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Question: If the function f is defined by \(\frac{\sum_{r = 1}^{n}{(r^{3} - r^{2})}}{n^{4}}\), then at what po...

If the function f is defined by ∑r=1n(r3−r2)n4\frac{\sum_{r = 1}^{n}{(r^{3} - r^{2})}}{n^{4}}, then at what points f is differentiable

A

Everywhere

B

Except at lim⁡n→∞\lim_{n \rightarrow \infty}

C

Except at f(x)f(y)\frac{f(x)}{f(y)}

D

Except at Limn→∞nCx(mn)x(1−mn)n−x{\underset{n \rightarrow \infty}{Lim}}^{n}C_{x}\left( \frac{m}{n} \right)^{x}\left( 1 - \frac{m}{n} \right)^{n - x} or ± 1

Answer

Everywhere

Explanation

Solution

We have,;

Lf′(0)=lim⁡h→0f(−h)−f(0)−hL f ^ { \prime } ( 0 ) = \lim _ { h \rightarrow 0 } \frac { f ( - h ) - f ( 0 ) } { - h } = lim⁡h→0−h1+h−0−h\lim _ { h \rightarrow 0 } \frac { \frac { - h } { 1 + h } - 0 } { - h } = 1

Rf′(0)=lim⁡h→0f(h)−f(0)hR f ^ { \prime } ( 0 ) = \lim _ { h \rightarrow 0 } \frac { f ( h ) - f ( 0 ) } { h } = lim⁡h→0h1+h−0h\frac { \lim _ { h \rightarrow 0 } \frac { h } { 1 + h } - 0 } { h } = lim⁡h→011+h\lim _ { h \rightarrow 0 } \frac { 1 } { 1 + h } = 1

So, Lf′(0)=Rf′(0)=1L f ^ { \prime } ( 0 ) = R f ^ { \prime } ( 0 ) = 1

So, f(x)f ( x ) is differentiable at x=0x = 0; Also f(x)f ( x ) is differentiable at all other points.

Hence, f(x)f ( x ) is everywhere differentiable.