JEE Mathematics 2023 Question with Solution

Given: $f(x) = |x^2 - 5x + 6| - 3x + 2$ on $[-1, 3]$.

Find: The sum of the absolute maximum and minimum values.

From the solution, the function is treated as

$$f(x) = x^2 - 5x + 6 - 3x + 2 = x^2 - 8x + 8$$

Then,

$$f'(x) = 2x - 8$$

Setting the derivative equal to zero,

$$2x - 8 = 0$$

so

$$x = 4$$

Since $x = 4$ is outside $[-1, 3]$, there are no critical points within the interval according to the provided solution.

Now evaluate at the endpoints:

$$f(-1) = (-1)^2 - 8(-1) + 8 = 1 + 8 + 8 = 17$$ $$f(3) = 3^2 - 8(3) + 8 = 9 - 24 + 8 = -7$$

Therefore, the absolute maximum value is $17$ and the absolute minimum value is $-7$. Their sum is

$$17 + (-7) = 10$$

However, the solution also states "The Correct Option is B," while the computed result is $10$, which matches option A. Using the working shown, the defensible answer is A.

Given: $f(x) = |x^2 - 5x + 6| - 3x + 2$.

Find: Which option matches the sum of the absolute maximum and minimum values.

The extracted solution does not split the absolute value into intervals, although its hint mentions that this should be done. Instead, it directly uses

$$f(x) = x^2 - 8x + 8$$

and proceeds by endpoint evaluation after checking the derivative.

Using that extracted working:

1. Differentiate to get

$$f'(x) = 2x - 8$$

2. Solve

$$f'(x) = 0 \implies x = 4$$

3. Since $4 \notin [-1, 3]$, check only the endpoints used in the solution.
4. Compute

$$f(-1) = 17, \qquad f(3) = -7$$

5. Add maximum and minimum values:

$$17 + (-7) = 10$$

So the numerical result obtained from the shown working is $10$, which corresponds to option A.