QJEEM22Ma1c5e: JEE Main 2022 Mathematics - Class 12 JEEM22Ma1c5e Solution: The sum of absolute maximum and absolute minimum values of the…

Question

Verified Solution • 2025–26 Session moderate

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

Accepted Answer

f(x)=|(2 x-1)(x+2)|+\frac{\sin 2 x}{2}

0 \leq x&lt;\frac{1}{2} \quad f(x)=(1-2 x)(x+2)+\frac{\sin 2 x}{2}

f^{\prime}(x)=-4 x-3+\cos 2 x&lt;0

For

x \geq \frac{1}{2}: \quad f^{\prime}(x)=4 x+3+\cos 2 x&gt;0

So, minima occurs at

x=\frac{1}{2}

\begin{aligned}\left.f(x)\right|_{\min } &amp;=\left|2\left(\frac{1}{2}\right)^{2}+\frac{3}{2}-2\right|+\sin \left(\frac{1}{2}\right) \cdot \cos \left(\frac{1}{2}\right) \\\\&amp;=\frac{1}{2} \sin 1\end{aligned}

So, maxima is possible at

x=0

or

x=1

Now checking for

x=0

and

x=1

, we can see it attains its maximum value at

x=1

\begin{aligned}\left.f(x)\right|_{\max }=&amp;|2+3-2|+\frac{\sin 2}{2} \\\\&amp;=3+\frac{1}{2} \sin 2\end{aligned}

Sum of absolute maximum and minimum value

=3+\frac{1}{2}(\sin 1+\sin 2)

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

=

3 + {1 \over 2}(1 + 2\cos (1))\sin (1)

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