Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 6 will help students in understanding the difficulty level of the exam.

## CBSE Sample Papers for Class 12 Maths Term 2 Set 6 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.

A problem of mathematics is given to 3 students whose chances of solving are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\). What is the probability that the problem will be solved?

Answer:

Let A, B, C be the respective events of solving the problem.

Then, P(A) = \(\frac{1}{2}\), P(B) = \(\frac{1}{3}\) and P(C) = \(\frac{1}{4}\)

Clearly A, B, C are independent events and the problem will be solved if at least one student solves it.

Required probability P (A âˆª B âˆª C)

= 1 – \(P(\bar{A}) P(\bar{B}) P(\bar{C})\)

= 1 – (1 – \(\frac{1}{2}\))(1 – \(\frac{1}{3}\))(1 – \(\frac{1}{4}\))

= 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)

Question 2.

If A(1, 2, – 2) and B(3, – 4, 5) are two points, find the direction cosines of OA, OB and AO.

Answer:

The direction cosines of OA are proportional to 1 – 0, 2 – 0, – 2 – 0 or 1, 2, – 2.

Hence, the actual direction cosines of OA are:

\(\frac{1}{\sqrt{1^{2}+2^{2}+(-2)^{2}}}, \frac{2}{\sqrt{1^{2}+2^{2}+(-2)^{2}}}, \frac{-2}{\sqrt{1^{2}+2^{2}+(-2)^{2}}}\)

or

\(\frac{1}{3}, \frac{2}{3}, \frac{-2}{3}\)

Therefore, the direction cosines of AO will be \(\frac{-1}{3}, \frac{-2}{3}, \frac{2}{3}\)

Similarly, the direction cosines of OB will be

\(\frac{3}{\sqrt{3^{2}+(-4)^{2}+5^{2}}}, \frac{-4}{\sqrt{3^{2}+(-4)^{2}+5^{2}}}, \frac{5}{\sqrt{3^{2}+(-4)^{2}+5^{2}}}\)

or

\(\frac{3}{5 \sqrt{2}}, \frac{-4}{5 \sqrt{2}}, \frac{1}{\sqrt{2}}\)

Question 3.

Find the solution of the differential equation:

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

Answer:

we have,

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

Integrating both sides

Question 4.

Find âˆ«e^{x}\(\left(\frac{\sin 4 x-4}{1-\cos 4 x}\right)\)dx

OR

Evaluate âˆ«\(\frac{d x}{1+\tan x}\)

Answer:

Question 5.

If \(\vec{a}, \vec{b}\) and \(\vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), then find the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\).

Answer:

Question 6.

If P(not A) = 0.7, P(B) = 0.7 and P(B/A) = 0.5, then find P(A/B) and P(A âˆª B).

Answer:

We have, P(not A) = 0.7 or P(\(\overline{\mathrm{A}}\)) = 0.7

1 – P(A) = 0.7

â‡’ P(A) = 0.3

Section – B

Question 7.

Find the cartesian equation of the plane passing through points A(0, 0, 0) and B(3, -1, 2), parallel to the \(\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\)

OR

Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

\(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\) and \(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\)

Answer:

Equation of plane passing through the point A(0,0,0) is

a(x – 0) + b(y- 0) + c(z – 0) = 0

â‡’ ax + by + cz = 0 …………(1)

Since, the plane (1) passing through the point B(3, -1,2)

Put x = 3, y = -1, z = 2 in equation (1), We get

3a – b + 2c = 0 ………….(2)

Also, the plane (1) is parallel to the line

\(\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\)

If the plane is parallel to the line, then normal to the plane is perpendicular to the line.

âˆ´ a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

âˆ´ a(1) + b(-4) + c(7) = 0

â‡’ a – 4b + 7c = 0

Now, multiplying equation (3) by 3 and subtracting it from equation (2), we get.

Putting b = \(\frac{19}{11}\) c in equation (2), we get

â‡’ 3a – \(\frac{19 c}{11}\) + 2c = 0

â‡’ 3a + \(\frac{-19 c+22 c}{11}\) = 0

â‡’ 3a + \(\frac{3 c}{11}\) = 0

â‡’ 3a = \(\frac{-3 c}{11}\)

â‡’ a = \(\frac{- c}{11}\)

Now, puthng a = \(\frac{-c}{11}\) and b = \(\frac{19c}{11}\) in equatIon (1), we get the required equation of plane as

\(\frac{-c}{11}\)x + \(\frac{19c}{11}\)y + cz = 0

â‡’ \(\frac{-x}{11}+\frac{19 y}{11}\) + z = 0

â‡’ -x + 19y + 11z = 0

â‡’ x – 19y – 11z = 0

OR

Let the required line be parallel to the vector \(\vec{b}\) given by \(\vec{b}\) = b_{1}iÌ‚ + b_{2}jÌ‚ + b_{3}kÌ‚. The position vector of the point(1, 2, -4) is a = iÌ‚ + 2jÌ‚ – 4kÌ‚.

The equation of the line passing through (1, 2, -4) and parallel to \(\vec{b}\) is \(\vec{r}=\vec{a}+\lambda \vec{b}\) =

â‡’ r = (iÌ‚ + 2jÌ‚ – 4kÌ‚) + (b_{1}iÌ‚ + b_{2}jÌ‚ + b_{3}kÌ‚) ………..(1)

The equations of the lines are

\(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\) ……….(2)

\(\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}\) ………..(3)

Line (1) and line (2) are perpendicular to each other. Therefore,

3b_{1} – 16b_{2} + 7b_{3} = O

Also, line (1) and line (3) are perpendicular to each other. Therefore,

3b_{1} + 8b_{2} – 5b_{3} = 0

From equation (4) and (5), we have

â‡’ Direction ratios of \(\vec{b}\) are 2, 3 and 6 or \(\vec{b}\) = 2iÌ‚ +3jÌ‚ + 6kÌ‚,

Substituting \(\vec{b}\) = 2iÌ‚ +3jÌ‚ + 6kÌ‚ in equation (1), we have \(\vec{r}\)(iÌ‚ + 2jÌ‚ – 4kÌ‚) + k(2iÌ‚ + 3jÌ‚ + 6kÌ‚)

Hence, this is the equation of the required line.

Question 8.

Find the area of parallelogram, whose diagonals are \(\vec{a}\) = 3iÌ‚ + jÌ‚ – 2kÌ‚ and \(\vec{b}\) = iÌ‚ – 3jÌ‚ + 4kÌ‚

Answer:

Let ABCD be a parallelogram and its diagonals are

Hence, Area of required parallelogram = \(\frac{1}{2}|\overrightarrow{A C} \times \overrightarrow{B C}|\)

= \(\frac{1}{2}\) Ã— 10âˆš3

= 5âˆš3 sq. units

Question 9.

Solve the differential equation:

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

OR

Find a particular solution of the differential equation (x + 1)\(\frac{d y}{d x}\) = 2e^{-y} – 1, given y = 0, when x = 0.

Answer:

We have, y + x\(\frac{d y}{d x}\) = x – y\(\frac{d y}{d x}\)

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

â‡’ \(\frac{d y}{d x}=\frac{x-y}{x+y}\) ………(i)

Which is a homogeneous differential equation.

Putting y = Vx â‡’ \(\frac{d y}{d x}\) = V + x\(\frac{d V}{d x}\)

Putting t = 1 – 2V – V^{2}

â‡’ dt = (- 2 – 2V) dV

â‡’ –\(\frac{1}{2}\)dt = (1 + V)dV

Now, equation (ii) becomes,

= – \(\frac{1}{2t}\)dt = \(\frac{dx}{x}\)

On integrating above equation, we get

– \(\frac{1}{2}\)âˆ«\(\frac{1}{t}\)dt = âˆ«\(\frac{1}{x}\)dx

– \(\frac{1}{2}\) log|t| = log|x| + log C

– \(\frac{1}{2}\) log |1 – 2V – V^{2}| = log {C|x|}

log {C|x|} + log|1 – 2V – V^{2}| = 0

C|x|\(\left(1-\frac{2 y}{x}-\frac{y^{2}}{x^{2}}\right)^{1 / 2}\) = 0

C(x^{2} – 2xy – y^{2})^{1/2} = 0.

OR

We have,

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

â‡’ -log|2 – e^{y}| = log|x + 1| + C [âˆµâˆ«\(\frac{f^{\prime}(x)}{f(x)}\)dx = log[f(x)] + C]

It is given that y = 0 when x = 0

â‡’ – log 1 = log 1 + C [âˆµ log1 = 0]

â‡’ C = 0

âˆ´ -log|2 – e^{y}| = log|x+1|

â‡’ log|\(\frac{1}{2-e^{y}}\)| = log|x+1|

â‡’ \(\frac{1}{2-e^{y}}\) = x + 1

â‡’ 2 – e^{y} = \(\frac{1}{x+1}\)

â‡’ e^{y} = 2 – \(\frac{1}{x+1}\)

â‡’ e^{y} = \(\frac{2 x+1}{x+1}\)

Taking log both sides we get,

y = log_{e} \(\left|\frac{2 x+1}{x+1}\right|\)

Question 10.

Evaluate: âˆ«\(\frac{d x}{2 \sin ^{2} x+5 \cos ^{2} x}\)

Answer:

Let I = âˆ«\(\frac{d x}{2 \sin ^{2} x+5 \cos ^{2} x}\)

Dividing numerator and denominator by cos^{2}x, we get

Section – C

Question 11.

Evaluate: âˆ«^{4}_{2}\(\frac{x}{x^{2}+1}\)dx

Answer:

Consider,

I = âˆ«^{4}_{2}\(\frac{x}{x^{2}+1}\)dx

and let x^{2} + 1^{2} = t

â‡’ 2x dx = dt

â‡’ x dx = \(\frac{1}{2}\)dt

Where x =2,

t = (2)^{2} + 1 = 5

and when x = 4, t = (4)^{2} + 1 = 17

Question 12.

Using integration, find the area bounded by the curve x^{2} = 4y and the line x = 4y – 2.

OR

Find the area of the region:

{(x, y): y^{2} â‰¤ 4x, 4x^{2} + 4y^{2} â‰¤ 9}

Answer:

Given, x^{2} = 4y is a parabola, symmetric to y-axis and x = 4y – 2 is a line.

Put x =4y – 2 in x^{2} = 4y

(4y – 2)^{2} = 4y

â‡’ 16y^{2} + 4 – 16y – 4y = 0

â‡’ 16y^{2} – 20y + 4 = 0

â‡’ 4y^{2} – 5y + 1 = 0

â‡’ 4y(y – 1) – 1(y – 1) = 0

â‡’ (y – 1)(4y – 1) = 0

âˆ´ y = 1, \(\frac{1}{4}\)

When y = 1, x = 2

When y = \(\frac{1}{4}\) x = -1

OR

The area bounded by the curves,

{(x, y): y^{2} â‰¤ 4x, 4x^{2} + 4y^{2} â‰¤ 9} is represented in figure

The points of intersection of both the curves are

(\(\frac{1}{2}, \sqrt{2}\)) and (\(\frac{1}{2},-\sqrt{2}\))

The required area is given by OABCO.

It can be observed that area OABCO is symmetrical about x-axis

Area OABCO = 2 Ã— Area OBCO

Area OBCO = Area OMC + Area MBC

Question 13.

Find the vector and cartesian equation of the plane containing the two lines

\(\vec{r}\) = 2Ã® + Äµ – 3kÌ‚ + Î»(Ã® + 2Äµ + 5kÌ‚)

and \(\vec{r}\) = 3Ã® + 3Äµ – 7kÌ‚ + Î¼(3Ã® – 2Äµ + 5kÌ‚)

Answer:

Here \(\vec{a}\) = 2Ã® + Äµ – 3kÌ‚

\(\overrightarrow{a^{\prime}}\) = 3Ã® + 3Äµ – 7kÌ‚

\(\vec{b}\) = Ã® + 2Äµ + 5kÌ‚

\(\overrightarrow{b^{\prime}}\) = 3Ã® – 2Äµ + 5kÌ‚

Vector equation of plane is

[\(\vec{r}\) – (2Ã® + Äµ – 3kÌ‚)]. [(Ã® + 2Äµ + 5kÌ‚) Ã— (3Ã® – 2Äµ + 5kÌ‚)] = 0

\(\vec{b} \times \overrightarrow{b^{\prime}}\) = \(\left|\begin{array}{rrr}

\hat{i} & \hat{j} & \hat{k} \\

1 & 2 & 5 \\

3 & -2 & 5

\end{array}\right|\) = 20Ã® + 10Äµ – 8kÌ‚

âˆ´ [\(\vec{r}\) – (2Ã® + Äµ – 3kÌ‚)].(20Ã® + 10Äµ – 8kÌ‚) = 0

â‡’ \(\vec{r}\).(20Ã® + 10Äµ – 8kÌ‚) – (40 + 10 + 24) = 0

â‡’ \(\vec{r}\).(20Ã® + 10Äµ – 8kÌ‚) = 74

or \(\vec{r}\).(10Ã® + 5Äµ – 4kÌ‚) = 37 ………..(i)

is the required vector equation.

Let, \(\vec{r}\) = xÃ® + yÄµ + zkÌ‚

Using this in equation (i), we get

10x + 5y – 4z = 37

as the equation of plane in Cartesian form.

Question 14.

In a family there are four children. All of them have to work in their family business to earn their livelihood at the age of 18.

Based on the above information, answer the following questions:

(i) Find the probability that all children are girls, if it is given that elder child is a boy.

(ii) Find the probability that two middle children are boys, if it is given that eldest child is a girl.

Answer:

Let B and G denotes the boy and girl respectively. If a family has 4 children then each of four children can either boy or girl.

Sample space is given by;

S = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG}

âˆ´ n(S) = 16

(i) Let E = All children are girls

âˆ´ E = {GGGG}

i.e, n(E) = 1

F = Elder child is boy.

âˆ´ F = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG, BGGB, BGGG}

i.e., n( F) = 8

Now, n(E âˆ© F) = 0

âˆ´ P(E âˆ© F) = \(\frac{n(\mathrm{E} \cap \mathrm{F})}{n(\mathrm{~S})}=\frac{0}{16}\) = 0

âˆ´ P(E/F) = \(\frac{P(E \cap F)}{P(F)}\) = 0

Thus, there is no probability that all children are girls, if it is given that elder child is a boy.

(ii) Let E = Two middle children are boys

âˆ´ E = {BBBB, BBBG, GBBB, GBBG} i.e., n(E) = 4

F = Elder child is girl.

âˆ´ F = {GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG}

i.e., n( F) = 8

Now, n( E âˆ© F) = 2

âˆ´P(E/F) = \(\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{2}{8}\)

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

Thus, there is probability \(\frac{1}{4}\) that two children are boys, if it is given that eldest child is a girl.