A) \[-\frac{1}{2}\]
B) \[\frac{1}{2}\]
C) \[-\frac{5}{8}\]
D) \[\frac{3}{8}\]
Correct Answer: C
Solution :
Given \[f(x)=\int_{1}^{x}{|t|}dt,\left[ -\frac{1}{2},\frac{1}{2} \right]\] On differentiating, we get \[f'(x)=|x|.1\] \[=|x|\left\{ \begin{matrix} x, & 0\le x\le \frac{1}{2} \\ -x, & -\frac{1}{2}\le x<0 \\ \end{matrix} \right.\] At \[x=0,f'(x)=0\] But\[f'\,'(0)\]is not defined. Now, \[f(1/2)=\int_{1}^{1/2}{|t|}\,dt\] \[=-\int_{1/2}^{1}{t}\,dt=-\frac{1}{2}[{{t}^{2}}]_{\frac{1}{2}}^{1}\] \[=-\frac{1}{2}\left( 1-\frac{1}{4} \right)=-\frac{3}{8}\] \[f\left( -\frac{1}{2} \right)=\int_{1}^{-1/2}{|t|}dt=-\int_{1/2}^{1}{|t|}dt\] \[=-\int_{-1/2}^{0}{(-t)}dt-\int_{0}^{1}{t\,dt}\] \[=\left[ \frac{{{t}^{2}}}{2} \right]_{-1/2}^{0}-\left[ \frac{{{t}^{2}}}{2} \right]_{0}^{1}\] \[=-\frac{1}{8}-\frac{1}{2}=-\frac{5}{8}\] Hence, maximum value of\[f(x)\]is\[-\frac{5}{8}\]
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