Class 12 Mathematics: CBSE Sample Question Paper- Term II (2021-22)- 5 | 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. Q.14 is a case-based problem having 2 sub parts of 2 marks each.

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Section - A

Q.1. Find the value of 
OR
Evaluate :

Let I =
= +
When f(x) is an even function, then,

and if f(x) is an odd function, then
= 0
∴ I =
 = π
OR
Let,
I =  
=
Also, put e^x = t, ⇒ e^xdx = dt
⇒ I =
= tan^-1t + C
= tan^-1(e^x) + C

Q.2. Show that  = 0 is the solution of y= e^–x (A cos x + B sin x).

Given that, y = e^–x (A cos x + B sin x)
On differentiating both sides w.r.t., x we get
dy/dx = -e^-x(A cos x + B sin x) + e^–x (-A sin x + B cos x)
dy/dx = -y + e^-x(-A sin x + B cos x)
Again, differentiating both sides w.r.t. x, we get
 (-A sin x + B cos x)
⇒
⇒
⇒
⇒  = 0

(Hence Proved 1)

Q.3. Find the projection of vector  on the vector .

= 4 + 6 + 2 = 12

or p =    = 3
= 12/3 = 4

Q.4. If the lines  and  are perpendicular to each other, then find the value of p.

Using formula for perpendicular condition,
l₁l₂ + m₁m₂ + n₁n₂ = 0
or – 8p + 6p – 28 = 0
or – 2p = 28
∴ p = 14

Q.5. If P(A) = 0.4, P(B) = 0.8 and P(B/A) = 0.6, then P(A ∪ B).

Here,
P(A) = 0.4, P(B) = 0.8 and P(B/A) = 0.6
∵
⇒ P(B ∩ A) =
= 0.6 x 0.4 = 0.24
∵ P(A ∪ B) = P(A) + P(B) - P(B ∩ A)
= 0.4 + 0.8 - 0.24
= 1.2 - 0.24
= 0.96

Q.6. Find the probability distribution of X, the number of heads is a simultaneous toss of two coins.

Let X be the number of heads
Possible values of X are 0, 1, 2.
P(x = 0) = 1/4, P(x = 1) = 1/2, P(x = 2) = 1/4
The probability distribution of X is :

Section - B

Q.7.Evaluate:

I =        ..(i)
Apply the property
I =
or I =
or I =         ..(ii)
Adding eqn. (i) and (ii), we get
or 2I =
So, I =

Q.8. Solve  = 0 subject to the initial condition y(0) = 0.
OR
Find the particular solution of the following differential equation :
 ; y = 0 when x = 0

Given differential equation can be written as: 

Comparing with

⇒ P = , Q =
I.F. (Integrating factor)
=
=
= 1 + x²
∴ General solution is :
y(1 + x²) =
or, y(1 + x²) =
Putting x = 0 and y = 0, we get C = 0
∴ Solution is:
y =
OR
Given equation can be written as

⇒
⇒ -log|2 - e^y| + logc = log|x + 1|
⇒ (2 - e^y)(x + 1) = c
When x= 0 , y = 0 ⇒ c =  1
∴ The required solution is (2 - e^y) ( x +1) = 1

Q.9. Find the area of the parallelogram whose diagonals are represented by the vectors  and.
OR
Find λ and μ if

The vector equation for diagonals are  and
Now,
=
 = √4 + 16 + 16 = √36 = 6
Area of the parallelogram
=  = 3 sq. units
OR
 = 0
or
or = 0
or 3μ + 9λ = 0 ...(i)
or μ – 27 = 0 ...(ii)
or – λ – 9 = 0 ...(iii)
From eqn. (ii) and (iii),
μ = 27
and λ = 9

Q.10. Find the value of λ, so that the lines  and 

are at right angles. Also, find whether the lines are intersecting or not.

Given lines are :
and
As lines are perpendicular,
= 0
⇒ λ =7
So, lines are

Consider
Δ =  
=
= -63
Since, as ∆ ≠ 0 ⇒ lines are not intersecting.

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Section - C

Q.11. Show that: = 

I =                       ...(i)
By applying property

I =
I =                       ...(ii)
Adding eqn. (i) & (ii)
∴ 2I =
=
⇒ 2I =
=
=
I =
I =

Q.12. Find the area of the region bounded by the parabola y² = x and the line 2y = x.
OR
Find the area of the region bounded by the parabola y² = 16x and the line x = 4.

When y² =x  and 2y = x
Solving we get y² =2y
⇒ y = 0, 2  and when y = 2, x = 4 and y = 0
⇒ x = 0
So, points of intersection are (0, 0) and (4, 2).
Graphs of parabola y² = x and 2y = x are as shown in the adjoining figure :

From the figure, area of the shaded region,
A =
=
=
= 4/3 sq. units
Parabola y2 = 16x and line x = 4
at x = 4, y² = 64 ⇒ y = ±8
Hence, the point of intersection (4, 8) and (4, –8)
OR

Area =
=
=

=
= 16/3 x 8
= 128/3 sq. units

Q.13. If  are unit vectors such that   and the angle between  and 

 is π/6, then prove that: 

As given, both  and
 are unit vectors
or
Let,
then
or  = |λ|
or  |λ| =
= 2
∴ λ = ±2
∴

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Case-Based/Data Based

Let E₁ = Bag I is chosen, E₂ = Bag II is chosen, E₃ = Bag III is chosen, A = The two balls drawn from the chosen bag are one white and one red.
P(E₁) = 1/3
= P(E₂) = P(E₃),
,

(i) By Bayes’ Theorem, Required probability
=
=  x
=
=
= 64/199
(ii) By Bayes' theorem Required probability
=
=
=
=
= 54/199