CBSE Sample Papers for Class 12 Maths Term 2 Set 8 with Solutions

Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 8 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Maths Term 2 Set 8 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.
Evaluate : ∫\(\frac{3 x^{2}}{x^{6}+1}\) dx
OR
Find the value of integral: ∫\(\frac{\tan ^{3} x}{\cos ^{3} x}\)dx
Answer:
Let, I = ∫\(\frac{3 x^{2}}{x^{6}+1}\) dx = ∫\(\frac{3 x^{2}}{\left(x^{3}\right)^{2}+1}\)dx
Put x³ = t
⇒ 3x³dx = dt
∴ I = ∫\(\frac{d t}{t^{2}+1}\) = tan^-1 t + C
= tan^-1(x³) + C

OR

Now, let, cos x = t
⇒ -sin x dx = dt
⇒ sin x dx = dt

Substituting t = cos x, we get
⇒ I = \(\frac{1}{5 \cos ^{5} x}-\frac{1}{3 \cos ^{3} x}\) + C

Therefore, ∫\(\frac{\tan ^{3} x}{\cos ^{3} x}\) dx = \(\frac{1}{5 \cos ^{5} x}-\frac{1}{3 \cos ^{3} x}\) + C

Question 2.
If θ is the angle between two vectors î – 2ĵ + 3k̂ and 3î – 2ĵ + k̂ find sin θ.
Answer:
\(\vec{a}\) = î – 2ĵ + 3k̂
\(\vec{b}\) = 3î – 2ĵ + k̂

\(\vec{a} \times \vec{b}\) = \(\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
3 & -2 & 1
\end{array}\right|\) = 4î + 8ĵ + 4k̂

Question 3.
Solve the following differential equation :
\(\frac{d y}{d x}\) = x³ cosec y, given that y(0) = 0
Answer:
\(\frac{d y}{d x}\) = x³ cosec y, given that y(0) = 0
⇒ ∫\(\frac{d y}{\operatorname{cosec} y}\) = ∫x³ . dx
⇒ ∫ sin y dy = ∫x³ . dx
⇒ – cos y = \(\frac{x^{4}}{4}\) + C
Puffing x = 0, y = 0
⇒ -1 = 0 + c
⇒ c = -1
∴ – cos y = \(\frac{x^{4}}{4}\) – 1
⇒ cos y = 1 – \(\frac{x^{4}}{4}\)

Question 4.
A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in the number less than 4.
Answer:
S = {(1, 1), (1, 2) ……………. (6, 6)}
∴ n(s) = 36
A = Red die resulted in a number less than 4.
= {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)}
n(A) = 18
B = Sum of number is 8.
B = {(4, 4), (6, 2), (2, 6), (5, 3), (3, 5)}
n(B) = 5
A ∩ B = ((5, 3), (6, 2))
n(A ∩ B) = 2
∴ P(B/A) = Probability of sum of number 8 when red die resulted in a number less than 4
= \(\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})}=\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~A})}=\frac{2}{18}=\frac{1}{9}\)

Question 5.
Find the direction cosines of the line, which is perpendicular to the lines whose direction consines are proportional to 1, – 1, 2 and 2,1, -1.
Answer:
Let l,m, n be the direction cosines of the required line.
Then l(1) + m(-1) + n(2) = 0
and l(2) + m(1)+ n(-1)=O
l – m + 2n = 0 ……….(1)
and 2l + m – n = 0 ………..(2)
On solving (1) and (2) by cross multiplication, we get

Question 6.
A bag contains 5 green and 12 blue marbles. A second bag contains 7 green and 9 blue marbles. One marble is picked at random from the first bag and mixed up with marbles in the second bag. Then, a marble is randomly picked from it. Find the probability that the picked marble is green.
Answer:
Let us consider the events :
X marble picked from first bag is green
Y = marble picked from first bag is blue and
A = marble picked from second bag is green
Now, P(X) = \(\frac{5}{17}\), P(Y) = \(\frac{12}{17}\)

Also, probability of picking up a green marble from second bag, if X has already occured.
P\(\left(\frac{\mathrm{A}}{\mathrm{X}}\right)\) = \(\frac{8}{17}\)
Similarly P\(\left(\frac{\mathrm{A}}{\mathrm{Y}}\right)\) = \(\frac{7}{17}\)

∴ By the law of total probability, we get
P(A) = P(X)P\(\left(\frac{\mathrm{A}}{\mathrm{X}}\right)\) + P(Y)P\(\left(\frac{\mathrm{A}}{\mathrm{Y}}\right)\)
= \(\frac{5}{17} \times \frac{8}{17}+\frac{12}{17} \times \frac{7}{17}\)
= \(\frac{40}{289}+\frac{84}{289}\)
= \(\frac{124}{289}\)

Section – B

Question 7.
Show that the direction consines of a vector equally inclined to the axes OX, OY and OZ are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)
Answer:
Let a vector be equally inclined to axes OX, OY and OZ at angle a.
Then, the direction cosines of the vector are cos a, cos a and cos a.
Now, cos² a + cos² a + cos² a = 1
⇒ 3 cos² a = 1
⇒ cos a = \(\frac{1}{\sqrt{3}}\)
Hence, the direction cosines of the vector which is equally inclined to the axes are \(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\)

Question 8.
Find the value of λ, so that following lines are perpendicular to each other.
\(\frac{x+5}{5 \lambda+2}=\frac{2-y}{5}=\frac{1-z}{-1}\) and \(\frac{x}{1}=\frac{2 y+1}{4 \lambda}=\frac{1-z}{-3}\)
OR
Find the distance between the point P(6, 5, 9) and the plane determined by the point A(3, – 1, 2), B(5, 2, 4) and C(- 1, -1, 6).
Answer:
Given equation of lines are

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

OR

Let A, B, C be the three points in the plane. D is the foot of perpendicular drawn from a point P to the plane. PD is the required distance to be determined, which is the projection of \(\overrightarrow{\mathrm{AP}}\) on \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}\).

Hence, PD = the dot product of \(\overrightarrow{\mathrm{AP}}\) with the unit vector along \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}\)
So, \(\overrightarrow{\mathrm{AP}}\) = 3i + 6j + 7k

Question 9.
Evaluate ∫^3/2 ₀|x cos πx| dx.
Answer:
0 < x < \(\frac{1}{2}\) ⇒ x cos πx > 0
\(\frac{1}{2}\) < x < \(\frac{3}{2}\) ⇒ x cos π x < 0

Question 10.
Find the general solution of the following differential equation :
xdy – (y + 2x²)dx = 0
OR
Find the particular solution of the differential equation.
\(\frac{d y}{d x}\) + y cot x = 2x + x² cot x(x ≠ 0), where y\(\left(\frac{\pi}{2}\right)\) = 0
Answer:
xdy – (y + 2x²)dx = 0
⇒ \(\frac{d y}{d x}-\frac{y}{x}\) = 2x

Comparing with \(\frac{d y}{d x}\) +Py = Q
P = – \(\frac{1}{x}\), Q = 2x
∴ I.F = e^∫Pdx = e^{∫1/x dx} = e^{-log x}
= x^-1 = \(\frac{1}{x}\)

Solution is given by
y(I.F) = ∫Q(I.F) dx
⇒ \(\frac{y}{x}\) = ∫(2x × \(\frac{1}{x}\)) dx
⇒ \(\frac{y}{x}\) = 2x +C
⇒ y = 2x² + Cx

OR

The given equation is a linear differential equation of the type \(\frac{dy}{dx}\) + Py = Q.
where P = cot x and Q = 2x + x² cot x
Therefore, I.F. = e^{∫cot x} dx = e^{∫l0g sin x} = sin x

Hence, the solution of the differential equation is given by
y sinx = ∫(2x + x² cotx) sin x dx + C
or y sinx = ∫2x sinx dx + ∫ x² cosx dx + C
or y sinx = sin x\(\left(\frac{2 x^{2}}{2}\right)\) – ∫cosx\(\left(\frac{2 x^{2}}{2}\right)\) dx + ∫x² cos x dx + C
or y sinx = x² sinx – ∫x² cosx dx + ∫x²cos x dx + C
or y sinx = x²sin x + C
Substituting y = 0 and x = \(\frac{π}{2}\) in equation (1), we get
0 = \(\left(\frac{\pi}{2}\right)^{2}\) sin \(\left(\frac{\pi}{2}\right)\) + C
C = \(\frac{-\pi^{2}}{4}\)
or
Substituting the value of C in equation (1), we get
y sin x = x² sin x – \(\frac{\pi^{2}}{4}\)
or
y = x² – \(\frac{\pi^{2}}{4 \sin x}\) (sin x ≠ 0)
which is the particular solution of the given differential equation.

Section – C

Question  11.
A person is known to speak lie 2 out of 5 times. He throws a die and report that it is a number greater than 3. Find the probability that it is actually a number greater than 3.
Answer:
Let A be the event that a number greater than 3 and B be the event that a number is not greater than 3. Sample space, S = (1, 2, 3, 4, 5, 6}
Then, A = {4, 5, 6} and B = {1, 2, 3}
P(A) = P(Number greater than 3)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)
and p(B) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Let E be the event that the person throws a die and report that it is number greater than 3.
∴ \(P\left(\frac{E}{A}\right)=\frac{3}{5}\)
If he speaks lie 2 out of 5 times, then it means he speaks truth 3 out of 5 times.
\(P\left(\frac{E}{B}\right)\) = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)
By baye’s theorem, required probability

Question 12.
Find the perpendicular distance of the point (2, 3, 4) from the line \(\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}\) Also, find coordinates of foot of perpendicular.
OR
Find the coordinates of the point where the line through the points (3, – 4, – 5) and (2, – 3, 1) crosses the plane determined by the points (1, 2, 3), (4, 2, – 3) and (0, 4, 3).
Answer:
The given equation of line is \(\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}\) which can be written in standard form as :
\(\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}\) ……… (i)
Let T be any random point on equation (i), then its coordinates are calculated as follows:

\(\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}\) = λ (say)
⇒ \(\frac{x-4}{-2}\) = λ, \(\frac{y}{b}\) = λ, \(\frac{z-1}{-3}\) = λ
⇒ x= – 2λ + 4, y = 6λ, z = – 3λ + 1
Now, direction ratios of line PT = (- 2λ + 4 – 2, 6λ – 3, – 3λ + 1 – 4)
= (- 2λ + 2, 6λ – 3, – 3λ – 3)
∵ PT ⊥ AB
∴ a₁a₂ + b₁b₂ + c₁c₂ = 0
where a₁ = – 2λ + 2, b₁ = 6λ – 3, c₁ = – 3λ – 3
a₂ = – 2, b₂ = 6, c₂ = – 3
∴ – 2(- 2λ + 2) + 6(6λ – 3) – 3(- 3λ – 3) = 0
⇒ 4λ – 4 + 36λ – 18 + 9λ + 9 = 0
⇒ 49λ – 13 = 0
⇒ λ = \(\frac{13}{49}\)

OR

Equation of the line through the points (3, – 4, – 5) and (2, – 3, 1) is given by
\(\frac{x-3}{-1}\) = \(\frac{y+4}{1}\) = \(\frac{z+5}{6}\) = k
[∵ Equation of line passing through two points is \(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)]
⇒ x = 3 – k, y = k – 4, z = 6k – 5 is a point on the line(i)
⇒ Equation of the plane with point (1, 2, 3) is given as
A(x – 1) + B(y – 2) + C(z – 3) = 0 ……… (ii)
∵ (4, 2, – 3) lies on the plane (ii),
∴ 3A + 0B – 6C = 0
⇒ 3A = 6C
⇒ A = 2C …….. (iii)
Also, (0, 4, 3) lies on the plane (ii)
⇒ A + 2B + 0C = 0
⇒ A = 2B
∴ A = 2B = 2C
B = C [using (iii)]
∴ Equation (ii) becomes
⇒ 2C(x – 1) + C(y – 2) + C(z – 3) = 0
⇒ 2x – 2 + y – 2 + z – 3 = 0
⇒ 2x + y + z – 7 = 0
⇒ 2x + y + z = 7 ……. (iv)
Since line (i) crosses the plane (iv)
∴ 2(3 – k) + (k – 4) + (6k – 5) = 7
⇒ 6 – 2k + k – 4 + 6k – 5 = 7
⇒ 5k – 3 = 7
⇒ 5k = 10
⇒ k = 2
Hence, the required point is (3 – 2, 2 – 4, 12 – 5) i.e., (1, – 2, 7).

Question 13.
Evaluate:

Answer:

Again putting tan θ = t, then sec²θ dθ = dt
where θ = 0, then t = 0
and θ = \(\frac{\pi}{6}\), then t = \(\frac{1}{\sqrt{3}}\)

Question 14.
Two roads are to be constructed along the curve y² = 2x and x² = 2y which are intersected with each other. The common portion is to be covered by the green grass and planting colour flower plants shown below:

Attempt the following questions on the basis of the above case study :
(i) Find the points at which both the curves intersect with each other.
(ii) Which area is to be planted?
Answer:
Given curves are
y² = 2x
⇒ y = √2x
and x² = 2y
Putting the value of y in (i),
\(\left(\frac{x^{2}}{2}\right)^{2}\) = 2x
⇒ \(\frac{x^{4}}{4}\) = 2x
⇒ x⁴ = 8x
⇒ x⁴ – 8x = 0
⇒ x(x³ – 8) = 0
⇒ x = 0 or x = 2.
Thus, x = 0 or x = 2.
When x = 0, then y = 0
When x = 2, then y = 2
Hence points of intersection are (0, 0) and (2, 2)

(ii) Given function will be integrated w.r.t x from 0 to 2