Verified Solution • 2025–26 Session moderate

QJEEM24Mab917: JEE Main 2024 Mathematics - Class 12 JEEM24Mab917 Solution: Let and . If , then is equal to ____________.

Let f:(0,∞)→Rf:(0, \infty) \rightarrow \mathbf{R} and F(x)=∫0xtf(t)dt\mathrm{F}(x)=\int\limits_0^x \mathrm{t} f(\mathrm{t}) \mathrm{dt} . If F(x2)=x4+x5\mathrm{F}\left(x^2\right)=x^4+x^5 , then ∑r=112f(r2)\sum\limits_{\mathrm{r}=1}^{12} f\left(\mathrm{r}^2\right) is equal to ____________.

Step-by-Step Verified Solution

F(x)=∫0xt⋅f(t)dtF(x)=\int\limits_0^x t \cdot f(t) d t

amp;F′(x)=xf(x)amp;F(x2)=x4+x5, let x2=tamp;F(t)=t2+t5/2amp;F′(t)=2t+5/2t3/2amp;t⋅f(t)=2t+5/2t3/2amp;f(t)=2+5/2r1/2\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}

amp;∑r=112f(r2)=∑r=1122+52ramp;=24+5/2[12(13)2]amp;=219\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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