Students can access the CBSE Sample Papers for Class 11 Maths with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.

## CBSE Sample Papers for Class 11 Maths Term 2 Set 2 with Solutions

Time: 2 Hours

Maximum Marks: 40

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 marls 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.

A wheel makes 360 revolutions in 1 min. Through how many radians does it turn in 1 second.

OR

Find the degree measure of the angle subtended at the centre of circle of radius 100 cm by an arc of 22 length 22 cm. (Use π = \(\frac{22}{7}\))

Answer:

No. of revolution in one minute = 360

∴ No. of revolutions m one second = \(\frac{360}{2}\)

= 6

∴ 6 revolutions = 6 × 2π = 12π radians

Hence, no. of radians turned in one second = 12π

OR

Given, l = length of arc = 22 cm

r = radius of circle = 100 cm

Let θ be the angle subtended at the centre, then

Question 2.

If parabola y^{2} = px passes through point (2, -3), find the length of latus rectum. (2)

Answer:

Given equation of parabola is y^{2} = px …(i)

Since (i) passes through (2, -3).

∴ (-3)^{2} = p.2

⇒ p = \(\frac{9}{2}\)

∴ y^{2} = \(\frac{9}{2}\)x

⇒ y^{2} = 4. \(\frac{9}{8}\)x

Comparing above by y^{2} = 4ax

a = \(\frac{9}{8}\)

∴Length of latus rectum = 4a = \(\frac{9}{2}\) units

Commonly Made Error:

Students make mistakes while comparing the given equation with the general form of parabola.

Answering Tip:

Learn the equations of parabola and the terms released it. Understand the equation and its properties with the help of its diagram.

Question 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?

OR

Determine the number of 5 card combining out of deck of 52 cards if each selection of 5 cards has exactly one king. (2)

Answer:

Number of ways to fill the III place = 8

Number of ways to fill the IV place = 7

Number of ways to fill the V place = 6

Hence, total number of ways = 8 × 7 × 6 = 56 × 6 = 336

OR

1 king out of 4 kings can be selected in ^{4}C_{1} ways

Remaining 4 cards out of remaining 48 cards can be selected in ^{4}C_{4} ways.

Hence, total number of ways

= ^{4}C_{1} × ^{48}C_{4} = 4 × ^{48}C_{4}

Question 4.

Given P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\). Find P(A ∪ B) if A and B are mutually exclusive events. (2)

Answer:

P(A) = \(\frac{3}{5}\), P(B) = \(\frac{1}{5}\)

A and B are mutually exclusive events

∴ P(A ∩ B) = 0

Now P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= \(\frac{3}{5}+\frac{1}{5}\) – 0

P(A ∪ B) = \(\frac{4}{5}\) = 0.8

Question 5.

Solve the inequality: -3 ≤ 4 – \(\frac{7x}{2}\) ≤ 18 (2)

Answer:

The given inequality -3 ≤ 4 – \(\frac{7x}{2}\) ≤ 18 Adding (-4) to each term,

-3 – 4 ≤ 4 – \(\frac{7x}{2}\) – 4 ≤ 18 – 4

-7 ≤ – \(\frac{7x}{2}\) ≤ 14

Multiplying by (\(\frac{-2}{7}\)) to each term

⇒ 2 ≥ x ≥ -4

⇒ 4 ≤ x ≤ 2

∴ Solution set = [-4, 2]

Question 6.

Find the derivatives of 2x – \(\frac{3}{4}\) (2)

Answer:

Let f(x) = 2x – \(\frac{3}{4}\)

∴ f'(x) = \(\frac{d}{d x}\)(2x – \(\frac{3}{4}\))

= 2\(\frac{d}{d x}\)x – \(\frac{d}{d x}\)(\(\frac{3}{4}\))

= 2 × 1 – 0

= 2

Section – B

Question 7.

Find the centre and radius of the circle 2x^{2} + 2y^{2} – x = 0. (3)

Answer:

Given equation of the circle is

2x^{2} + 2y^{2} – x = 0

Question 8.

Prove that: \(\frac{\cos 4 x \sin 3 x-\cos 2 x \sin x}{\sin 4 x \cdot \sin x+\cos 6 x \cdot \cos x}\) = tan 2x (3)

OR

Prove that:

tan α. tan (60° – α) . tan (60° + a) = tan 3α (3)

Answer:

Commonly Made Error:

Many students get confused among the formula of trigonometry. They must learn all the formulae properly and practise the derivation of the formula to avoid mistake.

Answering Tip:

Simplification of trigonometric equations needs sufficient practice. Give ample practice such type of questions.

OR

L.H.S = tan α. tan (60° – α). tan (60° + a)

Question 9.

From a class of 40 students, in how many ways can five students be chosen for an excursion party. (3)

Answer:

From a dass of 40 students, five students can be chosen for an excursion party in ^{4}C_{5} ways

= 658,008

Question 10.

Find the value of the expression

3[sin^{4}(\(\frac{3 \pi}{2}\) – α) + sin^{4}(3π + α)] – 2[sin^{6}(\(\frac{\pi}{2}\) + α) + sin^{6}(5π – α)]

Answer:

Given 3[sin^{4}(\(\frac{3 \pi}{2}\) – α) + sin^{4}(3π + α)] – 2[sin^{6}(\(\frac{\pi}{2}\) + α) + sin^{6}(5π – α)]

= 3[cos^{4}α + sin^{4}(π + α)] – 2[cos^{6} α + sin^{6}(π – α)]

= 3(cos^{6}α + sin^{6}α] – 2[cos^{6} α + sin^{6}α]

= 3(cos^{4}α + sin^{4} α + 2sin^{2}α cos^{2}α – 2sin^{2}α cos^{2}α] – 2(cos^{2}α + sin^{2}α)^{3} – 3cos^{2}α sin^{2}α(cos^{2}α + sin^{2}α)]

= 3[(cos^{2}α + sin^{2}α)^{2} – 2sin^{2}αcos^{2}α] -2[1 – 3cos^{2}αsin^{2}α]

= 3(1 – 2sin^{2}αcos^{2}α – 2 + 6cos^{2}αsin^{2}α]

= 3 – 6sin^{2}αcos^{2}α – 2 + 6sin^{2}αcos^{2}α

= 1

Section – C

Question 11.

Using first principle find the derivative of x cos x function: (4)

Answer:

Let f(x) = x cos x

∴ f(x + h) = (x + h)cos(x + h)

Question 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?

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 = \(\frac{9}{5}\)C +32? (4)

Answer:

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 add content in the (460 + x) L of mixture

= (460 × \(\frac{9}{100}\) + x × \(\frac{3}{100}\))L

It is given that add content in the resulting mixture must be more them 5% but less than 7%.

Therefore,

⇒ 2300 + 5x < 4140 + 3x < 3220 + 7x

Taking first two inequalities,

2300 + 5x < 4140 + 3x

⇒ 5x – 3x < 4140 – 2300

⇒ 2x < 1840

⇒ x < \(\frac{1840}{2}\)

⇒ x < 920 …(i)

Taking last two inequalities,

4140 + 3x < 3220 + 7x

⇒ 3x – 7x < 3220 – 4140

⇒ -4x < -920 ⇒ 4x > 920

⇒ x > \(\frac{920}{4}\)

⇒ x > 230 …(ii)

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

OR

Let the required temperature be x° F.

Given that, F = \(\frac{9}{5}\)C + 32

⇒ 5F = 9C + 32 × 5

⇒ 9C = 5F – 32 × 5

∴ C = \(\frac{5 F-160}{9}\)

Since, temperature in degree Celsius lies between 40°C to 45°C.

Therefore, 40 < \(\frac{5 F-160}{9}\) < 45

40 < \(\frac{5 F-160}{9}\) < 45

⇒ 40 × 9< 5x – 160 < 45 × 9 [multiplying throughout by 9]

⇒ 360 < 5x – 160 < 405 [adding 160 throughout]

⇒ 360 + 160 <5x< 405+160

⇒ 520 < 5x < 565 1

⇒ \(\frac{520}{5}\) < x < \(\frac{565}{5}\) [divide throughout by 5]

⇒ 104 < x < 113

Hence, the range of temperature in degree Fahrenheit is between 104 °F to 113°F.

Question 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 PA^{2} + PB^{2} = k^{2}, where k is constant. (4)

Answer:

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

PA^{2} = (x – 3)^{2} + (y – 4)^{2} + (z – 5)^{2}

= x^{2} + 9 – 6x + y^{2} + 16 – 8y + z^{2} + 25 – 10z

= x^{2} + y^{2} + z^{2} – 6x – 8y – 10z + 50

PB^{2} = (x + 1)^{2} + (y – 3)^{2} + (z + 7)^{2}

= x^{2} + 1 + 2x + y^{2} + 9 – 6y + z^{2} + 49 + 14z

= x^{2} + y^{2} + z^{2} + 2x – 6y + 14z + 59

Now, if PA^{2} + PB^{2} = k^{2}, then

(x^{2} + y^{2} + z^{2} – 6x – 8y – 10z + 50) + (x^{2} + y^{2} + z^{2} + 2x – 6y + 14z + 59)

⇒ 2x^{2} + 2y^{2} + 2z^{2} – 4x – 14y + 4z + 109 = k^{2}

⇒ 2(x^{2} + y^{2} + z^{2} – 2x – 7y + 2z) + 109 = k^{2}

⇒ 2(x^{2} + y^{2} + z^{2} – 2x – 7y + 2z) = k^{2} -109

⇒ x^{2} + y^{2} + z^{2} – 2x – 7y + 2z = \(\frac{k^{2}-109}{2}\)

Thus, the required equation is

x^{2} + y^{2} + z^{2} – 2x – 7y + 2z = \(\frac{k^{2}-109}{2}\)

Case-Based/Data Based

Question 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.

(i) Find the probability that lost piece is triangle. (2)

Answer:

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

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) = \(\frac{F(E)}{T(E)}\)

= \(\frac{8}{18}=\frac{4}{9}\)

(ii) Find the probabilities that lost piece is square of blue colour & triangle of red colour. (2)

Answer:

Number of favourable outcomes for the events that lost figure is square of blue colour, i.e., F(E) – 6 and T(E) = 18

∴ P(getting a blue square),

P(E) = \(\frac{F(E)}{T(E)}\)

= \(\frac{6}{18}=\frac{1}{3}\)

Number erf 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) = \(\frac{5}{18}\)