Question

For any real number 'b', let f(2) denote the maximum of the function $\left| \sin x + \frac{2}{3 + \sin x} + b \right|$ over all x Î R, then the

minimum of f(2) over all b Î R is –

• A. ¼
• B. 2/4
• C. ¾
• D. 1/8

Answer 1

Let y = 3 + sin x, y Î [2, 4] and assumes all value there in.

Also, let g(y) = y +$\frac{2}{y}$, this function is increasing on [2, 4]

So, g(2) £ g(y) £ g(4)

Thus 3 £ g(y) £ $\frac{9}{2}$and both extreme values are attained.

It now follows that the minimum of

f(2) = max ( |g(y) + b – 3|) is 3/4, which is attained at

b = – 3/4 for if b > – $\frac{3}{4}$,

then choose x = $\frac{\pi}{2}$,

So y = 4 and then g(y) + b – 3 > 3/4

While if b < $\frac{- 3}{4}$, then chose x = – $\frac{\pi}{2}$,

So y = 2 and g(y) + b – 3 = – 3/4

On the other hand, range for g(y) is

$\frac{- 3}{4}$ £ g(y) + b – 3 £ $\frac{3}{4}$ for b = –3/4