JEE Mains · Maths · STD 11 - 8. sequence and series
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.
- A \(32\)
- B \(64\)
- C \(128\)
- D \(256\)
Answer & Solution
Correct Answer
(C) \(128\)
Step-by-step Solution
Detailed explanation
\(\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{x^{2}}{2}+\frac{\alpha}{2 x^{5}}+\frac{\alpha}{2 x^{5}}\) \(\geq 7\left(\frac{\alpha^{2}}{2^{7}}\right)^{\frac{1}{7}}\) \(\frac{7 \cdot(\alpha)^{2 / 7}}{2}=14\) \(\left(\alpha^{2}\right)^{1 / 7}=2^{2}\)…
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