Category: Class 11 Mathematics Objective Questions

Mathematics online test on “Trigonometric Functions of Sum and Difference of Two Angles-2”.

1. If sin x=1/2 and cos x=(sqrt{3})/2, then find sin 2x.
a) (sqrt{3})/2
b) 1/2
c) 1/(sqrt{2})
d) 1
Answer: a
Clarification: We know, sin2x = 2 sin x cos x
sin 2x = 2*1/2*(sqrt{3})/2
sin 2x = (sqrt{3})/2.

2. If tan x=1/(sqrt{3}), then sin 2x =___________________
a) 1/(sqrt{2})
b) 1/2
c) (sqrt{3})/2
d) 1
Answer: c
Clarification: We know, sin 2x = (2 tan x) / (1 + tan²x)
sin 2x=(2*1/(sqrt{3})) / (1+1/3) = (2*3)/4*(sqrt{3}) = (sqrt{3})/2.

3. If tan x = 1/(sqrt{3}), then tan2x =_________________
a) 1
b) (sqrt{3})
c) 1/(sqrt{3})
d) 0
Answer: b
Clarification: We know, sin 2x = (2 tan x) / (1 – tan²x)
tan 2x = (2*1/(sqrt{3})) / (1-1/3) = (2*3)/2*(sqrt{3}) = (sqrt{3}).

4. Which is correct?
a) sin 3x = 3sinx – 4sin³x
b) sin 3x = 4sinx – 3sin³x
c) sin 3x = 3sin³x – 4sinx
d) sin 3x = 4sin³x – 3sinx
Answer: a
Clarification: sin 3x = sin (2x + x)
= sin 2x cos x + cos 2x sin x
= (2 sin x cos x) cos x + (1 – 2sin² x) sin x
= 2 sin x (1 – sin² x) + sin x – 2 sin³ x
= 2 sin x – 2 sin³ x + sin x – 2 sin³ x
= 3 sin x – 4 sin³ x.

5. Which is correct?
a) cos 3x = 3cosx – 4cos³x
b) cos 3x = 4cosx – 3cos³x
c) cos 3x = 3cos³x – 4cosx
d) cos 3x = 4cos³x – 3cosx
Answer: d
Clarification: cos 3x = cos (2x +x)
= cos 2x cos x – sin 2x sin x
= (2cos² x – 1) cos x – 2sin x cos x sin x
= (2cos² x – 1) cos x – 2cos x (1 – cos² x)
= 2cos³ x – cos x – 2cos x + 2 cos³ x
= 4cos³ x – 3cos x.

6. If sin x= 1/2, then sin 3x =_______________
a)(sqrt{3})/2
b) 1/2
c) 1/(sqrt{2})
d) 1
Answer: d
Clarification: We know, sin 3x = 3sin x – 4sin³x
= 3(1/2) – 4(1/2)³
= 3/2 – 4/8 = 3/2 – 1/2 = 1.

7. If cos x=1/2, then cos 3x =_______________
a) 0
b) -1
c) 1/(sqrt{2})
d) 1
Answer: b
Clarification: We know, cos 3x=4cos³x – 3cos x
= 4(1/2)³–3(1/2)
= 4/8 – 3/2 = 1/2 – 3/2 = -1.

8. Find tan 3x if tan x= 1.
a) 1
b) -1
c) 1/(sqrt{3})
d) (sqrt{3})
Answer: b
Clarification: We know, tan 3x = (3tanx – tan³x)/(1-3tan²x)
= (3*1 – 1)/(1-3)
= (2)/(-2) = -1.

9. cos 75° + cos 15° =___________________
a) (frac{sqrt{3}}{sqrt{2}})
b) (frac{sqrt{2}}{sqrt{3}})
c) (frac{sqrt{3}}{2})
d) (frac{1}{sqrt{2}})
Answer: a
Clarification: We know, cos x + cos y = 2 cos (x+y)/2 cos (x-y)/2
cos 75° + cos 15° = 2 cos (75°+15°)/2 cos(75°-15°)/2
= 2 * cos 45° * cos30°
= 2*(1/(sqrt{2}))*((sqrt{3})/2)
= (sqrt{3})/(sqrt{2}).

10. cos 75° – cos 15° =___________________
a) (frac{sqrt{3}}{sqrt{2}})
b) (frac{sqrt{2}}{sqrt{3}})
c) (frac{1}{sqrt{2}})
d) –(frac{1}{sqrt{2}})
Answer: d
Clarification: cos x – cos y = – 2 sin (x+y)/2 sin (x-y)/2
cos 75° – cos 15° = – 2 sin (75°+15°)/2 sin(75°-15°)/2
= – 2 sin 45° sin 30°
= – 2*1/(sqrt{2})*1/2= – 1/(sqrt{2}).

11. sin 75° + sin 15° = _________________
a) (frac{sqrt{3}}{sqrt{2}})
b) (frac{sqrt{2}}{sqrt{3}})
c) (frac{sqrt{3}}{2})
d) (frac{1}{sqrt{2}})
Answer: a
Clarification: We know, sin x + sin y = 2 sin (x+y)/2 cos (x-y)/2
sin 75° + sin 15° = 2 sin (75°+15°)/2 cos(75°-15°)/2
= 2 sin 45° cos 30°
= 2*1/(sqrt{2})*(sqrt{3})/2 = (sqrt{3})/(sqrt{2}).

12. sin 75° – sin 15° =_________________
a) (frac{sqrt{3}}{sqrt{2}})
b) (frac{sqrt{2}}{sqrt{3}})
c) (frac{sqrt{3}}{2})
d) (frac{1}{sqrt{2}})
Answer: d
Clarification: We know, sin x – sin y = 2 cos (x+y)/2 sin (x-y)/2
sin 75° – sin 15° = 2 cos (75°+15°)/2 sin(75°-15°)/2
= 2 cos 45° sin 30°
= 2*1/(sqrt{2})*1/2 = 1/(sqrt{2}).

13. 2 sin 75° sin15° =_____________________________
a) (frac{sqrt{3}}{2})
b) (frac{sqrt{2}}{sqrt{3}})
c) (frac{1}{2})
d) (frac{1}{sqrt{2}})
Answer: c
Clarification: We know, 2 sin x sin y = cos (x – y) – cos (x + y)
2 sin 75° sin15° = cos (75°-15°) – cos (75°+15°)
= cos 60° – cos 90° = 1/2 – 0 = 1/2.

14. 2 cos 75° cos 15° =____________________
a) (frac{sqrt{3}}{sqrt{2}})
b) (frac{1}{2})
c) (frac{sqrt{3}}{2})
d) (frac{1}{sqrt{2}})
Answer: b
Clarification: We know, 2 cos x cos y = cos (x + y) + cos (x – y)
2 cos 75° cos 15° = cos (75°+15°) + cos (75°-15°)
= cos 90° + cos 60°
= 0 + 1/2 = 1/2.

all areas of Mathematics for online tests,

Mathematics Multiple Choice Questions on “Combinations”.

1. Order matters in combination.
a) True
b) False
Answer: b
Clarification: Combination means selection and selection does not require ordering. So, order does not matter in combination.

2. ⁿC_r = ________________
a) n!
b) (frac{n!}{r!})
c) (frac{n!}{(n-r)!})
d) (frac{n!}{(n-r)! r!})
Answer: d
Clarification: Permutation is known as selection. ⁿC_r means arranging r objects out of n.
ⁿC_r = (frac{n!}{(n-r)! r!}).

3. ⁿC₀ = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
Answer: b
Clarification: We know, ⁿC_r = (frac{n!}{(n-r)! r!}).
ⁿC₀ = (frac{n!}{(n-0)! 0!} = frac{n!}{n!(1)}) = 1.

4. ⁿC_n = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
Answer: b
Clarification: We know, ⁿC_r = (frac{n!}{(n-r)! r!}).
ⁿC_n = (frac{n!}{(n-n)! n!} = frac{n!}{(0)! n!}) = 1/1 = 1.

5. ⁿP_r = ⁿC_r * ______________
a) r!
b) 1/r!
c) n!
d) 1/n!
Answer: a
Clarification: We know, ⁿP_r = (frac{n!}{(n-r)!}) and ⁿC_r = (frac{n!}{(n-r)! r!}).
=> ⁿP_r / ⁿC_r = (frac{frac{n!}{(n-r)!}}{frac{n!}{(n-r)! r!}}) = r!
=> ⁿP_r = ⁿC_r * r!

6. Is ⁿC_r = ⁿC_n-r true?
a) True
b) False
Answer: a
Clarification: We know, ⁿC_r = (frac{n!}{(n-r)! r!}).
Replacing r by n-r, we get ⁿC_n-r = (frac{n!}{(n-(n-r))!(n-r)!} = frac{n!}{r! (n-r)!}) = ⁿC_r
=> ⁿC_r = ⁿC_n-r

7. If ⁿC₂ = ⁿC₃ then find n.
a) 2
b) 3
c) 5
d) 6
Answer: c
Clarification: We know, ⁿC_p = ⁿC_q
=>either p = q or p + q = n.
Here p=2 and q=3 so, p ≠ q.
Hence p + q = n => n = p + q = 2 + 3 = 5.

8. If ⁶C₂ = ⁶C_X then find possible values of x.
a) 2
b) 4
c) 2 and 4
d) 3
Answer: c
Clarification: We know, ⁿC_p = ⁿC_q
=>either p = q or p + q = n.
=> either x=2 or x+2=6
=> either x=2 or x=4.

9. Determine n if ²ⁿC₃: ⁿC₃ = 9:1.
a) 7
b) 14
c) 28
d) 32
Answer: b
Clarification: We know, ⁿC_r = (frac{n!}{(n-r)! r!}).
²ⁿC₃ / ⁿC₃ = (frac{frac{n!}{3! (2n-3)!}}{frac{n!}{3! (n-3)!}})
= (frac{2n! (n-3)!}{n! (2n-3)!})
9/1 = (frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)})
9 = (frac{2(2n-1)2}{(n-2)})
9n-18 = 8n-4
=>n=14.

10. If ¹⁴C_r = 14 and ¹⁵C_r = 15. Find the value of ¹⁴C_r-1.
a) 1
b) 14
c) 15
d) 3
Answer: a
Clarification: We know, ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r
Substituting n=14 and r=4, we get ¹⁴C_r + ¹⁴C_r-1 = ¹⁵C_r
=>¹⁴C_r-1 = ¹⁵C_r – ¹⁴C_r = 15-14 = 1.

11. Out of a group of 5 persons, find the number of ways of selecting 3 persons.
a) 1
b) 5
c) 10
d) 15
Answer: c
Clarification: Out of a group of 5 persons, we have to select 3 persons. This can be done in ⁵C₃ ways.
ⁿC_r = (frac{n!}{(n-r)! r!})
⁵C₃ = (frac{5!}{(5-3)! 3!} = frac{5*4*3!}{(2)! 3!} = frac{20}{2})
i.e. 10 ways.

12. In a family, 5 males and 3 females are there. In how many ways we can select a group of 2 males and 2 females from the family?
a) 3
b) 10
c) 30
d) 40
Answer: c
Clarification: Out of 5 males, 2 males can be selected in ⁵C₂ ways.
⁵C₂ = (frac{5!}{(5-2)! 2!} = frac{5*4*3!}{(3)! 2!} = frac{20}{2}) = 10 ways.
Out of 3 females, 2 females can be selected in ³C₂ ways .
³C₂ = (frac{3!}{(3-2)! 2!} = frac{3*2!}{(1)! 2!} = frac{3}{1}) = 3 ways.
So, by the fundamental principle of counting we can select a group of 2 males and 2 females from the family in 10*3 = 30 ways.

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

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

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

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

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

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

6. What is major axis length for ellipse ((frac{x}{25})^2+(frac{y}{16})^2)=1?
a) 5 units
b) 4 units
c) 8 units
d) 10 units
Answer: d
Clarification: Comparing the equation with ((frac{x}{a})^2+(frac{y}{b})^2)=1, we get a=5 and b=4.
Major axis length = 2a = 2*5 =10 units.

7. What is minor axis length for ellipse ((frac{x}{25})^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=5 and b=4.
Minor axis length = 2b = 2*4 = 8 units.

8. What is length of latus rectum for ellipse ((frac{x}{25})^2+(frac{y}{16})^2)=1?
a) 25/2
b) 32/5
c) 5/32
d) 8/5
Answer: b
Clarification: We know, length of latus rectum = 2b²/a.
So, length of latus rectum of given ellipse = 2*42/5 = 32/5.

9. What is equation of latus rectums of ellipse ((frac{x}{25})^2+(frac{y}{16})^2)=1?
a) x=±3
b) y=±3
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=5 and b=4.
For ellipse, c²=a²-b²=25-16=9 => c=3.
Equation of latus rectum x=±c i.e. x=±3.

10. If length of major axis is 10 and minor axis is 8 and major axis is along x-axis then find the equation of ellipse.
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: b
Clarification: Given, 2a=10 => a=5 and 2b=8 => b=4.
Equation of ellipse with major axis along x-axis is ((frac{x}{a})^2+(frac{y}{b})^2)=1.
So, equation of given ellipse is ((frac{x}{5})^2+(frac{y}{4})^2)=1.

11. If foci of an ellipse are (0, ±3) and length of semimajor axis is 5 units, then find the equation of ellipse.
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, a=5 and c=3 => b²=a²-c² = 5²-3²=4² => b=4.
Equation of ellipse with major axis along y-axis is ((frac{x}{b})^2+(frac{y}{a})^2)=1.
So, equation of given ellipse is ((frac{x}{4})^2+(frac{y}{5})^2)=1.

Mathematics Multiple Choice Questions on “Probability – Random Experiments-1”.

1. Probability is __________________________
a) Number of outcomes in favour of event
b) Total number of possible outcomes
c) Ratio of number of outcomes in favour to total number of outcomes
d) Ratio of total number of outcomes to number of outcomes in favour
Answer: c
Clarification: Probability is chance of an outcome to appear. It is the ratio of number of outcomes in favour to total number of outcomes.

2. Probability of getting head on an unbiased coin is ________
a) 1/4
b) 1
c) 0
d) 1/2
Answer: d
Clarification: An unbiased coin can have head or tail as outcome i.e. there are two possible outcomes.
So, probability of getting head on an unbiased coin is 1/2.

3. Probability of getting tail on an unbiased coin is ________
a) 1/4
b) 1
c) 0
d) 1/2
Answer: d
Clarification: An unbiased coin can have head or tail as outcome i.e. there are two possible outcomes.
So, probability of getting tail on an unbiased coin is 1/2.

4. Probability of getting an even number on dice is ___________
a) 1
b) 1/2
c) 1/3
d) 0
Answer: b
Clarification: There are six possible outcomes on dice i.e. 1 to 6.
Even numbers on dice are 2,4,6 i.e. three outcomes in favour of the event.
So, probability of getting an even number on dice is 3/6 = 1/2.

5. Probability of getting an odd number on dice is ___________
a) 1
b) 1/2
c) 1/3
d) 0
Answer: b
Clarification: There are six possible outcomes on dice i.e. 1 to 6.
Odd numbers on dice are 1,2,3 i.e. three outcomes in favour of the event.
So, probability of getting an odd number on dice is 3/6 = 1/2.

6. Probability of getting prime number on dice is ___________
a) 1/2
b) 1/4
c) 1/3
d) 1
Answer: a
Clarification: There are six possible outcomes on dice i.e. 1 to 6.
Prime numbers on dice are 2,3,5 i.e. three outcomes in favour of the event.
So, probability of getting a prime number on dice is 3/6 = 1/2.

7. Probability of getting composite number on dice is ___________
a) 1/2
b) 1/4
c) 1/3
d) 1
Answer: c
Clarification: There are six possible outcomes on dice i.e. 1 to 6.
Composite numbers on dice are 4,6 i.e. two outcomes in favour of the event.
So, probability of getting a composite number on dice is 2/6 = 1/3.

8. Probability of getting 7 on a dice is ________
a) 1/2
b) 0
c) 1
d) 1/3
Answer: b
Clarification: There are six possible outcomes on dice i.e. 1 to 6.
7 does not appear on a dice so probability of getting 7 on a dice is zero.

9. If two coins are tossed simultaneously what are total number of possible outcomes?
a) 2
b) 4
c) 6
d) 8
Answer: b
Clarification: If two coins are tossed simultaneously total number of possible outcomes are 4.
{HH, TT, HT, TH}

10. If two coins are tossed simultaneously what is the probability of getting exactly one head?
a) 1/2
b) 1/3
c) 1/4
d) 3/4
Answer: a
Clarification: If two coins are tossed simultaneously total number of possible outcomes are 4.
{HH, TT, HT, TH} out of which {HT, TH} favour the event.
So, probability of getting exactly one head = 2/4 = 1/2.

11. If two coins are tossed simultaneously what is the probability of getting exactly one tail?
a) 1/2
b) 1/3
c) 1/4
d) 3/4
Answer: a
Clarification: If two coins are tossed simultaneously total number of possible outcomes are 4.
{HH, TT, HT, TH} out of which {HT, TH} favour the event.
So, probability of getting exactly one tail = 2/4 = 1/2.

12. If two coins are tossed simultaneously what is the probability of getting at least one head?
a) 1/2
b) 1/3
c) 1/4
d) 3/4
Answer: d
Clarification: If two coins are tossed simultaneously total number of possible outcomes are 4.
{HH, TT, HT, TH} out of which {HH, HT, TH} favour the event.
So, probability of getting at least one head = 3/4.

13. If two coins are tossed simultaneously what is the probability of getting at most one head?
a) 1/2
b) 1/3
c) 1/4
d) 3/4
Answer: d
Clarification: If two coins are tossed simultaneously total number of possible outcomes are 4.
{HH, TT, HT, TH} out of which {HT, TH, TT} favour the event.
So, probability of getting at most one head = 3/4.

14. If two coins are tossed simultaneously what is the probability of getting all heads?
a) 1/2
b) 1/3
c) 1/4
d) 3/4
Answer: c
Clarification: If two coins are tossed simultaneously total number of possible outcomes are 4.
{HH, TT, HT, TH} out of which {HH} favour the event.
So, probability of getting all heads = 1/4.

15. If two coins are tossed simultaneously what is the probability of getting no heads?
a) 1/2
b) 1/3
c) 1/4
d) 3/4
Answer: c
Clarification: If two coins are tossed simultaneously total number of possible outcomes are 4.
{HH, TT, HT, TH} out of which {TT} favour the event.
So, probability of getting no heads = 1/4.

Mathematics Multiple Choice Questions on “Equal Sets”.

1. Which of the following two sets equal?
a) {1,2,3} and {2,3,4}
b) {1,3,5} and {1,3,5,7}
c) {3,4,7} and {7,4,3}
d) {1,2,7} and {2,7,1,4}
Answer: c
Clarification: Two sets A and B are said to be equal if every element of set A is in set B and vice-versa.
{3,4,7} and {7,4,3} are equal.

2. Let A be set of prime numbers less than 6 and B be the set of prime factors of 30. Set A and B are ____________
a) Infinite
b) Empty
c) Singleton
d) Equal
Answer: d
Clarification: Prime numbers less than 6 are 2,3,5. A={2,3,5}
30=2*3*5 => 2,3,5 are prime factors of 30. B={2,3,5}
Since every element of A is in B and every element of B is in A so they are equal sets.

3. A={0} and B={}. Are sets A and B are equal?
a) True
b) False
Answer: b
Clarification: Set A has one element 0 while set B has no element in it and is empty so they both can’t be equal.

4. Let X be the set of letters of the word ALLOY and Y be the set of letters of the word LOYAL. Are the sets equal?
a) True
b) False
Answer: a
Clarification: X={A,L,O,Y} and Y={L,O,Y,A} .
Since both sets have same elements so they are equal.

5. Which of the sets are equal?
X={x : x-4=0 and x is a natural number}, Y={ x : x²=16 and x is a natural number}, Z={x : x>4 and x<16, x is a natural number}
a) X and Y
b) Y and Z
c) X and Z
d) X, Y and Z
Answer: a
Clarification: x-4=0 => x=4 X={4}
x²=16 => x²-16=0 => (x-4)(x+4)=0 => x=4,-4 but -4 is not a natural number so Y={4}
x>4 and x<16 can’t be true simultaneously so Z={ }.
Hence sets X and Y are equal.

6. Which of the following sets are equal?
X={x : x is letter of word LIFE}, Y={x : x is letter of the word WIFE}, Z={x : x is letter of the word FILE}
a) X and Y
b) Y and Z
c) X and Z
d) X, Y and Z
Answer: c
Clarification: X={L,I,F,E}, Y={W,I,F,E}, Z={F,I,L,E}
So, sets X and Z are equal.

7. Which of the following sets are equal?
a) {a, b, c, d} and {d, c, b, a}
b) {4,8,12,16} and {8,12,16,18}
c) {x : x is a multiple of 10} and {10,20,30}
d) {2,4,6,8} and {x : x is an even number}
Answer: a
Clarification: {a, b, c, d} and {d, c, b, a} are equal sets.
{4,8,12,16} and {8,12,16,18} are not equal their elements are not matching.
{x : x is a multiple of 10} = {10,20,30,………} which is infinite set and not equal to {10,20,30}.
{x : x is an even number} = {2,4,6,………….} which is an infinite set and not equal to {2,4,6}.

8. Which of the following sets are equal?
X={1,-1}, Y={-1,1}, Z={x : x is root of x²-1=0 and x is an integer}
a) X and Y
b) Y and Z
c) X and Z
d) X, Y and Z
Answer: d
Clarification: X={1,-1} which is equal to Y{-1,1}.
x²-1=0 => (x-1)(x+1)=0 => x=1,-1. Z={1,-1} so set X,Y and Z are equal.

9. Equal sets _______ number of elements.
a) Must have same
b) May have same
c) Can’t have same
d) Shouldn’t have different
Answer: a
Clarification: Equal sets are the sets having same number of elements with value of each element set A and B to be equal.

10. If two sets have equal number of elements, then they ___________
a) Are equal
b) Are not equal
c) May be equal
d) Are finite
Answer: c
Clarification: Equal sets are the sets having same number of elements with value of each element set A and B to be equal. So, if the two sets have same number of elements, then they may or may not be equal.

11. If A = {1,10,15,17} and B = {1,5,7,8} then which of the following is the correct statement about A and B?
a) “A” and “B” are equal sets
b) “A” and “B” are equivalent sets
c) “A” and “B” are empty sets
d) “A” and “B” are infinite sets
Answer: b
Clarification: Two sets are equal if each element in A is in B and vice versa. If a number of elements in both sets are the same then they are equivalent sets.

12. If A = {0}, B = {x: x is a non-negative root of x²+2x=0}, C= {x: x>10 and x<5}, D = {x: x²=36} then choose the correct option.
a) A=C
b) A=D
c) B=C
d) A=B
Answer: d
Clarification: Here for set B the value of x = 0 since -2 is a negative root, C is a null set, also solving x²=36 gives x=6 and x =-6 Since A and B have same elements in them and also an equal number of elements, Therefore, A is equal to B.

13. If A= {1,2,3} and B= {x∈R: x³-6x²+11x-6=0} then A and B are equal sets?
a) True
b) False
Answer: a
Clarification: The roots of x³-6x²+11x-6=0 are 1,2,3 hence A and B are the same sets so the statement is true.

14. If A={1,2} and B= {x ∈ R: x²-3x+2=0} then?
a) “A” and “B” are only equivalent sets and not equal
b) “A” and “B” are equal
c) “A” and “B” are not Equivalent
d) “A” and “B” are infinite sets
Answer: b
Clarification: The roots of the equation x²-3x+2=0 are 1 and 2 also A has the same elements, therefore, A and B are equal and equivalent as well as they have the same number of elements.

15. Which of the following statements are correct?
a) Equal and Equivalent sets are actually the same
b) Equivalent sets have a different number of elements
c) Equal sets have the same elements
d) Two null sets are not equal
Answer: c
Clarification: If two sets A and B are equal then they are equivalent as well because all elements of A are in B and all elements of B are in A which means a number of elements in both sets are same.

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).