If f (x) = {cases} x² ( x)/(1+x²), & x ≠ 0 \\ 0, & x = 0 {cases} then f is

Correct Answer is : differentiable at zero

continuous but not differentiable at zero

not continuous and not differentiable at zero

Correct Answer is : differentiable at zero

Test Index

TS EAMCET 4 May 2018 Shift 1 Solved Paper

Section: Mathematics

Question : 64 of 160

Marks: +1, -0

If f (x) =

\begin{cases} \frac{x^2 \log(\cos x)}{\log(1+x^2)}, & x \neq 0 \\ 0, & x = 0 \end{cases}

then f is

Solution:

Consider the function.

f (x) =

\begin{cases} \frac{x^2 \log(\cos x)}{\log(1+x^2)}, & x \neq 0 \\ 0, & x = 0 \end{cases}

\text{LHL}

x=0

is,

\lim\limits_{h \to 0} \frac{\log(\cos x)}{\frac{\log(1+x^2)}{x^2}}

= \lim\limits_{h \to 0} \frac{\log(\cos(-h))}{\frac{\log(1+h^2)}{h^2}}

= \frac{\log(1)}{1}

=0

\text{RHL}

x=0

is,

\lim\limits_{h \to 0} \frac{\log(\cos x)}{\frac{\log(1+x^2)}{x^2}}

= \lim\limits_{h \to 0} \frac{\log(\cos(-h))}{\frac{\log(1+h^2)}{h^2}}

= \frac{\log(1)}{1}

=0

Therefore, the function

f(x)

is continuous at

x=0

Now,

\text{LHD}

x=0

is,

\lim\limits_{h\to 0} \frac{ \frac{(0-h)^2 \log(\cos(0-h))}{\log(1+h^2)} - 0 }{-h}

= \lim\limits_{h\to 0} \frac{ \frac{h^2 \log(\cos h)}{\log(1+h^2)} }{-h}

= \lim\limits_{h\to 0} \frac{ h \times \frac{\log(\cos h)}{\log(1+h^2)} }{-h^2}

= 0 \times \frac{\log(1)}{-1}

=0

Now,

\text{RHD}

x=0

is,

\lim\limits_{h\to 0} \frac{ \frac{(0+h)^2 \log(\cos(0+h))}{\log(1+h^2)} - 0 }{h}

= \lim\limits_{h\to 0} \frac{ \frac{h^2 \log(\cos h)}{\log(1+h^2)} }{h}

= \lim\limits_{h\to 0} \frac{ h \times \frac{\log(\cos h)}{\log(1+h^2)} }{h^2}

= 0 \times \frac{\log(1)}{-1}

=0

Therefore, the function is differentiable at

x=0

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