Students must start practicing the questions from RBSE 12th Maths Model Papers Set 6 with Answers in English Medium provided here.

## RBSE Class 12 Maths Board Model Paper Set 6 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 f: R’ → R be given by f(x) = (3 – x^{3})^{\(\frac{1}{3}\)}, then fof(x) is : (1)

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

(b) x^{3}

(c) x

(d) (3- x^{3})

Answer:

(c) x

(ii) The simplest form of cot^{-1}\(\left(\frac{1}{\sqrt{x^{2}-1}}\right)\), x >1 is: (1)

(a) tan^{-1}x

(b) sec^{-1}x

(c) cosec^{-1}x

(d) sin^{-1}x

Answer:

(b) sec^{-1}x

(iii) The simplest form of the number of all possible matrices of order 2 × 2 with each entry 1, 2 or 3 : (1)

(a) 81

(b) 27

(c) 64

(d) 8

Answer:

(a) 81

(iv) Minor of an element of a determinant of order (n ≥ 2) is a determinant of order : (1)

(a) n

(b) n – 1

(c) n + 1

(d) n + 2

Answer:

(b) n – 1

(v) If x = a(θ – sin θ) and y = a(1 + cos θ), then \(\frac{dy}{dx}\) at θ = \(\frac{\pi}{3}\) is : (1)

(a) √2

(b) √3

(c) – √3

(d) 1

Answer:

(c) – √3

(vi) ∫\(\frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}\) is equal to: (1)

(a) tanx + cotx + C

(b) tanx + cosecx + C

(c) -tanx + cotx + C

(d) tanx + secx + C

Answer:

(a) tanx + cotx + C

(vii) The sum of the order and degree of the differential equation

\(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{3}\) + x^{4} = 0. (1)

(a) 2

(b) 3

(c) 5

(d) 4

Answer:

(d) 4

(viii) If \(\vec{a}\) is a non-zero vector of magnitude ‘a’ and λ a non-zero scalar, then λ \(\vec{a}\) is unit vector if : (1)

(a) λ = 1

(b) λ = – 1

(c) a = |λ|

(d) a = 1/|λ|

Answer:

(d) a = 1/|λ|

(ix) If P(Ā) > 0.7, P(B) = 0.7 and P(\(\frac{B}{A}\)) = 0.5, then P(A ∪ B) is : (1)

(a) 0.15

(b) 0.85

(c) 3/14

(d) 5/12

Answer:

(b) 0.85

(x) The simplest form of cos θ\(\left[\begin{array}{rr}

\cos \theta & \sin \theta \\

-\sin \theta & \cos \theta

\end{array}\right]\) + sin θ \(\left[\begin{array}{rr}

\sin \theta & -\cos \theta \\

\cos \theta & \sin \theta

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

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

1 & 0 \\

0 & 1

\end{array}\right]\)

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

0 & 1 \\

1 & 0

\end{array}\right]\)

(c) 1

(d) 2

Answer:

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

1 & 0 \\

0 & 1

\end{array}\right]\)

(xi) If v = 5 cos x – 3 sin x, then \(\frac{d^{2} y}{d x^{2}}\) is: (1)

(a) – 5 sin x – 3 cos x

(b) – 5 cos x + 3 sin x

(c) 3 sin x – 5 cos x

(d) 3 cos x + 5 sin x

Answer:

(b) – 5 cos x + 3 sin x

(xii) The unit vector perpendicular to each of the vertors (1)

(a) \(\frac{7 \hat{i}-6 \hat{j}+10 \hat{k}}{\sqrt{185}}\)

(b) \(\frac{7 \hat{i}+6 \hat{j}+10 \hat{k}}{\sqrt{185}}\)

(c) \(\pm\left(\frac{7 \hat{i}-6 \hat{j}-10 \hat{k}}{\sqrt{185}}\right)\)

(d) \(\pm\left(\frac{6 \hat{i}-7 \hat{j}+10 \hat{k}}{\sqrt{185}}\right)\)

Answer:

(c) \(\pm\left(\frac{7 \hat{i}-6 \hat{j}-10 \hat{k}}{\sqrt{185}}\right)\)

Question 2.

Fill in the blanks :

(i) If * is a binary operation on N given by a * b = LCM (a, b) for all a, b e N, then 5 * 7 is ____________. (1)

Answer:

35

(ii) The value of tan^{-1}(tan\(\frac{7 \pi}{6}\)) is ____________. (1)

Answer:

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

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

1 & 5 \\

6 & 7

\end{array}\right]\), then (A + A’) is ____________. (1)

Answer:

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

2 & 11 \\

11 & 14

\end{array}\right]\)

(iv) If y = tan^{-1}\(\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\) then \(\frac{d y}{d x}\) is ____________. (1)

Answer:

\(\frac{3}{1+x^{2}}\)

(v) The value of ∫^{3}_{2}\(\frac{d x}{x^{2}-1}\) is ____________. (1)

Answer:

\(\frac{1}{2}\)log\(\frac{3}{2}\)

(vi) If \(\vec{a}\) and \(\vec{b}\) are perpendicular vectors, \(|\vec{a}+\vec{b}|\) = 13 and |\(\vec{a}\)| = 5, then the value ____________ (1)

Answer:

12

Question 3.

Very Short Answer Type Questions

(i) Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor translitive. (1)

Answer:

R is reflexive, since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric, as (1, 2) ∈ R but (2, 1) ∉ R. Similarly, R is not transitive as (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R.

(ii) Find the value of tan^{-1}[2 cos(2 sin^{-1}\(\frac{1}{2}\))] (1)

Answer:

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

Answer:

Given, a_{ij} = \(\frac{(i+2 j)^{2}}{2}\)

Then

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

5 & 3 & 8 \\

2 & 0 & 1 \\

1 & 2 & 3

\end{array}\right|\), write the cofactor of element a_{32}. (1)

Answer:

Cofactor of elements a_{32}

= (-1)^{3+2}\(\left|\begin{array}{ll}

5 & 8 \\

2 & 1

\end{array}\right|\) = (-1)(5 – 16) = 11

(v) Differentiate tan^{-1}\(\left(\frac{1+\cos x}{\sin x}\right)\) with respect to x. (1)

Answer:

Let y = tan^{-1}\(\left(\frac{1+\cos x}{\sin x}\right)\)

(vi) Evaluate : ∫cos 4x cos x dx. (1)

Answer:

Let I = ∫cos 4x cos x dx

[cos(A + B) + cos(A – B) – 2cos A cos B]

∴ ∫cos 4x.cos x dx

= \(\frac{1}{2}\)∫ 2cos4xcosx dx

= \(\frac{1}{2}\)∫ cos (4x + x) + cos (4x – x) dx

= \(\frac{1}{2}\)∫(cos5x + cos3x) dx

= \(\frac{1}{2}\)∫ cos 5x dx + \(\frac{1}{2}\)∫ cos 3x dx

= \(\frac{1}{2} \times \frac{1}{5}\)sin5x + \(\frac{1}{2} \times \frac{1}{3}\)sin3x + C

= \(\frac{\sin 5 x}{10}+\frac{\sin 3 x}{6}\) + C

(vii) Find the general solution of the differential equation \(\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}\) (1)

Answer:

The given differential equation is:

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

Since, 1 + y^{2} ≠ 0, therefore, separating the variables, the given differential equation can be written as

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

Integrating both sides of equation (ii), we get:

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

⇒ tan^{-1}y = tan^{-1}x + C

which is the general solution of equation (i).

(viii) Find the position vector of the mid-point of. the vector joining the points P(2, 3, 4) and Q(4, 1, – 2). (1)

Answer:

Let origin is O, then position vector of P and Q with respect to O are \(\overrightarrow{O P}\) and \(\overrightarrow{O Q}\).

Now \(\overrightarrow{O P}\) = 2î + 3ĵ + 4k̂ and \(\overrightarrow{O Q}\) = 4î + ĵ – 2k̂

Let R is the mid point of PQ, then

Hence, the required mid-point is 3î + 2ĵ + k̂.

(ix) Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black. (1)

Answer:

There are 26 black cards in a deck of 52 cards.

Let P(A) be the probability of getting a black card in the first draw.

∴ P(A) = \(\frac{26}{52}=\frac{1}{2}\)

Let P(B) be the probability of get¬ting a black card on the second draw. Since the card is not replaced,

P(B) = \(\frac{25}{51}\)

Thus, probability of getting both the cards black

= \(\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}\)

(x) For what value of x, the matrix \(\left[\begin{array}{cc} (1)

2 x+4 & 4 \\

x+5 & 3

\end{array}\right]\) is a singular matrix?

Answer:

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

2 x+4 & 4 \\

x+5 & 3

\end{array}\right]\)

If matrix A is singular, then |A| = 0

⇒ \(\left[\begin{array}{cc}

2 x+4 & 4 \\

x+5 & 3

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

⇒ (2x + 4) × 3 – (x + 5) × 40

⇒6x + 12 – 4x – 20 = 0 ⇒ 2x = 8

Thus, x = 4

(xi) Form a differential equation representing the family of curve \(\frac{x}{a}+\frac{y}{b}\) = 1 by eliminating arbitrary constants a and b. (1)

Answer:

The given equation is :

\(\frac{x}{a}+\frac{y}{b}\) = 1 ……..(i)

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

Hence, equation (iii) is the required differential equation.

(xii) Find the area of a triangle having the points A (1, 1, 1), B (1, 2, 3) and C (2, 3,1) as its vertices. (1)

Answer:

We have

\(\overrightarrow{A B}\)= ĵ + 2k̂ and \(\overrightarrow{A C}\) = î + 2ĵ

The area of the given triangle is \(\frac{1}{2}\)||

Thus, the required area is \(\frac{1}{2}\)\(\sqrt{21}\)

Section – B

Short Answer Type Questions

Question 4.

Consider a bijective function f: R_{+} → (7, ∞) given by f(x) = 16x^{2} + 24x + 7, where R_{+} is the set of all positive real numbers. Find the inverse function of f. (2)

Answer:

The function is given by f(x) = 16x^{2} + 24x + 7

Let y = f(x)

⇒ y = 16x^{2} + 24x + 7

⇒ y = 16x^{2} + 24x + 7 + 2 – 2

⇒ y = 16x^{2} + 24x + 9 – 2

⇒ y = (4x + 3)^{2} – 2

⇒ (4x + 3)^{2} = y + 2

⇒ 4x + 3 = \(\sqrt{(y+2)}\)

⇒ x = \(\frac{\sqrt{(y+2)}-3}{4}\)

Now, y = f(x)

Thus, f^{-1} = x = \(\frac{\sqrt{(y+2)}-3}{4}\)

Question 5.

If F(x) = \(\left[\begin{array}{ccc}

\cos x & -\sin x & 0 \\

\sin x & \cos x & 0 \\

0 & 1 & 1

\end{array}\right]\), show that : F(x) F(y) = F(x + y). (2)

Answer:

Question 6.

Find A^{-1}, where A = \(\left[\begin{array}{ccc}

1 & 2 & -3 \\

2 & 3 & 2 \\

3 & -3 & -4

\end{array}\right]\). Hence, solve the system of equations,

x + 2y – 3z = 4, 2x + 3y + 2z = 2 and 3x – 3y – 4z = 11. (2)

Answer:

We have A = \(\left[\begin{array}{ccc}

1 & 2 & -3 \\

2 & 3 & 2 \\

3 & -3 & -4

\end{array}\right]\)

∴ |A| = \(\left[\begin{array}{ccc}

1 & 2 & -3 \\

2 & 3 & 2 \\

3 & -3 & -4

\end{array}\right]\)

= 1(-12 + 6) – 2(-8 – 6) – 3(-6 – 9)

= -6 + 28 + 45 = 67 ≠ 0

So, A is invertible.

Cofactors of |A| are

Matrix formed by the cofactors of |A| is

The given system of equation is

x + 2y – 3z = – 4

2x + 3y + 2z = – 2

3x – 3y – 4z = 11

We know that AX = B

Thus, x = 3, y = – 2 and z = 1 is the required solution.

Question 7.

If f(x) = \(\sqrt{x^{2}+1}\), g(x) = \(\frac{x+1}{x^{2}+1}\) and h(x) = 2x – 3, then find f'(h(g'(x)). (2)

Answer:

Given, f(x) = \(\sqrt{x^{2}+1}\), g(x) = \(\frac{x+1}{x^{2}+1}\) and h(x) = 2x – 3

On differentiating above functions w.r.t. x, we get

Question 8.

Evaluate : ∫x\(\sqrt{1+2 x^{2}}\) dx (2)

Answer:

Let I = ∫x\(\sqrt{1+2 x^{2}}\) dx

Puffing 1 + 2x^{2} = t

⇒ 4xdx = dt

⇒ xdx = \(\frac{dt}{4}\)

∴ ∫x\(\sqrt{1+2 x^{2}}\) dx = ∫\(\frac{t^{1 / 2} d t}{4}\)

= \(\frac{1}{4}\)∫t^{1/2} = \(\frac{1}{4} \frac{t^{3 / 2}}{3 / 2}\) + C

= \(\frac{2}{4 \times 3}\) (1 + 2x^{2})^{3/2} + C

= \(\frac{1}{6}\)(1 + 2x^{2})^{3/2} +C

Question 9.

Find the probability distribution of X, the number of heads in a simultaneous toss of two coins. (2)

Answer:

When two coins are tossed, there may be 1 head, 2 heads or no head at all. Thus, the possible values of X are 0, 1, 2.

Now, P(X = 0)

= P (Getting no head) = P(TT) = \(\frac{1}{4}\)

P(X = 1) = P (Getting one head)

= P(HT or TH)

= \(=\frac{2}{4}=\frac{1}{2}\)

P(X = 2) = P (Getting two heads) = P(HH) = \(\frac{1}{4}\)

Thus, the required probability distribution of X is

Question 10.

Show that :

Answer:

Question 11.

Is the function defined by f(x) = x^{2} – sin x + 5 continuous at x = π? Verify. (2)

Answer:

Given function,

f(x) = x^{2} – sin x + 5

For continuity at x = π

Question 12.

By using properties of determinants show that: \(\left|\begin{array}{ccc}

x+4 & 2 x & 2 x \\

2 x & x+4 & 2 x \\

2 x & 2 x & x+4

\end{array}\right|\) = (5x + 4)(4 – x)^{2}

Answer:

Question 13.

Evaluate:

Answer:

Question 14.

Find the general solution of the differential equation x^{5} \(\frac{d y}{d x}\) = – y^{5}. (2)

Answer:

Given differential equation is

x^{5} \(\frac{d y}{d x}\) = – y^{5} or \(\frac{d y}{y^{5}}=-\frac{d x}{x^{5}}\) ………(i)

Integrating both sides of equation (i), we get

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

Question 15.

Show that the points A (2î + 3ĵ + 5k̂), B(î + 2ĵ + 3k̂) and C(7î – k̂) are collinear. (2)

Answer:

Here,

AB = (1 + 2)î + (2-3)ĵ + (3 – 5)k̂

= 3î – ĵ – 2k̂

BC = (7 – 1)î + (0 – 2)ĵ + (-1 – 3)k̂

= 6î – 2ĵ – 4k̂

AC =(7 + 2)î + (0-3)ĵ + (-1 – 5)k̂

= 9î – 3ĵ – 6k̂

|\(\overrightarrow{A B}\)| = \(\sqrt{14}\), |\(\overrightarrow{B C}\)| = 2\(\sqrt{14}\)

and |\(\overrightarrow{A C}\)| = 3\(\sqrt{14}\)

∴ \(|\overrightarrow{A C}|=|\overrightarrow{A B}|+|\overrightarrow{B C}|\)

Thus, the points A, B and C are collinear

Question 16.

A man is known to speak the truth 3 out of 5 times. He throws a die and reports that it is ‘1’. Find the probability that it is actually 1. (2)

Answer:

Let E_{1} = Event that 1 occurs in a die

E_{2} = Event that 1 does not occur in a die

A = Event that the man reports that 1 occur in a die

Then, P(E_{1}) = \(\frac{1}{6}\) and P(E_{2}) = \(\frac{5}{6}\)

∴ P(man reports that 1 occurs when 1 occur) = P\(\left(\frac{A}{E_{1}}\right)=\frac{3}{5}\)

and P (man reports that 1 occur but 1 does not occur) = P\(\left(\frac{A}{E_{2}}\right)=\frac{2}{5}\)

Thus, by Baye’s theorem, we get P(get actually 1 wfien he reports that 1 occur)

Section – C

Long Answer Type Questions

Question 17.

Show that : sin^{-1} \(\frac{12}{13}\) + cos^{-1} \(\frac{4}{5}\) + tan^{-1} \(\frac{63}{16}\) = π.

Or

Express tan^{-1} \(\frac{\cos x}{1-\sin x}\).where \(\frac{-3 \pi}{2}\) < x < \(\frac{\pi}{2}\) in the simplest form. (3)

Answer:

Let sin^{-1} \(\frac{12}{13}\) = x, cos^{-1} \(\frac{4}{5}\) = y,

tan^{-1} \(\frac{63}{16}\) = z

⇒ sin x = \(\frac{12}{13}\)

cos y = \(\frac{4}{5}\)

tan z = \(\frac{63}{16}\)

Thus, tan(x + y) = – tan z

⇒ tan(x + y) = tan (- z)

⇒ tan(x + y) = tan (π – z)

Therefore, x + y = – z or

x + y = π – z

Since, x, y and z are positive, x + y ≠ – z

Thus, x + y + z = π

sin^{-1}\(\frac{12}{13}\) + cos^{-1}\(\frac{4}{5}\) + tan^{-1}\(\frac{63}{16}\)

Question 18.

Examine the applicability of Mean Value Theorem for the function f(x) = x^{2} – 1, x ∈ [1, 2]

Or

Differentiate the following function w.r.t. x : x^{x} + x^{a} + a^{x} + a^{a}, ∀ a > 0 and x > 0. (3)

Answer:

It is polynomial function which is continuous in interval [1, 2] and is differentiable in interval (1, 2).

Again f(1)= 1^{2} – 1 = 0 and

f(2)= 2^{2} – 1 = 4 – 1 = 3

f'(x) = 2x exists in interval (1, 2)

f(x) = x^{2} – 1 is differentiable in interval (1, 2).

Then, in interval (1,2) a point c exists such that

f'(c) = \(\frac{f(b)-f(a)}{b-a}\)

i.e., f'(c) = \(\frac{f(2)-f(1)}{2-1}=\frac{3-0}{1}\) = 3

f'(c) = 3 ⇒ 2c = 3

⇒ c = \(\frac{3}{2}\) ∈ (1, 2)

Thus c = \(\frac{3}{2}\) ∈ (1, 2) such that

f'(c) = \(\frac{f(2)-f(1)}{2-1}\)

Hence, mean value theorem is verified.

Question 19.

Evaluate : ∫ e^{2x} sin x dx

0r

Find : ∫ \(\sqrt{x^{2}+3 x}\) dx (3)

Answer:

Let

= e^{2x} ∫ sin x dx

= ∫{\(\frac{d}{d x}\)e^{2x} ∫sin x dx} dx

= e^{2x} (- cos x) – ∫ 2e^{2x} (- cos x) dx

= – e^{2x} cos x + 2 ∫ e^{2x} cos x dx

⇒ I = -e^{2x} cos x + 2I_{1} .. .(i)

⇒ I_{1} = e^{2x} ∫cos x dx – ∫{\(\frac{d}{d x}\)e^{2x}∫cosx dx} dx

⇒ I_{1} = e^{2x} sin x – ∫ 2e^{2x} sin x dx + C_{1}

Putting the value of I, in equation (i), we get

I = – e_{2x} cos x + 2[e_{2x} sin x – 2 ∫e_{2x} sin x dx + C_{1}

⇒ I = – e_{2x} cos x + 2e_{2x} sin x – 4 ∫e_{2x} sin x dx + 2C_{1}

⇒ 1 = – e_{2x} cos x + 2e_{2x} sin x – 4I + 2C_{1}

⇒ I + 4I = – e_{2x} cos x + 2e_{2x} sin x + 2C_{1}

⇒ 5I = – e2* cos x + 2e2x sin x + 2C_{1}

⇒ I = –\(\frac{1}{5}\)e cosx + \(\frac{2}{3}\)e sinx + \(\frac{2}{5}\)C_{1}

⇒ I = \(\frac{1}{5}\)e_{2x} cosx + \(\frac{2}{5}\) e_{2x} sinx + C_{1}

(Where C = \(\frac{2}{5}\)C_{1})

I = \(\frac{e^{2 x}}{5}\)(2sin x – cos x) + C

Question 20.

Show that the points A(1, 2, 7), B(2, 6, 3) and C (3, 10, – 1) are collinear. (3)

Or

Let \(\vec{a}\) = î + 4ĵ + 2k̂, \(\vec{b}\) = 3î – 2ĵ + 7k̂ and \(\vec{c}\) = 2î – ĵ + 4k̂. Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\), and = 15. (3)

Answer:

Let O be the origin. Then w.r.t. O Position vector of point A

\(\overrightarrow{O B}\) = î + 2ĵ + 7k̂

Position vector of point B

\(\overrightarrow{O B}\) = 2î + 6ĵ + 3k̂

Position vector of point C

\(\overrightarrow{O C}\) = 3î + 10ĵ – k̂

Now, \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}\)

= 2 î + 6ĵ + 3k̂- (î + 2ĵ + 7k̂)

= (2-1)î + (6-2)ĵ + (3-7)k̂

\(\overrightarrow{A B}\) = î + 4ĵ – 4k̂

Then \(\overrightarrow{B C}=\overrightarrow{O C}-\overrightarrow{O B}\)

= (3î +10 ĵ- k̂) – (2î + 6ĵ + 3k̂)

= (3 – 2)î + (10 – 6)ĵ + (-1 – 3)k̂

or \(\overrightarrow{B C}\) = î + 4ĵ – 4k̂

\(\overrightarrow{A B}\) = î + 4ĵ – 4k̂ = 5C

i.e., \(\overrightarrow{A B}\) and \(\overrightarrow{B C}\) represent the same vector. So, A, B and C are the points of this vector. Thus, A, B and C are collinear.

Section – D

Essay Type Questions

Question 21.

Evaluate :

Or

Evaluate ;

Answer:

Putting tan x = t

⇒ sec^{2} x dx = dt

When x = 0, then t = 0,

When x = π/2, then t = ∞

Question 22.

Solve the differential equation : 2x^{2} \(\frac{dy}{dx}\) – 2xy + y^{2} = 0. (4)

Or

Shpw that the family of curves for which the slope of the tangent at any point (x, y) on it is \(\frac{x^{2}+y^{2}}{2 x y}\), is given by x^{2} – y^{2} = cx. (4)

Answer:

Given differential equation is

2x^{2} \(\frac{dy}{dx}\) – 2xy + y^{2} = 0

⇒ \(\frac{d y}{d x}=\frac{y}{x}-\frac{y^{2}}{2 x^{2}}\) ………(i)

which is a homogeneous differential equation as

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

On putting y = vx ⇒ \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) in Eq. (i), we get

which is the required solution.

Question 23.

Find the mean number of heads in three tosses of a fair coin. (4)

Or

Let X denote the number of hours you study during a randomly selected school day. The probability that X can take the values x, has the following form, where k is some unknown constant.

(i) Find the value of k.

(ii) What is the probability that you study at least two hours ? Exactly two hours ? At most two hours ? (4)

Answer:

Let X denote the success of getting heads.

Therefore, the sample space is S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that X can take the value of 0,1, 2, or 3.

Therefore, the required probability distribution is as follows :

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