Solveeit Logo
Login

Question

Question: If \(f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}\) then...

If f(x)=1−sin⁡x+cos⁡x1+sin⁡x+cos⁡xf(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x} then

A

f is continuous but not derivable at x=π.x = \pi.

B

f(π),f(\pi),

C

f(x)f(x)

D

f is not derivable at x=πx = \pi

Answer

f is not derivable at x=πx = \pi

Explanation

Solution

Here,

Thus, f(x)=sgn⁡x3=sgn⁡xf ( x ) = \operatorname { sgn } x ^ { 3 } = \operatorname { sgn } x which is neither continuous nor derivable at 0.

Note : f′(0+)=lim⁡h→0+f(0+h)−f(0)hf ^ { \prime } \left( 0 ^ { + } \right) = \lim _ { h \rightarrow 0 ^ { + } } \frac { f ( 0 + h ) - f ( 0 ) } { h } =lim⁡h→0+1−0h→∞= \lim _ { h \rightarrow 0 ^ { + } } \frac { 1 - 0 } { h } \rightarrow \infty and

f′(0−)=lim⁡h→0−f(0−h)−f(0)hf ^ { \prime } \left( 0 ^ { - } \right) = \lim _ { h \rightarrow 0 ^ { - } } \frac { f ( 0 - h ) - f ( 0 ) } { h } =lim⁡h→0−−1−0h→∞= \lim _ { h \rightarrow 0 ^ { - } } \frac { - 1 - 0 } { h } \rightarrow \infty.

∴ f′(0+)≠f′(0−)f ^ { \prime } \left( 0 ^ { + } \right) \neq f ^ { \prime } \left( 0 ^ { - } \right), ∴ f is not derivable at x=0x = 0.