Question

If $f\left(x\right)=a\sec x-b\tan x,a⟩b⟩0$ , then the minimum value of f(x) is

• A. $\sqrt{a^2+b^2}$
• B. $2\sqrt{a^2-b^2}$
• C. $\sqrt{a^2-b^2}$
• D. $a-b$

Answer 1

The correct option is C $\sqrt{a^2-b^2}$

$f\left(x\right)=a\sec x-b\tan x{f}^{{}^{\prime }}\left(x\right)=0a\sec x\tan x-b{\sec }^{2}x=0\therefore a\tan x=b\sec x\therefore b=a\sin xf\left(x\right)=a\sec x-b\tan x=\frac{a}{\cos x}-\frac{a{\sin }^{2}x\cos x}{\cos x}=a\cos x$

Minimum value=$a\cos x$

$b=a\sin x\sin x=\frac{b}{a}\cos x=\frac{\sqrt{a^2-b^2}}{a}$

$\therefore$ Minimum value=$a\cos x=\sqrt{a^2-b^2}$