JEE Main 2023 — Application Of Derivatives Question with Solution

The sum of the absolute maximum and minimum values of the function $$f(x)=\left|x^2-5x+6\right|-3x+2$$ in the interval $$[-1,3]$$ can be found by finding the maximum and minimum values of $$f(x)$$ in this interval and then adding them.

First, let's find the critical points of $$f(x)$$. To do this, we will find the zeros of the expression inside the absolute value:

$$x^2-5x+6=0$$

Solving this quadratic equation, we find that the zeros are $$x=2$$ and $$x=3$$. These are the critical points of $$f(x)$$.

Next, we evaluate $$f(x)$$ at these critical points and at the endpoints of the interval $$[-1,3]$$:

$$f(-1)=\left|(-1{)}^2-5(-1)+6\right|-3(-1)+2=17$$

$$f(2)=\left|(2{)}^2-5(2)+6\right|-3(2)+2=-4$$

$$f(3)=\left|(3{)}^2-5(3)+6\right|-3(3)+2=-7$$

So, the minimum value of $$f(x)$$ is $$-7$$ and the maximum value is $$17$$.

So, the sum of the absolute maximum and minimum values of the function is $$-7+17=10$$.

Note : The absolute maximum and minimum values of a function are the largest and smallest values that the function takes on a given interval, respectively. These values are also called the "extrema" of the function. The absolute maximum value is the highest point on the graph of the function, and the absolute minimum value is the lowest point.