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

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.

Evaluate ∫\(\frac{2^{x+1}-5^{x-1}}{10^{x}}\) dx. (2)

Answer:

Let I = ∫\(\frac{2^{x+1}-5^{x-1}}{10^{x}}\) dx

Question 2.

A line makes angle α, β, γ with the coordinate axes. If α + β = 90°, then find the value of γ. (2)

Answer:

If a tine makes angles α, β, γ with coordinate axes, then we have,

cos^{2} α + cos^{2}β + cos^{2}γ = 1

Since, α + β = 90°

⇒ α = 90° – β

⇒ cos α = cos (90° – β)

⇒ cos α = sin β

⇒ cos^{2}α = sin^{2}β

= 1 – cos^{2}β

⇒ cos^{2}α + cos^{2}β = 1

From (i) and (ii), we get

⇒ 1 + cos^{2}γ = 1

⇒ cos^{2}γ = 0

Hence, γ = \(\frac{\pi}{2}\) = 90°

Question 3.

Find the probability of the occurrence of a number greater than 2 in a throw of a die, if it is known that only even numbers can occur.

OR

An unbaised die is thrown. If a random variable X is defined as

Write the probability distribution of X. (2)

Answer:

Let A be the event of “occurrence of an even number”, and B be the event of “occurrence of a number greater than 2.

∴ P(A) = \(\frac{3}{6}=\frac{1}{2}\)

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

[∵ 4 and 6 are even numbers from 1 to 6 that are greater than 2]

Now,

Required probability = P(B/A)

= \(\frac{P(B \cap A)}{P(A)}\)

= \(\frac{1 / 3}{1 / 2}=\frac{2}{3}\)

OR

Here, X takes the values 0 and 1.

Also, P(X = 0) = P(an odd number = \(\frac{3}{6}=\frac{1}{2}\)

And, P(X = 1) = P(an even number) = \(\frac{3}{6}=\frac{1}{2}\)

Thus, the probability distribution of X is:

Question 4.

If \(\vec{a}\) = 5î – ĵ – 3k̂ and \(\vec{b}\) = î + 3ĵ – 5k̂, then show that \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) are orthogonal. (2)

Answer:

We have,

\(\vec{a}+\vec{b}\) = (5î – ĵ – 3k̂) + (î + 3ĵ – 5k̂)

= 6î + 2ĵ – 8k̂

and \(\vec{a}-\vec{b}\) = (5î – ĵ – 3k̂) – (î + 3ĵ – 5k̂)

= 4î – 4ĵ + 2k̂

Now, \((\vec{a}+\vec{b})\). \((\vec{a}-\vec{b})\) = (6î + 2ĵ – 8k̂) . (4î – 4ĵ + 2k̂)

= 6 × 4 + 2 × (- 4) + (- 8) × 2

= 24 – 8 – 16

= 0

Hence, \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) are orthogonal i.e., perpendicular to each other.

Question 5.

Check whether the differential equation (x – y) \(\frac{d y}{d x}\) = x + 2y is homogenous or not. (2)

Answer:

We have,

Since, the power of λ is zero.

∴ The given differential equation is homogeneous.

Question 6.

Bag I contains 2 white and 4 red balls. Bag II contains 3 white and 3 red balls. One of the bags is selected at random and a ball is drawn from it. If the ball drawn is white, what is the probability that it is drawn from Bag I? (2)

Answer:

Let E_{1} and E_{2} be the events of selecting Bag I and Bag II respectively.

Also, let W be the event of drawing a white ball. Then,

P(E_{1})= \(\frac{1}{2}\) and P(E_{2}) = \(\frac{1}{2}\)

Section – B

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

Question 7.

Evaluate ∫tan^{-1}(secx + tanx)dx.

OR

Evaluate ∫_{0}^{1} \(\frac{1}{\sqrt{a x-x^{2}}}\) dx. (3)

Answer:

Let ∫tan^{-1}(secx + tanx)dx.

[Dividing numerator and denominator by \(\frac{x}{2}\)]

Question 8.

Find the general solution of y^{2} dx + (x^{2} – xy + y^{2}) dy = 0. (3)

Answer:

Given, diffrentiaL equation is

y^{2}dx + (x^{2} – xy + y^{2})dy =0

⇒ y^{2}dx = -(x^{2} – xy + y^{2})dy

⇒ y^{2} \(\frac{d x}{d y}\) = -(x^{2} – xy + y^{2})

⇒ \(\frac{d x}{d y}=-\left(\frac{x^{2}}{y^{2}}-\frac{x}{y}+1\right)\) ……(i)

This is a homogeneous differential equation.

Let x = vy …….(ii)

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

Substituting these values in equation (i), we get

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

⇒ y\(\frac{d v}{d y}\) = -v^{2} – 1

⇒ \(\frac{d v}{d y}\)

On integrating both sides, we get

tan^{-1}(y) = -log y + c

⇒ tan^{-1}\(\) + log y = c [From (ii)]

Caution:

Here, we are differentiating x w.r.t y so we will use x = vy for substitution.

Question 9.

Find the distance between the parallel planes 6x + 2y – 3z = 12 and 12x + 4y – 6z = 17. (3)

Answer:

Given: Equation of ptanes ore:

6x + 2y – 3z = 12 ………(i)

and 12x + 4y – 6z = 17

or 6x + 2y – 3z = \(\frac{17}{2}\)

∵ Planes (1) and (ii) are paralleL

∴ Distance between them

Related Theory:

Distance between two parallel planes ax + by + cz = d_{1} and ax + by + cz = d_{2} is given as \(\left|\frac{d_{1}-d_{2}}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)

Question 10.

Let the vectors \(\vec{a}, \vec{b}, \vec{c}\) be given as a_{1}x î + a_{2}ĵ + a_{3} k̂, b_{1}î + b_{2}ĵ + b_{3}k̂ and c_{1}î + c_{2}ĵ + c_{3}k̂ respectively. Then show that \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a}+\vec{b})+(\vec{a} \times \vec{c})\)

OR

Find the area of the triangle whose vertices are A (1,1,2), B (2, 3, 5) and C (1, 5, 5). (3)

Answer:

We have,

\(\vec{b}+\vec{c}\) = (b_{1}î + b_{2}ĵ + b_{3}k̂) + (c_{1}î + c_{2}ĵ + c_{3}k̂)

= (b_{1} + c_{1})î + (bsub>2 + csub>2)ĵ + (b_{3} + c_{3}) k̂

= {a_{2} (b_{3} + c_{3}) – a_{3} (b_{2} + c_{2})}î – {a_{1} (b_{3} + c_{3}) – a_{3}(b_{1} + c_{1}}ĵ + {a_{1}(b_{2} + c_{2}) – a_{2}(b_{1} + c_{1})k̂

So \((\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\) = (a_{2}b_{3} – a_{3}b_{2})î – (a_{1}b_{3} – a_{3}b_{1})ĵ + (a_{1}b_{2} – a_{2}b_{1}) k̂ + (a_{1}c_{3} – a_{3}c_{1}) î -(a_{1}c_{3} – a_{3}c_{1})ĵ + (a_{1}c_{2} – a_{2}c_{1}) k̂

= {a_{2}(b_{3} + c_{1}) – a_{3}(b_{2} + c_{2})} î – {a_{1}(b_{3} + c_{3}) – a_{3}(b_{1} + c_{1}}ĵ + {i (b_{2} + c_{2}) – 2 (b_{1} + c_{1})} k̂ …(iv)

From (i) and (iv), we get

\(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a}+\vec{b})+(\vec{a} \times \vec{c})\)

Hence proved.

OR

We have, \(\overrightarrow{\mathrm{AB}}\) = (2 – 1)î + (3 – 1)ĵ + (5 – 2) k̂

= î + 2ĵ + 3k̂

and \(\overrightarrow{\mathrm{AC}}\) = (1 – 1)î + (5 – 1)ĵ +(5 – 2)k̂

= 4ĵ +3k̂

We know,

Area of ∆ABC = \(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|\)

= (6 – 12)î – (3 – 0)ĵ + (4 – 0)k̂ – 6î – 3ĵ + 4k̂

∴\(|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{(-6)^{2}+(-3)^{2}+4^{2}}\)

= \(\sqrt{36+9+16}\)

= \(\sqrt{61}\)

∴ Area of ∆ABC = \(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|=\frac{\sqrt{61}}{2}\)sq. units

Section – C

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

Question 11.

Evaluate ∫\(\frac{1}{3+2 \sin x+\cos x}\) dx

OR

Evaluate ∫_{0}^{π/2}\(\frac{\cos x}{1+\cos x+\sin x}\) dx. (4)

Answer:

Let I = ∫\(\frac{1}{3+2 \sin x+\cos x}\) dx

Put tan \(\frac{x}{2}\) = t

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

⇒ sec^{2}\(\frac{x}{2}\) dx = 2dt

∴ I = ∫\(\frac{d t}{t^{2}+2 t+2}\)

= ∫\(\frac{d t}{(t+1)^{2}+1}\)

= tan^{-1}(t + 1) + C

[∵∫\(\frac{1}{x^{2}+a^{2}}\)dx = \(\frac{1}{a}\)tan^{-1}\(\frac{x}{a}\)

= tan^{-1}(tan\(\frac{x}{2}\) + 1) + C

OR

Put 1 + tan\(\frac{x}{2}\) = t

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

or sec^{2} \(\frac{x}{2}\)dx = 2dt

Also, when x = 0, t = 1

when x = \(\frac{π}{2}\) t =2

∴ I_{1} = ∫_{1}^{2}\(\frac{d t}{t}\)

= [log t]_{1}^{2}

= log 2

Substituting the value of I_{1} in (iii), we get

2I = \(\frac{π}{2}\) – Log 2

⇒ I = \(\frac{π}{2}\) – \(\frac{1}{2}\)Log 2

Question 12.

Find the area bounded by the curve y = √x, x = 2y + 3 in the first quadrant and x-axis. (4)

Answer:

Given: y= √x and x = 2y + 3 in the first quadrant

Putting x = 2y + 3 in y = √x we get

y = \(\sqrt{2 y+3}\)

⇒ y^{2} = 2y+3

⇒ y^{2} – 2y – 3 = 0

⇒ y^{2} – 3y + y – 3 = 0

⇒ y(y – 3) + 1(y – 3) = 0

⇒ (y + 1)(y – 3) = 0

⇒ y = -1, 3

∴ x = 1.9

∴ Points of intersection of the given curve and the line are (1, -1) and (9, 3).

∴ Required area of shaded region

Question 13.

If î + ĵ + k̂, 2î +5ĵ, 3î + 2 ĵ – 3k̂ and î – 6ĵ – k̂ are the position vectors of points A, B, C and D respectively, then find the angle between the vectors \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{CD}}\). (4)

Answer:

We have, \(\overrightarrow{\mathrm{AB}}\) = (2 – 1)î + (5 – 1)ĵ + (0 – 1)k̂

= î + 4ĵ – k̂

And

\(\overrightarrow{\mathrm{CD}}\) = (1 – 3)î +(-6 – 2) + (-1 + 3)k̂

= -2î – 8ĵ + 2k̂

Let θ be the angle between the vectors \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{CD}}\)

Hence the angle between the vectors \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{CD}}\) is π.

Case-Based/Data-Based

Question 14.

In an examination, an examinee either guesses or copies or knows the answer of MCQs with four choices. The probability that he makes a guess is \(\frac{1}{3}\), and the probability that he copies answer is \(\frac{1}{6}\). The probability that his answer is correct given that he copied it, is \(\frac{1}{8}\)

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

(A) What is the probability that he answers the question correctly? (2)

Answer:

Let E_{1}, E_{2}, E_{3,} and E be the events defined as follows:

E_{1}: Examinee guesses the answer.

E_{2}: Examinee copies the answer.

E_{3}: Examinee knows the answer.

E: Examinee answers the question correctly.

∴ P(E_{1}) = \(\frac{1}{3}\)

P(E_{2}) = \(\frac{1}{6}\)

P(E_{3}) = 1 – \(\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{1}{2}\)

Also, = (Since, a question has four choices, out of which only one is correct.)

P\(\left(\frac{E}{E_{2}}\right)=\frac{1}{8}\)

P\(\left(\frac{E}{E_{3}}\right)\) = 1

(When the student knows the answer, the probability of his answer being correct is 100%.)

(A) P(E) = P(E_{1}) × P\(\left(\frac{E}{E_{1}}\right)\) + P\(\left(\frac{E}{E_{2}}\right)\) + P(E_{3}) × P\(\left(\frac{E}{E_{3}}\right)\)

(B) Given that he answered the question correctly, what is the probability that he copied the answer? (2)

Answer: