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

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

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: ∫\(\frac{1}{\cos ^{2} x(1-\tan x)^{2}}\) dx
OR
Evaluate: ∫\(\frac{d x}{1-\tan x}\)
Answer:
We have given

Put cos x – sin x = t
Differentiating both sides, we get
⇒ (- sin x – cos x) dx = dt
⇒ (sin x + cos x) dx = – dt
∴ I = \(\frac{-1}{2} \int \frac{d t}{t}+\frac{1}{2} \int d x\)
⇒ I = \(\frac{-1}{2}\)log |t| + \(\frac{x}{2}\) + C
⇒ I = \(\frac{-1}{2}\) log |cos x – sin x| + \(\frac{x}{2}\) + C

Question 2.
Determine the order and degree of the following differential equation \(\frac{d^{2} y}{d x^{2}}\) = cos 3x + sin 3x
Answer:
⇒ \(\frac{d^{2} y}{d x^{2}}\) – cos 3x – sin 3x = 0
The highest order derivative present in the differential equation is \(\frac{d^{2} y}{d x^{2}}\). Therefore, its order is two.
It is a polynomial equation in \(\frac{d^{2} y}{d x^{2}}\) and the power raised to \(\frac{d^{2} y}{d x^{2}}\) is 1.
Hence, its degree is one.

Question 3.
Find a vector r equally inclined to the three axes and whose magnitude is 3√3 units.
Answer:
We have
|\(\vec{r}\)| = 3√3
Since, \(\vec{r}\) is equally inclined to the three axes, direction consines of the unit vector \(\vec{r}\) will be same.
i.e. l = m = n
Now we know that,
l² + m² + n² = 1
⇒ l² + l² + l² = 1 (∴ l = m = n)
⇒ 3l² = 1
⇒ l² = \(\frac{1}{3}\)

Question 4.
Find the value of X such that the line \(\frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{-4}\) is perpendicular to the plane 3x – y – 2z = 7.
Answer:
Given that line \(\frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{-4}\) is perpendicular to plane 3x – y – 2z = 7
∴ \(\frac{6}{3}=\frac{\lambda}{-1}=\frac{-4}{-2}\)
When a line is perpendicular to a plane, their direction ratios are proportional]
⇒ 2 = – λ = 2
⇒ – λ = 2
⇒ λ = – 2

Question 5.
Three persons A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4, 0.3 and 0.2 respectively. Find the probability of two hits.
Answer:
Here,
P(A) = 0.4, P(Ā) = 0.6, P(B) = 0.3,
P(B̄) = 0.7, P(C) = 0.2 and P(C̄) = 0.8
∴ Probability of two hits = P(A).P(B).P(C̄) + P(A).P(B̄) + P(C) + P(Ā).P(B).P(C)
= 0.4 × 0.3 × 0.8 + 0.4 × 0.7 × 0.2 + 0.6 × 0.3 × 0.2
= 0.096 + 0.056 + 0.036 = 0.188

Question 6.
A die is rolled. Consider the events
A = {2, 4, 6}, B = {4, 5}, C = (3, 4, 5, 6}
Find P[A ∪ B/C].
Answer:
Given that, A = {2, 4, 6), B = {4, 5}, C = {3, 4, 5, 6}
Now, A ∪ B = {2, 4, 6} ∪{4, 5} = {2, 4, 5, 6}
So, P(A ∪ B) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
Now, (A ∪ B) ∩ C = {2, 4, 5, 6} ∩ {3, 4, 5, 6} = {4, 5, 6}
So, P[(A ∪ B) ∩ C] = \(\frac{3}{6}=\frac{1}{2}\)
Also P(C) = \(\frac{4}{6}=\frac{2}{3}\)
Required probability = P[A ∪ B)/C] = \(\frac{\mathrm{P}[(\mathrm{A} \cup \mathrm{B}) \cap \mathrm{C}]}{\mathrm{P}(\mathrm{C})}\)
= \(\frac{1 / 2}{2 / 3}=\frac{1}{2} \times \frac{3}{2}=\frac{3}{4}\)

Section – B

Question 7.
Evaluate:

Answer:

Question 8.
Solve the following differential equation :
\(\frac{d y}{d x}\) + y = cos x – sin x
OR
Given that \(\frac{d y}{d x}\) = e^-2y and if x = 5, then y = 0. When y = 3, then find the value of x.
Answer:
Given differential equation is dy
\(\frac{d y}{d x}\) + y = cos x – sin x ……..(1)
Here P = 1, Q = cos x – sin x.
I.F. = e^{∫1 dx} = e^x
Solution of given equation will be given by
y.I.F. = ∫Q × I.F.dx + C
⇒ y.e^x = ∫e^x(cos x – sin x) dx + C}
⇒ y.e^x = ∫e^x cos x dx – ∫e^x sin x dx + C
⇒ y.e^x = – e^x sin x – ∫e^x.(- sin x)dx – ∫e^x sin x dx + C
⇒ y.e^x = – e^x sin x + C
⇒ y = – sin x + ce^{– x}

Given:
\(\frac{d y}{d x}\) = e^-2y
⇒ \(\frac{d y}{e^{-2 y}}\) = dx
On integrating both sides
⇒ ∫e^2ydy = ∫dx
⇒ \(\frac{e^{2 y}}{2}\) = x + C
When x = 5 and y = 0, then putting these values in equation (1),
\(\frac{e^{0}}{2}\) = 5 + C
⇒ \(\frac{1}{2}\) = 5 + C
⇒ C = \(\frac{1}{2}\) – 5 = \(\frac{-9}{2}\)

Question 9.
Using vectors, find the area of a triangle with vertices A (1, 1, 2), B (2, 3, 5) and C(1, 5, 5).
Answer:
A(1, 1, 2), B(2, 3, 5) and C (1, 5,5) are the vertices of the given triangle.
\(\overrightarrow{\mathrm{AB}}\) = (2 – 1) î + (3 – 1)ĵ + (5 – 2)k̂
∴ \(\overrightarrow{\mathrm{AB}}\) = î + 2ĵ + 3k̂
and \(\overrightarrow{\mathrm{AC}}\) = (1 – 1)î + (5 – 1)ĵ + (5 – 2)k̂
∴ \(\overrightarrow{\mathrm{AC}}\) = 0î + 4ĵ + 3k̂
Area of ∆ ABC = \(\frac{1}{2}\)

Question 10.
Find the shortest distance between the following pair of lines:
\(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\) and \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\)
OR
Find the cartesian equation of the line passing through the point A (1, 2, -4) and perpendicular to the lines:
Answer:
Given lines are
\(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
and \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\)
Here, (a₁, b₁, c₁) = (1, -2, 1); x₁ = 3, y₁ = 5, z₁ = 7
and (a₂, b₂, c₂) = (7, -6, 1); x₂ = -1, y₂ = -1, z₂ = -1
Now, we know that shortest distance between two lines is given by

Hence, the required shortest distance is 2√29 units.

Equation of line passing through A(1, 2, – 4) and parallel to \(\vec{b}\) = (b₁î + b₂ĵ + b₃k̂)

Since line (1) is perpendicular to the lines (2) and (3).
∴ 2b₁ + 3b₂ + 4b₃ = 0
and b₁ – 3b₂ + 5b₃ = 0
On solving (4) and (5), we get

Section – C

Question 11.
Evaluate: ∫\(\frac{3 x+4}{(x-1)(x+2)(x-3)}\)dx.
Answer:

⇒ (3x + 4) = A(x + 2) (x – 3)+ B (x – 3) (x – 1) + (x + 2) (x – 1) ……… (2)
Put x = 1 in (2)
⇒ 7 = A(3) (- 2) + B (0) + C(0)
⇒ 7 = – 6A
⇒ A = –\(\frac{7}{6}\)
Put x = 3 in (2)
⇒ 13 = 0 + 0 + C (5)(2)
⇒ C = \(\frac{13}{10}\)
Put x = – 2 in (2)
⇒ – 2 = B(- 3) (- 5)
⇒ B = \(\frac{-2}{15}\)
Putting these values in (1)

Question 12.
Find the area of the region enclosed between the parabola 4y = 3x² and the line 3x – 2y + 12 = 0.
OR
Find the area of the region quadrant enclosed by x-axis, line x = √3y and the circle x² + y² = 4.
Answer:
Given 4y = 3x²
and y = \(\frac{3}{2}\)x + 6
Solving (1) and (2), we get
x = – 2 or x = 4

The area of region bounded by the circle, x² + y² = 4, x = √3y, and the x-axis is the area OAB.
The point of intersection of the line and the circle in the first quadrant is √3, 1
Area ∆ OAB = Area of ∆ OCA + Area of ACB
∴ Area of ∆ OCA = \(\frac{1}{2}\) × OC × AC = \(\frac{1}{2}\) × √3 × 1 = \(\frac{\sqrt{3}}{2}\)

Therefore area enclosed by x-axis, the line x = √3y, and the circle x² + y² = 4 in the first quadrant
= \(\frac{\sqrt{3}}{2}+\frac{\pi}{3}-\frac{\sqrt{3}}{2}=\frac{\pi}{3}\) units.

Question 13.
Find the distance of point – 2î + 3ĵ – 4k̂ from the line \(\vec{r}\) = î + 2ĵ – k̂ + λ( î + 3ĵ – 9k̂) measured parallel to the plane x – y + 2z – 3 = 0.
Answer:
Let the given point be A (- 2î + 3ĵ – 4k̂)
Equation of line is,
\(\vec{r}\) = î + 2ĵ – k̂ + λ( î + 3ĵ – 9k̂) …….. (i)
Let B be the point on the line such that AB is parallel to plane
x-y + 2z – 3 = 0
In cartesian form, equation (1) can be written as,
So, \(\frac{x-1}{1}=\frac{y-2}{3}=\frac{z+1}{-9}\) = λ(say)
x = λ + 1,
y = 3λ + 2
z = – 9λ – 1
Let co-ordinates of B be (λ, + 1, 3λ + 2, – 9λ -1)
Now, equation of line AB can be written as,

The line AB is parallel to the given plane, so it will be perpendicular to plane’s normal.
Direction ratios of AB = (λ + 3, 3λ – 1, – 9λ + 3)
Direction ratio of plane = (1, – 1, 2)
∴ (λ + 3)(1) + (3λ – 1)(-1) + (- 9λ + 3) (2) = 0
⇒ λ + 3 – 3λ + 1 – 18λ + 6 = 0
⇒ – 20λ + 10 = 0
⇒ λ = \(\frac{1}{2}\)
So, point B is

Thus, Distance AB = \(\sqrt{\left(\frac{3}{2}+2\right)^{2}+\left(\frac{7}{2}-3\right)^{2}+\left(\frac{-11}{2}+4\right)^{2}}\)
= \(\frac{\sqrt{59}}{2}\) Units.

Question 14.
In an office three employees Vinay, Sonia and Iqbal process incoming copies of a certain from. Vinay processes 50% of the forms. Sonia processes 20% and Iqbal processes the remaining 30% of the forms. Vinay has an error rate of 0.06%, Sonia has an error rate of 0.04% and Iqbal has an error rate of 0.03%.

Based on the above information answer the following:
(i) The total probability of committing an error in processing the form is?
(ii) The manager of the company wants to do a quality check. During inspection he selects a form at random from the days output of processed forms. If the form selected at random has an error, the probability that the form is NOT processed by Vinay is?
Answer:
p(V) = 50% = \(\frac{50}{100}\)
p(S) = 50% = \(\frac{20}{100}\)
p(I) = 50% = \(\frac{30}{100}\)
Let E be the event that error occurred.
P(E/V) = \(\frac{6}{100}\)
P(E/S) = \(\frac{4}{100}\)
P(E/I) = \(\frac{3}{100}\)

(i) P(E) = P(V) P(E/V) + P(S) P(E/S) + P(I) P(E/I)

(ii)