JEE MAIN - Mathematics (2022 - 25th July Evening Shift - No. 18)
The sum of the maximum and minimum values of the function $$f(x)=|5 x-7|+\left[x^{2}+2 x\right]$$ in the interval $$\left[\frac{5}{4}, 2\right]$$, where $$[t]$$ is the greatest integer $$\leq t$$, is ______________.
Answer
15
Explanation
$$f(x) = |5x - 7| + [{x^2} + 2x]$$
$$ = |5x - 7| + [{(x + 1)^2}] - 1$$
Critical points of
$$f(x) = {7 \over 5},\sqrt 5 - 1,\,\sqrt 6 - 1,\,\sqrt 7 - 1,\,\sqrt 8 - 1,\,2$$
$$\therefore$$ Maximum or minimum value of $$f(x)$$ occur at critical points or boundary points
$$\therefore$$ $$f\left( {{5 \over 4}} \right) = {3 \over 4} + 4 = {{19} \over 4}$$
$$f\left( {{7 \over 5}} \right) = 0 + 4 = 4$$
as both $$|5x - 7|$$ and $${x^2} + 2x$$ are increasing in nature after $$x = {7 \over 5}$$
$$\therefore$$ $$f(2) = 3 + 8 = 11$$
$$\therefore$$ $$f{\left( {{7 \over 5}} \right)_{\min }} = 4$$ and $$f{(2)_{\max }} = 11$$
Sum is $$4 + 11 = 15$$
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