JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(x, y, z\) be positive real numbers such that \(x + y + z = 12\) and \(x^3y^4z^5 = (0. 1 ) (600)^3\). Then \(x^3 + y^3 + z^3\) is equal to
- A \(342\)
- B \(216\)
- C \(258\)
- D \(270\)
Answer & Solution
Correct Answer
(B) \(216\)
Step-by-step Solution
Detailed explanation
\(x+y+z=12\) \(AM \ge GM\) \(\frac{{3\left( {\frac{x}{3}} \right) + 4\left( {\frac{y}{4}} \right) + 5\left( {\frac{z}{5}} \right)}}{{12}} \ge \) \(\sqrt[{12}]{{{{\left( {\frac{x}{3}} \right)}^3}{{\left( {\frac{y}{4}} \right)}^4}{{\left( {\frac{z}{5}} \right)}^5}}}\)…
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