JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 9)
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 :
$$3 + {{\sin (1){{\cos }^2}\left( {{1 \over 2}} \right)} \over 2}$$
$$3 + {1 \over 2}(1 + 2\cos (1))\sin (1)$$
$$5 + {1 \over 2}(\sin (1) + \sin (2))$$
$$2 + \sin \left( {{1 \over 2}} \right)\cos \left( {{1 \over 2}} \right)$$
Explanation
$f(x)=|(2 x-1)(x+2)|+\frac{\sin 2 x}{2}$
$0 \leq x<\frac{1}{2} \quad f(x)=(1-2 x)(x+2)+\frac{\sin 2 x}{2}$
$f^{\prime}(x)=-4 x-3+\cos 2 x<0$
For $x \geq \frac{1}{2}: \quad f^{\prime}(x)=4 x+3+\cos 2 x>0$
So, minima occurs at $x=\frac{1}{2}$
$$ \begin{aligned} \left.f(x)\right|_{\min } &=\left|2\left(\frac{1}{2}\right)^{2}+\frac{3}{2}-2\right|+\sin \left(\frac{1}{2}\right) \cdot \cos \left(\frac{1}{2}\right) \\\\ &=\frac{1}{2} \sin 1 \end{aligned} $$
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 $x=1$
$$ \begin{aligned} \left.f(x)\right|_{\max }=&|2+3-2|+\frac{\sin 2}{2} \\\\ &=3+\frac{1}{2} \sin 2 \end{aligned} $$
Sum of absolute maximum and minimum value
$=3+\frac{1}{2}(\sin 1+\sin 2)$
$=3+\frac{1}{2}(\sin 1+2\sin 1\cos 1)$
= $$3 + {1 \over 2}(1 + 2\cos (1))\sin (1)$$
$0 \leq x<\frac{1}{2} \quad f(x)=(1-2 x)(x+2)+\frac{\sin 2 x}{2}$
$f^{\prime}(x)=-4 x-3+\cos 2 x<0$
For $x \geq \frac{1}{2}: \quad f^{\prime}(x)=4 x+3+\cos 2 x>0$
So, minima occurs at $x=\frac{1}{2}$
$$ \begin{aligned} \left.f(x)\right|_{\min } &=\left|2\left(\frac{1}{2}\right)^{2}+\frac{3}{2}-2\right|+\sin \left(\frac{1}{2}\right) \cdot \cos \left(\frac{1}{2}\right) \\\\ &=\frac{1}{2} \sin 1 \end{aligned} $$
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 $x=1$
$$ \begin{aligned} \left.f(x)\right|_{\max }=&|2+3-2|+\frac{\sin 2}{2} \\\\ &=3+\frac{1}{2} \sin 2 \end{aligned} $$
Sum of absolute maximum and minimum value
$=3+\frac{1}{2}(\sin 1+\sin 2)$
$=3+\frac{1}{2}(\sin 1+2\sin 1\cos 1)$
= $$3 + {1 \over 2}(1 + 2\cos (1))\sin (1)$$
Comments (0)
