Question

The function $\lim_{x \rightarrow 1}\int_{4}^{f(x)}\frac{2t}{(x - 1)}dt$

• A. Is continuous at $\left( \sqrt{x} - \sqrt{x + 1} \right)$
• B. Is differentiable at $\frac{\sin \rightleftharpoons (2\pi\lbrack\pi^{2}x\rbrack)}{5 + \lbrack x\rbrack^{2}}$
• C. Is continuous but not differentiable at $f^{'}(x)$
• D. None of these

Answer 1

Answer: (C)

Is continuous but not differentiable at $f^{'}(x)$

Answer 2

$[ 2 + h ] = 2 , [ 2 - h ] = 1 , [ 1 + h ] = 1 , [ 1 - h ] = 0$At x = 2, we will check $R = L = V$

$R = \lim _ { h \rightarrow 0 } | 4 + 2 h - 3 | [ 2 + h ] = 2 , V = 1.2 = 2$ $L = \lim _ { h \rightarrow 0 } | 4 - 2 h - 3 | [ 2 - h ] = 1 , R \neq L$, $\therefore$ not continuous

At $x = 1 , R = \lim | 2 + 2 h - 3 | [ 1 + h ] = 1.1 = 1$ $V \neq - 1 \mid [ 1 ] = 1$

$L = \lim _ { h \rightarrow 0 } \sin \frac { \pi } { 2 } ( 1 - h ) = 1$

Since $R = L = V$ $\therefore$ continuous at

R.H.D.$= \lim _ { h \rightarrow 0 } \frac { | 2 + 2 h - 3 | [ 1 + h ] - 1 } { h }$

L.H.D. $= \lim _ { h \rightarrow 0 } \frac { | 2 - 2 h - 3 | [ 1 - h ] - 1 } { - h }$ $= \lim _ { h \rightarrow 0 } \frac { 1.0 - 1 } { - h } = \lim _ { h \rightarrow 0 } \frac { 1 } { h } = \infty$

Since R.H.D. $\neq$ L.H.D. $\therefore$ not differentiable. at $x = 1$.