Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 7 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Maths Term 2 Set 7 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:
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 1
OR
Evaluate:
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 2
Answer:
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 3

CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with Solutions

Question 2.
A refrigerator box contains 2 milk chocolates and 4 dark chocolates. Two chocolates are drawn at random. Find the probability distribution of the number of milk chocolates. What is the most likely outcome? Find the distance between the two planes 2x + 3y + 4z = 4 and 4x + 6y + 8z = 16.
Answer:
Number of milk chocolates = 2
Number of dark chocolates = 4
Number of chocolate drawn = 2
Let X be the random variable representing no. of milk chocolates drawn, then x can be 0, 1, 2.
p(X = 0) = \(\frac{{ }^{4} C_{2}}{{ }^{6} C_{2}}\) = \(\frac{12}{30}\) = \(\frac{6}{15}\)
p(X = 1) = \(\frac{{ }^{2} C_{1} \times{ }^{4} C_{1}}{{ }^{6} C_{2}}\) = \(\frac{16}{30}\) = \(\frac{8}{15}\)
p(X = 2) = \(\frac{{ }^{2} \mathrm{C}_{2}}{{ }^{6} \mathrm{C}_{2}}\) = \(\frac{2}{30}\) = \(\frac{1}{15}\)

Question 3.
Find the distance between the two planes 2x + 3y + 4z = 4 and 4x + 6y + 8z = 16.
Answer:
Given planes are,
2x + 3y + 4z = 4
and 4x + 6y + 8z = 16
or 2x + 3y + 4z = 8
From the above equations, it can be seen that given planes are parallel.
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 4

Question 4.
Find the sum of order and degree of the differential equation:
\(\left(\frac{d^{4} y}{d x^{4}}\right)^{2}\) = \(\left[x+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
Answer:
Since, \(\left(\frac{d^{4} y}{d x^{4}}\right)^{2}\) = \(\left[x+\left(\frac{d y}{d x}\right)^{2}\right]^{3}\)
\(\left(\frac{d^{4} y}{d x^{4}}\right)^{2}\) = x3 + 3x2\(\left(\frac{d y}{d x}\right)^{2}\) + 3x\(\left(\frac{d y}{d x}\right)^{4}\)
The equation represents a fourth order differential equation.
The highest power raised to \(\frac{d^{4} y}{d x^{4}}\) is 2.
Hence, degree of the differential equation is 2.
So, the sum of order and degree of differential equation is
= 4 + 2
= 6.

CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with Solutions

Question 5.
Ram can find a solution of 60% of the problems given in mathematics examination paper and Shyam can find out 80% of it. What is the probability that atleast one of them will solve the problem that is selected at random?
Answer:
Let us consider the events:
A = Ram finds the solution
B = Shyam finds the solution
Since, A and B are independent events.
So P(A) = \(\frac{60}{100}=\frac{3}{5}\) and P(B) = \(\frac{80}{100}=\frac{4}{5}\)
∴ Required probability = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B) [∵ A and B are independent events]
= P(A) + P(B) – P(A) P(B) [.∴ P(A ∩ B) = P(A) P(B)]
= \(\)
= \(\frac{7}{5}-\frac{12}{25}\)
= \(\frac{23}{25}\).

Question 6.
Find the magnitude of each of the two vectors \(\vec{a}\) and \(\vec{b}\), having the same magnitude such that the angle between them is 60° and their scalar product is \(\frac{9}{2} \cdot\).
Answer:
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 5

CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with Solutions

Section – B

Question 7.
If \(\vec{a}\) = iÌ‚ + jÌ‚ + kÌ‚, \(\vec{b}\) = 2iÌ‚ – jÌ‚ + 3 kÌ‚ and \(\vec{c}\) = iÌ‚ – 2 jÌ‚ + kÌ‚, find a unit vector parallel to the vector 2 \(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\).
Answer:
Given, \(\vec{a}\) = iÌ‚ + jÌ‚ + kÌ‚, \(\vec{b}\) = 2iÌ‚ – jÌ‚ + 3 kÌ‚ and \(\vec{c}\) = iÌ‚ – 2 jÌ‚ + kÌ‚
Now,
2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\) = 2(iÌ‚ + jÌ‚ + kÌ‚) – (2iÌ‚ – jÌ‚ + 3kÌ‚) + 3(iÌ‚ – 2jÌ‚ + kÌ‚)
= 3iÌ‚ – 3jÌ‚ + 2kÌ‚
∴ |2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\)| = \(\sqrt{(3)^{2}+(-3)^{2}+(2)^{2}}\)
= √22
Hence, the unit vector along 2\(\vec{a}\) – \(\vec{b}\) + 3\(\vec{c}\) is
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 6

Question 8.
Find the vector equation of the plane which passes through the point (5, 2, – 4) and perpendicular to the line with direction ratios 2, 3, -1.
OR
From the point P(1, 2,4), a perpendicular is drawn on the plane 2x + y – 2z + 3 = 0. Find the equation, the length and co-ordinates of the foot of the perpendicular.
Answer:
The position vector of point (5, 2, – 4) is \(\vec{a}\) = 5iÌ‚ + 2jÌ‚ – 4kÌ‚
The normal vector \(\vec{N}\) perpendicular to the plane is
\(\vec{N}\) = 2iÌ‚ + 3jÌ‚ – kÌ‚
The vector equation of plane is given by
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 7
Hence, it is the vedor equation of the required plane.

OR

Draw PM perpendicular upon the given plane.
Equation of the line PM through P(1, 2, 4) and perpendicular to the plane 2x + y – 2z + 3 = 0 is,
\(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-4}{-2}\) = λ (say)
x = 2λ + 1, y = λ + 2, z = – 2λ
∴ Any point on it is given by
M(2λ + 1, λ + 2, – 2λ + 4)
This point lies on the given plane.
∴ 2(2λ + 1) + (λ + 2) – 2(4 – 2λ) + 3 = 0
⇒ 4λ + 2 + λ + 2 – 8 + 4 + 3 = 0
⇒ 9λ = 1
⇒ λ = \(\frac{1}{9}\)
∴ M is \(\left(\frac{2}{9}+1, \frac{1}{9}+2,4-\frac{2}{9}\right)\).
i.e., M\(\left(\frac{11}{9}, \frac{19}{9}, \frac{34}{9}\right)\).
∴ Length of the perpendicular.
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 8

CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with Solutions

Question 9.
Evaluate: ∫\(\frac{\sin ^{6} x+\cos ^{6} x}{\sin ^{2} x \cdot \cos ^{2} x}\) dx.
Answer:
Let I = ∫\(\frac{\sin ^{6} x+\cos ^{6} x}{\sin ^{2} x \cdot \cos ^{2} x}\) dx
Here, Numerator = sin6 x + cos6 x
= (sin2 x)3 + (cos2 x)3
= (sin2 x + cos2 x) (sin4 x + cos4 x – sin2 x cos2 x) [∵ a3 + b3 = (a + b) (a2 + b2 – ab)]
= 1.[(sin2 x + cos2 x)2 – 2 sin2 x cos2 x – sin2 x cos2 x]
= 1 – 3 sin2 x cos2 x
∴ I = ∫\(\frac{1-3 \sin ^{2} x \cdot \cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x\)
= ∫\(\frac{1}{\sin ^{2} x \cos ^{2} x}\) dx – 3∫1 dx + C
Multiplying and dividing denominator by cos2 x in first integral
= ∫\(\frac{\sec ^{4} x}{\tan ^{2} x}\) dx – 3x + C
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 9

Put tan x = t
⇒ sec2 dx = dt
⇒ I = ∫\(\frac{\left(t^{2}+1\right)}{t^{2}}\) dt – 3x + C
⇒ ∫\(\left[1+\frac{1}{t^{2}}\right]\) dt – 3x + C
= t – \(\frac{1}{t}\) – 3x + C
= tan x – cot x – 3x + C.

Question 10.
Solve the following differential equation:
(x2 – 1)\(\frac{d y}{d x}\) + 2xy = \(\frac{1}{x^{2}-1}\): |x| ≠ 1
Solve the following differential equation:
(x2 + 1)\(\frac{d y}{d x}\) + 2xy = \(\sqrt{x^{2}+4}\)
Answer:
Given differential equation is
(x2 – 1)\(\frac{d y}{d x}\) + 2xy = \(\frac{1}{x^{2}-1}\)
Divide by (x2 – 1)
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 10
which is the required solution.

OR

Given differential equation is.
(x2 + 1)\(\frac{d y}{d x}\) + 2xy = \(\sqrt{x^{2}+4}\)
Divide by (x2 + 1)
\(\frac{d y}{d x}+\frac{2 x}{x^{2}+1} y\) = \(\frac{\sqrt{x^{2}+4}}{x^{2}+1}\)
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 11
It is the required solution of given differential equation.

CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with Solutions

Section – C

Question 11.
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and found to be both clubs. Find the probability of the lost card being a club.
Answer:
Let, E1 be the event that the lost card is club.
Let, E2 be the event that the lost card is not club.
Let, A be the event that both two cards drawn are clubs.
Then,
p(E1) = \(\frac{13}{52}=\frac{1}{4}\), p(E2) = \(\frac{39}{52}=\frac{3}{4}\),
When one card is lost, number of remaining cards in the pack = 51.
When E1 has occured i.e., a card of clubs is lost, then the probability of drawing 2 cards of clubs from the remaining pack.
i.e., \(\left(\frac{\mathrm{A}}{\mathrm{E}_{2}}\right)\) = \(\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{13 \times 12}{51 \times 50}=\frac{26}{425}\)
When E2 has occurred i.e., when a card is not lost, then the probability of drawing 2 cards of clubs from the remaining pack.
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 12

Question 12.
Find the area of the region bounded by the curves x2 + y2 = 4, y = √3x and x-axis in the first quadrant.
OR
Find the area of ellipse x2 + 9y2 = 36 using integration.
Answer:
x2 + y2 = 4
It is a circle with centre (0, 0) and r = 2.
y = √3x
From equation (1) and (2),
x2 + 3x2 = 4
⇒ 4x2 = 4
⇒ x2 = 1
⇒ x = ±1
When x = 1, y = √3
When x = – 1, y = – √3
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 13
∴ Points of intersection of (1) and (2), are (1, √3) and (- 1, -√3).
From(1), y = \(\sqrt{2^{2}-x^{2}}\)
∴ Required area = Shaded area
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 14

OR

CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 15

CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with Solutions

Question 13.
Evaluate:
∫\(\frac{2 x+5}{\sqrt{7-6 x-x^{2}}}\)dx.
Answer:
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 16

Question 14.
A badminton hall is to be constructed in the form of a cube in a sports complex in 3-D as given below:
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 17
Length of side of cube be V units, four diagonals are OP, MC, AN and BL.
From the above information given, answer the following questions :
(i) Find the direction cosines of diagonal AN.
Answer:
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 18

CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with Solutions

(ii) Write the cartesian equation of the side BN.
Answer:
Cartesian equation of side BN where B(0, a, 0), N(0, a, a) is
CBSE Sample Papers for Class 12 Maths Term 2 Set 7 with solutions 19
This is the required equation.