CBSE Sample Papers for Class 12 Maths Term 2 Set 3 with Solutions

Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Maths Term 2 Set 3 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 bag contains 5 red, 7 green and 4 white balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, green and green respectively.
Answer:
A bag has 5 red, 7 green and 4 white balls.
So, Total balls = 5 + 7 + 4 = 16
P(White, green and green) = P(White). P(green/white).P(green/white and green)
\(\frac{4}{16} \times \frac{7}{15} \times \frac{6}{14}=\frac{1}{20}\)

Question 2.
Find the direction cosines of the line passing through the two points (-1, 2, 3) and (2, -3, 5).
Answer:
We know that the direction cosines of a line joining two points P and Q are given by

Thus, the direction cosines of the line joining two points is
\(\frac{3}{\sqrt{38}}, \frac{-5}{\sqrt{38}}, \frac{2}{\sqrt{38}}\)

Question 3.
Find: ∫ cos 6x \(\sqrt{1+\sin 6 x}\) dx.
OR
Find ∫\(\left(x+\frac{1}{x}\right)^{2}\)dx
Answer:
We have I = ∫ cos 6x \(\sqrt{1+\sin 6 x}\) dx.
Put t = 1 + sin 6x, so that dt = 6 cos 6x dx

Therefore ∫ cos 6x \(\sqrt{1+\sin 6 x}\) dx = \(\frac{1}{2}\)∫t^1/2 dt
= \(\frac{1}{6} \times \frac{2}{3}\)(t)^3/2 + C
= \(\frac{1}{9}\)(1 + sin6x)^3/2 + C

OR

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

Question 4.
From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being queens?
Answer:
Total number of ways of getting two cards,
n(S) = ⁴C₂ = \(\frac{52 \times 51}{2 \times 1}\) = 1326

Let E be the event of getting two queens out of 4.

Question 5.
\(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are unit vectors such that 2\(\vec{a}\) + 3\(\vec{b}\) + 4\(\vec{c}\) = 0 and the angle between \(\vec{a}\) and \(\vec{b}\) is θ, find the value of (4cosθ – 1).
Answer:
2\(\vec{a}\) + 3\(\vec{b}\) + 4\(\vec{c}\) = 0
2\(\vec{a}\) + 3\(\vec{b}\) = – 4\(\vec{c}\)

Taking dot product with itself

⇒ 4 + 9 + 12.cos θ = 16
⇒ 12 cos θ = 3
⇒ cos θ = \(\frac{1}{4}\)
Now, 4 cos θ – 1 = 4 × \(\frac{1}{4}\) – 1
= 0

Question 6.
Find the order and degree of the following differential equation:
\(\left\{1+\left(\frac{d y}{d x}\right)^{2}\right\}^{3 / 2}=k \frac{d^{2} y}{d x^{2}}\)
Answer:
The highest derivative occuring in the differential equation:

Which is of order 2. Hence the order of the differential equation is 2.
The equation

as a polynomial in derivative.
Raising both sides of the powre 2 we have,

The exponent of the highest order derivative, i.e.,\(\frac{d^{2} y}{d x^{2}}\) is 2. Thus the degree of the equation is 2.

Section – B

Question 7.
Find the shortest distance between the following pairs of lines:
\(\frac{x-8}{3}=\frac{y-5}{-2}=\frac{z-10}{7}\) and \(\frac{x-15}{3}=\frac{y+1}{8}=\frac{z-5}{-5}\)
Answer:
Given equation of two lines
\(\frac{x-8}{3}=\frac{y-5}{-2}=\frac{z-10}{7}\) ……….(1)
and
\(\frac{x-15}{3}=\frac{y+1}{8}=\frac{z-5}{-5}\) …….(2)

Line (1) passes through (8, 5, 10) and has direction ratios proportional to 3, -2, 7, So its vector equation is
\(\overrightarrow{r_{1}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}\)

Where \(\overrightarrow{a_{1}}\) = 8î + 5ĵ + 10k̂ and \(\overrightarrow{b_{1}}\) = 3î + 2ĵ + 7k̂
Line (2) passes through (15, – 1, 5) and has direction ratios 3, 8, -5, so its vector equation is
\(\overrightarrow{r_{2}}=\overrightarrow{a_{2}}+\lambda \overrightarrow{b_{2}}\)

Where \(\overrightarrow{a_{2}}\) = 15î – ĵ + 5k̂ and \(\overrightarrow{b_{2}}\) = 3î + 8ĵ – 5k̂

The shortest distance between the lines

OR

Given equation of lines are
\(\frac{x+5}{5 \lambda+2}=\frac{2-y}{5}=\frac{1-z}{-1}\)
and \(\frac{x}{1}=\frac{2 y+1}{4 \lambda}=\frac{1-z}{-3}\)

Above equation can be written as

Comparing equations (i) and (ii) with one point form of line \(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\) we get
a₁ = 5λ + 2, b₁ = -5, c₁ = 1
and a₂ = 1, b₂ = 2λ, c₂ = 3

Since, two lines are perpendicular,
∴ a₁a₂ + b₁b₂ + c₁c₂ = 0
⇒ 1(5λ + 2) + 2λ(-5) + 3(1) = 0
⇒ 5λ + 2 – 10λ + 3 = 0
⇒ -5λ + 5 = 0
⇒ 5λ = 5
⇒ λ = 1

Question 8.
Evaluate ∫\(\frac{(3 x+4)}{x^{2}+12 x+32}\) dx
Answer:
Let I = ∫\(\frac{(3 x+4)}{x^{2}+12 x+32}\) dx
Let 3x + 4 = A(\(\frac{d}{d x}\)(x² + 12x + 32) + B
⇒ 3x + 4 = A(2x + 12) + B

On comparing the coefficients,
2A = 3
⇒ A = \(\frac{3}{2}\)
and 12A + B = 4
⇒ B = 4 – 12.\(\frac{3}{2}\)
= 4 – 18 = – 14

Putting x² + 12x + 32 = t in first integral

Question 9.
If \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = ĵ – k̂, then find \(\vec{c}\) such that \(\vec{a} \times \vec{c}=\vec{b}\) and \(\vec{a} \cdot \vec{c}\) = 3
Answer:
Let \(\vec{a}\) = xî + yĵ + zk̂
and \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = ĵ – k̂

Given \(\vec{a} \times \vec{c}=\vec{b}\)
\(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
x & y & z
\end{array}\right|\) = ĵ – k̂

⇒ î(z – y) – ĵ(z – x) + k̂(y – x) = ĵ – k̂
∴ z – y = 0 ……..(2)
x – y = 1 ………(3)

\(\vec{a} \cdot \vec{c}\) = 3

Again given,
(î + ĵ + k̂). (xî + yĵ + zk̂) = 3 ……….(4)
⇒ x + y + z = 3 ………..(5)

On adding equation (2) and (3), we get
2x – y – z = 2

Solving equation (4) and (5), we get
x = \(\frac{5}{3}\), y = \(\frac{2}{3}\) and z = \(\frac{2}{3}\)

Now, \(\vec{c}\) = \(\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\)
= \(\frac{1}{3}\)(5î + 2ĵ + 2k̂)

Question 10.
Solve the differential equation:
x\(\frac{d y}{d x}\) – y + sin \(\left(\frac{y}{x}\right)\) = 0

OR

Find a particular solution:
\(\frac{d y}{d x}\) = y tanx; y = 1 when x = 0.
Answer:
The given differential equation is,
x\(\frac{d y}{d x}\) – y + sin \(\left(\frac{y}{x}\right)\) = 0

Intergrating both sides, we get

Intergrating both sides, we get
∫\(\frac{d y}{y}\) = ∫ tan x dx

⇒ log y = log(sec x) + log C
⇒ log y = log(C × sec x)
⇒ y = C × sec x ………(1)

Now, y = 1, when x = 0
⇒ 1 = C × sec 0
⇒ C = 1

Substituting C = 1 in equation (1), we get
y = sec x.

Section – C

Question 11.
Find ∫\(\frac{\sec x}{1+{cosec} x}\) dx
Answer:


For t = -1,
⇒ – 1 = A[1 + (- 1)] [1 – (- 1)] + B[1 – (-1)] + C(1 + (-1)]²
⇒ -1 = A × 0 + B × 2 + C(0)²
⇒ -1 = 2B
⇒ B = \(\frac{1}{2}\)

For t = 1,
⇒ 1 = A(1 + 1) (1 – 1) +B(1 – 1) +C (1 + 1)²
⇒ 1 = A × 0 + B × 0 + C (2)²
⇒ 1 = 4C
⇒ C = \(\frac{1}{4}\)

For t = 0,
⇒ 0 = A(1 + 0)(1 – 0) + B(1 – 0) + C(1 + 0)²
⇒ 0 = A(1)(1) + B(1) + C(1)²
⇒ 0 = A + B + C
⇒ 0 = A + \(\left(\frac{-1}{2}\right)+\frac{1}{4}\)
⇒ A = \(\frac{1}{4}\)

From (1), we get

Question 12.
Find the area of the region enclosed between the parabola y² = x and the line x + y = 2.
OR
Using integration, find the area of a triangle ABC, coordinates of whose vertices are A(4,l), B(6,6), C(8,4).
Answer:
Given, curve is y² = x ……….(1)
It is a right hand parabola whose vertex is (0, 0)

Given x + y = 2
⇒ y = 2 – x

From equation (1) and (2)
(2 – x)² = x
⇒ 4+ x² – 4x = x
⇒ x² – 5x + 4 = 0
⇒ x² – 4x – x + 4 = 0
⇒ x(x – 4) – 1(x – 4) = 0
∴ x = 1, 4.
Putting in equation (2), we get
if, x = 1 then y = 1 and
if, x = 4 then y = -2

OR

The given points are A(4, 1), B(6, 6), and C(8, 4)
Equation of AB is given by
⇒ y – 1 = \(\frac{6-1}{6-4}\)(x – 4)
⇒ y – 1 = \(\frac{5}{2}\)(x – 4)
⇒ y = \(\frac{5}{2}\)x – 9 ………….(1)

Equation of BC is given by
y – 6 = 1(x – 6)
⇒ y = -x + 12

Similarly, equation of AC is given by
y – 1 = \(\frac{3}{4}\)(x – 4)
⇒ y = \(\frac{3}{4}\)x – 2

Question 13.
Find the equation of the plane passing through the points (1, 0, – 2), (3, – 1, 0) and perpendicular to the plane 2x – y + z = 8. Also find the distance of the plane thus obtained from the origin.
Answer:
Given points, P(1, 0, -2), Q (3, -1, 0)
Given plane 2x – y + z = 8

Normal vector of given plane,
\(\overrightarrow{n_{1}}\) = 2î – ĵ + k̂

Now,
\(\overrightarrow{P Q}\) = 2î – ĵ + 2k̂

Normal vector of required plane,
\(\overrightarrow{n_{2}}=\left(\overrightarrow{n_{1}} \times \overrightarrow{\mathrm{PQ}}\right)\)

= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
2 & -1 & 2
\end{array}\right|\)
= – 1.î – 2ĵ + 0k̂ = -i – 2ĵ

Required equation of plane,
-1(x – 1) + (-2)(y – 0) + 0(z + 2) = 0
⇒ -x + 1 – 2y + 0 = 0
⇒ x + 2y = 1

Therefore, the required equation of the plane is, x + 2y = 1.
Now, distance from origin (0,0,0) is,

Therefore, distance from origin is \(\frac{1}{\sqrt{5}}\) units.

Question 14.
Sanjay, Ajay and Vijay are three best friends. After completing their MBA from IIM Lucknow. They apply for the job in the same company for the post of manager in finance department. Chances of selection of Sanjay, Ajay and Vijay are in the same ratio of 1 : 2 : 4. The probabilities that Sanjay can increase the profits of the company by his efforts is 0.8 whereas probabilities for the sam task of Ajay and Vijay are 0. 5 and 0.3 respectively.
From the above information attempt the following questions:
(i) A be event of increase the profit then find if it is due to Ajay.
(ii) If E₁, E₂ and E₃, be events of not increase the profits, then find the probability that is due to the appointment of Vijay?
Answer:
Let E₁ be event of selecting Sanjay, E₂ be event of selecting Ajay and E₃ be event of selecting Vijay.
∴ P(E₁) = \(\frac{1}{7}\)
P (E₂) = \(\frac{2}{7}\)
P(E₃) = \(\frac{4}{7}\)

(i) A be event of increasing profits
P(\(\frac{\mathrm{A}}{\mathrm{E}_{1}}\))
P(\(\frac{\mathrm{A}}{\mathrm{E}_{2}}\))
P(\(\frac{\mathrm{A}}{\mathrm{E}_{3}}\))

(ii) Let P be the event of not increases in profits.