Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced
14 Questions MCQ Test Maths 35 Years JEE Main & Advanced Past year Papers | Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced
is equal to (1984 )
is equal to (1984 )We have,
(1 + cosπ/ 8) (1 + cos 3π/8) 1 + cos 5π/8) (1 + cos 7π/8)
= (1 + cosπ/8) (1 + cos 3π/8)(1 + cos(p - 3π/8)) (1 + cos(π-π/8))
= (1 + cos π/8) (1 + cos3π/8) 1 - cos3π/8 (1 - cosπ/8)
= (1 - cos2π/8) (1 - cos2 3π/8) = sin 2 π/8 sin2 3π/8
∴ (c) is the correct answer.
The expression 3 
- 
is equal to
- is equal toThe given expression is
The number of all possible triplets (a1, a2, a3) such that a1 + a2 cos(2x) + a3sin2(x) = 0 for all x is (1987 - 2 Marks)
Since a1 + a2 cos 2x + a3 sin2x = 0 for all x
Putting x = 0 and x = π/2 , we get a1 + a2 = 0 ....(1)
and a1 – a2+ a3 = 0 ....(2)
⇒ a2 = – a1 and a3 = – 2a1
∴ The given equation becomes
a1– a1 cos 2x – 2a1 sin2 x = 0, ∀ x
⇒ a1 (1– cos 2x – 2 sin2 x) = 0, ∀ x
⇒ a1 (2 sin2 x– 2 sin2 x) = 0, ∀ x
The above is satisfied for all values of a1.
Hence infinite number of triplets (a1, – a1, – 2a1) are possible.
The values of θ lying between θ = 0 and θ = π/2 and satisfying the equation (1988 - 2 Marks)
= 0 are
(a,c) We have
Operating 
Operating 
Expanding along R1 we get [1 + 4 sin 4θ + 1]= 0
Let 2sin2x + 3sinx – 2 > 0 and x2 – x – 2 < 0 (x is measured in radians). Then x lies in the interval (1994)
2 sin2 x + 3 sin x – 2 > 0
(2 sin x – 1 ) (sin x + 2 ) > 0
⇒ 2 sin x – 1 > 0 (∵ – 1 ≤ sin x ≤ 1)
⇒ sin x > 1/2 ⇒ x ∈ (π/6, 5π/6) ....(1)
Also x2 – x – 2 < 0
⇒ (x – 2) (x + 1) < 0 ⇒ – 1 < x < 2 ....(2)
Combining (1) and (2) x ∈ (π/6, 2).
The minimum value of the expression sin α + sin β + sin γ , where α, β, γ are real numbers satisfying α + β + γ = π is
sin α + sin β + sin γ
∴ Each 
lies between –1
and 1, therefore 
∴ min value = – 2.
The number of values of x in the interval [0, 5π] satisfying the equation 3 sin2 x – 7 sin x + 2 = 0 is (1998 - 2 Marks)
3 sin2 x – 7 sin x + 2 = 0, put sin x = s
⇒ (s – 2) (3s – 1) = 0 ⇒ s = 1/3
(s = 2 is not possible)
Number of solutions of sin x =
from the following graph is 6 between [0, 5π]
Which of the following number(s) is/are rational? (1998 - 2 Marks)
We know that sin 
(irrational)
(irrational)
sin 15° . cos 15° =
(2 sin 15° cos 15°)
sin 15° cos 75° = sin 15° cos (90 – 15°)
= sin 15° sin 15° = sin2 15° 
(irrational)
For a positive integer n, let (1999 - 3 Marks) 
Then
Similarly, 
If 
then (2009)
Given that
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒ 
Also 
For 
the solution (s) of 
is (are)
(2009)
We have
⇒ 
⇒
⇒
⇒
⇒
⇒
⇒ 
Let 
be such that 
then φ cannot satisfy (2012)
As tan 
∴ θ ∈ II or IV quadrant ...(1)
And 
⇒ θ ∈ III or IV quadrant ...(2)
Also θ ∈ [0, 2π] ...(3)
Combining above three equations (1), (2) and (3); θ ∈ IV
quadrant and more precisely 
Now, 
But 
The number of points in 
for which x2 – x sin x – cos x = 0, is (JEE Adv. 2013)
(c) Let f(x) = x2 – x sin x – cos x
∴ f' (x) = 2x – x cos x = x(2 – cos x)
∴ f is increasing on 
and decreasing on 
∴ y = f(x) meets x-axis twice.
i.e., f(x) = 0 has two points in 
Let f(x) = x sin πx, x > 0. Then for all natural numbers n, f'(x) vanishes at (JEE Adv. 2013)
We have f(x) = x sin πx, x > 0
We observe, from graph of y = tanπx and y = –πx that they intersect at unique point in the intervals
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