Question

For any real number b, let f(2) denotes 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. $\frac{1}{4}$
• B. $\frac{2}{4}$
• C. $\frac{3}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

$\frac{3}{4}$

Answer 2

Let y = 3 + sin x, y Î [2, 4] and assumes all values 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 external values are attained.

It now follows that the minimum of

f(2) = max(|g(y) + b – 3|) is $\frac{3}{4}$,

which is attained at b = $- \frac{3}{4}$

For if b > $- \frac{3}{4}$, then choose x = $\frac{\pi}{2}$, so y = 4 and then

g(y) + b – 3 > $\frac{3}{4}$

While if b < $- \frac{3}{4}$, then choose x = $- \frac{\pi}{2}$, so y = 2 and

g(y) + b – 3 = $- \frac{3}{4}$

On the other hand range for g(y) is

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

\ Minimum value of f(2) is $\frac{3}{4}$.