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

Correct Answer is : differentiable

continuous but not differentiable

not continuous, but differentiable

Correct Answer is : differentiable

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AP EAPCET 04-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 64 of 160

Marks: +1, -0

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

x=0, f(x)

Solution: 👈: Video Solution

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

Now,

\lim\limits_{x \to 0} \frac{f(x)-f(0)}{x-0}

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

=\lim\limits_{x \to 0} \frac{x \log[1-(1-\cos x)]}{1-\cos x} \cdot \frac{1-\cos x}{\log(1+x)}.

=\lim\limits_{x \to 0} \frac{\log[1-(1-\cos x)]}{1-\cos x} \cdot \frac{2 \sin^2 \left(\frac{x}{2}\right)}{4 \left(\frac{x}{2}\right)^2}, x^2 \frac{x}{\log(1+x)}=0

So,

f

is differentiable hence it is also continuous.

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