Category: Class 11 Mathematics Objective Questions

Mathematics Multiple Choice Questions on “Trigonometric Equations – 1”.

1. The solutions of a trigonometric equation for which ___________ are called principal solutions.
a) 0 < x < 2π
b) 0 ≤ x < π
c) 0 ≤ x < 2π
d) 0 ≤ x < nπ
Answer: c
Clarification: The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solutions. x should lie between 0 and 2π except 2π.

2. The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the principal solution.
a) True
b) False
Answer: b
Clarification: The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution. n is used in general solution only.

3. Find the principal solutions of the equation cos x = 1/2.
a) π/6, 5π/6
b) π/3, 5π/3
c) π/3, 2π/3
d) 2π/3, 5π/3
Answer: b
Clarification: Since cos x is positive in 1^st and 4^th quadrant, so there are two principal solutions to above equation i.e. x = π/3, 2π-π/3. So, x = π/3, 5π/3.

4. Find the principal solutions of the equation tan x = – √3.
a) π/6, 5π/6
b) π/3, 5π/3
c) π/3, 2π/3
d) 2π/3, 5π/3
Answer: d
Clarification: Since tan x is negative in 2^nd and 4^th quadrant so, there are two principal solutions to above equation i.e. x = π – π/3, 2π – π/3. So, x=2π/3, 5π/3.

5. Find the principal solutions of the equation sec x = -2.
a) 2π/3, 4π/3
b) 2π/3, 5π/3
c) 4π/3, 5π/3
d) π/3, 5π/3
Answer: a
Clarification: Since sec x is negative in 2^nd and 3^rd quadrant so, there are two principal solutions to above equation i.e. x = π – π/3, π + π/3. So, x=2π/3, 4π/3.

6. Find the principal solutions of the equation sin x = -1/√2.
a) π/4, 3π/4
b) 3π/4, 5π/4
c) 3π/4, 7π/4
d) 5π/4, 7π/4
Answer: d
Clarification: Since sin x is negative in 3^rd and 4^th quadrant so, there are two principal solutions to above equation i.e. x = 2π – π/4, 2π + π/4. So, x=5π/4, 7π/4.

7. Find the principal solutions of the equation cosec x=2.
a) π/6, 5π/6
b) π/6, 7π/6
c) 5π/6, 11π/6
d) 5π/6, 7π/6
Answer: a
Clarification: Since cosec x is positive in 1^st and 2^nd quadrant, so there are two principal solutions to above equation i.e. π/6, π – π/6. So, x=π/6, 5π/6.

8. Find the principal solutions of the equation cot x=1/√3.
a) π/3, 4π/3
b) 2π/3, 5π/3
c) 4π/3, 5π/3
d) π/3, 5π/3
Answer: a
Clarification: Since cot x is positive in 1^st and 3^rd quadrant, so there are two principal solutions to above equation i.e. π/3, π + π/3. So, x=π/3, 4π/3.

9. Find general solution to equation sin x = 1/2.
a) x = nπ + (-1)ⁿ π/3
b) x = nπ + (-1)ⁿ π/6
c) x = nπ + (-1)ⁿ 2π/3
d) x = nπ + (-1)ⁿ 5π/6
Answer: d
Clarification: sin x= 1/2
sin x = sin π/6
x = nπ + (-1)ⁿ π/6.

10. Find general solution to equation cos x = – 1/√2.
a) x = 2nπ±7π/4
b) x = 2nπ±3π/4
c) x = 2nπ±π/4
d) x = 2nπ±π/3
Answer: c
Clarification: cos x = – 1/√2
cos x = – cos π/4 = cos (π- π/4) = cos 3π/4
So, x = 2nπ±π/4.

11. Find general solution to equation cot x = √3.
a) x = nπ + π/3
b) x = nπ + π/6
c) x = nπ + 2π/3
d) x = nπ + 5π/6
Answer: b
Clarification: cot x = √3
tan x = 1/√3
tan x = tan π/6
x = nπ + π/6.

12. Solve: tan x = cot x
a) x = nπ/2 + (π/4)
b) x = nπ + (3π/4)
c) x = nπ + (π/4)
d) x = nπ/2 + (3π/4)
Answer: a
Clarification: tan x = cot x
tan x = cot (π/2 – x)
x = nπ + (π/2 – x)
2x = nπ + (π/2)
x = nπ/2 + (π/4).

Mathematics online quiz on “Binomial Theorem for Positive Integral Index”.

1. What is the coefficient of x²y² in (x + 1)² . (x + 1)³?
a) 1
b) 5
c) 2
d) 10
Answer: a
Clarification: We know that (a + b)² = a² + 2ab + b²
(a + b)³ = a³ + 3ab² + 3a²b + b²
Using these formulae, we get
P(x) = (x² + 2xy + y²)(x³ + 3xy² + 3x²y + y²)
P(x) = 3xy⁴ + 9x²y³ + 10x³y² + 5x⁴y + x⁵ + y⁴ + 2xy³ + x²y²
The coefficient of x²y² in (x + 1)² . (x + 1)³ is 1.

2. What is the remainder when 8⁴⁸ is divided by 63?
a) 4
b) 2
c) 1
d) 7
Answer: c
Clarification: 8⁵⁸ can be written as (8²)²⁴.
8⁴⁸ = (64)²⁴
8⁴⁸ = (63 + 1)²⁴
We know that (60 + 1)²⁴ = (Sigma_{r = 0}^{r = 24})(24Cr 63^{24 – r} 1^r)
= 24C₀ 63²⁴ 4⁰ + 24C₁ 63²³ 4¹ +….+24C₂₃ 63¹ 4²³ + 24C₂₄ 63⁰ 1²⁴
= 63 x k + 1
Therefore, the remainder will be 1.

3. What is the remainder when 4¹⁰³ is divided by 17?
a) 10
b) 14
c) 13
d) 16
Answer: d
Clarification: 4¹⁰³ = 4 x 4¹⁰²
4¹⁰³ = 4 x (4²)⁵¹
4¹⁰³ = 4 x (16)⁵¹
4¹⁰³ = 4 x (17 – 1)⁵¹
4¹⁰³ = 4 x (Sigma_{r = 0}^{r = 51})(51Cr 17^{24 – r} (-1)^r
4¹⁰³ = 4 x [51C₀ 17⁵¹ (-1)⁰ + 51C₁ 17⁵¹ (-1)¹ +….+ 51C₅₀ 17¹ (-1)⁵⁰ + 51C₅₁ 17⁰ (-1)⁵¹]
4¹⁰³ = 4 x 17 x k – 1
The remainder = 17 – 1
Remainder = 16.

4. What is the integral part of (√3 + 1)⁸?
a) 1558
b) 1551
c) 1552
d) 1556
Answer: c
Clarification: By binomial expansion,
(√3 + 1)⁷ = (Sigma_{r = 0}^{r = 7})(7Cr √3^{7 – r} (1)^r)
Whenever, r is an even number, 8 – r will also be even. Then √3 will also have an even power and thereby be integral.
Integral parts = 8C₀ (√3)⁰ + 8C₂ (√3)² + 8C₄ (√3)⁴ + 8C₆ (√3)⁶ + 8C₈ (√3)⁸
Integral parts = 1 + 28 x 3 + 70 x 9 + 28 x 27 + 1 x 81
Integral part = 1552.

5. What is the expansion of (x + y)¹⁰⁰⁰?
a) (Sigma_{r = 0}^{r = 1000})(1000Cr x^{r – 1000} y^r)
b) (Sigma_{r = 0}^{r = 1000})(100Cr x^{1000 – r} y^r)
c) (Sigma_{r = 0}^{r = 999})(1000Cr x^{r – 1000} y^r)
d) (Sigma_{r = 0}^{r = 999})(1000Cr x^{1000 – r} y^r)
Answer: b
Clarification: The expansion can be done using binomial theorem.
(x + y)¹⁰⁰⁰ = 1000C₀ x¹⁰⁰⁰ y⁰ + 1000C₁ x⁹⁹⁹ y¹ +….+ 1000C₉₉₉ x¹ y⁹⁹⁹ + 1000C₁₀₀₀ x⁰ y¹⁰⁰⁰
This can also be written as,
(x + y)¹⁰⁰⁰ = (Sigma_{r = 0}^{r = 1000})(100Cr x^{1000 – r} y^r).

6. What is the real part of (11 + i)³?
a) 1331
b) 1332
c) 1328
d) 1329
Answer: c
Clarification: (11 + i)³ = 11³ + 3.11².i +3.i².11 +i³
= 1331 + 363i – 3 – i
= 1328 + 365i.

7. What are the coefficients of the first and the last term of (a + b)ⁿ?
a) 2
b) 1
c) Coefficients depend on n
d) 3
Answer: b
Clarification: The coefficient of the first term and last term is same. The first term is nC₁ aⁿ and the last term is nC₀ bⁿ unless, a and b are numbers that change the value of the coefficient.

8. What is the remainder when (4)^{2n + 1} is divided by 5?
a) 4
b) 1
c) 2
d) 3
Answer: a
Clarification: The powers of four follow the given order:
4¹ = 4
4² = 16
4³ = 64
4⁴ = 256
4⁵ = 1024 and so on.
Odd powers of 4, have the number 4 in the units place. When 5 divides the nearest ten, 4 will be obtained as the remainder each time.

9. What is the expansion of the series (xy + 2)²?
a) x² + y² + 4
b) xy² + 4 +2xy
c) x²y² + 2xy + 4
d) x²y² + 4xy + 4
Answer: d
Clarification: (a + b)² can be expanded using binomial theorem to get:
(a + b)² = a² + 2ab + b²
Here, a = xy and b = 2
Therefore, (xy + 2)² = (xy)² + 2(xy)(2) + (2)²
(xy + 2)² = x²y² + 4xy + 4.

10. What is the answer of (frac{x^2+y^2+2xy}{x^2-y^2})?
a) (x – y) (x + y)^-2
b) (x + y) (x – y)^-2
c) (x + y) (x – y)^-1
d) (x – y) (x + y)^-1
Answer: c
Clarification: x² + y² + 2xy is the expansion of (x + y)²
x² – y² can be written as (x – y)(x + y)
Substituting in the fraction we get, (frac{(x + y)^2}{(x – y)(x + y)}).
After cancelling the terms we get, (frac{x^2+y^2+2xy}{x^2-y^2}) = (x + y) (x – y)^-1.

11. What is the value of (frac{7^3+2^3+84}{7^2-2^2}) ?
a) 9 (frac{1}{2})
b) 9 (frac{2}{3})
c) 9 (frac{1}{3})
d) 9 (frac{1}{4})
Answer: b
Clarification: Using binomial theorem we know that (a + b)³ = a³ + 3ab² + 3a²b + b³
Therefore, (7 + 2)³ = 7³ + 2³ + (3 x 7 x 2²) + (3 x 2 x 7²)
(9)³ = 7³ + 2³ + 84 + (3 x 2 x 7²)
729 = 7³ + 2³ + 84 + 294
7³ + 2³ + 84 = 435
Also 7² – 2² = (7 – 2)(7 + 2)
7² – 2² = (5)(9)
7² – 2² = 45
So (frac{7^3 + 2^3 + 84}{7^2-2^2}) = 435 / 45
= 9 (frac{30}{45})
= 9 (frac{2}{3}).

12. What is the value of (frac{101^3-99^3+2969703–3029697}{ 101^2 – 99^2})?
a) 1
b) 1/200
c) 1/100
d) 1/50
Answer: d
Clarification: The numerator when simplified is of the form (101 – 99)³
The denominator can be simplified as (101 – 99)(101 + 99)
When we substitute in the numerator and denominator we get (2 x 2 x 2) / (2 x 200)
= 1/50.

13. What is the quotient when x⁴ + 4x³y + 6x²y + 4xy³ + y⁴ is divided by (x + y)?
a) (x + y)³
b) x² + y²
c) (x + y)²
d) (x + y)
Answer: a
Clarification: Using binomial expansions properties, x⁴ + 4x³y + 6x²y + 4xy³ + y⁴ can be written as
= 4C₀x⁴y⁰ + 4C₁x³y¹ + 4C₂x²y² + 4C₃x¹y² + 4C₄x⁰y⁴
= (x + y)⁴
When divided by (x + y), we get (x + y)³.

14. What is the real part of (9 + 3i)²?
a) 81
b) 90
c) 54
d) 72
Answer: d
Clarification: Using binomial theorem (9 + 3i)² = 81 + 54i + 9i²
We know that i² = –1
Therefore, (9 + 3i)² = 81 + 54i – 9
(9 + 3i)² = 72 + 54i
Real part = 72.

all areas of Mathematics for online Quizzes,

Mathematics Multiple Choice Questions on “Conic Sections – Hyperbola”.

1. A hyperbola has ___________ vertices and ____________ foci.
a) two, one
b) one, one
c) one, two
d) two, two
Answer: d
Clarification: A hyperbola has two vertices lying on each end and two foci lying inside the hyperbola.
If P is a point on hyperbola and F₁ and F₂ are foci then |PF₁-PF₂| remains constant.

2. The center of hyperbola is the same as a vertex.
a) True
b) False
Answer: b
Clarification: No, center and vertex are different for hyperbola.
Hyperbola has one center and two vertices.

3. Find the coordinates of foci of hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1.
a) (±5,0)
b) (±4,0)
c) (0,±5)
d) (0,±4)
Answer: a
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
For hyperbola, c²=a²+b²=9+16=25 => c=5.
So, coordinates of foci are (±c,0) i.e. (±5,0).

4. Find the coordinates of foci of hyperbola ((frac{y}{16})^2-(frac{x}{9})^2)=1.
a) (±5,0)
b) (±4,0)
c) (0,±5)
d) (0,±4)
Answer: c
Clarification: Comparing the equation with ((frac{y}{a})^2-(frac{x}{b})^2)=1, we get a=4 and b=3.
For hyperbola, c²=a²+b²= 16+9=25 => c=5.
So, coordinates of foci are (0,±c) i.e. (0,±5).

5. What is eccentricity for ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) 2/5
b) 3/5
c) 15
d) 5/3
Answer: d
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
For hyperbola, c²=a²+b²= 9+16=25 => c=5.
We know, for hyperbola c=a*e
So, e=c/a = 5/3.

6. What is transverse axis length for hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) 5 units
b) 4 units
c) 8 units
d) 6 units
Answer: d
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
Transverse axis length = 2a = 2*3 =6 units.

7. What is conjugate axis length for hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) 5 units
b) 4 units
c) 8 units
d) 10 units
Answer: c
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
Conjugate axis length = 2b = 2*4 =8 units.

8. What is length of latus rectum for hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) 25/2
b) 32/3
c) 5/32
d) 8/5
Answer: b
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
We know, length of latus rectum = 2b²/a.
So, length of latus rectum of given hyperbola = 2*42/3 = 32/3.

9. What is equation of latus rectums of hyperbola ((frac{x}{9})^2-(frac{y}{16})^2)=1?
a) x=±5
b) y=±5
c) x=±2
d) y=±2
Answer: a
Clarification: Comparing the equation with ((frac{x}{a})^2-(frac{y}{b})^2)=1, we get a=3 and b=4.
For hyperbola, c²=a²+b²= 9+16=25 => c=5.
Equation of latus rectum x=±c i.e. x= ±5.

10. If length of transverse axis is 8 and conjugate axis is 10 and transverse axis is along x-axis then find the equation of hyperbola.
a) ((frac{x}{4})^2-(frac{y}{5})^2)=1
b) ((frac{x}{5})^2-(frac{y}{4})^2)=1
c) ((frac{x}{10})^2-(frac{y}{8})^2)=1
d) ((frac{x}{8})^2-(frac{y}{10})^2)=1
Answer: a
Clarification: Given, 2a=8 => a=4 and 2b=10 => b=5.
Equation of hyperbola with transverse axis along x-axis is ((frac{x}{4})^2-(frac{y}{5})^2)=1.
So, equation of given hyperbola is ((frac{x}{4})^2-(frac{y}{5})^2)=1.

11. If foci of a hyperbola are (0, ±5) and length of semi transverse axis is 3 units, then find the equation of hyperbola.
a) ((frac{x}{4})^2-(frac{y}{3})^2)=1
b) ((frac{x}{3})^2-(frac{y}{4})^2)=1
c) ((frac{x}{10})^2+(frac{y}{8})^2)=1
d) ((frac{x}{8})^2-(frac{y}{6})^2)=1
Answer: a
Clarification: Given, a=3 and c=5 => b²=c²-a² = 5²-3²=4² => b=4.
Equation of hyperbola with transverse axis along y-axis is ((frac{y}{a})^2-(frac{x}{b})^2)=1.
So, equation of given hyperbola is ((frac{y}{3})^2-(frac{x}{4})^2)=1.

12. A hyperbola in which length of transverse and conjugate axis are equal is called _________ hyperbola.
a) isosceles
b) equilateral
c) bilateral
d) right
Answer: b
Clarification: A hyperbola in which length of transverse and conjugate axis is equal is called equilateral hyperbola. In this type of hyperbola, a=b i.e. 2a=2b or length of transverse and conjugate axis are equal.

Mathematics Assessment Questions for IIT JEE Exam on “Probability – Random Experiments-2”.

1. If n coins are tossed simultaneously what is the total number of outcomes?
a) 2^n-2
b) 2^n-1
c) 2ⁿ
d) 2ⁿ⁺¹
Answer: c
Clarification: If n coins are tossed simultaneously total number of possible outcomes are 2ⁿ.
For 2 coins, it is 4. For 3 coins it is 8 etc.

2. If four dice are rolled simultaneously then what is the total number of possible outcomes?
a) 6
b) 36
c) 216
d) 1296
Answer: d
Clarification: If four dice are rolled simultaneously then total number of possible outcomes is 6⁴=1296.

3. If two dice are simultaneously rolled then what is probability of getting sum 4?
a) 1/6
b) 1/4
c) 1/12
d) 1/9
Answer: c
Clarification: If two dice are simultaneously rolled then total number of possible outcomes is 6*6 =36.
We can get 4 as sum in these cases {(1,3), (2,2), (3,1)}. So, probability of getting sum 4 is 3/36 = 1/12.

4. If two dice are simultaneously rolled then what is probability of getting sum 11?
a) 1/6
b) 1/4
c) 1/9
d) 1/18
Answer: d
Clarification: If two dice are simultaneously rolled then total number of possible outcomes is 6*6 =36.
We can get 11 as sum in these cases {(5,6), (6,5)}. So, probability of getting sum 11 is 2/36 = 1/18.

5. A coin is tossed and a dice is rolled simultaneously what is total number of possible outcomes?
a) 2
b) 6
c) 12
d) 24
Answer: c
Clarification: If a coin is tossed and a dice is rolled simultaneously then total number of possible outcomes will be 2*6 =12.

6. A coin is tossed and if head come then a red dice is rolled and if tail come then a blue dice is rolled then what is the possible number of outcomes?
a) 6
b) 12
c) 24
d) 72
Answer: b
Clarification: If head come on coin then red dice can have 6 outcomes so total 1*6 = 6 outcomes. And if tail come on coin then blue dice can have 6 outcomes so total 1*6 = 6 outcomes.
So, total number of possible outcomes = 6 + 6 = 12.

7. If 4 bulbs are there each of which can be defective or non-defective then what is the total number of possible outcomes?
a) 8
b) 16
c) 4
d) 2
Answer: b
Clarification: Each bulb can have two cases i.e. either defective or non-defective. Then total number of outcomes will be 2⁴ = 16.

8. If 4 bulbs are there each of which can be defective or non-defective then what is the probability that all the bulbs are defective?
a) 1/8
b) 1/16
c) 1/4
d) 1/2
Answer: b
Clarification: Each bulb can have two cases i.e. either defective or non-defective. Then total number of outcomes will be 2⁴ = 16.
So, the probability that all the bulbs are defective is 1/16.

9. If 4 bulbs are there each of which can be defective or non-defective then what is the probability that at least one bulb is non- defective?
a) 7/8
b) 15/16
c) 3/4
d) 1/2
Answer: b
Clarification: Each bulb can have two cases i.e. either defective or non-defective. Then total number of outcomes will be 2⁴ = 16.
Probability that all the bulbs are defective is 1/16.
So, the probability that at least one bulb is non-defective is 1-1/16 = 15/16.

10. The probability can be negative.
a) True
b) False
Answer: b
Clarification: Since probability is the ratio of number of outcomes in favour to total number of possible outcomes and number (counting) cannot be negative so, probability cannot be negative.

Mathematics Assessment Questions for IIT JEE Exam,

Mathematics Multiple Choice Questions on “Subsets – 1”.

1. If every element of set X is in set Y then_____________
a) X⊂Y
b) Y⊂X
c) X=Y
d) X≠Y
Answer: a
Clarification: If every element of set X is in set Y then X is called subset of Y. X⊂Y.
But every element of Y may or may not be the element of X so we can’t say Y⊂X and hence we can’t decide equality.

2. If set A is equal to set B then ____________
a) A⊂B
b) B⊂A
c) A⊂B and B⊂A
d) neither A⊂B nor B⊂A
Answer: c
Clarification: If set A is equal to set B then every element of set A is in set B i.e. A⊂B and every element of set B is in set A i.e. B⊂A. Hence A⊂B and B⊂A.

3. Let X= {1,2,3}, Y= {}, Z= {1,2,3}, then which of the following is true?
a) X⊂Y
b) Only Y⊂X and Y⊂Z
c) Z⊂Y
d) Y⊂X and Y⊂Z and X⊂Z
Answer: d
Clarification: Null set is the subset of every set so Y⊂X and Y⊂Z.
Since set X is equal to set Z so, Z⊂X and X⊂Z.

4. Let A= {2,3,5} and B= {3,5,7}. Which of the following is true?
a) A⊂B
b) B⊂A
c) A=B
d) A⊂A
Answer: d
Clarification: Since every set is subset of itself so A⊂A and B⊂B.
Since every element of set A is not in set B so, A is not a subset of B. Also, every element of set B is not in set A so, B is not a subset of A. Hence, A≠B.

5. Let X be set of rational numbers. Which of the following is superset of X?
a) Set of real numbers
b) Set of natural numbers
c) Set of whole numbers
d) Set of integers
Answer: a
Clarification: If X is subset of Y then Y is called superset of X. Set of rational numbers is subset of set of real numbers so, set of real numbers is called superset of X.

6. Let X be set of rational numbers. Which of the following is not subset of X?
a) Set of real numbers
b) Set of natural numbers
c) Set of whole numbers
d) Set of integers
Answer: a
Clarification: Set of rational numbers { x : x=p/q where p and q are integers and q≠0}.
Set of real number is not a subset of X. Set of natural numbers, whole numbers, integers are subset of X.

7. Let A = {1, 3}, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Then ___________
a) A⊂B
b) B⊂A
c) C⊂B and A⊂C
d) B⊂C and A⊂C
Answer: d
Clarification: Here A has two elements 1 and 3. They both belongs to C so, A⊂C. Here B has three elements 1, 5 and 9. They all belongs to C so, B⊂C.

8. If an element x∈A and A⊂B then x∈B.
a) True
b) False
Answer: a
Clarification: If A⊂B then every element of A is in set B. Since x is an element of A so, x also belong to B. x∈B is true.

9. If X∈A and A⊂B then X⊂B.
a) True
b) False
Answer: b
Clarification: Let X = {1,2}. A= {{1,2},3}, B= {{1,2},3,4}. Since elements of X does not belongs to set B so, X is not a subset of B. X⊂B is false.

10. Let A = {1, 2, {3, 4}, 5}. Which of the following is true?
a) {3, 4} ⊂ A
b) {3, 4} ∈ A
c) {{3, 4}} ⊂ A
d) {1, 2, 5} ⊂ A
Answer: a
Clarification: Here A has elements 1,2, {3,4} and 5. So, {3, 4} ∈ A and {{3, 4}} ⊂ A. {1, 2, 5} ⊂ A. {3, 4} is not subset of A.