Please refer to the Class 11 Mathematics Sample Paper below. These CBSE Sample Papers for Class 11 Mathematics have been prepared based on the latest guidelines and examination patterns issued for the current academic year. We have provided Term 1 and Term 2 sample papers with solutions. You can click on the links below and access the free latest CBSE Sample Papers for Mathematics for Standard 11. All guess papers for Mathematics Class 11 have been prepared by expert teachers. You will be able to understand the type of questions which are expected to come in the examinations and get better marks.

## CBSE Sample Papers for Class 11 Mathematics

Term 2 Sample Papers |

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

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

**Section – A**

**1. If 1/6! + 1/8! = x/9!, find x****Answer :**

**OR**

**If ^{n}P_{4} : ^{n}P_{2} = 12 : 1, find n**

**Answer :**

**2. Write the eccentricity of the hyperbola, 9y ^{2} – 4x^{2} = 36**

**Answer :**

**3. Differentiate: (3x + 4) / (5x ^{2} – 7x + 9)**

**Answer :**

**4. If sin A = 3/5 and π/2 < A < π . Find cos A, tan 2A.****Answer : **Since, sin A > 0 and π/2 < A < π, then A is in II Quadrant.

**5. Solve for x : (2x-3)/4 + 9 ≥ 3 + 4x/3 , x ∈ R.****Answer : **

**6. If the letters of the word ‘ALGORITHM’ are arranged at random in a row what is the probability the letter ‘GOR’ must remain together as a unit ?****Answer :** Number of letters in the word ‘ALGORITHM’ is 9

If ‘GOR’ remain together, then considered it as 1 number

∴ Number of letters = 6 + 1 = 7

Number of words, if ‘GOR’ remain together = 7!

Total number of words from the letters of the word ‘ALGORITHM’ = 9!

∴ Reθuired probability

= 7!/9! = 7!/(9×8×7!) = 1/(9×8) = 1/72

**OR**

**A card is selected from a pack of 52 cards.**

(i) What is the sample space?

(ii) Calculate the probability that card is an ace of spade?**Answer :** (i) Sample space i.e. n(S) = 52

(ii) Number of ace of spade in a pack = 1

i.e. n(E) = 1

Probability that card is an ace = n(E) / n(S) = 1/52

**Section – B**

**7. If tan θ + sin θ = m and tan θ – sin θ = n, then prove that : m ^{2} – n^{2} = 4sinθ⋅tan θ**

**Answer :**Given, tanθ + sinθ = m and tanθ – sinθ = n,

To prove, m

^{2}– n

^{2}= 4 sin θ tan θ

L.H.S.= m

^{2}– n

^{2}

= (m + n)(m – n)

= [(tan θ + sin θ) + (tan θ – sin θ)]

[(tan θ + sin θ) – (tan θ – sin θ)]

(2tanθ )(2sinθ)

= 4sinθ tanθ

= R.H.S

**Hence proved**

**8. Find the co-ordinate of the vertices, foci eccentricity and length of latus rectum of the hyperbola**

x^{2}/25 – y^{2}/4 = 1**Answer **:

**OR**

**Find the distance between the directrices of the hyperbola x ^{2} – y^{2} = 8.**

**Answer :**

**9. If tan (A + B) = p, tan (A – B) = θ, then show that : tan2A = (p+θ) / (1-pθ)****Answer :** Given, tan (A + B) = p, and tan (A – B) = q,

L.H.S. = tan 2 A = tan (A + B + A – B)

= tan [(A + B) + (A – B)]

= ( tan (A + B) + tan (A – B) ) / 1 – tan (A + B)·tan (A – B)

= (p + q) / 1 – pq

= R.H.S. **Hence proved.**

**10. From a group of 7 boys and 5 girls, a team consisting of 4 boys and 2 girls is to be made. In how many different ways it can be done?****Answer :** From a group of 7 boys, 4 boys can be choose in ^{7}C_{4} ways.

From a group of 5 girls, 2 girls can be choose in ^{5}C_{2} ways.

∴ Required number of ways to choose a team consisting 4 boys and 2 girls are

= ^{7}C_{4} × ^{5}C_{2}

= 7! / (3!4!) × 5! / (3!2!)

= (7·6·5 / 3·2·1) × (5·4/2·1)

= 35 × 10

= 350

**Section – C**

**11. Find the derivative of the functions a/x ^{4} – b/x^{2} + cosx where a and b are non-zero constant.**

**Answer :**

**12. Find the eθuation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x – 1.****Answer :**

**OR**

**Find the eθuation of a circle whose centre is (3, –1) and which cuts off a chord length 6 units on the line 2x – 5y + 18 = 0.****Answer :**

**13. Solve the following system of ineθualities graphically**

2x + y ≤ 24,

x + y ≥ 11,

2x + 5y ≤ 40, x, y ≥ 0.**Answer :** Inequality (x ≥ 0) represents the region on the right of Y-axis and Y-axis itself. Inequality (y ≥ 0) represents the region above X-axis and X-axis itself.

2x + y ≤ 24 …(i)

x + y ≥ 11 …(ii)

2x + 5y ≤ 40, x, y ≥ 0 …(iii)

We first draw the graph of lines

2x + y = 24, x + y = 11 and 2x + 5y = 40

Now, 2x + y = 24, passes through A(12, 0) and B(0, 24) 1

Again, x + y = 11, passes through C(11, 0) and D(0, 11)

Further 2x + 5y = 40, passes through E(20, 0) and F(0, 8)

Shaded area PQAC is the solution area

**Case-Based/Data Based**

**14.** One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Conseθuently the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}. You are told that the chances of John’s promotion is same as that of Gurpreet, Rita’s chances of promotion are twice as likely as Johns. Aslam’s chances are four times that of John.

**Based on the given information, answer the following θuestions:**

(i) What is the probability that John got promoted?

(ii) If A = {John promoted or Gurpreet promoted}, Find P(A).**Answer :** Let Event :

J = John promoted

R = Rita promoted

A = Aslam promoted

G = Gurpreet promoted

Given sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}

i.e., S = {J, R, A, G)

It is given that, chances of John’s promotion is same as that of Gurpreet.

P(J) = P(G)

Rita’s chances of promotion are twice as likely as John.

P(R) = 2P(J)

and Aslam’s chances of promotion are four times that of John.

P(A) = 4P(J)

(i) P(J) + P(R) + P(A) + P(G) = 1

⇒ P(J) + 2P(J) + 4P(J) + P(J) = 1

⇒ 8P(J) = 1

P(J) = P(John Promoted) = 1/8

(ii) A = John promoted or Gurpreet promoted

∴ A = J ∪ G

P(A) = P (J ∪ G)

∴ P(A) = P(J) + P(G) – P (J ∩ G)

P(A) = 1/8 + 1/8 – 0 [∵ P (J ∩ G) = 0]

= 1/4