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Question: Let f(x) = a sec x – b tanx, a \> b \> 0. Minimum value of f(x) Is...

Let f(x) = a sec x – b tanx, a > b > 0. Minimum value of f(x)

Is

A

a2+ b2

B

(a2− b2)

C

a2+b2\sqrt { a ^ { 2 } + b ^ { 2 } }

D

a2−b2\sqrt { a ^ { 2 } - b ^ { 2 } }

Answer

a2−b2\sqrt { a ^ { 2 } - b ^ { 2 } }

Explanation

Solution

f(x) = a−bsin⁡xcos⁡x\frac { a - b \sin x } { \cos x }

⇒ f(x) = asin⁡x−bcos⁡2x\frac { a \sin x - b } { \cos ^ { 2 } x }f '(x) = 0

= x = sin⁡−1(ba)\sin ^ { - 1 } \left( \frac { b } { a } \right) or x = π−sin⁡−1(ba)\pi - \sin ^ { - 1 } \left( \frac { b } { a } \right)

Sign scheme of f'(x) is

Clearly x = sin⁡−1(ba)\sin ^ { - 1 } \left( \frac { b } { a } \right) is the point of minima for

y = f(x). When x = sin⁡−1(ba)\sin ^ { - 1 } \left( \frac { b } { a } \right), secx = aa2−b2\frac { a } { \sqrt { a ^ { 2 } - b ^ { 2 } } }

tan x = ba2−b2\frac { b } { \sqrt { a ^ { 2 } - b ^ { 2 } } } ⇒ f(x) = a2−b2\sqrt { a ^ { 2 } - b ^ { 2 } }