If f(x)={cases} x² ≤ft| π/x |, & x ≠ 0 \\ 0, & x=0 {cases}, then at x=2, f(x) is

Correct Answer is : Continuous but not differentiable

Continuous but not differentiable

Correct Answer is : Continuous but not differentiable

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TG EAMCET 4 May 2025 Shift 2 Paper

Question : 63 of 160

Marks: +1, -0

f(x)=\begin{cases} x^2 \left| \cos\frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x=0 \end{cases}

x=2, f(x)

Solution:

\;f(x)=x^2 \left| \cos\frac{\pi}{2} \right|, x \neq 0

\;=0, \;\; x=0

\;f(2^{+}) =4 \left| \cos\frac{\pi}{2} \right| =0

\;f(2^{-}) =4 \left| \cos\frac{\pi}{2} \right| =0

\;f(2)=0

Continuous act

=2

\;f'(2^{+}) =\lim\limits_{h \to 0^{+}} \frac{f(2+h)-f(2)}{h}

\;=\lim\limits_{h \to 0^{+}} \frac{(h+2)^2 \left| \cos \frac{\pi}{2+h} \right|}{h}

\;=\lim\limits_{h \to 0^{+}} \frac{ h+2^2 \cos \left(\frac{\pi}{2+h}\right) }{h}

\;=\lim\limits_{h \to 0^{+}} 2(h+2) \cos \left(\frac{\pi}{h+2}\right) + (h+2)^2

\;x - \sin \left(\frac{\pi}{2+h}\right) \times \frac{-\pi}{(2+h)^2}

\;=4 \cos \frac{\pi}{2} + \left(4 \times -\sin \frac{\pi}{2} \times \frac{-\pi}{4}\right) = \pi

\;f'(2) \lim\limits_{h \to 0^{-}} \frac{f(2+h)-f(2)}{h}

\;=\lim\limits_{h \to 0^{-}} \frac{(2+h)^2 \left| \cos \frac{\pi}{2+h} \right|}{h}

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

\; \; \; -2(2+h) \cos \frac{\pi}{2+h}-(2+h)^2

\;=\lim\limits_{h \to 0^{-}} \frac{ \times \sin \left(\frac{\pi}{2+h}\right) \times \frac{\pi}{(2+h)^2} }{1}

\;=0-\pi =-\pi

\;f'(2^{+}) \neq f'(2)

So,

f

is not differentiable

a t=2

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