Question

$0<a<b<\frac{\pi }{2};f(a,b)=\frac{\tan b-\tan a}{b-a}$ then

• A. $f(a,b)<\frac{1}{2}$
• B. $f(a,b)\ge 1$
• C. $f(a,b)\le 1$
• D. $f(a,b)<0$

Answer 1

The correct option is B $f(a,b)\ge 1$

According to mean value theorem

Let f(x)=tanx

$x\in (0,\frac{\pi }{2}$)

$f^{\prime }\left(x\right)=se{c}^{2}x$

There will exist a c at which

$f^{\prime }\left(c\right)=\frac{f\left(a\right)-f\left(b\right)}{a-b}$

$se{c}^{2}c=\frac{tana-tanb}{a-b}$

so a<c<b

$c=\frac{\pi }{2}$

$f(a,b)\ge se{c}^2\frac{\pi }{2}\to f(a,b)\ge 1$