Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.

## CBSE Sample Papers for Class 12 Maths Term 2 Set 4 with Solutions

Maximum Marks : 40

Time : 2 Hours

General Instructions:

- This question paper contains three sections – A, B and C. Each part is compulsory.
- Section – A has 6 short answer type (SA1) questions of 2 marks each.
- Section – B has 4 short answer type (SA2) questions of 3 marks each.
- Section – C has 4 long answer type questions (LA) of 4 marks each.
- There is an internal choice in some of the questions.
- Q14 is a case-based problem having 2 sub parts of 2 marks each.

Section – A

Question 1.

Determine order and degree of given differential equation:

\(\left(\frac{d \mathrm{~S}}{d t}\right)^{4}\) + 3S\(\frac{d^{2} S}{d t^{2}}\) = 0

Answer:

The given differential equation

\(\left(\frac{d \mathrm{~S}}{d t}\right)^{4}\) + 3S\(\frac{d^{2} S}{d t^{2}}\) = 0

The highest order derivative presents in the given differential equation is \(\frac{d^{2} S}{d x^{2}}\) Therefore, its order is two. It is a polynomial equation in \(\frac{d^{2} S}{d t^{2}}\) and \(\frac{d \mathrm{~S}}{d t}\). The power raised to \(\frac{d^{2} S}{d t^{2}}\) is 1. Hence, its degree is one.

Question 2.

Find the coordinates of the foot of the perpendicular from 0(0, 0, 0) to the plane 3x – 4y + 5z – 10 = 0.

Answer:

Let the coordinates of foot of perpendicular be V(x_{1}, y_{1}, z_{1}) from point 0(0, 0, 0). Equation of plane in normal from is given by

\(\frac{3}{\sqrt{50}}\)x – \(\frac{4}{\sqrt{50}}\)y + \(\frac{5}{\sqrt{50}}\)z = âˆš2

[Dividing through by \(\sqrt{3^{2}+(-4)^{2}+5^{2}}=\sqrt{50}\)]

The direction cosines of the normal drawn from the origin to the given plane are

l = \(\frac{3}{5 \sqrt{2}}\)

m = \(\frac{-4}{5 \sqrt{2}}\)

n = \(\frac{5}{5 \sqrt{2}}\)

as OP = âˆš2

So coordinate of foot of perpendicular P(\(\frac{3}{5}, \frac{-4}{5}\), 1)

Question 3.

Find the angle between unit vectors \(\widehat{\mathrm{a}}\) and \(\widehat{\mathrm{b}}\) so that \(\sqrt{3a}\) – b is also a unit vector.

Answer:

a and b are unit vectors and \(\sqrt{3a}\) -b is also unit vector

To find angle between a and b.

Suppose angle between a and b is Î¸.

\(\vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}|\) cos Î¸ (Dot product of two vectors)

\(\vec{a} \cdot \vec{b}\) = cos Î¸

As \(\vec{a}\) and \(\vec{b}\) are unit vector so, \(|\vec{a}|=|\vec{b}|\) = 1

\(\sqrt{3a}\) – b is also unit vector i.e. \(|\sqrt{3} \vec{a}-\vec{b}|\) = 1

Squaring both sides, we get

(\( \sqrt{3} \vec{a}-\vec{b}\))^{2} = 1

â‡’ (âˆš3)2 \(|\vec{a}|^{2}+|\vec{b}|^{2}-2 \sqrt{3} \vec{a} \cdot \vec{b}\) = 1 [since, \(\vec{a} \cdot \vec{b}\) = cos Î¸]

â‡’ 3.1 + 1 – 2.âˆš3.cosÎ¸ = 1

â‡’ 4 – 2âˆš3.cosÎ¸ = 1

â‡’ 2âˆš3. cos Î¸ = 3

â‡’ cos Î¸ = \(\frac{\sqrt{3}}{2}\)

â‡’ Î¸ = \(\frac{\pi}{6}\)

Therefore, the angle between the two unit vectors is \(\frac{\pi}{6}\)

Question 4.

Ten cards numbered 1 to 10 are placed in box, mixed up throughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?

Answer:

Let A be the event the number on the card drawn is even and B be the event the number on the card drawn is greater than 3.

Now, the sample space of the experiment is

S = {1, 2, 3, 4, 5, 6, 7, 8, 9,10}

Then, A = {2,4, 6, 8,10}, B = {4, 5, 6, 7, 8, 9,10}

and A âˆ© B = {4, 6, 8,10}

P(A) = \(\frac{5}{10}\) P(B) = \(\frac{7}{10}\) = and P(A âˆ© B) = \(\frac{4}{10}\)

Then, P(A/B) = \(\frac{P(A \cap B)}{P(B)}=\frac{4 / 10}{7 / 10}=\frac{4}{7}\)

Question 5.

Find âˆ«\(\frac{\sin x-\cos x}{\sqrt{1+\sin 2 x}}\) dx, 0 < x < Ï€/2.

OR

Find âˆ«(log x)^{2}dx

Answer:

According to given

Let I = âˆ«\(\frac{\sin x-\cos x}{\sqrt{1+\sin 2 x}}\) dx, 0 < x < Ï€/2.

Let sin x + cos x = t

â‡’ (cos x – sin x) dx = dt

âˆ´ I = âˆ«\(\frac{-1}{t}\)dt

= -log t + C

= log\(\left(\frac{1}{t}\right)\) + C

â‡’ I = log\(\left(\frac{1}{\sin x+\cos x}\right)\) + C

OR

â‡’ I = x(logx)^{2} – âˆ«\(\frac{2 x \log }{x}\) dx

â‡’ I= x.(logx)^{2} – I_{1} + C_{1} ……..(1)

â‡’ I_{1} = âˆ«2.log x dx

â‡’ I_{1} =2x.log – 2âˆ«\(\frac{x}{x}\) dx

â‡’ I_{1} = 2x.log – 2x + C_{2}

â‡’ I = x.(logx)^{2} – 2x.log x + 2x + C_{1} – C_{2}

â‡’ I = x.(log x)^{2} – 2x. Iogx + 2x + C (where C = C_{1} – C_{2})

Question 6.

A bag contains 8 black and 5 blue balls three balls are drawn at random without replacements. What is the probability that all drawn balls are black colours?

Answer:

1st black ball drawn at random the probability = \(\frac{8}{13}\)

2nd black ball drawn at random the probability = \(\frac{7}{12}\)

3rd black ball drawn at random the probability = \(\frac{6}{11}\)

P = \(\frac{8}{13} \times \frac{7}{12} \times \frac{6}{11}\)

So, the total probability = \(\frac{28}{13 \times 11}\) = 0.195

Section – B

Question 7.

Find the scalar components of a unit vector which is perpendicular to each of the vectors Ã® +2Äµ – kÌ‚ and 3Ã® – Äµ – 2kÌ‚.

Answer:

Let, \(\vec{a}\) = Ã® +2Äµ – kÌ‚

and \(\vec{b}\) = 3Ã® – Äµ – 2kÌ‚

We know that a unit vector perpendicular to each of vector \(\vec{a}\) and \(\vec{b}\) is given as \(\left[\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\right]\)

âˆ´ Sclar components are \(\frac{3}{\sqrt{83}}, \frac{-5}{\sqrt{83}}, \frac{-7}{\sqrt{83}}\)

Question 8.

Find the shortest distance between the lines and l2 whose vector equations are \(\vec{r}\) = Ã® + jÌ‚ + Î»(2iÌ‚ – jÌ‚ + kÌ‚) and \(\vec{r}\) = 2iÌ‚ + jÌ‚ – kÌ‚ + Î¼(3iÌ‚ – 5jÌ‚ + 2kÌ‚).

OR

Find cartesian equation of a line where the direction ratios of the parallel vector to it are (2, -1, 3) and passes through (5, -2, 4).

Answer:

We have given two vector equations

\(\vec{r}\) = Ã® + jÌ‚ + Î»(2iÌ‚ – jÌ‚ + kÌ‚) …….(1)

and \(\vec{r}\) = 2iÌ‚ + jÌ‚ – kÌ‚ + Î¼(3iÌ‚ – 5jÌ‚ + 2kÌ‚)

Comparing (1) and (2), with \(\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}\) and \(\vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}\) respectively,

We get, \(\overrightarrow{a_{1}}\) = Ã® + jÌ‚ , \(\overrightarrow{b_{1}}\) = 2Ã® – jÌ‚ + kÌ‚

\(\overrightarrow{a_{2}}\) = 2Ã® + jÌ‚ – kÌ‚ and \(\overrightarrow{b_{1}}\) = 3Ã® – 5jÌ‚ + 2kÌ‚

Therefore, \(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\) = Ã® – kÌ‚

and \(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\) = (2iÌ‚ – jÌ‚ + kÌ‚) Ã— (3iÌ‚ – 5jÌ‚ + 2kÌ‚)

OR

Given, direction ratios of the parallel vector = (2, -1, 3)

âˆ´ The direction ratios of required line are (2kÌ‚ – kÌ‚, 3kÌ‚)

Also, the given point is (x_{1}, y_{1}, z_{1}) = (5, -2, 4)

Cartesian equation of line passes through (x_{1}, y_{1}, z_{1}) and with direction ratios a, b, c is given by

\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)

âˆ´ The equation of required lines is,

\(\frac{x-5}{2 k}=\frac{y+2}{-k}=\frac{z-4}{3 k}\)

i.e, \(\frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3}\)

Question 9.

Find the general solution of the following differential equation :

xdy – (y + 2x^{2})dx = 0

OR

Find the particular solution of the differential equation e^{x} tan y dx + (2 – e^{x}) sec^{2} y dy = 0, given that y = Ï€/4 when x = 0.

Answer:

xdy – (y + 2x^{2})dx = 0

â‡’ \(\frac{d y}{d x}=\frac{y+2 x^{2}}{x}\)

â‡’ \(\frac{d y}{d x}-\frac{y}{x}\) = 2x

Comparing \(\frac{d y}{d x}\) +Py = Q

P = \(\frac{-1}{x}\), Q = 2x

I.F = e^{âˆ«Pdx} = e^{-âˆ«1/xdx} = e^{-logx}

= x^{-1} = \(\frac{1}{x}\)

Solution is given by

y(I.F) = âˆ«Q(I.F) dx

â‡’ \(\frac{y}{x}\) = âˆ«(2x Ã— \(\frac{1}{x}\))dx

â‡’ \(\frac{y}{x}\) = 2x +C

â‡’ y = 2x^{2} +Cx

OR

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

â‡’ e^{x} tan y dx – (e^{x} – 2)sec^{2}y dy

âˆ´ âˆ«\(\frac{e^{x}}{e^{x}-2}\)dx = âˆ«\(\frac{\sec ^{2} y}{\tan y}\)dx

â‡’ log |e^{x} – 2| = log |tan y| + log C

â‡’ log e^{x} – 2|= log (C + tan y)

â‡’ e^{x} – 2 = C tany

Given, x = 0, y = \(\frac{\pi}{4}\)

âˆ´ e^{0} – 2 = C tan\(\left(\frac{\pi}{4}\right)\)

â‡’ 1 – 2 = C tan\(\left(\frac{\pi}{4}\right)\)

â‡’ -1 = C Ã— 1

â‡’ C = -1

âˆ´ e^{x} – 2 = -tany

or

e^{x} – 2 + tany = 0.

Question 10.

Find: âˆ«\(\frac{x^{2}+1}{\left(x^{2}+2\right)\left(x^{2}+3\right)}\)dx

Answer:

I = âˆ«\(\frac{x^{2}+1}{\left(x^{2}+2\right)\left(x^{2}+3\right)}\)dx

Let x^{2} = y

Section – C

Question 11.

Find the coordinates of foot of perpendicular drawn from the point (0, 2, 3) on the line = \(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\). Also, find the length of perpendicular.

Answer:

Given equation of line is

\(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\)

Any random point T on the given line is calculated as follows

\(\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\) = Î»(say)

âˆ´ coordinates of point T are (5Î» -3, 2Î» + 1, 3Î» – 4).

Let point is P (0, 2, 3)

Now, Direction ratios of line PT = (5Î», – 3 – 0, 2Î» + 1 – 2, 3Î» – 4 – 3)

= (5Î» – 3, 2Î» – 1, 3Î» – 7)

âˆµ PT is perpendicular to the given line.

â‡’ a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

a_{1} = 5Î»- 3, b_{1} = 2Î» – 1, c_{1} = 3Î» – 7

a_{2} = 5, b_{2} = 2, c_{2} = 3

â‡’ 5(5Î» – 3) + 2 (2Î» – 1) + 3 (3Î» – 7) = 0

â‡’ 25Î» – 15 + 4Î» – 2 + 9Î» – 21 =0

â‡’ 38Î» – 38 = 0

â‡’ Î» = 1

âˆ´ The foot of perpendicular T = (5L- 3,2X +1,3X – 4)

= (2, 3, -1) (Put Î» = 1)

Also, length of perpendicular, PT = Distance between point P and T

PT = \(\sqrt{(0-2)^{2}+(2-3)^{2}+(3+1)^{2}}\)

= \(\sqrt{4+1+16}\)

= \(\sqrt{21}\) units

Question 12.

Find the area under the curve y = 2âˆšx and between the lines x = 0 and x = 1.

OR

Find the area of parabola y^{2} = 4ax bounded by its latus rectum.

Answer:

Given y = 2

OR

The graph of required parabola and its latus rectum is given alongside. The vertex of parabola y^{2}= 4ax is at (O, O). The equation of latus rectum LSL’ is x = a

Also, parabola is symmetric about the X-axis.

âˆ´ The required area of the region OLL’O

= 2(area of region OLSO)

= 2âˆ«^{a}_{0}ydx = 2âˆ«^{a}_{0}\(\sqrt{4 a x}\) dx

= 2 Ã— 2âˆšaâˆ«^{a}_{0} âˆšx dx

= 4âˆša\(\frac{2}{3}\)[x^{3/2}]^{a}_{0} = \(\frac{8}{3}\)âˆša(a^{3/2})

= \(\frac{8}{3}\) a^{2} sq. units

Question 13.

Evaluate âˆ«\(\frac{x^{9}}{\left(4 x^{2}+1\right)^{6}}\)dx

Answer:

Let I = âˆ«\(\frac{x^{9}}{\left(4 x^{2}+1\right)^{6}}\)dx

Question 14.

A doctor is going to visit a patient. From the past experience, it is known that the probabilities that he will come by cab, metro, bike or by other means of transport are respectively 0.3, 0.2, 0.1 and 0.4. The probabilities that he will be late are 0.25, 0.3, 0.35 and 0.1, if he comes by cab, metro, bike and other means of transport respectively.

Based on the above information, answer the following questions :

(i) When the doctor arrives late, what is the probability that he comes by metro?

(ii) When the doctor arrives late, what is the probability that he comes by other means of transport?

Answer:

Let E be the event that the doctor visit the patient late and let A_{1}, A_{2}, A_{3}, A_{4} be the events that the doctor comes by cab, metro, bike and other means of transport respectively.

P(A_{1}) = 0.3, P(A_{2}) = 0.2, P(A_{3}) = 0.1, P(A_{4}) = 0.4

P(E/A_{1}) = Probability that the doctor arriving late when he comes by cab = 0.25

Similarly, P(E/A_{2}) = 0.3,

P(E/A_{3}) = 0.35

And P(E/A_{4}) = 0.1

(i) P(A_{2}/E) = Probability that the doctor arriving late when he comes by metro

P(A_{2})P(E/A_{2})

= \(\frac{\mathrm{P}\left(\mathrm{A}_{2}\right) \mathrm{P}\left(\mathrm{E} / \mathrm{A}_{2}\right)}{\Sigma \mathrm{P}\left(\mathrm{A}_{i}\right) \mathrm{P}\left(\mathrm{E} / \mathrm{A}_{i}\right)}\)

= \(\frac{(0.2)(0.3)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)

= \(\frac{0.06}{0.075+0.06+0.035+0.04}\)

= \(\frac{0.06}{0.21}\)

= \(\frac{2}{7}\)

Thus, the probability that the doctor arriving late when he comes by metro is \(\frac{2}{7}\).

(ii) P(A_{4}/E) = Probability that the doctor arriving late when he comes by any means of transport

Thus, the probability that the doctor arriving late when he comes by any means of transport is \(\frac{4}{21}\).