# Class 11 Mathematics Sample Paper

We have provided Class 11 Mathematics Sample Paper as per the latest CBSE examination pattern for the current academic year. The following CBSE Sample Papers for Class 11 Mathematics has been prepared based on the guess papers issued recently. Students will be able to practice these papers and get good marks in upcoming Mathematics exams for Class 11.

## Class 11 Mathematics Sample Paper Term 2 With Solutions Set A

Section – A

1. A wheel makes 360 revolutions in 1 min. Through how many radians does it turn in 1 second.
No. of revolution in one minute = 360
∴ No. of revolutions in one second = 360/60 = 6
∴ 6 revolutions = 6 × 2p = 12p radians
Hence, no. of radians turned in one second = 12p

OR

Find the degree measure of the angle subtended at the centre of circle of radius 100 cm by an arc of length 22 cm. ( Use π = 22/7 )
Given, l = length of arc = 22 cm
r = radius of circle = 100 cm
Let q be the angle subtended at the centre, then
= (11/50 x 180 /π)0 = (11x18x7 / 5×22)0 = (63/5)0
= 120 + (3/5)0      (∴10 = 60′)
= 12036′

2. If parabola y2 = px passes through point (2, −3), find the length of latus rectum.
Given equation of parabola is y2 = px …(i)
Since (i) passes through (2, −3).
(-3)2 = p.2
⇒ p = 9/2
∴ y2 = 9/2x
⇒ y2 = 4.9/x
comparing above by y2 = 4ax
a = 9/8
∴ Length of latus rectum = 4 = 9/2 units

3. How many 5 digit telephone numbers can be constructed using the digits 0 to 9, if each number starts with 67 and no digit appears more than once?
Number of ways to fill the III place = 8
Number of ways to fill the IV place = 7   (Image 178)
Number of ways to fill the V place = 6
Hence, total number of ways
= 8 x 7 x 6 = 56 x 6 = 336

OR

Determine the number of 5 card combining out of deck of 52 cards if each selection of 5 cards has exactly one king.
1 king out of 4 kings can be selected in 4C1 ways
Remaining 4 cards out of remaining 48 cards can be selected in 48C4 ways.
Hence, total number of ways
= 4C1 X 48C4 = 4X C4

4. Given P(A) = 3 /5 and P(B) = 1/5 . Find P(A ∪ B) if A and B are mutually exclusive events.

5. Solve the inequality: -3 ≤ 4 – 7x/2 ≤ 18.
The given inequality -3 ≤ 4 – 7x/2 ≤ 18 Adding (−4) to each term,

6. Find the derivatives of 2x – 3/4
Let f(x) = 2x – 3/4
f'(x) = d/dx(2x-3/4)
= 2×1-0
= 2

Section – B

7. Find the centre and radius of the circle 2x2 + 2y2 – x = 0.
Given equation of the circle is
⇒ 2x2 + 2y2 – x = 0
⇒ x2 + y2 – x/2 = 0

8. Prove that : cos4x sin3x – cos2x – sinx / sin4x . sinx + cos6x .cosx = tan2x

OR

Prove that:
tanα.tan (600 – α) – tan (600 +α) = tan 3α

9. From a class of 40 students, in how many ways can five students be chosen for an excursion party.
From a class of 40 students, five students can be chosen for an excursion party in 40C5 ways
= 40! / 5(40 -5)!
= 40! / 5! 35! = 40.39.38.37.36.35! / 5.4.3.2.1.35!
= 658 , 008

10. Find the value of the expression
3[sin4 (3π/2 – α) + sin4 (3π + α)] – 2[sin6 (π/2 + α)+ sin6 (5π – α)]

Section – C

11. Using first principle find the derivative of x cos x function:

12. A solution of 9% acid is to be diluted by adding 3% acid solution to it. The resulting mixture is to be more than 5% but less than 7% acid. If there is 460 L of the 9% solution, how many litres of 3% solution will have to be added?
Let x L of 3% solution be added to 460 L of 9% solution of acid.
Then, total quantity of mixture = (460 + x) L
Total acid content in the (460 + x) L of mixture
= (460 x 9/100 + x x 3/100)L
It is given that acid content in the resulting mixture must be more than 5% but less than 7%.
Therefore,

Hence, the number of litres of the 3% solution of acid must be more than 230 L and less than 920 L.

OR

A solution is to be kept between 40°C and 45°C. What is the range of temperature in degree Fahrenheit, if the conversion formula is F = C + 9/5 C + 32?

13. If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 +PB2 = k2,where k is constant.
The coordinates of points A and B are given as (3, 4, 5) and (–1, 3, –7) respectively. Let the co-ordinates of point P be (x, y, z).
On using distance formula, we obtain

PA2 = (x – 3)2 + (y – 4)2 + (z – 5)2 1
= x2 + 9 – 6x + y2 + 16 – 8y + z2 + 25 – 10z
= x2 + y2 + z2 – 6x – 8y – 10z + 50
PB2 = (x + 1)2 + (y – 3)2 + (z + 7)2
= x2 + 1 + 2x + y2 + 9 – 6y + z2 + 49 + 14z
= x2 + y2 + z2 + 2x – 6y + 14z + 59
Now, if PA2 + PB2 = k2, then
(x2 + y2 + z2 – 6x – 8y – 10z + 50) + (x2 + y2 + z2 + 2x – 6y + 14z + 59)
⇒ 2×2 + 2y2 + 2z2 – 4x – 14y + 4z + 109 = k2
⇒ 2(x2 + y2 + z2 – 2x – 7y + 2z) + 109 = k2
⇒ 2(x2 + y2 + z2 – 2x – 7y + 2z) = k2 – 109
⇒ x2 + y2+ z2 – 2x – 7y + 2z = (k2 – 109) / 2
Thus, the required equation is
x2 + y2+ z2 – 2x – 7y + 2z = (k2 – 109) / 2

Case-Based/Data Based

14. A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. (Image 166)
(i) Find the probability that lost piece is triangle.
(ii) Find the probabilities that lost piece is square of blue colour & triangle of red colour.
(i) Since, total no. of triangles = 8
Triangles with blue colour = 3
Triangles with red colour = 8 – 3 = 5
and total no. of squares = 10
Squares with blue colour = 6
Squares with red colour = 10 – 6
= 4 1
Number of favourable outcomes for the event that lost figure is triangle,
i.e., F (E) = 8
Total figures (square and triangle)
= 8 + 10 = 18
i.e., T(E) = 18
Probability (getting a triangle),
P(E) = F(E) / T(E)
= 8/18 = 4/9

(ii) Number of favourable outcomes for the events that lost figure is square of blue colour, i.e., F(E) = 6 and T(E) = 18
T P(getting a blue square),
P(E) = F(E) / T(E)
= 6/18 = 1/3
Number of favourable outcomes for the event that lost figure is triangle of red colour = 5,
i.e.,  F(E) = 5 and
T(E) = 18
∴ P(lost figure is red triangle),
P(E) = 5/18