JEE MAIN - Mathematics (2022 - 28th July Morning Shift - No. 13)

If the minimum value of $$f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0$$, is 14 , then the value of $$\alpha$$ is equal to :

128

256

$$f(x) = {{5{x^2}} \over 2} + {\alpha \over {{x^5}}}\,\,\{ x > 0\} $$

$$f'(x) = 5x - {{5\alpha } \over {{x^6}}} = 0$$

$$ \Rightarrow x = {(\alpha )^{{1 \over 7}}}$$

JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Application of Derivatives Question 56 English Explanation

$$f{(x)_{\min }} = {{5{{(\alpha )}^{{2 \over 7}}}} \over 2} + {\alpha \over {{\alpha ^{{5 \over 7}}}}} = 14$$

$${5 \over 2}{\alpha ^{{2 \over 7}}} + {\alpha ^{{2 \over 7}}} = 14$$

$${7 \over 2}{\alpha ^{{2 \over 7}}} = 14$$

$$\alpha = 128$$