CBSE Sample Papers for Class 12 Maths Term 2 Set 5 with Solutions

Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 5 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Maths Term 2 Set 5 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:

OR
Evaluate: ∫ sin³ x cos³ x dx
Answer:

OR

Let, I = ∫sin³x.cos³x dx
= ∫sin³ x.cos² x.cosx dx
= ∫sin³ x.(1 – sin² x).cosx dx
Let, sin x = t,
⇒ cos x dx = df
∴ I = ∫t³(1 – t²)dt
= ∫(t³ – t⁵)dt
= \(\frac{t^{4}}{4}-\frac{t^{6}}{6}\) + C
= \(\frac{1}{4}\) sin⁴x – \(\frac{1}{6}\)sin⁶x + C

Question 2.
Determine the sum of order and degree of the given differential equation.
(y”‘)² + (y”)³ + (y’)⁴+ y⁵ = 0.
Answer:
The given differential equation is
(y”‘)² + (y”)³ + (y’)⁴+ y⁵ = 0.
The highest order derivative present in the differential equation is y”‘ Therefore, its order is three.
The given differential equation is a polynomial equation in y”’, y” and y’.
The highest power raised to y'” is 2.
Hence, its degree is 2.
Now, we know order is 3 and degree is 2.
So, Sum of order and degree of given differential equation is 3 + 2 = 5.

Question 3.
Write a vector of magnitude 15 units in the direction of vector î – 2ĵ + 2k̂.
Answer:
Given vectors is î – 2ĵ + 2k̂
Unit vector in the direction of î – 2ĵ + 2k̂ is given by

So, Vector of magnitude 15 units in the direction of î – 2ĵ + 2k̂ will be:
15â = 15\(\left(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{3}\right)\)
= 5(î – 2ĵ + 2k̂)
= 5î – 10ĵ + 10k̂

Question 4.
If the line has the direction ratio – 18, 12, – 4, then what are its direction consines?
Answer:
The direction ratios of the lines are -18,12, – 4
Direction cosines of the lines are

Hence, direction cosines of line are \(\frac{-9}{11}, \frac{6}{11}, \frac{-2}{11}\).

Question 5.
Mother, father and son line up at random for a family photos. If A and B are two events given by A = Son on one end, B = Father in the middle, find P(B/A).
Answer:
5. If mother (M), father (F), and son (S) lineup for the family picture, then the sample space will be
S = (MFS, MSF, FMS, FSM, SMF, SFM)
A = (MFS, FMS, SMF, SFM),
∴ B = (MFS, SFM)
And A ∩ B = (MFS, SFM)
∴ P(A ∩ B) = \(\frac{2}{6}=\frac{1}{3}\),
P(B) = \(\frac{2}{6}=\frac{1}{3}\)
P(A) = \(\frac{4}{6}=\frac{2}{3}\)
∴ P(B/A) = \(\frac{P(A \cap B)}{P(A)}=\frac{1 / 3}{2 / 3}=\frac{1}{2}\)
⇒ P(B/A) = \(\frac{1}{2}\)

Question 6.
A coin is tossed 5 times. Find the probability of getting (i) at least 4 heads, and (ii) at most 4 heads.
Answer:
Total number of probability of tossing a coin 5 times is 32
(i) Probability of getting atleast 4 heads.

(ii) Probability of getting at most 4 heads.

Section – B

Question 7.
Evaluate: ∫x sin x dx.
Answer:
Let, Using integration by parts,
∫f(x)g(x)dx = f(x)∫g(x)dx -∫f'(x)[∫g(x)dx]dx
So, we take
f(x) = x and g(x) = sin x
So, I = x∫sin x dx – ∫\(\frac{d(x)}{d x}\) (∫sin x dx) dx
⇒ I = x (- cos x) – ∫1.(- cosx) dx
⇒ I = – x cos x + ∫cos x dx
⇒ I = – x cos x + sin x + C

Question 8.
Solve the differential equation
\(\frac{d y}{d x}\) = 1 + x + y² + xy², if y = 0 when x = 0
OR
Solve the differential equation:
x log x \(\frac{d y}{d x}\) + y = 2 log x.
Answer:
Given differential equation is,
\(\frac{d y}{d x}\) = 1 + x + y² + xy²
⇒ \(\frac{d y}{d x}\) = (1 + x) + y²(1 + x)
⇒ \(\frac{d y}{d x}\) = (1 + x) (1 + y²)
⇒ \(\frac{1}{1+y^{2}}/latex]dy = (1 + x)dx
Integrating on both sides
∫[latex]\frac{1}{1+y^{2}}/latex] = ∫(1 + x)dx
⇒ tan^-1y = x + [latex]\frac{x^{2}}{2}\) + C
⇒ y = tan \(\left(x+\frac{x^{2}}{2}+C\right)\)
It is given that, y = 0, when x = 0
∴ 0 = tan (0 + C)
⇒ C = 0
∴ y = tan \(\left(x+\frac{x^{2}}{2}\right)\)

OR

Given differential equation is,

Question 9.
If \(\vec{a}\) = \(2 \vec{i}+4 \hat{j}-5 \hat{k}\) and \(\vec{b}\) = î + 2ĵ + 3k̂, find the unit vector in the direction \(\vec{a}\) + \(\vec{a}\)
Answer:
We have,
\(\vec{a}+\vec{b}\) = (2î + 4ĵ – 5k̂) + (î + 2ĵ + 3k̂)
= 3î + 6ĵ – 2k̂
⇒ |\(\vec{a}+\vec{b}\)| = \(\sqrt{(3)^{2}+(6)^{2}+(-2)^{2}}\)
= \(\sqrt{9+36+4}\) = √49
= 7
Hence, the unit vector in the direction of \(\vec{a}+\vec{b}\)

Question 10.
Find the foot of perpendicular from the point (1, 6, 3) on the line;
x = \(\frac{y-1}{2}=\frac{z-2}{2}\)
OR
If vector \(\vec{a}\) = 2î + 2ĵ + 3k̂, \(\vec{b}\) = – î + 2ĵ + k̂ and \(\vec{c}\) = 3î + ĵ are such that \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c}\), then find the value of λ.
Answer:
Given line is,
x = \(\frac{y-1}{2}=\frac{z-2}{3}\)
It can be written as
\(\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{3}\) = λ(say)
So, x = λ, y = 2λ + 1 and z = 3λ + 2.
∴ Coordinates of any general point which lines on the given line are (λ, 2λ + 1, 3λ + 2).
∴ Coordinates of foot of perpendicular are (λ, 2λ + 1,3λ + 2).
Now, direction ratios of line joining (1,6,3) and foot of perpendicular (λ, 2λ + 1, 3λ + 2) will be,
λ – 1, 2λ + 1 – 6, 3λ + 2 – 3 = λ – 1, 2λ – 5, 3λ – 1.
Also, direction ratios of given line are 1, 2, 3.
Since, line joining (1, 6, 3) and foot of perpendicular is perpendicular to given line.
∴ 1(λ – 1) + 2 (2λ – 5) + 3 (3λ – 1) = 0
⇒ λ – 1 + 4λ – 10 + 9λ – 3 = 0
⇒ 14λ, -14 = 0
⇒ λ = 1
∴ Co-ordinates of foot of perpendicular
= (λ, 2λ + 1, 3λ + 2)
= (1, 2(1) + 1, 3(1)+ 2)
= (1, 3, 5)

OR

Given that \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c}\).
\((\vec{a}+\lambda \vec{b}) \cdot \vec{c}\) = o …… (1)
Now, given that
\(\vec{a}\) = 2î + 2ĵ + 3k̂
\(\vec{b}\) = – î + 2ĵ + k̂
and \(\vec{c}\) = 3î + ĵ
∴ \(\vec{a}\) + λ \(\vec{a}\) = (2 z + 2 y+3 fc) + A,(~ z + 2 j + λ)
\(\vec{a}\) + λ\(\vec{b}\) = i(2 – λ) + j (2 + 2λ) + k(3 + λ)
From equation (1), we get
[î(2 – λ) + ĵ(2 + 2λ) + k̂(3 + λ)].[3 î + ĵ] = 0
⇒ 3(2 – λ) + 1(2 + 2λ) =0
⇒ 6 – 3λ + 2 + 2λ = 0
⇒ 8 – λ = 0
λ = 8

Section – C

Question 11.
Evaluate:

Answer:
Given,

Question 12.
Find the area of region bounded by the curve y² = 2y – x and the Y – axis.
OR
Find the area of region bounded by the curves:
x² = y and y = |x|.
Answer:

Given equation of curve is,
y² = 2y – x
⇒ y² – 2y = – x
⇒ y² – 2y + 1 = – x + 1
⇒ (y – 1)² = – (x – 1)
It is left handed parabola with vertex (1, 1).
At Y – axis, x = 0
∴ y² – 2y – 0 (from(1))
⇒ y (y – 2) = 0
⇒ y = 0
or y = 2

Given curves are
y = x²
and y = |x|
⇒ x² – |x| = 0
⇒ |x| {|x| – 1} = 0
⇒ |x| = or |x| = 1
∴ x = 0, x = – 1 or + 1
If x = 0 y = 0
If x = ± 1 y = 1

Question 13.
Find the shortest distance between the lines
\(\overrightarrow{r_{1}}\) = 3î + 2ĵ – 4k̂ + λ(î + 2ĵ + 2k̂)
and \(\overrightarrow{r_{1}}\) = 5î – 2ĵ + µ(3î + 2ĵ + 6k̂)
If the line intersect find their point of intersection.
Answer:
Given line are:

⇒ \(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\) = 8î – 4k̂
∴ \(\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\) = (2î – 4ĵ + 4k̂).(8î – 4k̂)
= 16 – 16 = 0
∴ Shortest distance between given lines is
\(\mid \frac{\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\vec{b}_{1} \times \overrightarrow{b_{2}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\) = 0 units.
Hence, the lines are intersecting for points of intersection.
Equation (1) and (2)
3î + 2ĵ – 4k̂ + λ(î + 2ĵ + 2k̂) = 5î – 2ĵ + µ(3î + 2ĵ + 6k̂)
∴ 3 + 2λ = 5 + 3µ
⇒ λ = 3µ + 2
2 + 2λ = – 2 + 2µ
λ = µ – 2
and – 4 + 2λ = 6µ
∴ λ = 3µ + 2
⇒ 2µ = – 4
∴ µ = – 2
and λ = – 4
Putting in (1), we get
\(\overrightarrow{r_{1}}\) = 3î + 2ĵ – 4k̂ – 4(î + 2ĵ + 2k̂)
⇒ \(\overrightarrow{r_{1}}\) = – î – 6ĵ – 12k̂
∴ Point of intersection is (- 1, – 6, – 12)

Question 14.
A glass jar contains twenty white balls of plastic numbered from 1 to 20, ten red balls of plastic numbered from 1 to 10, forty yellow balls of plastic numbered from 1 to 40 and ten blue balls of plastic numbered from 1 to 10. If these 80 balls of plastic are thoroughly shuffled so that each ball has the same probability of being drawn.

Based on the above information answer the following:
(i) Determine the probabilities of drawing a ball of plastic that is red or yellow and numbered 1, 2, 3 or 4.
(ii) Discuss the probabilities of drawing a plastic ball which is numbered 5, 15, 25 or 35.
Answer:
As we have given;
Total number of balls in jar = 80
20 white balls of numbered 1 to 20
i.e., 1 white ball contains one number,
10 Red balls of numbered 1 to 10
i.e., 1 Red ball contains one number,
Similarly, Yellow ball contains one number per ball and
Blue ball contains one number per ball.

(1) Number of Red balls of numbered 1,2,3, and 4 = 4
Number of Yellow balls of numbered 1,2,3 and 4 = 4
P(Red or yellow and numbered 1, 2, 3 or 4) = P(Red numbered 1, 2, 3 or 4) + P(Yellow numbered 1, 2, 3 or 4)
= \(\frac{4}{80}+\frac{4}{80}\)
= \(\frac{8}{80}\)
= \(\frac{1}{10}\)

(ii) Number of White balls of numbered 5, 15 = 2,
Number of Red balls of numbered 5 = 1,
Number of Yellow balls of numbered 5, 15, 25 and 35 = 4,
Number of Blue balls of numbered 5 = 1
P(numbered 5, 15, 25 or 35) = P(5) + P(15) + P(25) + P(35)
= P(White) + P(RecI) + P(Yellow) + P(Blue)
= \(\frac{2}{80}+\frac{1}{80}+\frac{4}{80}+\frac{1}{80}\)
= \(\frac{8}{80}\)
= \(\frac{1}{10}\)