JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 25)
Let $f:(0, \infty) \rightarrow \mathbf{R}$ and $\mathrm{F}(x)=\int\limits_0^x \mathrm{t} f(\mathrm{t}) \mathrm{dt}$. If $\mathrm{F}\left(x^2\right)=x^4+x^5$, then $\sum\limits_{\mathrm{r}=1}^{12} f\left(\mathrm{r}^2\right)$ is equal to ____________.
Answer
219
Explanation
$F(x)=\int\limits_0^x t \cdot f(t) d t$
$\begin{aligned} & F'(x)=x f(x) \\\\ & F\left(x^2\right)=x^4+x^5, \quad \text { let } x^2=t \\\\ & F(t)=t^2+t^{5 / 2} \\\\ & F^{\prime}(t)=2 t+5 / 2 t^{3 / 2} \\\\ & t \cdot f(t)=2 t+5 / 2 t^{3 / 2} \\\\ & f(t)=2+5 / 2 r^{1 / 2}\end{aligned}$
$\begin{aligned} & \sum_{r=1}^{12} f\left(r^2\right)=\sum_{r=1}^{12} 2+\frac{5}{2} r \\\\ & =24+5 / 2\left[\frac{12(13)}{2}\right] \\\\ & =219\end{aligned}$
$\begin{aligned} & F'(x)=x f(x) \\\\ & F\left(x^2\right)=x^4+x^5, \quad \text { let } x^2=t \\\\ & F(t)=t^2+t^{5 / 2} \\\\ & F^{\prime}(t)=2 t+5 / 2 t^{3 / 2} \\\\ & t \cdot f(t)=2 t+5 / 2 t^{3 / 2} \\\\ & f(t)=2+5 / 2 r^{1 / 2}\end{aligned}$
$\begin{aligned} & \sum_{r=1}^{12} f\left(r^2\right)=\sum_{r=1}^{12} 2+\frac{5}{2} r \\\\ & =24+5 / 2\left[\frac{12(13)}{2}\right] \\\\ & =219\end{aligned}$
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