and P(T) = \(\frac{1}{4}\)

P(X = 0) = P(HH) = P(H).P(H)

= \(\left(\frac{3}{4}\right)^{2}=\frac{9}{16}\)

P(X = 1)= P(one tail and one head)

= P(HT, TH) = P(HT) + P(TH)

= P(H) . P(T) + P(T). P(H)

= \(\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}\)

= \(\frac{3}{16}+\frac{3}{16}=\frac{6}{16}=\frac{3}{8}\)

P(X = 2)= P(two tails) = P(TT)

= P(T).P(T)

= \(\left(\frac{1}{4}\right)^{2}=\frac{1}{16}\)

Therefore, the required probability distribution is as follows

Now, mean = ΣX.P(X)

= \(0 \times \frac{9}{16}+1 \times \frac{3}{8}+2 \times \frac{1}{16}\)

= \(0+\frac{3}{8}+\frac{1}{8}=\frac{4}{8}=\frac{1}{2}\)

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