Students must start practicing the questions from RBSE 12th Maths Model Papers Set 7 with Answers in English Medium provided here.

## RBSE Class 12 Maths Board Model Paper Set 7 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 f:R – \(\left\{-\frac{4}{3}\right\}\) → R be a function defined as f(x) = \(\frac{4 x}{3 x+4}c\). The inverse of f is the map g: Range f → R – \(\left\{-\frac{4}{3}\right\}\) given by: (1)

(a) g(y) = \(\frac{3 y}{3-4 y}\)

(b) g(y) = \(\frac{4 y}{4-3 y}\)

(c) g(y) = \(\frac{4 y}{3y-4}\)

(d) g(y) = \(\frac{4 y}{3-4 y}\)

Answer:

(b) g(y) = \(\frac{4 y}{4-3 y}\)

(ii) tan^{-1}\(\left(\frac{x}{y}\right)\) – tan^{-1}\(\left(\frac{x-y}{x+y}\right)\) is equal to (1)

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

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

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

(d) \(\frac{3 \pi}{4}\)

Answer:

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

(iii) If 2A – 3B + 5C = 0, where B = \(\left[\begin{array}{rrr}

-2 & 2 & 0 \\

3 & 1 & 4

\end{array}\right]\) and C = \(\left[\begin{array}{rrr}

2 & 0 & -2 \\

7 & 1 & 6

\end{array}\right]\) then the value of matrix A is : (1)

(a) \(\left[\begin{array}{ccc}

8 & 3 & 5 \\

13 & -1 & 9

\end{array}\right]\)

(b) \(\left[\begin{array}{ccc}

-8 & -3 & 5 \\

-13 & -1 & 9

\end{array}\right]\)

(c) \(\left[\begin{array}{ccc}

8 & 3 & 5 \\

13 & 1 & 9

\end{array}\right]\)

(d) \(\left[\begin{array}{ccc}

-8 & 3 & 5 \\

-13 & -1 & -9

\end{array}\right]\)

Answer:

(d) \(\left[\begin{array}{ccc}

-8 & 3 & 5 \\

-13 & -1 & -9

\end{array}\right]\)

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

x & 2 \\

18 & x

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

6 & 2 \\

18 & 6

\end{array}\right|\), then x is equal to: (1)

(a) 6

(b) ± 6

(c) – 6

(d) 0

Answer:

(b) ± 6

(v) If x = at^{2} and y = 2at, then \(\frac{d y}{d x}\) is: (1)

(a) 2 at

(b) 2a

(c) t

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

Answer:

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

(vi) ∫\(\frac{d x}{x^{2}+16}\) is equal to: (1)

(a) tan^{-1}\(\left(\frac{x}{4}\right)\) + C

(b) \(\frac{1}{4}\)tan^{-1}\(\left(\frac{x}{4}\right)\) + C.

(c) cot^{-1}(x) + C

(d) \(\frac{1}{4}\) cot^{-1}(x) + C

Answer:

(b) \(\frac{1}{4}\)tan^{-1}\(\left(\frac{x}{4}\right)\) + C.

(vii) Degree of differential equation \(\frac{d^{4} y}{d x^{4}}\) – sin \(\left(\frac{d^{3} y}{d x^{3}}\right)\) = o is: (1)

(a) 0

(b) 3

(c) 4

(d) not defined

Answer:

(d) not defined

(viii) Find \(\vec{a}\) = î – 7ĵ + 7k̂ and \(\vec{b}\) = 3î – 2ĵ + 2k̂, then \(|\vec{a} \times \vec{b}|\) is: (1)

(a) \(|\vec{a} \times \vec{b}|\)

(b) √2

(c) 19

(d) √3

Answer:

(a) \(|\vec{a} \times \vec{b}|\)

(ix) If P(B) = 0.5 and P(A ∩ B) = 0.32. then P(A/B) is: (1)

(a) 1/2

(b) 8/15

(c) 16/25

(d) 32/45

Answer:

(c) 16/25

(x) If A and B are two matrices such that AB exists, then BA: (1)

(a) may exist

(b) may not exist

(c) may or may not exist

(d) none of these

Answer:

(c) may or may not exist

(xi) If y = tan^{-1}x,then \(\frac{d^{2} y}{d x^{2}}\) is: (1)

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

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

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

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

Answer:

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

(xii) The value of \(\vec{a}\), for vectors 2î – 3ĵ + 4k̂ and aî + 6ĵ – 8k̂ are collinear, is: (1)

(a) – 3

(b) – 4

(c) 6

(d) – 8

Answer:

(b) – 4

Question 2.

Fill in the blanks: .

(i) If R = {(x, y) / x, y ∈ I, x^{2} + y^{2} ≤ 4} is a relation in I, then domain of R is ____________ . (1)

Answer:

{- 2, – 1, 0, 1, 2},

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

Answer:

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

(iii) If A = \(\left[\begin{array}{lll}

1 & 2 & 3 \\

2 & 3 & 1

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

3 & -1 & 3 \\

-1 & 0 & 2

\end{array}\right]\), then 2A – B is ____________ . (1)

Answer:

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

(iv) If f (x) = x|x|, then f ‘(x) = ____________ . (1)

Answer:

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

-1 & 5 & 3 \\

5 & 6 & 0

\end{array}\right]\)

(v) The value of:

is ____________ . (1)

Answer:

1

(vi) The sum of the vectors \(\vec{a}\) = î – 2ĵ + k̂, \(\vec{b}\) = – 2î + 4ĵ + 5k̂ and \(\vec{c}\) = î – 6ĵ – 7k̂ is ____________ . (1)

Answer:

– 4ĵ – k̂

Question 3.

Very Short Answer Type Questions

(i) Show that the function f: R → R, defined as f(x) = x^{2}, is neither one-one nor onto. (1)

Answer:

Since, f(- 1) = 1 = f(1), f is not one-one. Also, the element -2 in the co-domain R is not image of any element x in the domain R.

Therefore, f is onto.

(ii) Prove that: 3 sin^{-1}x = sin^{-1}(3x – 4x^{3}), x ∈ \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)

Answer:

Let sin^{-1} x = θ ⇒ sin θ = x

∵ sin 3θ = 3 sin θ – 4 sin^{3} θ

⇒ sin 3θ = 3x – 4x^{3}

⇒ 3θ = sin^{-1} (3x – 4x^{3})

Thus, 3 sin^{-1} x = sin^{-1} (3x – 4x^{3})

(iii) Find the values of x, y and z from the equation: \(\left[\begin{array}{c}

x+y+z \\

x+z \\

y+z

\end{array}\right]\) = \(\left[\begin{array}{l}

9 \\

5 \\

7

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

Answer:

Given,

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

x+y+z \\

x+z \\

y+z

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

9 \\

5 \\

7

\end{array}\right]\)

On comparing corresponding elements, we get

x + y + z = 9 …(i)

x + z = 5 …(ii)

y + z = 7 ….. (iii)

From equation (i) and (ii)

y = 9 – 5 = 4

From equation (i) and (iii)

x = 9 – 7 = 2

In equation (i) putting y = 4, x = 2

2 + 4 + z = 9 ⇒ z = 9 – 6

∴ z = 3

Thus, x = 2, y = 4, z = 3

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

1 & 2 \\

4 & 2

\end{array}\right]\), then show that: |2A| = 4 |A|

Answer:

= 4(1 × 2 – 4 × 2)

= 4(2 – 8) = 4 × (- 6) = – 24

From (i) and (ii), we get

|2A| = 4|A|

(v) Show that the function defined by f (x) = sin (x^{2}) is a continuous function. (1)

Answer:

The function f may be thought of as a composition g o h of the two functions g and h, where g(x) = sin x and h(x) = x^{2}. Since, both g and h are continuous functions, it can be deduced that f is a continuous function.

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

Answer:

Let I = \(\int \frac{2-3 \sin x}{\cos ^{2} x}\)

= ∫\(\frac{2}{\cos ^{2} x}\) dx – 3∫\(\frac{\sin x}{\cos ^{2} x}\) dx

= 2∫sec^{2}x dx – 3∫sec x tan x dx

= 2 tan x – 3 sec x + C

(vii) Verify that the function y = e^{-3x} is a solution of the differential equation \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}\) – 6y = 0.

Answer:

(vii) Given function is y = e^{-3x}

Differentiating both sides w.r.t.

\(\frac{d y}{d x}\) = – 3e^{– 3x} ………. (i)

Now, differentiating (i) w.r.t. x, we have

\(\frac{d^{2} y}{d x^{2}}\) = 9e^{-3x}

Substituting the values of \(\frac{d^{2} y}{d x^{2}}, \frac{d y}{d x}\) and y in the given differential equation, we get

L.H.S. = 9 e^{-3x} + (- 3e^{-3x}) – 6.e^{-3x}

= 9e^{-3} – 9e^{-3x} = 0 = R.H.S.

Thus, the given function is a solution of the given differential equation.

(viii) Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (- 5, 7). (1)

Answer:

Let initial point and terminal point of the vector are A and B.

Then, co-ordinate of A (2,1) and, co-ordinate of B = (- 5, 7)

Now, from formula

\(\overrightarrow{A B}\) = (x_{2} – x_{1})î + (y_{2} – y_{1})ĵ + (z_{2} – z_{1})k̂

\(\overrightarrow{A B}\) = (- 5 – 2)î + (7 – 1)ĵ

= -7î + 6ĵ

∴ Scalar components of \(\overrightarrow{A B}\) are – 7 and 6 and vector components of \(\overrightarrow{A B}\) are – 7î and 6ĵ.

(ix) In a school, there are 1000 students out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl? (1)

Answer:

Let E is event randomly selected students read in XII and F denotes event randomly selected student is girl

Now P(F) = \(\frac{430}{1000}\) = 0.43

and P(E ∩ F) = \(\frac{43}{1000}\) = 0.043

Thus, P(E/F) = \(\frac{P(E \cap F)}{P(F)}\)

= \(\frac{0.043}{0.43}\) = 0.1

(x) if three vertices of a triangle are respectively (- 4, 5), (- 4, – 2) and (k, – 5), then find value of k for which area of triangle is \(\frac{63}{2}\) sq. units. (1)

Answer:

Let ∆ is area of triangle. Then area of triangle is

(xi) Write the differential equation obtained by eliminating the arbitrary constant C in the equation representing the family of curves xy = C cos x. (1)

Answer:

Given, equation of family of curves is

On differentiating both sides w.r.t. x, we get xy = C cos x.

1.y + x\(\frac{d y}{d x}\) = C (- sin x) .. .(i)

⇒ y + x\(\frac{d y}{d x}\) = –\(\left(\frac{x y}{\cos x}\right)\) sin x

[from Eq. (i)]

Thus, y + x\(\frac{d y}{d x}\) + xy tan x = 0,which is the required differential equation.

(xii) If a unit vector \(\vec{a}\) makes angle \(\frac{\pi}{3}\) with î, \(\frac{\pi}{4}\) with and an acute angle θ with k̂ then find the value of θ. (1)

Answer:

We have

Section – B

Short Answer Type Questions

Question 4.

Consider f: R_{+} → [4, ∞) given b f(x) = x^{2} + 4 Show that f is invertible with the inverse f^{-1} of given by f^{-1} (y) = \(\sqrt{y-4}\), where R_{+} is the set of all non-negative real numbers. (2)

Answer:

Let x_{1}, x_{2} ∈ R, then

f(x_{1}) = f(x_{2}) ⇒ x_{1}^{2} + 4 = x_{2}^{2} + 4

⇒ x_{1}^{2} = x_{2}^{2} ⇒ x_{1} = x_{2}

Thus, f is one-one.

Again, let y ∈ [4, ∞) is an arbitrary element then x ∈ R_{+} will be such element for which y = f(x)

Now y = f(x)

⇒ y = x^{2} + 4

⇒ x^{2} = y – 4

⇒ x = \(\sqrt{y-4}\), y>4

For each y ≥ 4, \(\sqrt{y-4}\) ∈ R_{+}

Now f(x) = f(\(\sqrt{y-4}\))

= (\(\sqrt{y-4}\))^{2} + 4

= y – 4 + 4

∴ f(\(\sqrt{y-4}\)) = y

Thus, f is onto.

Now, f is one-one onto then it will be invertible. Thus,/is invertible.

Now, for f^{-1} inverse of f

fof^{-1}(x) = x

⇒ f(f^{-1}x)) = x

⇒ [f^{-1}(x)]^{2} + 4 = x

⇒ (f^{-1}(x))^{2} = x – 4

⇒ f^{-1}(x) = \(\sqrt{x-4}\)

⇒ f^{-1}(y) = \(\sqrt{y-4}\)

Question 5.

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

4 & -3 \\

2 & -1

\end{array}\right]\) as a sum of a symmetric and a skew-symmetric matrix. (2)

Answer:

So, P is a symmetric matrix and Q is a skew-symmetric matrix.

Now, P + Q = \(\frac{1}{2}\)(A + A’) + \(\frac{1}{2}\)(A – A’)

Thus, matrix A is expressed as the sum of symmetric matrix and skew-symmetric matrix.

Question 6.

Solve the following system of linear equations, using matrix method:

2x + 3y + 3z = 5

x – 2y + z = -4

3x – y – 2z = 3

Answer:

Given system of equations is :

2x + 3y + 3z = 5

x – 2y + z = -4

3x – y – 2z = 3

Writing it in matrix form

AX = B

= 2(4 + 1) – 3(- 2 – 3) + 3(- 1 + 6)

= 2 × 5 – 3 × (- 5) + 3 × 5

= 10 + 15 + 15 = 40

∴ |A| = 40 ≠ 0

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

If A_{ij} is cofactor of a_{ij} in A, then

A_{11} = + [- 2 × (- 2) – (- 1) × 1] = 4 + 1 = 5

A_{12} = (- 1) [- 2 – 3] = (- 1) (- 5) = 5

A_{13} = + [1 × (- 1) – 3 × (- 2)] = – 1 + 6 = 5

A_{21} = (- 1)[- 6 + 3] = (- 1) (- 3) = 3

A_{22} = + [2 × ( – 2) – 3 × 3] = – 4 – 9 = – 13

A_{23} = (- 1) [- 2 – 9] = (- 1) (- 11) = 11

A_{31} = +[3 × 1 – (- 2) × 3] = 3 + 6 = 9

A_{32} = (- 1) [2 – 3] = – 1(- 1) = 1

A_{33} = + [2 × ( – 2) – 1 × 3] = – 4 – 3 = – 7

Question 7.

Show that the function defined by f(x) = cos x^{2} is a continuous function.

Answer:

Given function, f(x) = cos x^{2}

Let x = c ∈ R be any arbitrary real number, then

value of function at x c

∴ Function is continuous at x C.

Since, c is arbitrary real number.

∴ f is continuous for all real numbers.

Thus, function cos x^{2} is continuous.

Question 8.

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

Answer:

Let I = ∫\(\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}\)dx

Putting cos^{-1}x = t ⇒ x = cos t

⇒ – \(\frac{1}{\sqrt{1-x^{2}}}\)dx = dt

∴ I = ∫(cos t).t dt

I = – [t ∫cos t dt – ∫{\(\frac{d}{d t}\)t∫ cos t dt} dt]

= – [t sin t – ∫1.sin t dt]

= – t sin t + ∫ sin t dt

= – t sin t + (- cos t) + C

= – t sin t – cos t + C

= – t\(\sqrt{1-\cos ^{2} t}\) – cos t + C

= – (cos^{-1} x)\(\sqrt{1-x^{2}}\) – x + C

Question 9.

A family has two children. What is the probability that both the children are boys given that at least one of them is a boy? (2)

Answer:

Let b stand for boy ang g for girl.

The sample space of the experiment is

S = {(b, b), (g, b), (b, g), (g, g)}

Let E and F denote the following events:

E: ‘both the children are boys’

F: ’at least one of the child isa boy’

∴ E = {(b, b)}

and F = {(b, b), (g, b), (b, g)}

Now E ∩ F = {(b,b))

∴ P(F) = \(\frac{3}{4}\) and P(E ∩ F) = \(\frac{1}{4}\)

Thus,

P(E/F) = \(\frac{P(E \cap F)}{P(F)}\) = \(\frac{\frac{1}{4}}{\frac{3}{4}}\) = \(\frac{1}{3}\)

Question 10.

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

1 & 2 & 2 \\

2 & 1 & x \\

-2 & 2 & -1

\end{array}\right]\) is a matrix satisfying AA’ = 9I, find x.

Answer:

We have,

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

1 & 2 & 2 \\

2 & 1 & x \\

-2 & 2 & -1

\end{array}\right]\)

Also, AA’ = 9I

On equating the corresponding elements, we get

4 + 2x = 0 and 5 + x^{2} = 9

⇒ x = – 2 and x^{2} = 4

⇒ x = – 2 and x = ±2

Thus, the value of x is -2.

Question 11.

If e^{x} + e^{y} = e^{x + y}, then prove that \(\frac{d y}{d x}\) + e^{y-x} = 0.

Answer:

Given,

e^{x} + e^{y} = e^{x + y}

Dividing Eq. (i) by e^{x + y}, we get

e^{-y} + e^{-x} = 1 …(ii)

Differentiating both sides of Eq. (ii) w.r.t.’x’, we get

Question 12.

Solve the following: \(\left|\begin{array}{ccc}

4-x & 4+x & 4+x \\

4+x & 4-x & 4+x \\

4+x & 4+x & 4-x

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

Answer:

On expanding along R_{1}, we get

(12 + x)[1((- 2x) (- 2x) – 0} – 0 + 0] =0

⇒ 4(12 + x)x^{2} = 0

Either 12 + x = 0 ⇒ x = – 12

or x^{2} = 0 ⇒ x = 0

Thus, x = 0, – 12

Question 13.

Evaluate:

Answer:

Question 14.

For the following differential equation, find the general solution: (2)

e^{x} tan y dx + (1 – e^{x}) sec^{2} y dy = 0

Answer:

Given differential equation is

e^{x} tan y dx + (1 – e^{x}) sec^{2} y dy = 0 ….. (i)

Dividing equation (i) by (1 – e^{x}) tan y, we get

\(\left(\frac{e^{x}}{1-e^{x}}\right)\) dx + \(\frac{\sec ^{2} y}{\tan y}\) dy = o ……. (ii)

Integrating equation (ii), we get

∫\(\frac{e^{x}}{1-e^{x}}\) dx + ∫\(\frac{\sec ^{2} y}{\tan y}\) dy = 0

or – log |1 – e^{x}| + log |tan y| = log C

Let 1 – e^{x} = u

or – e^{x} dx = du

or e^{x} dx = du

∴ ∫\(\frac{e^{x}}{1-e^{x}}\) = – ∫\(\frac{d u}{u}\)

= – log |u|

= log |1 – e^{x}| and tan y = V

sec^{2}y dy = dV

∫\(\frac{\sec ^{2} y}{\tan y}\) = – ∫\(\frac{d V}{V}\)

= log |V| = log |tan y|

= log |tan y|

= log C + log |1 – e^{x}|

⇒ log |tan y| = log C(1 – e^{x})

⇒ tan y = C(1 – e^{x}) …(iii)

Equation (iii) is the required solution of given differential equation.

Question 15.

If \(\vec{a}\) = î + ĵ + k, \(\vec{b}\) = 2î – ĵ + 3k and \(\vec{c}\) î – 2ĵ + k̂, find a unit vector parallel to the vector 2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\). (2)

Answer:

Given

\(\vec{a}\) = î + ĵ + k, \(\vec{b}\) = 2î – ĵ + 3k, \(\vec{c}\) î – 2ĵ + k̂

Then 2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\)

= 2(î + ĵ + k̂) – (2î – ĵ + 3k̂) + 3(î – 2ĵ + k̂)

= 2î + 2ĵ + 2k̂- 2î + ĵ – 3k̂+ 3î – 6ĵ + 3k̂

= (2 – 2 + 3)î + (2 + 1 – 6)ĵ + (2 – 3 + 3)k̂

= 3î -3ĵ +2k̂

Question 16.

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that: (2)

(i) both balls are red.

(ii) first ball is black and second is red.

Answer:

Total number of balls = 18

Number of red balls = 8

Number of black balls = 10

(i) Probability of getting a red ball in the first draw

= \(\frac{8}{18}\) = \(\frac{4}{9}\)

The ball is replaced after the first draw.

∴ Probability of getting a red ball in the second draw

= \(\frac{8}{18}\) = \(\frac{4}{9}\)

Therefore, probability of getting both the balls red

= \(\frac{4}{9} \times \frac{4}{9}=\frac{16}{81}\)

(ii) Probability of getting first ball black

= \(\frac{10}{18}=\frac{5}{9}\)

The ball is replaced after the first draw.

Probability of getting second ball as red

= \(\frac{8}{18}=\frac{4}{9}\)

Therefore, probability of getting first ball as black and second ball as red

= \(\frac{5}{9} \times \frac{4}{9}=\frac{20}{81}\)

Section – C

Long Answer Type Questions

Queestion 17.

If sin^{-1}(1 – x) – 2 sin^{-1}x = \(\frac{\pi}{2}\), then find the value of x. (3)

Or

Prove that: tan^{-1}\(\frac{1}{5}\) + tan^{-1}\(\frac{1}{7}\) + tan^{-1}\(\frac{1}{3}\) + tan^{-1}\(\frac{1}{8}\) = \(\frac{\pi}{4}\) (3)

Answer:

sin^{-1}(1 – x) – 2 sin^{-1} x = \(\frac{\pi}{2}\)

⇒ sin^{-1}(1 – x) – 2 sin^{-1}x

= sin^{-1}(1 – x) + cos ^{-1}(1 – x)

⇒ – 2 sin^{-1} x

= cos^{-1}(1 – x)

[∴ sin^{-1} (1 – x) + cos^{-1} (1 – x) ⇒ \(\frac{\pi}{2}\)]

Let sin^{-1}x = θ

then sin θ = x

Let sin^{-1}x = θ

⇒ sin θ = x

∴ – 2θ = cos^{-1}(1 – x)

⇒ cos(- 2θ) = 1 – x

⇒ cos 2θ = 1 – x …………(i)

Since, we know that

cos 2θ ⇒ 1 – 2 sin^{2} θ

⇒ cos 2θ ⇒ 1 – 2x^{2} …………. (ii)

From equations (i) and (ii), we have

1 – x = 1 – 2x^{2}

⇒ 2x^{2} – x = θ ⇒ x (2x – 1) = 0

⇒ x = 0, x = \(\frac{1}{2}\).

But x = \(\frac{1}{2}\) does not satisfies given equation.

∴ x = 0

Question 18.

If y = cos^{-1}\(\left[\frac{2 x-3 \sqrt{1-x^{2}}}{\sqrt{13}}\right]\), then find \(\frac{d y}{d x}\). (3)

Or

For what value of λ is the function defined by

continuous at x = 0? What about continuity at x = 1? (3)

Answer:

Question 19.

Evaluate: ∫\(\int \frac{1-\sqrt{x}}{1+\sqrt{x}} d x\) (3)

Or

Evaluate: ∫\(\frac{d x}{\cos (x+a) \cos (x+b)}\) (3)

Answer:

Question 20.

Show that the points A(1, – 2, – 8), B(5, 0, – 2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC. (3)

Or

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (î + 2ĵ – k̂) and (- î + ĵ + k̂) respectively, in the ratio 2 : 1 (i) internally (ii)externally (3)

Answer:

Let O be origin, then with respect to O

Position vector of A

= \(\overrightarrow{O A}\) = î – 2ĵ – 8k̂

Position vector of B

Section – D

Essay Type Questions

Question 21.

Evaluate:

Or

Evaluate:

Answer:

Question 22.

For the differential equation xy\(\frac{d y}{d x}\) (x +2) (y + 2), find the solution curve passing through the point (1, – 1). (4)

Or

Show that the differential equation (x – y)\(\frac{d y}{d x}\) = x + 2y is homogeneous and solve it. (4)

Answer:

Given differential equation is

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

⇒ \(\frac{y d y}{y+2}\) = \(\frac{(x+2)}{x}\) dx ……. (i)

Integrating both sides of equation (i), we get

⇒ y – 2 log|y + 2| ⇒ x + 2 log x + C ……(ii)

Putting x = 1, y = -1 in equation (ii)

⇒ – 1 – 2 log |- 1 + 2|

⇒ 1 + 2 log 1 + C

⇒ – 1 – 2 log 1 = 1 + 2 × 0 + C

⇒ – 1 – 0 = 1 + C

⇒ C = – 1 – 1 = – 2 ⇒ C = – 2

Putting the value of C in equation (ii)

y – 2 log (y + 2) = x + 2 log x – 2

⇒ y – x + 2 = 2 log x + 2 log (y + 2)

⇒ y – x + 2 = log x^{2} + log(y + 2)^{2}

⇒ y – x + 2 = log{x^{2}(y + 2)^{2}}

which is the required equation of the curve.

Question 23.

Suppose a girl throws a die, if she gets I or 2, she tosses a corn three times and notes the number of tails. if she gets 3, 4, 5 or 6, she losses a coin once and notes whether a ‘head’ or ‘tail’ is obtained, if she obtained exactly one ‘tail’, what is the probability that she threw 3, 4, 5 or 6 with the die? (4)

Or .

Three rotten apples are mixed with seven fresh apples. Find the probability distribution of the number of rotten apples, il’ three apples are drawn one by one with replacement. Find the mean of the number of rotten apples. (4)

Answer:

Let E_{1} be the event that the girl gets 1 or 2, E2 be the event that the girl gets 3, 4, 5 or 6 and A be the event that the girl gets exactly a ‘tail’.

Then, P(E_{1}) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

and P(E_{2}) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

\(P\left(\frac{A}{E_{1}}\right)\) = P (getting exactly one tail when a coin is tossed three times = \(\frac{3}{8}\)

\(P\left(\frac{A}{E_{2}}\right)\) = P (getting exactly a tail when a coin is tossed once) = \(\frac{1}{2}\)

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