Let f ( x ) = 0 x t ( t 2 – 9 t + 20 ) d t , 1 x 5 . If the range of f is , then 4 equals : [2025]

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Solution

Q.
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 : [2025]

1 253

2 154

3 157

4 125

Ans.
(3)

We have, $f (x) = \int_{0}^{x} t (t^{2} - 9 t + 20) d t$

$\Rightarrow f' (x) = x^{3} - 9 x^{2} + 20 x = x (x - 4) (x - 5)$

Where $1 \leq x \leq 5$

On integration, we get

$f (x) = \frac{x^{4}}{4} - \frac{9 x^{3}}{3} + \frac{20 x^{2}}{2}$

$= \frac{x^{4}}{4} - 3 x^{3} + 10 x^{2}$ , where $1 \leq x \leq 5$

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

$f (4) = \frac{{(4)}^{4}}{4} - 3 {(4)}^{3} + 10 {(4)}^{2} = 32 = β$

$f (5) = \frac{{(5)}^{4}}{4} - 3 {(5)}^{3} + 10 {(5)}^{2} = \frac{125}{4}$

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

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