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JEE Mains · Maths · STD 11 - 8. sequence and series

ને \(f(x)=\frac{5 x^{2}}{2}+\frac{\alpha}{x^{5}}, x>0\), ની ન્યૂનતમ કિંમત \(14\) હોય, તો \(\alpha\) ની કિંમત .......... છે.

  1. A \(32\)
  2. B \(64\)
  3. C \(128\)
  4. D \(256\)
Verified Solution

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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