JEE Main 2025 Jan 29 Shift 2, Mathematics Q18: Maxima & Minima

JEE Main2025Jan 29, Shift 2Mathematics

Let $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$ , $1 \leq x \leq 5$ . If the range of $f$ is $[α, β]$ , then $4 (α + β)$ equals:

A
$157$
B
$253$
C
$125$
D
$154$

Solution

By FTC, $f^{'} (x) = x (x^{2} - 9 x + 20) = x (x - 4) (x - 5)$ . On the interval $[1, 5]$ , signs are $f^{'} > 0$ for $1 < x < 4$ and $f^{'} < 0$ for $4 < x < 5$ , so $f$ has a maximum at $x = 4$ and minimum at an endpoint.

Integrate: $f (x) = \frac{x ^{4}}{4} - 3 x^{3} + 10 x^{2}$ .

$f (1) = \frac{1}{4} - 3 + 10 = \frac{29}{4}$ .

$f (4) = 64 - 192 + 160 = 32$ .

$f (5) = \frac{625}{4} - 375 + 250 = \frac{125}{4}$ .

So $α = min = f (1) = \frac{29}{4}$ and $β = max = f (4) = 32$ .

$4 (α + β) = 4 (\frac{29}{4} + 32) = 29 + 128 = 157$ .

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q18: Maxima & Minima

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