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The sum of absolute maximum and absolute minimum values of the function \mathrm{f(x)=\left|2 x^{2}+3 x-2\right|+\sin x \cos x} in the interval \mathrm{[0,1]} is:

Option: 1

\mathrm{3+\frac{\sin (1) \cos ^{2}(1 / 2)}{2}}


Option: 2

\mathrm{3+\frac{1}{2}(1+2 \cos (1)) \sin (1)}


Option: 3

\mathrm{5+\frac{1}{2}( \sin (1)+\sin (2)) }


Option: 4

\mathrm{2+\sin \left ( \frac{1}{2}\right )\cos \left ( \frac{1}{2} \right ) }


Answers (1)

best_answer

\mathrm{f\left ( x \right )= \left | 2x^{2}+3x-2 \right |+\sin x\, \cos x}

\mathrm{f\left ( x \right )= \left\{\begin{matrix} -\left ( 2\, x^{2}+3x-2 \right ) &+\sin \, x\cos x & ;0\leq x\leq \frac{1}{2}\\ 2\, x^{2}+3x-2 &+\sin x\, \cos x & ;\frac{1}{2}< x< 1 \end{matrix}\right.}

\mathrm{{f}'\left ( x \right )= \left\{\begin{matrix} -4\, x-3 &+\frac{\cos \, 2x}{4}& ;0\leq x< \frac{1}{2}\\ 4\, x+3&+\frac{\cos \, 2x}{4} & ;\frac{1}{2}< x\leq 1 \end{matrix}\right.}

\mathrm{{f}'\left ( x \right )< 0\quad \forall\: x\, \in\left [0,\frac{1}{2} \right )}
\mathrm{{f}'\left ( x \right )> 0\quad \forall\: x\, \in\left (\frac{1}{2},1 \right ]}

\mathrm{f\left ( 0 \right )= 2}
\mathrm{f\left ( 1 \right )= 3+\sin 1\, \cos 1}
\mathrm{f\left ( \frac{1}{2} \right )= \sin \frac{1}{2}\, \cos \frac{1}{2}}

\mathrm{\therefore }   Sum of Absolute maximum & minimum
         \mathrm{= f\left ( 1 \right ) +f\left ( \frac{1}{2} \right )}
         \mathrm{= 3+\sin \frac{1}{2}\, \cos \frac{1}{2}\left ( 1+2\, \cos 1 \right )}
         \mathrm{= 3+ \frac{1}{2}\sin 1\, \left ( 1+2\, \cos 1 \right )}

The correct answer is option (B)

Posted by

Shailly goel

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