Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.

## CBSE Sample Papers for Class 12 Maths Term 2 Set 1 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.

Find: âˆ«\(\frac{\log x}{(1+\log x)^{2}}\) dx [2]

OR

Find: âˆ«\(\frac{\sin 2 x}{\sqrt{9-\cos ^{4} x}}\) dx

Answer:

Question 2.

Write the sum of the order and the degree of the following differential equation: [2]

\(\frac{d}{d x}\left(\frac{d y}{d x}\right)=5\) = 5

Answer:

Here, \(\frac{d}{d x}\left(\frac{d y}{d x}\right)=5\) = 5

â‡’ \(\frac{d^{2} y}{d x^{2}}\) = 5 â‡’ \(\left[\frac{d^{2} y}{d x^{2}}\right]^{1}\) = 5

âˆ´ order = 2 and degree = 1

and sum = 2 + 1 = 3.

Question 3.

If and are unit vectors, then prove that |aÌ‚ + bÌ‚| = 2 cos \(\frac{\theta}{2}\), where Î¸ is the angle between them. [2]

Answer:

Question 4.

Find the direction cosines of the following line:

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

Answer:

Clearly direction ratios are < 1, 1, 4 > and its direction ratios (dr’s) are to be converted to direction cosines.

Question 5.

A bag contains 1 red and 3 white balls. Find the probability distribution of the number of red balls if 2 balls are drawn at random from the bag one-by-one without replacement. [2]

Answer:

Here, Sample space = 1 red ball + 3 white balls

Here P(not red) = \(\frac{3}{4} \times \frac{2}{3}\) = \(\frac{1}{2}\) â†’ for [x = 0]

P(1 red ball) = \(\frac{1}{4} \times \frac{3}{3}+\frac{3}{4} \times \frac{1}{3}\) = \(\frac{1}{4}+\frac{1}{4}\) = \(\frac{1}{2}\) â†’ for [x = 1]

âˆ´ P(X = 0) = \(\frac{1}{2}\) and P(X = 1) = \(\frac{1}{2}\), so probability distribution table will be

X | 0 | 1 |

P(X) | \(\frac{1}{2}\) | \(\frac{1}{2}\) |

Question 6.

Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card Jack?. [2]

Answer:

Total cards = 52

Total no. of red cards = 26 and red card without jack = 26 – 2 = 24

No. of Jacks = 4

Now drawing of cards one by one without replacement are independent events.

âˆ´ Probability of getting first card or a red card + second card a jack (without replacement)

= [getting a red jack + getting a jack] or [getting a red card (not a jack) + getting a jack]

= \(\left[\frac{2}{52} \times \frac{3}{51}\right]+\left[\frac{24}{52} \times \frac{4}{51}\right]=\frac{1}{26}.\)

Section – B

Question 7.

Find \(\int \frac{x+1}{\left(x^{2}+1\right) x}\) dx. [3]

Answer:

Here, one of the factors in the denominator (x^{2} + 1) is quadratic.

Question 8.

Find the general solution of the following differential equation: [3]

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

OR

Find the particular solution of the following differential equation, given that y = 0 when x = \(\frac{\pi}{4}\);

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

Answer:

Given, x\(\frac{d y}{d x}\) = y – x sin \(\left(\frac{y}{x}\right)\) ……….. (i)

x and y are of same degree (one).

so it is a homogeneous differential equation

put y = vx, on differentiating we get

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

Substituting in (i), we get

Integrating on both sides

âˆ«\(\frac{d v}{\sin v}\) = – âˆ«\(\frac{d x}{x}\)

log |cosec v – cot v| = – log |x| + log k, k > 0, log k is an arbitrary constant

or log |cosec v – cot v| + log |x| = log k

or log |(cosec v – cot v)x| = log k

= |(cosec v – cot v)x| =Â± k

which is the required general solution.

OR

The differential equation is a linear differential equation

I.F. = e^{âˆ«cot x dx} = e^{log sin x} = sin x

The general solution is given by

Question 9.

If \(\vec{a} \neq \overrightarrow{0}, \vec{a}, \vec{b}=\vec{a}, \vec{c}, \vec{a} \times \vec{b}=\vec{a} \times \vec{c}\), then show that \(\vec{b} \times \vec{c}\).

Answer:

Question 10.

Find the shortest distance between the following lines:

\(\vec{r}\) = (iÌ‚ + jÌ‚ – kÌ‚) + s(2iÌ‚ + jÌ‚ + kÌ‚)

\(\vec{r}\) = (iÌ‚ + jÌ‚ – 2kÌ‚) + t(4iÌ‚ + 2jÌ‚ + 2kÌ‚)

OR

Find the vector and the cartesian equations of the plane containing the point iÌ‚ + 2jÌ‚ – kÌ‚ and parallel to the lines \(\vec{r}\) = (iÌ‚ + 2jÌ‚ + 2kÌ‚) + s(2iÌ‚ – 3jÌ‚ + 2kÌ‚) = 0 and \(\vec{r}\) = (3iÌ‚ + jÌ‚ – 2kÌ‚) + t(iÌ‚ – 3jÌ‚ + k) = 0.

Answer:

Here, the lines are parallel. The shortest distance = \(\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \times \vec{b}\right|}{|\vec{b}|}\)

= \(\frac{(3 \hat{k}) \times(2 \hat{i}+\hat{j}+\hat{k}) \mid}{\sqrt{4+1+1}}\) ………. (i)

(3kÌ‚) Ã— (2iÌ‚ + jÌ‚ + kÌ‚) = \(\left|\begin{array}{lll}

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

0 & 0 & 3 \\

2 & 1 & 1

\end{array}\right|\) = – 3iÌ‚ + 6jÌ‚

âˆ´ |- 3kÌ‚ + 6jÌ‚| = \(\sqrt{3^{2}+6^{2}}\) = âˆš45 = 3âˆš5

Hence, the required shortest distance = \(\frac{3 \sqrt{5}}{\sqrt{6}}\) units from (i)

OR

Since, the plane is parallel to the given lines, the cross product of the vectors 2iÌ‚ – 3jÌ‚ + 2 kÌ‚ and iÌ‚ – 3jÌ‚ + kÌ‚ will be a normal to the plane

(2iÌ‚ – 3jÌ‚ + 2 kÌ‚) Ã— (iÌ‚ – 3jÌ‚ + kÌ‚) = \(\left|\begin{array}{rrr}

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

2 & -3 & 2 \\

1 & -3 & 1

\end{array}\right|\) = 3iÌ‚ – 3kÌ‚

The vector equation of the plane is

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

or \(\vec{r}\) . (iÌ‚ – kÌ‚) = 2

and the cartesian equation of the plane is x – z – 2 = 0.

Section – C

Question 11.

Evaluate:

Answer:

The given definite integral =

Here, x(x – 1) (x – 2) = 0 has three zeroes x = 0, x = 1 and x = 2

Now look at the figure:

For interval – 1 â‰¤ x â‰¤ 0

at x = – 1 x(x – 1) (x – 2) = (- 1) (- 1 – 1) (- 1 – 2) = – 6

at x = 0 x(x – 1)(x – 2) = 0 (0 – 1)(0 – 2) = 0

âˆ´ |x^{3} – 3x^{2} + 2x| = – (x^{3} – 3x^{2} + 2x) for – 1 â‰¤ x â‰¤ 0 ……. (i)

For interval 0 â‰¤ x â‰¤ 1

at x = 0

x(x – 1) (x – 2) = 0(0 – 1) (0 – 2) = 0

at x = \(\frac{1}{2}\) â‡’ 0 â‰¤ \(\frac{1}{2}\) â‰¤ 1

at x = 1

x(x – 1) (x – 2) = (1) (1 – 1) (1 – 2) = 0

âˆ´ |x^{3} – 3x^{2} + 2x| = – x^{3} – 3x^{2} + 2x for 0 < x < 1 ……. (ii)

For interval 1 â‰¤ x â‰¤ 2

at x = 1

x(x – 1) (x – 2) = 1(1 – 1) (1 – 2) = 0

at x = \(\frac{3}{2}\) â‡’ 1 â‰¤ \(\frac{3}{2}\) â‰¤ 2, we have

at x = 2, we have

x(x – 1) (x – 2) = (2) (2 – 1) (2 – 2) = 0

âˆ´ |x^{3} – 3x^{2} + 2x| = – (x^{3} – 3x^{2} + 2x) for 1 â‰¤ x â‰¤ 2 ……. (iii)

Question 12.

Using integration, find the area of the region in the first quadrant enclosed by the line x + y = 2, the parabola y^{2} = x and the x-axis.

OR

Using integration, find the area of the region {(x, y): 0 â‰¤ y â‰¤ âˆš3x, x^{2} + y^{2} â‰¤ 4} [4]

Answer:

Solving x + y = 2 and y^{2} = x simultaneously, we get the points of intersection as (1, 1) and (4, – 2).

The required area = the shaded area (OABO)

Solving y = âˆš3x and x^{2} + y^{2} = 4, we get the points of intersection as (1, âˆš3) and (- 1, -âˆš3).

The required area

= the shaded area (OABO)

Question 13.

Find the foot of the perpendicular from the point (1, 2, 0) upon the plane x – 3y + 2z = 9. Hence, find the distance of the point (1, 2, 0) from the given plane. [4]

Answer:

The equation of the line perpendicular to the plane and passing through the point (1, 2, 0) is

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

The coordinates of the foot of the perpendicular are (Î¼ + 1, – 3Î¼ + 2, 2Î¼) for some Î¼.

These coordinates will satisfy the equation of the plane. Hence, we have

Î¼ + 1 – 3(- 3Î¼ + 2) + 2(2Î¼) = 9

â‡’ Î¼ = 1

The foot of the perpendicular is (2, – 1, 2).

Hence, the required distance = \(\sqrt{(1-2)^{2}+(2+1)^{2}+(0-2)^{2}}\)

= âˆš14 units.

Case-Based/Data-Based

Question 14.

An insurance company believes that people can be divided into two classes: those who are accident prone and those who are not. The company’s statistics show that an accident-prone person will have an accident at sometime within a fixed one-year period with probability 0.6, whereas this probability is 0.2 for a person who is not accident prone. The company knows that 20 percent of the population is accident prone.

Based on the given information, answer the following questions.

(i) What is the probability that a new policyholder will have an accident within a year of purchasing a policy? [2]

(ii) Suppose that a new policyholder has an accident within a year of purchasing a policy. What is the probability that he or she is accident prone? [2]

Answer:

Let E_{1} = The policy holder is accident prone.

E_{2} = The policy holder is not accident prone.

E = The new policy holder has an accident within a year of purchasing a policy.

(i) P(E) = P(E_{1}) Ã— P(E/E_{1}) + P(E_{2}) Ã— P(E/E_{2})

= \(\frac{20}{100} \times \frac{6}{10}+\frac{80}{100} \times \frac{2}{10}=\frac{7}{25}\)

(ii) By Bayes’ Theorem,