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

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 given expression is

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.

(a,c)  We have

Operating  

Operating 

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

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).

sin α + sin β + sin γ

∴ Each   lies between –1

and 1, therefore  

∴   min value = – 2.

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]

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)

Similarly,  

 Given that

⇒

⇒

⇒

⇒

⇒

⇒

⇒ 

⇒ 

Also  

We have

⇒ 

⇒

⇒

⇒

⇒ 

⇒

⇒  

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 

(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 

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