250+ TOP MCQs on Trigonometric Functions & Answers | Class 11 Maths

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

1. If sin x=0 then x = ________
a) nπ
b) (2n+1) π/2
c) (n+1) π
d) nπ/2
Answer: a
Clarification: When know, sin x =0 whenever x is 0, π, 2π, 3π,….. i.e. all integral multiples of π so, x=nπ when sin x=0.

2. If cos x=0 then x = ________
a) nπ
b) (2n+1) π/2
c) (n+1) π
d) nπ/2
Answer: b
Clarification: When know, cos x =0 whenever x is π/2, 3π/2, 5π/2, ………… i.e. all odd integral multiples of π/2
so, x=(2n+1) π/2 when cos x=0.

3. If tan x = 0 then x = _________
a) nπ
b) (2n+1) π/2
c) (n+1) π
d) nπ/2
Answer: a
Clarification: We know, tan x = sin x / cos x. So, tan x will be zero wherever sin x is zero except the points where cos x is also zero. We know there is no point where sin x as well as cos x both are zero. So, tan x = 0 => x=nπ.

4. 1-sin245° = ___________
a) 1/2
b) 1
c) 0
d) √3 /2
Answer: a
Clarification: We know, sin245° + cos245°=1
So, 1- sin245° = cos245° = (1/√2)2 = 1/2.

5. 1-cos2x=_________
a) sin x
b) cos x
c) sin 2x
d) sin2x
Answer: d
Clarification: We know, sin2x+ cos2x=1
So, 1-cos2x=sin2x.

6. 1-sec2x=_________
a) cot2x
b) tan2x
c) -tan2x
d) -cot2x
Answer: c
Clarification: We know, sec2x – tan2x=1
So, 1-sec2x=-tan2x.

7. 1+ tan2x=_______________
a) sec2x
b) -sec2x
c) cosec2x
d) -cosec2x
Answer: a
Clarification: We know, sec2x – tan2x=1
So, 1+ tan2x=sec2x.

8. cot2x – cosec2x = __________
a) 1
b) -1
c) sin2x
d) cos2x
Answer: b
Clarification: We know, cosec2x – cot2x = 1
So, cot2x – cosec2x = -1.

9. cosec2x – 1 = ______________
a) cot2x
b) -cot2x
c) tan2x
d) -tan2x
Answer: a
Clarification: We know, cosec2x – cot2x = 1
So, cosec2x – 1 = cot2x.

10. tan x is not defined for_______
a) 0
b) nπ/2
c) (2n+1) π/2
d) nπ
Answer: c
Clarification: We know, tan x is not defined when cos x = 0.
cos x = 0 whenever x is π/2, 3π/2, 5π/2, ………… i.e. all odd integral multiples of π/2
so, x=(2n+1) π/2.

11. sin (-45°) = ______________
a) 1
b) -1
c) 1/√2
d) -1/√2
Answer: d
Clarification: We know, sin(-x) = sin x
So, sin (-45°) = -sin 45° = -1/√2.

12. cos (-60°) = ________________
a) -√3/2
b) 1/2
c) √3/2
d) -1/2
Answer: b
Clarification: We know, cos (-x) = cos x
So, cos(-60°) = cos 60°=1/2.

250+ TOP MCQs on Permutations-1 & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Permutations-1”.
1. Permutation is also known as selection.
a) True
b) False
View Answer
Answer: b
Clarification: Permutation is known as arrangement. Selection is another name for combinations.
It involves arrangement of letters, numbers, persons etc.
2. nPr = ________________
a) n!
b) (frac{n!}{r!})
c) (frac{n!}{(n-r)!})
d) (frac{n!}{(n-r)! r!})
View Answer
Answer: c
Clarification: Permutation is known as arrangement. nPr means arranging r objects out of n.
nPr = (frac{n!}{(n-r)!}).
3. 6! = _____________
a) 24
b) 120
c) 720
d) 8
View Answer
Answer: c
Clarification: We know, n! = n.(n-1).(n-2).(n-3)…..
6! = 6.5.4.3.2.1 = 720.
4. (frac{7!}{5!}) = ____________________
a) 7
b) 42
c) 230
d) 30
View Answer
Answer: b
Clarification: We know, n! = n.(n-1).(n-2).(n-3)…… = n(n-1)!
(frac{7!}{5!} = frac{7.6!}{5!} = frac{7.6.5!}{5!}) = 7.6 = 42.
5. (frac{100}{10!} = frac{1}{8!} + frac{x}{9!}). Find x.
a) 1
b) 2
c) 3
d) 4
View Answer
Answer: a
Clarification: (frac{100}{10!} = frac{1}{8!} + frac{x}{9!}).
We know, n! = n.(n-1). (n-2). (n-3) …………… = n(n-1)!
(frac{100}{10.9.8!} = frac{1}{8!} + frac{x}{9.8!})
=> (frac{100}{10.9} = frac{1}{1} + frac{x}{9})
=> (frac{10}{9} = 1 + frac{x}{9})
=> (frac{x}{9} = frac{10}{9} – 1 = frac{1}{9})
=> x=1.
6. nP0 = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
View Answer
Answer: b
Clarification: We know, nPr = (frac{n!}{(n-r)!}).
nP0 = (frac{n!}{(n-0)!} = frac{n!}{(n)!}) = 1.
7. nPn = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
View Answer
Answer: a
Clarification: We know, nPr = (frac{n!}{(n-r)!}).
nPn = (frac{n!}{(n-n)!} = frac{n!}{(0)!}) = n!
8. The number of permutations of n different objects taken r at a time, where repetition is allowed is _______________
a) n!
b) r!
c) nPr
d) nr
View Answer

Answer: d
9. Find the number of permutations of word DEPENDENT.
a) 132400
b) 1512500
c) 1663200
d) 1723400
View Answer
Answer: c
Clarification: There are total 9 letters out of which 1T, 2N, 2D, 3E, 1P.
Total number of permutations are (frac{9!}{3!2!2!} = frac{9.8.7.6.5.4.3.2.1}{6*2*2} = frac{362880}{24}) = 15120.
10. Find the number of 5 letter words which can be formed from word IMAGE without repetition using permutations.
a) 20
b) 60
c) 120
d) 240
View Answer
Answer: c
Clarification: IMAGE is a 5 letters word. We have to arrange all 5 letters of the word IMAGE without repetition. So, total permutations are nPr = 5P5 = 5! = 5.4.3.2.1 = 120.
11. Find the number of 5 letter words that can be formed from word IMAGE using permutations if repetition is allowed.
a) 25
b) 120
c) 125
d) 3125
View Answer
Answer: d
Clarification: IMAGE is a 5 letters word. We have to arrange all 5 letters of the word IMAGE with repetition allowed. So, total permutations are nr = 55 = 3125.

250+ TOP MCQs on Conic Sections – Parabola-1 & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Conic Sections – Parabola-1”.

1. Find the focus of parabola with equation y2=100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)
Answer: c
Clarification: Comparing equation with y2=4ax.
4a=100 => a=25.
Focus is at (a, 0) i.e. (25, 0).

2. Find the focus of parabola with equation y2=-100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)
Answer: d
Clarification: Comparing equation with y2=-4ax.
4a=100 => a=25.
Focus is at (-a, 0) i.e. (-25, 0).

3. Find the focus of parabola with equation x2=100y.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)
Answer: a
Clarification: Comparing equation with x2=4ay.
4a=100 => a=25.
Focus is at (0, a) i.e. (0, 25).

4. Find the focus of parabola with equation x2=-100y.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)
Answer: b
Clarification: Comparing equation with x2=-4ay.
4a=100 => a=25.
Focus is at (0, -a) i.e. (0, -25).

5. Find the equation of latus rectum of parabola y2=100x.
a) x=25
b) x=-25
c) y=25
d) y=-25
Answer: a
Clarification: Comparing equation with y2=4ax.
4a=100 => a=25. Line passing through focus perpendicular to axis is latus rectum.
Equation of latus rectum is x=a => x=25.

6. Find the equation of latus rectum of parabola y2=-100x.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: b
Clarification: Comparing equation with y2=-4ax.
4a=100 => a=25. Line passing through focus perpendicular to axis is latus rectum.
Equation of latus rectum is x=-a => x=-25.

7. Find the equation of latus rectum of parabola x2=100y.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: d
Clarification: Comparing equation with x2=4ay.
4a=100 => a=25. Line passing through focus perpendicular to axis is latus rectum.
Equation of latus rectum is y=a => y=25.

8. Find the equation of latus rectum of parabola x2=-100y.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: c
Clarification: Comparing equation with x2=-4ay.
4a=100 => a=25. Line passing through focus perpendicular to axis is latus rectum.
Equation of latus rectum is y=-a => y=-25.

9. Find the equation of directrix of parabola y2=100x.
a) x=25
b) x=-25
c) y=25
d) y=-25
Answer: b
Clarification: Comparing equation with y2=4ax.
4a=100 => a=25. Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is x=-a => x=-25.

10. Find the equation of directrix of parabola y2=-100x.
a) x=25
b) x=-25
c) y=-25
d) y=25
Answer: a
Clarification: Comparing equation with y2=-4ax.
4a=100 => a=25. Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is x=a => x=25.

250+ TOP MCQs on Statistics – Mean Deviation & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Statistics – Mean Deviation”.

1. Find the mean deviation about the median of the scores of a batsman given below.

Innings Scores
1 20
2 56
3 0
4 84
5 11
6 120

a) 10
b)10.5
c) 11
d) 9
Answer: b
Clarification: Mean deviation = (frac{1}{n})[Σi = n |xI – A|], where A is median or AM
From the given data, Median, A = (20 + 56)/2 = 38
⇒ Mean deviation = 1/6 x (63) = 10.5.

2. What is the mean deviation from the mean for the following data?

a) 0
b) 3
c) 1
d) ½
Answer: a
Clarification: Mean = (117 + 156 + 206 + 198 + 223)/5 = 180

Xi 117 156 206 198 223
Xi – mean -63 -24 26 18 43

Mean deviation = (frac{1}{n})[Σi = 5 |xI – mean|] = 1/5 x [(-63) + (-24) + 26 + 18 + 43] = 1/5 x [0] = 0.

3. The mean deviation of an ungrouped data is 150. If each observation is increased by 3.5%, then what is the new mean deviation?
a) 153.5
b) 3.5
c) 155.25
d) 150
Answer: c
Clarification: If x1, x2, …, xn are the observations, then the new observations are (1.035) x1, (1.035) x2, ……, (1.035) xn.
Therefore, the new mean is (1.035) x̄
Now, Mean deviation = (frac{1}{n})[Σi = n |xI – mean|]
⇒ New mean deviation = (frac{1}{n})[Σi = n|(1.035)xI – (1.035) x̄|] = (1.035) × (frac{1}{n})[Σi = n |xI – mean|] = 1.035 x 150 = 155.25.

4. Find the mean deviation about mean from the following data:

xi 3 5 20 25 27
fi 5 12 20 8 15

a) 7.7
b) 15
c) 8.7
d) 6.2
Answer: a
Clarification: From the given data,

xi fi fixi |xi-18| fi|xi-18|
3 5 15 15 75
5 12 60 13 156
20 20 400 2 40
25 8 200 7 56
27 15 405 9 135
Σ fi = 60 Σ fixi = 1080 Σ fi|xi – 15| = 462

Now, Mean = (frac{1}{n}) Σ fixi = 1080/60 = 18
⇒ Mean deviation = (frac{1}{n}) Σ fi|xi – 18| = 462/60 = 7.7.

5. What is the geometric mean of 5,52, ….,5n?
a) 5n/2
b) 5(n+1)/2
c) 5n(n+1)/2
d) 5n
Answer: b
Clarification: Geometric Mean = (5 x 52 x …… x 5n)1/n = [5(1+2+…+n)]1/n = [5n(n+1)/2]1/n = 5(n+1)/2.

6. In a class there are 20 juniors, 15 seniors and 5 graduate students. If the junior averaged 65 in the midterm exam, the senior averaged 70 and the graduate students averaged 91, then what is the mean of the centre class approximately?
a) 71
b) 74
c) 70
d) 72
Answer: c
Clarification: Combined mean = (Σ xini)/(Σ ni) = (20 × 65 + 15 × 70 + 5 × 91)/(20 + 15 + 5) = 70.

7. Find the mean deviation from mean of the observations: a, a+d, …., (a+2nd).
a) n(n + 1)d2/3
b) n(n + 1)d2/2
c) a + n(n + 1)d2/2
d) n(n + 1)d/(2n + 1)
Answer: d
Clarification: Mean = (frac{1}{n}) Σ xi = (frac{1}{2n+1}) [a + (a + d) + … + (a + 2nd)] = a + nd
⇒ Mean Deviation = (frac{1}{2n+1}) [2 × d × (1 + 2 + … + n)] = [n (n + 1) (d)]/(2n + 1).

250+ TOP MCQs on Sets and their Representations & Answers | Class 11 Maths

Mathematics MCQs for Class 11 on “Sets and their Representations – 2”.

1. How to define a set?
a) A collection of well-defined objects or element
b) A collection of unordered objects or element
c) Any random elements
d) A collection of special characters
Answer: a
Clarification: Generally, a set is defined as a collection of well defined objects or elements.
Each element in a set is unique.
Say for example, if S a set it is represented as,
S = {x: 2x2 ᵾ x< 5 and x € N}
Then the elements present in the set will be
S = {2, 8, 18, 32}.

2. How is a set denoted?
a) ()
b) {}
c) []
d) **
Answer: b
Clarification: A set is represented by {}.
Usually, but not necessarily a set is denoted by a capital letter e.g. A, B……. V, W, X, Y, Z.
The elements are enclosed between { } denoted by small letters a, b, ……., y, z.

3. How will you define a set of all real numbers?
a) {x: -1 < x < 1}
b) [x: -∞ < x < ∞]
c) {x: -∞ < x < ∞}
d) {x: -Z < x < +Z}
Answer: c
Clarification: All the numbers whether it is an integer or rational number or irrational number is defined as Real Number. The range of the real number lies between in the range (-∞, +∞).

4. How will you define Union of two sets A and B?
a) {x: x € A or x € B}
b) {x: x € A or x € B (or both)}
c) {x: x € A and B}
d) {x: x € A – B}
Answer: b
Clarification: Union of two or more sets is the set of all elements that belongs to any of these sets.
The symbol used for this union of sets is ‘∪‘.
If A = {1, 2, 3, 4} and B = {2, 4, 5, 6} and C = {1, 2, 6, 8}
Then, A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}.

5. How will you define the difference of two sets B-A?
a) {x: x € A and x Ɇ B}
b) {x: x Ɇ A and x € B}
c) {x: x € A and x € B}
d) {x: x Ɇ A and x Ɇ B}
Answer: b
Clarification: The difference of a set A and B is denoted as A-B. A-B is a set of those elements that are in the set A but not in the set B. Similarly, the difference of a set B and A is denoted as B-A. It is a set of those elements that are in the set B but not in the set A.

6. What will be the set of the interval (a, b]?
a) {x: a < x < b}
b) {x: a ≤ x ≤ b}
c) {x: a < x ≤ b}
d) {x: a ≤ x < b}
Answer: c
Clarification: The symbol ( ) implies that the value will always be less than or greater than the x value i.e. end points are not included.
{ } implies that all the values that does not satisfy a given interval are included inside {}.
[ ] implies that the value will always be less than equal to or greater than equal to the x value i.e. end points are included. This is possible only when both a and b are finite.

7. How to define Wavy Curve Method f(x)?
a) (x-a1)n1 / (x-a2)n2 / (x-a3)n3 …………… / (x-ak)nk * (x-b1)m1 / (x-b2)m2 / (x-b3)m3 …….. /(x-bp)mp
b) (x-a1)n1 + (x-a2)n2 +(x-a3)n3 …………… + (x-ak)nk / (x-b1)m1 + (x-b2)m2 + (x-b3)m3 …….. + (x-bp)mp
c) (x-a1)n1 (x-a2)n2 (x-a3)n3 …………… (x-ak)nk / (x-b1)m1 (x-b2)m2 (x-b3)m3 …….. (x-bp)mp
d) (x-a1)n1 – (x-a2)n2 – (x-a3)n3 …………… – (x-ak)nk / (x-b1)m1 – (x-b2)m2 – (x-b3)m3 ……..- (x-bp)mp
Answer: c
Clarification: The method of intervals {or wavy curve} is used for solving inequalities of the form
f(x) = (x-a1)n1 (x-a2)n2 (x-a3)n3 …………… (x-ak)nk / (x-b1)m1 (x-b2)m2 (x-b3)m3 …….. (x-bp)mp > 0 (< 0, ≤ 0, or ≥ 0)
where, n1, n2, ,n3, …….. nk and m1, m2, m3, …….. , mp are natural numbers .
a1, a2, a3, ……..ak and b1, b2, b3, …….. bp are any real numbers such that ai ≠ bj where i = 1, 2, 3, ……. , k and j = 1, 2, 3, ….. , p.

8. How to solve for x, if |x-1| ≥ 3?
a) (-∞, -2) ∪ (4, ∞)
b) (-∞, -2] ∪ [4, ∞)
c) (0, -2] ∪ (4, 0)
d) (-∞, ∞) – {-2, 4}
Answer: b
Clarification: Given, |x-1| ≥ 3
= x-1 < -3 or x – 1 ≥ 3
= x ≤ -2 or x ≥ 4
Hence, x c (-∞, -2] ∪ [4, ∞).

9. What is the interval of f(x) = (x – 1)(x – 2)(x – 3)/(x3 + 6x2 + 11x + 6) where f(x) is positive?
a) (-∞, -3) ∪ (3, ∞)
b) (3, -2) ∪ (1, 1) ∪ (2, 3)
c) (-∞, -3) ∪ (2, -1) ∪ (1, 2) ∪ (3, ∞)
d) (-∞, ∞)
Answer: c
Clarification: f(x) = (x – 1)(x – 2)(x – 3)/(x3 + 6x2 + 11x+ 6)
After solving the cubic equation (x3 + 6x2 + 11x+ 6) we get (x+1)(x+2)(x+3)
Now, we can see that this implies f(x) = (x – 1)(x – 2)(x – 3)/(x + 1)(x + 2)(x + 3)
So, the critical points of x are, x = 1, 2, 3, -1, -2, -3
So, for f(x) > 0 ᵾ x € (-∞, -3) ∪ (2, -1) ∪ (1, 2) ∪ (3, ∞).

10. What is the interval of f(x) = (x – 1)(x – 2)(x – 3)/(x3 + 6x2 + 11x+ 6) where f(x) is negative?
a) (-∞, -3) ∪ (3, ∞)
b) (3, -2) ∪ (1, 1) ∪ (2, 3)
c) (-∞, -3) ∪ (2, -1) ∪ (1, 2) ∪ (3, ∞)
d) (-∞, ∞)
Answer: b
Clarification: f(x) = (x – 1)(x – 2)(x – 3)/(x3 + 6x2 + 11x+ 6)
After solving the cubic equation (x3 + 6x2 + 11x+ 6) we get (x+1)(x+2)(x+3)
Now, we can see that this implies f(x) = (x – 1)(x – 2)(x – 3)/(x + 1)(x + 2)(x + 3)
So, the critical points of x are, x = 1, 2, 3, -1, -2, -3
So, for f(x) < 0 ᵾ x € (3, -2) ∪ (1, 1) ∪ (2, 3).

11. What is the set of all x for which 1/(x – 1)(3 – x) ≤ 1?
a) (-∞, 1) ∪ (3, ∞)
b) (-∞, 1) ∪ (3, ∞) ∪ {2}
c) (-∞, 1) ∪ {2}
d) (3, ∞) ∪ {2}
Answer: b
Clarification: 1/(x – 1)(3 – x) ≤ 1
Now, on solving the equation further we get,
1/(x – 1)(3 – x) -1 ≤ 0
This also implies,
1- 1/(x – 1)(3 – x) ≥ 0
(x – 1)(3 – x) – 1/(x – 1)(3 – x) ≥ 0
So, (x – 2)2/(x – 1)(3 – x) ≥ 0.
This implies, (-∞, 1) ∪ (3, ∞) ∪ {2}.

12. Which one of the following is the correct representation of set A = {2,4,8,16….} in set builder form?
a) {x: x = 2n where n ∈ N}
b) {x: x = 2n where n ∈ N}
c) {x: x = 4n where n ∈ N}
d) {x: x = 2n+4 where n ∈ N}
Answer: b
Clarification: The sequence is a geometric progression with base 2 hence 2n is the correct answer.

13. A set can be a collection but a collection cannot be a set.
a) True
b) False
Answer: a
Clarification: A collection becomes a set when it is well defined for example a collection of good football players is not a set since the phrase “good football players” is vague and not defined.

250+ TOP MCQs on Trigonometric Functions & Answers | Class 11 Maths

Mathematics Questions and Answers for Entrance exams on “Trigonometric Functions – 2”.

1. cosec(-30°) =___________
a) -2
b) 2
c) 2/√3
d) -2/√3
Answer: a
Clarification: We know, cosec(-x) = cosec x
So, cosec(-30°) = -cosec 30°=-2.

2. sec(-45°) =_____________
a) 1
b) -1
c) √2
d) -√2
Answer: c
Clarification: We know, sec(-x) = sec x
So, sec(-45°)=sec 45°=1/(cos 45°)=√2.

3. cot x is not defined for_______
a) 0
b) nπ/2
c) (2n+1) π/2
d) nπ
Answer: d
Clarification: We know, cot x is not defined when sin x = 0.
sin x = 0 whenever x is 0, π, 2π, 3π, …. i.e. all integral multiples of π
so, x=nπ.

4. If sin x=-4/5 and x lies in 3rd quadrant, then find sec x.
a) 5/3
b) 3/5
c) -3/5
d) -5/3
Answer: d
Clarification: sin x=-4/5
We know, sin2x + cos2x=1
cos2x = 1-(-4/5)2 = 1-16/25=9/25
cos x=±3/5
cos x is negative in 3rd quadrant so, cos x=-3/5.
sec x = 1/cos x = 1/(-3/5) = – 5/3.

5. If cosec x = -5/12 and x lies in 2nd quadrant, then find cos x.
a) 12/13
b) 5/13
c) -13/5
d) 12/13
Answer: c
Clarification: cosec x = 5/12
sin x = 1/cosec x = 12/5
We know, sin2x+cos2x=1
cos2 x = 1-(-12/5)2 = 1+144/25 = 169/25
cos x = ±13/5
cos x is negative in 2nd quadrant so, cos x=-13/5.

6. If sec x = 13/5 and x lies in 4th quadrant, then find cot x.
a) 5/12
b) -5/12
c) 5/13
d) -5/13
Answer: b
Clarification: sec x = 13/5.
We know, sec2x – tan2x=1
tan2x = (13/5)2-1 = (169/25) – 1 = 144/25
tan x = ±12/5
tan x is negative in 4th quadrant so, tan x=-12/5
cot x = 1/tan x = 1/(-12/5) = – 5/12.

7. If tan x = -5/12 and x lies in 2nd quadrant, then find cosec x.
a) 12/5
b) 13/5
c) -13/5
d) -12/5
Answer: b
Clarification: tan x = -5/12
cot x = 1/tan x = 1/ (-5/12) = -12/5
We know, cosec2x – cot2x = 1
cosec2x =1+(-12/5)2 = 1+144/25 = 169/25
cosec x = ± 13/5
cosec x is positive in 2nd quadrant so, cosec x = 13/5.

8. sin (15π/6) =_____________
a) 1
b) -1
c) 0
d) 1/2
Answer: a
Clarification: sin(15π/6) = sin (2π + 3π/6) = sin (3π/6) {sin(2nπ+x)=sin x}
= sin (π/2) = 1.

9. cos (17π/3) =______________
a) 1/2
b) -1/2
c) √3/2
d) -√3/2
Answer: a
Clarification: cos (17π/3) = cos (2π*3 – π/3)
{cos (2nπ-x)=cos x}
= cos(π/3) = 1/2.

10. tan (19π/6) =____________________
a) √3
b) -1/√3
c) – √3
d) 1/√3
Answer: d
Clarification: tan (19π/6) = tan (2π + 7π/6) = tan 7π/6 {tan( 2nπ+x)=tan x}
= tan (π+π/6) = tan π/6 = 1/√3. {tan π+x = tan x}

11. cos ( -1500°) =______________
a) 1/2
b) -1/2
c) √3/2
d) -√3/2
Answer: a
Clarification: cos(-1500°) = cos(1500°) {cos(-x) = cos x}
= cos (4*360° + 60°) = cos 60° = 1/2. {cos (2nπ+x) = cos x}

12. sin 1710° =__________________
a) 1
b) -1
c) 0
d) 1/2
Answer: b
Clarification: sin 1710° = sin (360°*5 – 90°) {sin (2nπ-x)= – sin x}
=-sin 90° = -1.

13. tan 1560°=_________________
a) -√3
b) √3
c) 1/√3
d) -1/√3
Answer: a
Clarification: tan 1560° = tan (360°*4 + 120°) = tan 120° {tan (2nπ+x) = tan x}
= tan (180°-60°) = -tan 60° = – √3. {tan π-x = – tan x}

Mathematics for Entrance exams,