JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a , b , c\) and \(d\) be positive real numbers such that \(a+b+c+d=11\). If the maximum value of \(a^5 b^3 c^2 d\) is \(3750 \beta\), then the value of \(\beta\) is
- A \(90\)
- B \(110\)
- C \(55\)
- D \(108\)
Answer & Solution
Correct Answer
(A) \(90\)
Step-by-step Solution
Detailed explanation
\(\frac{5\left(\frac{ a }{5}\right)+3\left(\frac{ b }{3}\right)+2\left(\frac{ c }{2}\right)+ d }{11} \geq\left(\frac{ a ^5 b ^3 c ^2 d }{5^5 3^3 2^2}\right)^{1 / 11}\) \(1 \geq\left(\frac{a^5 b^3 c^2 d}{5^5 3^3 2^2}\right)^{1 / 11}\) \(\beta=90\)
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