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JEE MainMathematicsToughLimits, continuity and derivability

Let $f(x) = \left\{ \begin{matrix}
\int_{0}^{x}\mspace{2mu}\{ 1 + |1 - t|\} dt,\ \ \ x > 2 \\
\ \ \ \ 5x - 7,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \leq 2
\end{matrix} \right.\ $
Then f(2) is

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$f(x) = \left\{ \begin{matrix}
\int_{0}^{x}\mspace{2mu}\{ 1 + |1 - t|\} dt,\ \ \ \ x > 2 \\
5x - 7,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \leq 2
\end{matrix},f(x) = \int_{0}^{1}\mspace{2mu}(1 + (1 - t))dt + \int_{1}^{x}\mspace{2mu}(1 + (1 - t))dt \right.\ $
=[2t−t22]01+[t22]1x= \left\lbrack 2t - \frac{t^{2}}{2} \right\rbrack_{0}^{1} + \left\lbrack \frac{t^{2}}{2} \right\rbrack_{1}^{x}
=(2−12)+(x22−12)= \left( 2 - \frac{1}{2} \right) + \left( \frac{x^{2}}{2} - \frac{1}{2} \right)
∴f(x)=1+x22\therefore f(x) = 1 + \frac{x^{2}}{2}
$f(x) = \left\{ \begin{matrix}
5x - 7\ \ \ \ x \leq 2 \\
1 + \frac{x^{2}}{2}\ \ \ \ x > 2
\end{matrix} \right.\ $
L = R = V = 2 at x = 2 ∴\therefore continuous
L′=lim⁡h→0\mspace2muf(2−h)−f(2)−hL^{'} = \lim_{h \rightarrow 0}\mspace{2mu}\frac{f(2 - h) - f(2)}{- h}
=lim⁡h→0\mspace2mu5(2−h)−7−3−h=5= \lim_{h \rightarrow 0}\mspace{2mu}\frac{5(2 - h) - 7 - 3}{- h} = 5
=lim⁡h→0\mspace2mu5(2−h)−7−3−h=5= \lim_{h \rightarrow 0}\mspace{2mu}\frac{5(2 - h) - 7 - 3}{- h} = 5
R′=hth→0f(2+h)−f(2)h=2R^{'} = {ht}_{h \rightarrow 0}\frac{f(2 + h) - f(2)}{h} = 2
so not differentiable at x = 2

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

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