JEE Main 2023 — Application of Derivatives Question with Solution

Given,

$f\left(x\right)=\left|x^2-5x+6\right|-3x+2$

$\Rightarrow f\left(x\right)=\left|\left(x-3\right)\left(x-2\right)\right|-3x+2$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{cc}{x}^2-8x+8& ;x\in \left[-1,2\right]\\ -x^2+2x-4& ;x\in (2,3]\end{array}\right.$

$\Rightarrow f^{\text{'}}\left(x\right)=\left\{\begin{array}{cc}2x-8& ;x\in \left(-1,2\right)\\ -x+2& ;x\in \left(2,3\right)\end{array}\right.$

Now point of extrema will be $2x-8=0\Rightarrow x=4$which is not in domain and $-x+2=0\Rightarrow x=2$

Now for finding the value of global minima and maxima we will check the value of function at extrema points and boundary points,

So, $f\left(x\right)=x^2-8x+8$ will give, $f\left(-1\right)=17 \& f\left(2\right)=-4$

And $f\left(x\right)=-x^2+2x-4$ will give, $f\left(2\right)=4, f\left(3\right)=-7$

Hence, absolute minima is $-7$ and maxima is $17$

So, their sum is $-7+17=10$