Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 5 will help students in understanding the difficulty level of the exam.

## CBSE Sample Papers for Class 12 Maths Term 2 Set 5 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.

Evaluate:

OR

Evaluate: âˆ« sin^{3} x cos^{3} x dx

Answer:

OR

Let, I = âˆ«sin^{3}x.cos^{3}x dx

= âˆ«sin^{3} x.cos^{2} x.cosx dx

= âˆ«sin^{3} x.(1 – sin^{2} x).cosx dx

Let, sin x = t,

â‡’ cos x dx = df

âˆ´ I = âˆ«t^{3}(1 – t^{2})dt

= âˆ«(t^{3} – t^{5})dt

= \(\frac{t^{4}}{4}-\frac{t^{6}}{6}\) + C

= \(\frac{1}{4}\) sin^{4}x – \(\frac{1}{6}\)sin^{6}x + C

Question 2.

Determine the sum of order and degree of the given differential equation.

(y”‘)^{2} + (y”)^{3} + (y’)^{4}+ y^{5} = 0.

Answer:

The given differential equation is

(y”‘)^{2} + (y”)^{3} + (y’)^{4}+ y^{5} = 0.

The highest order derivative present in the differential equation is y”‘ Therefore, its order is three.

The given differential equation is a polynomial equation in y”â€™, y” and y’.

The highest power raised to y'” is 2.

Hence, its degree is 2.

Now, we know order is 3 and degree is 2.

So, Sum of order and degree of given differential equation is 3 + 2 = 5.

Question 3.

Write a vector of magnitude 15 units in the direction of vector iÌ‚ – 2jÌ‚ + 2kÌ‚.

Answer:

Given vectors is iÌ‚ – 2jÌ‚ + 2kÌ‚

Unit vector in the direction of iÌ‚ – 2jÌ‚ + 2kÌ‚ is given by

So, Vector of magnitude 15 units in the direction of iÌ‚ – 2jÌ‚ + 2kÌ‚ will be:

15aÌ‚ = 15\(\left(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\right)\)

= 5(iÌ‚ – 2jÌ‚ + 2kÌ‚)

= 5iÌ‚ – 10jÌ‚ + 10kÌ‚

Question 4.

If the line has the direction ratio – 18, 12, – 4, then what are its direction consines?

Answer:

The direction ratios of the lines are -18,12, – 4

Direction cosines of the lines are

Hence, direction cosines of line are \(\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\).

Question 5.

Mother, father and son line up at random for a family photos. If A and B are two events given by A = Son on one end, B = Father in the middle, find P(B/A).

Answer:

5. If mother (M), father (F), and son (S) lineup for the family picture, then the sample space will be

S = (MFS, MSF, FMS, FSM, SMF, SFM)

A = (MFS, FMS, SMF, SFM),

âˆ´ B = (MFS, SFM)

And A âˆ© B = (MFS, SFM)

âˆ´ P(A âˆ© B) = \(\frac{2}{6}=\frac{1}{3}\),

P(B) = \(\frac{2}{6}=\frac{1}{3}\)

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

âˆ´ P(B/A) = \(\frac{P(A \cap B)}{P(A)}=\frac{1 / 3}{2 / 3}=\frac{1}{2}\)

â‡’ P(B/A) = \(\frac{1}{2}\)

Question 6.

A coin is tossed 5 times. Find the probability of getting (i) at least 4 heads, and (ii) at most 4 heads.

Answer:

Total number of probability of tossing a coin 5 times is 32

(i) Probability of getting atleast 4 heads.

(ii) Probability of getting at most 4 heads.

Section – B

Question 7.

Evaluate: âˆ«x sin x dx.

Answer:

Let, Using integration by parts,

âˆ«f(x)g(x)dx = f(x)âˆ«g(x)dx -âˆ«f'(x)[âˆ«g(x)dx]dx

So, we take

f(x) = x and g(x) = sin x

So, I = xâˆ«sin x dx – âˆ«\(\frac{d(x)}{d x}\) (âˆ«sin x dx) dx

â‡’ I = x (- cos x) – âˆ«1.(- cosx) dx

â‡’ I = – x cos x + âˆ«cos x dx

â‡’ I = – x cos x + sin x + C

Question 8.

Solve the differential equation

\(\frac{d y}{d x}\) = 1 + x + y^{2} + xy^{2}, if y = 0 when x = 0

OR

Solve the differential equation:

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

Answer:

Given differential equation is,

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

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

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

â‡’ \(\frac{1}{1+y^{2}}/latex]dy = (1 + x)dx

Integrating on both sides

âˆ«[latex]\frac{1}{1+y^{2}}/latex] = âˆ«(1 + x)dx

â‡’ tan^{-1}y = x + [latex]\frac{x^{2}}{2}\) + C

â‡’ y = tan \(\left(x+\frac{x^{2}}{2}+C\right)\)

It is given that, y = 0, when x = 0

âˆ´ 0 = tan (0 + C)

â‡’ C = 0

âˆ´ y = tan \(\left(x+\frac{x^{2}}{2}\right)\)

OR

Given differential equation is,

Question 9.

If \(\vec{a}\) = \(2 \vec{i}+4 \hat{j}-5 \hat{k}\) and \(\vec{b}\) = iÌ‚ + 2jÌ‚ + 3kÌ‚, find the unit vector in the direction \(\vec{a}\) + \(\vec{a}\)

Answer:

We have,

\(\vec{a}+\vec{b}\) = (2iÌ‚ + 4jÌ‚ – 5kÌ‚) + (iÌ‚ + 2jÌ‚ + 3kÌ‚)

= 3iÌ‚ + 6jÌ‚ – 2kÌ‚

â‡’ |\(\vec{a}+\vec{b}\)| = \(\sqrt{(3)^{2}+(6)^{2}+(-2)^{2}}\)

= \(\sqrt{9+36+4}\) = âˆš49

= 7

Hence, the unit vector in the direction of \(\vec{a}+\vec{b}\)

Question 10.

Find the foot of perpendicular from the point (1, 6, 3) on the line;

x = \(\frac{y-1}{2}=\frac{z-2}{2}\)

OR

If vector \(\vec{a}\) = 2iÌ‚ + 2jÌ‚ + 3kÌ‚, \(\vec{b}\) = – iÌ‚ + 2jÌ‚ + kÌ‚ and \(\vec{c}\) = 3iÌ‚ + jÌ‚ are such that \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c}\), then find the value of Î».

Answer:

Given line is,

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

It can be written as

\(\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) = Î»(say)

So, x = Î», y = 2Î» + 1 and z = 3Î» + 2.

âˆ´ Coordinates of any general point which lines on the given line are (Î», 2Î» + 1, 3Î» + 2).

âˆ´ Coordinates of foot of perpendicular are (Î», 2Î» + 1,3Î» + 2).

Now, direction ratios of line joining (1,6,3) and foot of perpendicular (Î», 2Î» + 1, 3Î» + 2) will be,

Î» – 1, 2Î» + 1 – 6, 3Î» + 2 – 3 = Î» – 1, 2Î» – 5, 3Î» – 1.

Also, direction ratios of given line are 1, 2, 3.

Since, line joining (1, 6, 3) and foot of perpendicular is perpendicular to given line.

âˆ´ 1(Î» – 1) + 2 (2Î» – 5) + 3 (3Î» – 1) = 0

â‡’ Î» – 1 + 4Î» – 10 + 9Î» – 3 = 0

â‡’ 14Î», -14 = 0

â‡’ Î» = 1

âˆ´ Co-ordinates of foot of perpendicular

= (Î», 2Î» + 1, 3Î» + 2)

= (1, 2(1) + 1, 3(1)+ 2)

= (1, 3, 5)

OR

Given that \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c}\).

\((\vec{a}+\lambda \vec{b}) \cdot \vec{c}\) = o …… (1)

Now, given that

\(\vec{a}\) = 2iÌ‚ + 2jÌ‚ + 3kÌ‚

\(\vec{b}\) = – iÌ‚ + 2jÌ‚ + kÌ‚

and \(\vec{c}\) = 3iÌ‚ + jÌ‚

âˆ´ \(\vec{a}\) + Î» \(\vec{a}\) = (2 z + 2 y+3 fc) + A,(~ z + 2 j + Î»)

\(\vec{a}\) + Î»\(\vec{b}\) = i(2 – Î») + j (2 + 2Î») + k(3 + Î»)

From equation (1), we get

[iÌ‚(2 – Î») + jÌ‚(2 + 2Î») + kÌ‚(3 + Î»)].[3 iÌ‚ + jÌ‚] = 0

â‡’ 3(2 – Î») + 1(2 + 2Î») =0

â‡’ 6 – 3Î» + 2 + 2Î» = 0

â‡’ 8 – Î» = 0

Î» = 8

Section – C

Question 11.

Evaluate:

Answer:

Given,

Question 12.

Find the area of region bounded by the curve y^{2} = 2y – x and the Y – axis.

OR

Find the area of region bounded by the curves:

x^{2} = y and y = |x|.

Answer:

Given equation of curve is,

y^{2} = 2y – x

â‡’ y^{2} – 2y = – x

â‡’ y^{2} – 2y + 1 = – x + 1

â‡’ (y – 1)^{2} = – (x – 1)

It is left handed parabola with vertex (1, 1).

At Y – axis, x = 0

âˆ´ y^{2} – 2y – 0 (from(1))

â‡’ y (y – 2) = 0

â‡’ y = 0

or y = 2

Given curves are

y = x^{2}

and y = |x|

â‡’ x^{2} – |x| = 0

â‡’ |x| {|x| – 1} = 0

â‡’ |x| = or |x| = 1

âˆ´ x = 0, x = – 1 or + 1

If x = 0 y = 0

If x = Â± 1 y = 1

Question 13.

Find the shortest distance between the lines

\(\overrightarrow{r_{1}}\) = 3iÌ‚ + 2jÌ‚ – 4kÌ‚ + Î»(iÌ‚ + 2jÌ‚ + 2kÌ‚)

and \(\overrightarrow{r_{1}}\) = 5iÌ‚ – 2jÌ‚ + Âµ(3iÌ‚ + 2jÌ‚ + 6kÌ‚)

If the line intersect find their point of intersection.

Answer:

Given line are:

â‡’ \(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\) = 8iÌ‚ – 4kÌ‚

âˆ´ \(\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\) = (2iÌ‚ – 4jÌ‚ + 4kÌ‚).(8iÌ‚ – 4kÌ‚)

= 16 – 16 = 0

âˆ´ Shortest distance between given lines is

\(\mid \frac{\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\vec{b}_{1} \times \overrightarrow{b_{2}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\) = 0 units.

Hence, the lines are intersecting for points of intersection.

Equation (1) and (2)

3iÌ‚ + 2jÌ‚ – 4kÌ‚ + Î»(iÌ‚ + 2jÌ‚ + 2kÌ‚) = 5iÌ‚ – 2jÌ‚ + Âµ(3iÌ‚ + 2jÌ‚ + 6kÌ‚)

âˆ´ 3 + 2Î» = 5 + 3Âµ

â‡’ Î» = 3Âµ + 2

2 + 2Î» = – 2 + 2Âµ

Î» = Âµ – 2

and – 4 + 2Î» = 6Âµ

âˆ´ Î» = 3Âµ + 2

â‡’ 2Âµ = – 4

âˆ´ Âµ = – 2

and Î» = – 4

Putting in (1), we get

\(\overrightarrow{r_{1}}\) = 3iÌ‚ + 2jÌ‚ – 4kÌ‚ – 4(iÌ‚ + 2jÌ‚ + 2kÌ‚)

â‡’ \(\overrightarrow{r_{1}}\) = – iÌ‚ – 6jÌ‚ – 12kÌ‚

âˆ´ Point of intersection is (- 1, – 6, – 12)

Question 14.

A glass jar contains twenty white balls of plastic numbered from 1 to 20, ten red balls of plastic numbered from 1 to 10, forty yellow balls of plastic numbered from 1 to 40 and ten blue balls of plastic numbered from 1 to 10. If these 80 balls of plastic are thoroughly shuffled so that each ball has the same probability of being drawn.

Based on the above information answer the following:

(i) Determine the probabilities of drawing a ball of plastic that is red or yellow and numbered 1, 2, 3 or 4.

(ii) Discuss the probabilities of drawing a plastic ball which is numbered 5, 15, 25 or 35.

Answer:

As we have given;

Total number of balls in jar = 80

20 white balls of numbered 1 to 20

i.e., 1 white ball contains one number,

10 Red balls of numbered 1 to 10

i.e., 1 Red ball contains one number,

Similarly, Yellow ball contains one number per ball and

Blue ball contains one number per ball.

(1) Number of Red balls of numbered 1,2,3, and 4 = 4

Number of Yellow balls of numbered 1,2,3 and 4 = 4

P(Red or yellow and numbered 1, 2, 3 or 4) = P(Red numbered 1, 2, 3 or 4) + P(Yellow numbered 1, 2, 3 or 4)

= \(\frac{4}{80}+\frac{4}{80}\)

= \(\frac{8}{80}\)

= \(\frac{1}{10}\)

(ii) Number of White balls of numbered 5, 15 = 2,

Number of Red balls of numbered 5 = 1,

Number of Yellow balls of numbered 5, 15, 25 and 35 = 4,

Number of Blue balls of numbered 5 = 1

P(numbered 5, 15, 25 or 35) = P(5) + P(15) + P(25) + P(35)

= P(White) + P(RecI) + P(Yellow) + P(Blue)

= \(\frac{2}{80}+\frac{1}{80}+\frac{4}{80}+\frac{1}{80}\)

= \(\frac{8}{80}\)

= \(\frac{1}{10}\)