Question

What is the ratio of the greatest to the smallest value of 2 – 2 sin x – sin² x, 0 ≤ x ≤ (π/2)? 
1. -2
2. -1
3. 1
4. 2

Answer 1

Correct Answer - Option 1 : -2

Formula used:

(a + b)² = a² + 2ab + b²

Calculation:

Let 

y = 2 – 2 sin x – sin2 x

⇒ y = 3 - (sin2 x + 2 sin x + 1)

We know that,

(a + b)2 = a2 + 2ab + b2

⇒ y = 3 - (sin x + 1)²    ----(1)

We know that,

-1 ≤ sin θ ≤ 1

Also, we have 0 ≤ x ≤ (π/2). Hence,

sin x ∈ [0, 1]

For maxima, sin x = 0

From equation (1), 

(y)_max = 3 - (1)² = 2

For minima, sin x = 1

⇒ (y)_min = 3 - (1 + 1)² = -1

Hence, required ratio is

y_max/y_min = 2/-1 = -2

∴ ratio of the greatest to the smallest value is -2.