JEE MAIN - Mathematics (2024 - 31st January Morning Shift - No. 25)
Let $$S=(-1, \infty)$$ and $$f: S \rightarrow \mathbb{R}$$ be defined as
$$f(x)=\int_\limits{-1}^x\left(e^t-1\right)^{11}(2 t-1)^5(t-2)^7(t-3)^{12}(2 t-10)^{61} d t \text {, }$$
Let $$\mathrm{p}=$$ Sum of squares of the values of $$x$$, where $$f(x)$$ attains local maxima on $$S$$, and $$\mathrm{q}=$$ Sum of the values of $$\mathrm{x}$$, where $$f(x)$$ attains local minima on $$S$$. Then, the value of $$p^2+2 q$$ is _________.
Explanation
$$\mathrm{f}^{\prime}(\mathrm{x})=\left(\mathrm{e}^{\mathrm{x}}-1\right)^{11}(2 \mathrm{x}-1)^5(\mathrm{x}-2)^7(\mathrm{x}-3)^{12}(2 \mathrm{x}-10)^{61}$$
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Local minima at $$\mathrm{x}=\frac{1}{2}, \mathrm{x}=5$$
Local maxima at $$\mathrm{x}=0, \mathrm{x}=2$$
$$\therefore \mathrm{p}=0+4=4, \mathrm{q}=\frac{1}{2}+5=\frac{11}{2}$$
Then $$p^2+2 q=16+11=27$$
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