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MCQs (One or More Correct Option): Trigonometric Functions & Equations


MCQ Practice Test & Solutions: Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced (14 Questions)

You can prepare effectively for JEE Crack JEE with 35 Years of Previous Year Solved Papers with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced". These 14 questions have been designed by the experts with the latest curriculum of JEE 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 14

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*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 1

is equal to         (1984 )

Detailed Solution: Question 1

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.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 2

The expression 3     is equal to

Detailed Solution: Question 2

The given expression is

 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 3

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)

Detailed Solution: Question 3

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.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 4

The values of θ lying between θ = 0 and θ = π/2 and satisfying the equation (1988 - 2 Marks)

 = 0 are

Detailed Solution: Question 4

(a,c)  We have

Operating  

Operating 

Expanding along R1 we get [1 + 4 sin 4θ + 1]= 0

 

 

 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 5

Let 2sin2x + 3sinx – 2 > 0 and x2 – x – 2 < 0 (x is measured in radians). Then x lies in the interval (1994)

Detailed Solution: Question 5

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).

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 6

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

Detailed Solution: Question 6

sin α + sin β + sin γ

 

∴ Each   lies between –1

and 1, therefore  

∴   min value = – 2.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 7

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)

Detailed Solution: Question 7

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]

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 8

Which of the following number(s) is/are rational? (1998 - 2 Marks)

Detailed Solution: Question 8

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)

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 9

For a positive integer n, let (1999 - 3 Marks)       Then

Detailed Solution: Question 9

 

 

Similarly,  

Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 10

If     then (2009)

Detailed Solution: Question 10

 Given that

 

⇒ 

Checking the higher-power expressions

Evaluate:

So. option (a) is the correct answer.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 11

For   the solution (s) of     is (are)
(2009)

Detailed Solution: Question 11

We have

 

⇒ 

 

⇒  

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 12

   Let     be such that  then φ cannot satisfy (2012)​

Detailed Solution: Question 12

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 

 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 13

The number of points in for which x2 – x sin x – cos x = 0, is (JEE Adv. 2013)

Detailed Solution: Question 13

(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 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced - Question 14

Let f(x) = x sin πx, x > 0. Then for all natural numbers n, f'(x) vanishes at (JEE Adv. 2013)

Detailed Solution: Question 14

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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