JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 16)
Let $f(x)=\int\limits_0^x \mathrm{t}\left(\mathrm{t}^2-9 \mathrm{t}+20\right) \mathrm{dt}, 1 \leq x \leq 5$. If the range of $f$ is $[\alpha, \beta]$, then $4(\alpha+\beta)$ equals :
253
157
154
125
Explanation
$f^{\prime}(x)=x^3-9 x^2+20 x=x(x-4)(x-5)$
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$$\begin{aligned} & \therefore f(x)=\frac{x^4}{4}-\frac{9 x^3}{3}+\frac{20 x^2}{2} \\ & f(1)=\frac{1}{4}-3+10=\frac{29}{4}=\alpha \\ & \left.f(4)=\frac{256}{4}-3(64) \right\rvert\,+10(16)=32=\beta \\ & 4(\alpha+\beta)=4\left(\frac{29}{4}+32\right)=157 \end{aligned}$$
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