Courses

# 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

Description
This mock test of Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced for JEE helps you for every JEE entrance exam. This contains 14 Multiple Choice Questions for JEE Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced exercise for a better result in the exam. You can find other Test: MCQs (One or More Correct Option): Trigonometric Functions & Equations | JEE Advanced extra questions, long questions & short questions for JEE on EduRev as well by searching above.
*Multiple options can be correct
QUESTION: 1

### is equal to         (1984 )

Solution:

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
QUESTION: 2

### The expression 3 -     is equal to

Solution:

The given expression is

*Multiple options can be correct
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)

Solution:

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
QUESTION: 4

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

= 0 are

Solution:

(a,c)  We have

Operating

Operating

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

*Multiple options can be correct
QUESTION: 5

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

Solution:

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
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

Solution:

sin α + sin β + sin γ

∴ Each   lies between –1

and 1, therefore

∴   min value = – 2.

*Multiple options can be correct
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)

Solution:

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
QUESTION: 8

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

Solution:

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
QUESTION: 9

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

Solution:

Similarly,

*Multiple options can be correct
QUESTION: 10

If     then (2009)

Solution:

Given that

⇒

Also

*Multiple options can be correct
QUESTION: 11

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

Solution:

We have

⇒

⇒

*Multiple options can be correct
QUESTION: 12

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

Solution:

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

Now,

But

*Multiple options can be correct
QUESTION: 13

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

Solution:

(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
QUESTION: 14

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

Solution:

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