Please refer to Class 11 Mathematics Sample Paper with solutions provided below. All sample papers for Mathematics Class 11 have been designed as per the latest paper pattern issued by CBSE for the current academic year. Students should practice these guess papers for Class 11 Mathematics as it will help them to gain more understanding of the type of questions that are expected to be asked in upcoming Class 11 Mathematics exams. Please click on the links below to access free CBSE Sample Papers for Class 11 Mathematics.
Class 11 Mathematics Sample Paper Term 2 With Solutions Set A
1. If 5-4x/6 ≤ x/6 − 5, then x ∈
(a) [2, ∞)
(b) [–7, 7]
(c) [7, ∞)
(d) [7, ∞)
Answer
D
2. Find the centre and radius of circle x2 + y2 – 2x + 4y = 8.
(a) (1, 2), √13
(b) (1, –2), √13
(c) (–1, – 2), √13
(d) (2, –1), 13
Answer
B
3. If y = x2 + 2x, then dy/dx is
(a) x + 1
(b) 2x + 1
(c) 2x + 2
(d) None of these.
Answer
C
4. The probability that at least one of the events A and B occurs is 0.5. If A and B occur simultaneously with probability 0.3, then P(A) + P(B) is
(a) 0.4
(b) 0.8
(c) 1.2
(d) 1.6
Answer
C
Section – II
Case study-based question is compulsory. Attempt any 4 sub parts. Each sub-part carries 1 mark.
5. Riya had some empty space in her garden. So, Riya thought to buy some plants. For this, she went to the plant nursery to buy new plants. There she saw southern crabapple plant. She holds its one leaf and found that it has an elliptical shape having length of major axis 10 cm and that of minor axis 6 cm. If the vertical elliptical shaped leaf has centre at origin, then answer the following questions.
(i) The equation of curve formed by leaf be
(a) x2/9 + y2/16 =1
(b) x2/16 + y2/25 = 1
(c) x2/9 + y2/25 = 1
(d) x2/36 + y2/16 = 1
Answer
C
(ii) The eccentricity of ellipse is
(a) 3/5
(b) 4/5
(c) 5/3
(d) 2/5
Answer
B
(iii) The distance between foci is
(a) 8
(b) 6
(c) 12
(d) 16
Answer
A
(iv) Equation of its directrices are
(a) y = ± 9/4
(b) y = ± 25/4
(c) x = ± 5/4
(d) x = ± 25/3
Answer
B
(v) What are the ends of latus rectum?
(a) (±3,±9/5)
(b) (±5,3/5)
(c) (±4 , ±3/5)
(d) (±9/5, (±4)
Answer
D
PART – B
Section – III
6. If x+4/x-3 < 2, then find the interval in which x lies.
Answer: We have, x+4/x- 3 < 2 or x+4/x-3 -2<0
or -x+10/x-3 < 0 or x-10/x-3 > 0
⇒ {x – 10 > 0 and x – 3 > 0} or {x – 10 < 0 and x – 3 < 0}
⇒ {x > 10 and x > 3} or {x < 10 and x < 3}
⇒ x ∈ (–∞, 3) ∪ (10, ∞)
7. Find the equation of circle whose centre is (–1, 2) and radius is 4.
Answer: Equation of circle with centre (–1, 2) and radius 4
is [x – (–1)]2 + (y – 2)2 = (4)2
⇒ (x + 1)2 + (y – 2)2 = 16
⇒ x2 + 1 + 2x + y2 + 4 – 4y = 16
⇒ x2 + y2 + 2x – 4y – 11 = 0
8. If sinθ + cosecqθ= 2, then find the value of sin2θ + cosec2θ.
Answer: Given, sinq + cosecq = 2
Squaring on both sides, we get
sin2q + cosec2q + 2sinq · cosecq = 4
⇒ sin2q + cosec2q = 4 – 2 = 2
OR
Evaluate : cos 4q – cos 6q
Answer: Consider, cos 4q – cos 6q
= 2sin(4θ + 6θ/2) sin (6θ – 4θ/2) = 2 sin 5θ sin θ
9. Evaluate : n!/r!(n – r)! , when n = 6, r = 2.
Answer: Since n = 6, r = 2
We have 6!/2!(6! – 2)! = 6!/2! x 4! = 6×5/2 = 15
10. Find the derivative of a/x4 – b/x2 +cosx .
Answer: Let f(x) = a/x4 – b/x2 + cos x
⇒ f(x) = ax–4 – bx–2 + cos x … (i)
Differentiating (i) with respect to x, we get
d/dx[f(x)]= a (-4) x – 5− b(−2)x−3 + (−sinx)
= -4a/x5 + 2b/x3 – sin x
Section – IV
11. Evaluate : sin A · tan (A/2) + cos .
Answer: We have, sin A · tan (A/2) + 2 cos A
= sin A. sin A/2/+ 2cos A = 2sin A/2 cos A/2. sinA/2/cosA/2 + 2cos A
= 2 sin2 A/2 + 2cos A = 1 – cos A + 2 cos A = 1 + cos A
= 2 cos2 A/2 = 2/sec A/2 = 2/1+tan2 A/2
12. The probabilities that a student will receive an A, B, C and D grade are 0.30, 0.38, 0.22 and 0.10 respectively.
Find the probability that the student will receive
(i) at most C grade
(ii) B or C grade.
Answer: (i) P(at most C grade)
= P(C grade) + P(D grade) = 0.22 + 0.10 = 0.32
(ii) P(B or C grade)
= P(B grade) + P(C grade) = 0.38 + 0.22 = 0.60
OR
A young man visits a hospital for medical check-up. The probability that he has lungs problem is 0.45, heart problem is 0.29 and either lungs or heart problem is 0.47. What is the probability that he has both types of problems: lungs as well as heart?
Answer: Let L represents the lungs problem and H represents
the heart problem.
Given, P(L) = 0.45, P(H) = 0.29
P(L ∪ H) = 0.47
Q P (L ∪ H) = P(L) + P(H) – P(L ∩ H)
⇒ 0.47 = 0.45 + 0.29 – P(L ∩ H)
⇒ P(L ∩ H) = 0.45 + 0.29 – 0.47
⇒ P(L ∩ H) = 0.74 – 0.47 = 0.27
∴ The probability that he has both types of problems is 0.27.
13. If y = cos x/1+sinx , then find dy/dx.
Answer: We have, y = cos x/1+sinx
14. Find the derivative of f(x) = 2x-3/x+2 .
Answer:
Section – V
15. Two students Kartik and Kanika appeared in an examination. The probability that Kartik will qualify the examination is 0.06 and that Kanika will qualify the examination is 0.20. The probability that both will qualify the examination is 0.02. Find the probability that:
(i) both Kartik and Kanika will not qualify the examination.
(ii) at least one of them will not qualify the examination.
(iii) only one of them will qualify the examination.
Answer: Let E be the event that Kartik will qualify the
examination and F be the event that Kanika will qualify
the examination.
P(E) = 0.06, P(F) = 0.20 and P(E ∩ F) = 0.02
(i) Probability that both will not qualify the
examination
= P(E ∪F) = P(E∪F) = 1 – P(E ∪ F)
= 1 – [P(E) + P(F) – P(E ∩ F)]
= 1 – (0.06 + 0.20 – 0.02) = 1 – 0.24 = 0.72
(ii) Probability that atleast one of them will not qualify
the examination
= P(E′ ∪ F′) = P(E ∩ F)′ = 1 – 0.02 = 0.98
(iii) P (only E) = P(E) – P(E ∩ F)
= 0.06 – 0.02 = 0.04
P (only F) = P(F) – P(E ∩ F)
= 0.20 – 0.02 = 0.18
∴ Required probability = P (only E) + P(only F)
= 0.04 + 0.18 = 0.22
16. If | x +3| + x/x+2 > 1 , then find the interval in which x lies.
Answer:
OR
Show that the following system of inequalities x/2x+1 ≥ 1/4, 6x/4x-1 < 1/2 have no solution.
Answer:
Note that the common solution of (1) and (2) is null set.
Hence, the given system of inequalities has no solution.
17. Three consecutive vertices of a parallelogram ABCD are A(6, –2, 4), B(3, 5, –4) and C(–2, 2, 4). Find the coordinates of the fourth vertex.
Answer: Let the coordinates of the fourth vertex be D(x, y, z)
Section – VI
18. If the letters of the word ‘PATTERN’ are written in all possible orders and are arranged in dictionary order.
What is the rank of the word ‘PATTERN’?
Answer: Total letters in the word ‘PATTERN’ is 7.
∴ Total no. of words starting with ‘A’ = 6!/2!
Total no. of words starting with E = 6!/2!
Total no. of words starting with N = 6!/2!
Total no. of words starting with PAE = 4!/2!
Total no. of words starting with PAN = 4!/2!
Total no. of words starting with PAR = 4!/2!
Total no. of words starting with PATE = 3!
Total no. of words starting with PATN = 3!
Total no. of words starting with PATR = 3!
Total no. of words starting with PATTENR = 1
∴ Total no. of words before ‘PATTERN’
= 3 × 6!/2! + 3 × 4!/2! + 3 × 3! + 1 = 1135
Hence, rank of word ‘PATTERN’ = 1135 + 1 = 1136
OR
Find the number of arrangements of the letters of the word INDEPENDENCE when all the vowels always occur together.
Answer: There are 5 vowels in the given word, which are 4’s E and 1 I. Since, they have to always occur together, we treat them as a single object EEEEI for the time being.
This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3N’s and 2 D’s, can be rearranged in 8!3!2! ways.
Corresponding to each of these arrangements, the 5 vowels E, E, E, E and I can be rearranged in 5!/4! ways.
Therefore, by multiplication principle, the required number of arrangements = 8! /3! x 2! x 5! /4! = 16800
19. If cosθ = 4/5 and cos Φ = 12/13 , where q and f both lie in the fourth quadrant, then find the value of tan (q + Φ).
Answer: Given, cosθ = 4/5 and cos Φ = 12/13
Since q lies in the fourth quadrant, we have cos θ > 0, sin q < 0 and tan θ < 0
Again, since f lies in the fourth quadrant, we have cos Φ > 0, sin f < 0 and tan Φ < 0
OR
Evaluate : cos x + cos y + cos z + cos (x + y + z).
Answer:
