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

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

Time Allowed: 2 Hours

Maximum Marks: 40

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

(Section – A has 6 short answer type (SA-1) questions of 2 marks each.)

Question 1.

Let E and F be two events such that P(E) = \(\frac{1}{3}\), P(F) = \(\frac{1}{4}\), and P(E uF) = \(\frac{1}{2}\).Show that E and F are independent events. (2)

Answer:

We know,

P(E ∪ F) = P(E) + P(F) – P(E ∩ F)

⇒ P(E ∩ F) = P(E) + P(F) – P(E ∪ F)

= \(\frac{1}{3}+\frac{1}{4}-\frac{1}{2}=\frac{1}{12}\)

Also, P(E) × P(F) = \(\frac{1}{3} \times \frac{1}{4}=\frac{1}{12}\)

∵ P(E) × P(F) = P(E ∩ F)

∴ E and F are independent events.

Question 2.

Find the cosine of the angle between the vectors \(\vec{a}\) and \(\vec{b}\), where \(\vec{a}\) = 3î – 2ĵ – k̂ and \(\vec{b}\) = -2ĵ.

OR

Find λ and µ (2î + 6ĵ + 27k̂) × (î + j + pk̂) = 0 (2)

Answer:

Let the angle between the vector \(\vec{a}\) and \(\vec{b}\) be θ.

Then, cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)

⇒ cos θ = \(\frac{(3 \hat{i}-2 \hat{j}-\hat{k}) \cdot(-2 \hat{j})}{|3 \hat{i}-2 \hat{j}-\hat{k} \|-2 \hat{j}|}\) ………(i)

Here, (3î – 2ĵ – k̂).(-2ĵ) = -6î.ĵ + 4ĵ.ĵ + 2k̂.ĵ

= 0 + 4 + 0

= 4

⇒ (6µ – 27λ)î – (2µ – 27)ĵ +(2λ – 6)k̂ = 0

⇒ 6µ – 27λ = 0

⇒ 2µ – 27 = 0

⇒ 2λ – 6 = 0

⇒ λ = 3

µ = \(\frac{27}{2}\)

Question 3.

Evaluate ∫ log x dx. (2)

Answer:

Let I = ∫ log x dx

Using integrating formu[a for a product of two functions.

= logx.∫1 dx – ∫[\(\frac{d}{d x}\)(log x).∫1.dx]dx

= x log x – ∫\(\frac{1}{x}\) . x dx

= x log x – ∫1dx

= x tog x – x + C

Question 4.

if the foot of a perpendicular drawn from origin to a plane is (5, – 3, – 2), then find the equation of the plane. (2)

Answer:

We know, general equation of a plane is a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0 0(0,0,0)

From the figure, normal to the required plane is OP.

where, \(\vec{OP}\) = (5 – 0)î + (- 3 – 0) ĵ + (- 2 – 0)k̂

= 5î – 3ĵ – 2k̂

(a, b, c) = (5, -3, -2)

Also, the plane contains the point P(5, – 3, -2)

∴ Required equation of plane is:

5(x – 5) + (- 3) (y + 3) + (- 2) (z + 2) = 0

⇒ 5x – 3y – 2z – 38 = 0

Question 5.

Evaluate (2)

Answer:

= [2 tan^{-1}x – x]^{1}_{0}

= [(2 tan^{-1}1 – 1) — (2 tan^{-1} 0 – 0)]

= [(2.\(\frac{\pi}{4}\) – 1) – o]

= \(\frac{\pi}{2}\) – 1

Question 6.

Two cards are drawn successively without replacement from a well-shuffled pack of 52 playing cards. Find the probability distribution of the number of kings. (2)

Answer:

Let x denote the discrete random variable ‘the number of king1.

Then, x takes values 0, 1, and 2.

∴ P(X = 0) = P(both cards are not king)

= \(\frac{{ }^{48} C_{2}}{{ }^{52} C_{2}}=\frac{48 \times 47}{52 \times 51}=\frac{188}{221}\)

P(X = 1) = P(one card is king)

= \(\frac{{ }^{48} C_{1} \times{ }^{4} C_{1}}{{ }^{52} C_{2}}=\frac{48 \times 4 \times 2}{52 \times 51}=\frac{32}{221}\)

P(X = 2) = P(both ore king cards)

= \(\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}\)

So, the required probabiLity distribution of X is

Section – B

(Section – B has 4 short answer type (SA-2) questions of 3 marks each.)

Question 7.

Solve the differential equation

(1 + y^{2}) tan^{-1} x dx + 2y (1 + x^{2})dy = 0. (3)

Answer:

Given, differentiaL equation is

(1 + y^{2})tan^{-1}xdx + 2y(1 + x^{2})dy = 0

⇒ (1 + y^{2})tan^{-1}xdx = -2y(1 + x^{2})dy

⇒ \(\left(\frac{\tan ^{-1} x}{1+x^{2}}\right)\)dx = \(\left(\frac{-2 y}{1+y^{2}}\right)\)dy

On integrating both sides, we get

∫\(\frac{\tan ^{-1} x}{1+x^{2}}\)dx = ∫\(\frac{2 y}{1+y^{2}}\)dy

Put tan^{-1} x = t in LHS. we get

\(\frac{1}{1+x^{2}}\)dx = dt

Put 1 + y^{2} = u in RHS, we get

2y dy = du

∫t dt = ∫\(\frac{d u}{u}\)

⇒ \(\frac{t^{2}}{2}\) = – log u + c

⇒ (tan^{-1}x)^{2}= – log(1 + y^{2})+c

⇒ (tan^{-1} x)^{2} + log(1 + y^{2})=c

which is the required soLution of the given differential equation.

Question 8.

Three vectors \(\vec{a}, \vec{b}\) and \(\vec{c}, \vec{a}\) satisfy the condition \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) . Evaluate the quantity µ = \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\), if |\(\vec{a}\)|= 3, |\(\vec{b}\)|= 4 and |\(\vec{c}\)| = 2.

OR

Let \(\vec{a}\) = 2î + k̂, \(\vec{b}\) = î + ĵ + k̂ and \(\vec{c}\) = 4î – 3ĵ + 7 k̂. Find a vector \(\vec{r}\) which satisfies \(\vec{r} \times \vec{b}=\vec{c} \times \vec{b}\) and \(\vec{r} \cdot \vec{a}\) = 0. (3)

Answer:

Given

Similarly, \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}=-|\vec{b}|^{2}\) = -16 …(ii)

And, \(\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{c}=-|\vec{c}|^{2}\) =—4 ………..(iii)

Adding (i), (ii) and (iii), we have

2(\((\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)) = 9 – 16 – 4

= – 29

Since μ = \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\) (Given)

2 × μ = -29

= \(\frac{-29}{2}\)

Hence, the value of μ is \(\frac{-29}{2}\).

OR

Let the vector \(\vec{r}\) be given as xî + yĵ + zk̂

Now, \(\vec{r} \cdot \vec{a}\) = 0

⇒ (xî + yĵ + zk̂).(2î + k̂) = 0

⇒ 2x + z = 0 …(i)

Also, \(\vec{r} \times \vec{b}=\vec{c} \times \vec{b}\)

⇒ (y – z)î +(x – z)ĵ +(x – y)k̂ = – 10î – 3ĵ +7k̂

⇒ y – z = -10, x – z = -3. x – y = 7 ……….(ii)

Solving eq. (i) and (ii), we get

x = -1, y = -8, z = 2

∴ \(\vec{r}\) = -î – 8ĵ + 2k̂

Question 9.

Let a, b and c be three vectors such that \(\vec{a}\) ≠ 0 and \(\vec{a} \times \vec{b}=2 \vec{a} \times \vec{c}\), |\(\vec{a}\)| = |\(\vec{c}\)| = 1, |\(\vec{b}\)| = 4 and |\(\vec{b} \times \vec{c}\)| = \(\sqrt{15}\) , \(|\vec{b} \times \vec{c}|\) = λ\(\vec{a}\).Then find the value of λ. (3)

Answer:

Let angle between \(\vec{b}\) and \(\vec{c}\) be θ.

∵ \(|\vec{b} \times \vec{c}|\)

\(|\vec{b}| \cdot|\vec{c}|\) sin θ = \(\sqrt{15}\)

⇒ (4) (1) sin θ = \(\sqrt{15}\)

⇒ sin θ = \(\frac{\sqrt{15}}{4}\)

⇒ cos θ = \(\frac{1}{4}\)

Now, \(\vec{b}\) – 2\(\vec{c}\) = λ\(\vec{a}\)

Taking mode both sides, we get

⇒ 16 = λ^{2}

⇒ λ = 4

Hence, the value of λ is 4.

Question 10.

Evaluate ∫\(\frac{1}{x^{1 / 2}+x^{1 / 4}}\)dx

OR

Evaluate ∫^{2}_{-1}f(x) dx, where f(x) = |x + 1| + |x| + |x – 1|. (3)

Answer:

Let I = ∫\(\frac{1}{x^{1 / 2}+x^{1 / 4}}\)dx

Put x = t^{4}

⇒ dx = 4t^{3}

We have,

f(x) = |x + 1| + |x| + |x – 1|

Section – C

(Section – C has 4 long answer type questions (LA) of 4 marks each.)

Question 11.

Find the area of the region bounded by the curve y = cos x between x = 0 and x = π.

OR

Find the area of the region bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7. (4)

Answer:

The graph of cosine function is positive from 0 to \(\frac{\pi}{2}\) and negative from \(\frac{\pi}{2}\) to π.

= (1 + 0) + (0 + 1)

= 1 + 1 = 2 sq. units

OR

Given equations of the lines are x + 2y = 2, y – x = 1 and 2x + y = 7.

These lines intersect each other at points C(2, 3), D(0, 1) and E(4, – 1).

∴ Required area = ar(OACDO) + ar(ABCA) – ar(OADO) + ar(AEFBA) – ar(BEFB)

Question 12.

It is known that 40% of the students in a certain college are girls and 50% of the students are above the median height. If \(\frac{2}{3}\) of boys are above the median height, what is the probability that a randomly selected student who is below the median height is a girl? (4)

Answer:

Let the total number of students in the college be 100. Then, 40 are girls and 60 are boys.

Here, \(\frac{2}{3}\), of boys are above the median height

i.e., \(\frac{2}{3}\) × 60 = 40 boys are above the median height

Since, only 50 students are above the median height, so only 10 girls are above the median height.

Now, let the event G, B and M are described as:

G : A girl is selected.

B : A boy is.

M : Student selected is below the median height

Question 13.

Evaluate ∫_{0}^{1} tan^{-1}\(\left(\frac{2 x}{1-x^{2}}\right)\)dx (4)

Answer:

Let I = ∫_{0}^{1} tan^{-1}\(\left(\frac{2 x}{1-x^{2}}\right)\)dx

Put x = tan y

⇒ dx = sec^{2}y dy

For x = 0, y = tan^{-1} 0 = 0

For x = 1, y = tan^{-1} 1 = \(\frac{\pi}{4}\)

Case-Based/Data-Based

Question 14.

A football match is organized between students of class XII of two schools, say school A and school B. For each, a team from each school is chosen. Remaining students of Class XII of schools A and B are respectively sitting on the plane represented by the equations \(\vec{r}\)(î + ĵ + 2k̂) = 5 and \(\vec{r}\)(2î – ĵ + k̂) = 6, to cheer up the team of their respective school.

Based on the above information, answer the following two questions:

(i) Write the cartesian equation of the plane on which students of school A are seated. (2)

Answer:

Vector equation of plane on which students of school A are seated is:

\(\vec{r}\) .(î + ĵ + 2k̂) = 5

⇒ (xî + yĵ + zk̂).(î + ĵ + 2k̂) =5

⇒ x + y + 2z = 5

which is the required cartesian form.

(ii) Write the intercept form of the equation of the plane on which students of school B are seated. Also, find the distance of this plane from the origin. (2)

Answer:

Cartesian equation of the plane on which students of school B are seated is x – y + z = 6.

\(\frac{2 x}{6}-\frac{y}{6}+\frac{z}{6}\) = 1

\(\frac{x}{3}+\frac{y}{-6}+\frac{z}{6}\) = 1

Which is the required intercept form.

Its distance from the origin = \(\left|\frac{-6}{\sqrt{4+1+1}}\right|\)

= √6 units