Class 12 Mathematics: CBSE Sample Question Paper- Term II (2021-22)- 1 | Sample Papers for Class 12 Medical and Non-Medical PDF Download

Class-XII

Max. Marks: 40

General Instructions :

1. This question paper contains three section 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. Find
OR
Evaluate

Put, 1 – tan x = y
So that, –sec²x dx = dy

OR
Let
Since,


∴

= – [0 – (–1)] + (2 – 0)
= –1 + 2
= 1

Q.2. Show that the function y = ax + 2a² is a solution of the differential equation

y = ax + 2a² ⇒ (dy/dx) = a

From L.H.S. of differential equation

= 2(a)² + x(a) – (ax + 2a²) = 0
= R.H.S. 

Q.3. If  are mutually perpendicular unit vectors, then find value of

Given  are mutually perpendicular unit vectors, i.e.,

 ...(i)
and
 ...(ii)
Now,


[ ∵ Dot product is distributive over addition]

 = 4(1) + 1 + 1 = 6

Q.4. A line passes through the point with position vector  and makes angles 60°, 120°, and 45° with x, y and z-axis respectively. Find the equation of the line in the Cartesian form. 

D-Cosines of line are 1/2, -(1/2), 1/√2

Equation of line is :
or 2x – 4 = – 2y – 6 = √2 ( z -4).

Q.5. Find the probability distribution of X, the number of heads in a simultaneous toss of two coins.

Let X be the number of heads.

Possible v alues 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 :

Q.6. The probability that it will rain on any particular day is 50%. Find the probability that it rains only on first 4 days of the week.

P(rain on any particular day) = 50%

= 50/100 = 1/2
P(rain on first four days of week)
.

Section - B

Q.7. Prove that:  hence evaluate

Let

Put a - x = t ⇒ dx = dt

Now,

⇒

⇒

Put cos x = t ⇒ -sin x dx = dt
⇒
I is an even find in

Q.8. Find the general solution of the following differential equation: x dy – (y + 2x²)dx = 0
OR
Solve the following differential equation.

The given differential equation can be written as

⇒
Comparing above with (dy/dx) +Py = Q
Here P = -(1/x) and Q = 2x


Hence, the required solutions is:

⇒ (y/x) = 2x + c
⇒ y = 2x² + cx
OR
Given differential equation is
or
or
or
or
On integrating both sides, we get

On putting 1 + y² = t and 1 + x² = u²
or 2y dy = dt and 2x dx = 2u du
⇒ y dy  = dt/2 and x dx = u du

or
or  [put 1 + y² = t]
or  C
which is the required solution.

Q.9. Find the area of a rectangle having vertices A, B, C and D with position vectors   respectively.
The two vectors  represent the two sides AB and AC, respectively of DABC. Find the length of the median through A.

The position vectors of vertices A, B, C and D of rectangle ABCD are given as:

The adjacent sides  of the given rectangle are given as :



Now, it is known that the area of parallelogram whose adjacent sides are
Therefore, the area of the given rectangle is
OR

Now ABEC represent a parallelogram with AE as the diagonal.


Now,
∴

Q.10. Write the sum of intercepts cut off by the plane  on the three axes.

Given, equation of plane is
Put  we get
⇒ 2x + y - z = 5
or
On comparing it with standard equation of plane in intercept form
a = 5/2, b = 5 and c = -5
Now, required sum of intercepts cut off by the plane on the three axis = a + b + c
= (5/2) + 5 - 5 = 5/2 units.

Section - C

Q.11. Find :

Put, [sinx = t or cosxdx = dt]

 
Put , 1,
1 = 4C , i.e., C = 1/4 Put 0 ,
0 = A + B + C, which gives A = 3/4
Therefore the required integral
 
 + c

Q.12. Find the area bounded by lines x = 2y+  3, y – 1 = 0 and y + 1 = 0.
Or
Find the region bounded by the curve  y² =4x, y-axis and the line y = 3.

From the figure, area of the shaded region,


= [ 1 + 3 - 1 + 3]
= 6 sq. units
OR
The area bounded by the curve, y² = 4x,  y-axis, and y = 3 is represented as :

Area of OAB =

= (1/12) x 27
= 9/4 sq. units.

Q.13. Find the equation of plane passing through the points A(3, 2, 1), B(4, 2, –2) and C(6, 5, –1) and hence find the value of λ for which A(3, 2, 1), B(4, 2, –2), C(6, 5, –1) and D(λ, 5, 5) are coplanar.

A plane which passes through A(3, 2, 1), B(4, 2, –2) and C(6, 5, –1) is

⇒
⇒ (x - 3)(0 + 9)-(y - 2)(-2 + 9) + (z - 1)(3 - 0) = 0
⇒ 9x - 7(y - 2) + 3(z - 1) = 0
⇒ 9(x - 3) 9x - 7y + 3z = 16
Thus, plane passing through point A, B and C is 9x – 7y + 3z = 16

Now, given A, B, C and D (λ, 5, 5) are coplanar. So, D  lies on the plane passing through A, B and C
∴ 9λ -  7(5) + 3(5)=16
⇒ 9λ = 36
⇒ λ = 4

Case-Based/Data Based 

Q.14. Of the students in a school, it is known that 30% have 100% attendance and 70% students are irregular. Previous year results report that 70% of all students who have 100% attendance attain A grade and 10% irregular students attain A grade in their annual examination. At the end of the year, one student is chosen at random from the school and he was found to have an A grade. Let E1 and E2 be the events that selecting a student with 100% attendance and selecting a student who is not regular, respectively.
Based on the above information, answer the following questions:
(i) Find the values of P
(ii) What is the probability that the student has 100% attendance.

Let E₁ : Selecting a student with 100% attendance 

E₂ : Selecting a student who is not regular
A  : selected student attains A grade.

(i)
(ii)