Class 12 Mathematics: CBSE Sample Question Paper- Term II (2021-22)- 4 | Sample Papers for Class 12 Medical and Non-Medical PDF Download

Class-XII

Max. Marks: 40

General Instructions :

1. This question paper contains three sections A, B and C. Each part is compulsory.
2. Section - A has 6 short answer type (SA1) questions of 2 marks each.
3. Section - B has 4 short answer type (SA2) questions of 3 marks each.
4. Section - C has 4 long answer type questions (LA) of 4 marks each.
5. There is an internal choice in some of the questions.
6. Q14 is a case-based problem having 2 sub parts of 2 marks each.

Section - A

Q.1. Evaluate
OR
Find: .

Let I =
I =
=
=
Let cos x + sin x = t
⇒ (cosx - sinx)dx = dt
⇒ I =
= log|t| + C
= log|cosx + sinx| + C
OR

=
f(x) =
∴  
=

Q.2. Find the sum of the order and the degree of the following differential equations: 

Here,
 
Thus, order is 2 and degree is 3. So, the sum is 5.

Q.3. Find the position vector of a point which divides the join of points with position vectors  and  externally in the ratio 2 : 1.

Required vector = 
= 
= 

Q.4. Find the equation of line passing through (1, 1, 2) and (2, 3, –1).

Equation of line passes to (x₁, y₁, z₁) and (x₂, y₂, z₂)

So,

∴

Q.5. If A and B are two independent events, then prove that the probability of occurrence of at least one of A and B is given by 1 – P(A’) · P(B’).

Required probability
= P(A ∩ B)
= P(A) + P(B) – P(A) · P(B)
= P(A) [1 – P(B)] + 1 – P(B’)
= P(A) P(B’) – P(B’) + 1
= [1 – P(B’)] [1 – P(A)]
= 1 – P(A’) P(B’) (Hence Proved)

Q.6. One bag contains 3 red and 5 black balls. Another bag contains 6 red and 4 black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is red.

P (Red transferred and red drawn or black transferred and red drawn)
= 3/8 x 7/11 + 5/8 x 6/11
= 51/88

Section - B

Q.7. Find : .

Let

⇒ x + 1 = (Ax + B)x + C(x² + 1) (Anidentity)
Equating the coefficients, we get
B = 1, C = 1, A + C = 0
Hence, A  = –1, B = 1, C = 1
The given integral
=
=
=
=  + log|x| + c

Q.8. Find the general solution of the following differential equation:

OR
Find the particular solution of the following differential equation, given that y = 0 when x = π/4:

We have the differential equation

The equation is a homogeneous differential equation.
Putting y = vx
⇒
The differential equation becomes

⇒
⇒ cos v dv = -dx/x
Integrating both sides, we get
log|cosec v - cotv|= -log |x|+ logK,
K > 0 (Here, log K is an arbitrary constant.)
⇒ log|(cosec v - cotv)x| =  logK

⇒ |(cosec v - cotv)x| = K
⇒ (cosec v - cot v)x = ± K
⇒
which is the required general solution.
OR
The differential equation is a linear differential equation
⇒ IF =  = sin x
The general solution is given by

ysin x =

⇒ ysin x =
=
⇒ ysin x =
⇒ ysin x =  
⇒ ysin x =  
⇒ ysin x =
Given that y = 0, where x = π/4,
Hence, 0 =
⇒ c =
Hence, the particular solution is

y = -
Alternative method

The differential equation is a linear differential equations
∴ I.F. = =  sin x
The general solution is given by
ysin x =
=
=
ysin x = 2x -
ysin x = 2x -
ysin x = 2x - 2 tanx + 2secx + c
Given that
y = 0 when x = π/4
0 = π/2 - 2 + √2 + C
C =
Since, the particular solution is
ysin x =

Q.9. If , then show that .

We have

⇒
⇒
Also,
⇒
⇒
 cannot be both perpendicular to  and parallel to
Hence, .

Q.10. Find the shortest distance between the following lines:

OR
Find the vector and the cartesian equations of the plane containing the point  and parallel to the lines  and  = 0

Here, the lines are parallel. The shortest distance
=
=

Hence, the required shortest distance
=  units
OR
Since, the plane is parallel to the given lines, the cross product of the vector and 

 will be a normal to the plane
 =
=
The vector equation of the plane is

or,  = 2
and the cartesian equation of the plane is
x -z- 2 = 0.

Section - C

Q.11. Find:

Let,
I =
(On dividing Nr and Dr. by cos³ x)
=  
On substituting tan x = t  and sec²x dx = dt,
We get
I =
=
=
=
= +
= +
=  +
= ++

Q.12. Find the area of the region bounded by the curve x² = 4y and the straight-line x = 4y– 2
OR
Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x²+ y² = 32 

As x² = 4y and x = 4y – 2
So, x² = x + 2
x² - x - 2 = 0
(x -2)(x + 1) = 0
x = -1, 2
For x=−1,     y = 1/4  and for x = 2, y = 1
Points of intersection are (-1, 1/4) and (2, 1)
Graphs of parabola x² = 4y  and x = 4y − 2    are     shown in the following figure:

A=
=
=
= 9/8 sq. units
= 1 x 1/8 sq. units
OR
We have y = 0, y = x and the circle x² + y² = 32 in the first quadrant.
Solving y  = x with the circle
x² + x² = 32
x² = 16
x = 4 (In the first quadrant)
When x  = 4, y = 4 for the point of intersection of the circle with the x-axis.
Put y  = 0 in circle
x² + 0 = 32
x =
So, the circle intersects the x-axis at

From the above figure, area of the shaded region,
A =
= +
=
=
= 8 + [8π - 8 - 4π]
= 4π sq. units

Q.13. Find the shortest distance between the lines  and 

. If the lines intersect find their point of intersection.

We have
a₁ =
b₁ =
a₂ =
b₂ =

= +
=
∵  = 16 - 16 = 0
∴ The lines are intersecting and the shortest distance between the lines is 0.
Now for point of intersection
 =
⇒ 3 + λ = 5 + 3μ ....(i)
2 + 2λ = -2 + 2μ ....(ii)
-4 + 2λ = 6μ
Solving (i) and (ii) we get μ = –2 and λ= –4 Substituting in equation of line we get

=
Point of intersection is (–1, –6, 12)

Case-Based/Data Based

 

Let
E be the event = A solves the problem
F be the event = B solves the problem
G be the event = C solves the problem
H be the event = D solves the problem




(i) The required probability
= P(E ∪F∪G∪ H)
=
= 1 - 2/3 x 3/4 x 4/5 x 1/3
= 13/15
(ii) The required probability
=
=
=
= 7/15 + 1/30 + 2/45
= 42 + 3 + 4/90
= 49/90