JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(\mathrm{a} \gt 0\). If the function \(\mathrm{f}(\mathrm{x})=6 \mathrm{x}^3-45 \mathrm{a} \mathrm{x}^2+108 \mathrm{a}^2 \mathrm{x}+1\) attains its local maximum and minimum values at the points \(x_1\) and \(x_2\) respectively such that \(x_1 x_2=54\), then \(\mathrm{a}+\mathrm{x}_1+\mathrm{x}_2\) is equal to :-
- A \(15\)
- B \(18\)
- C \(24\)
- D \(13\)
Answer & Solution
Correct Answer
(B) \(18\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & f^{\prime}(x)=18 x^2-90 a x+108 a^2=0 \\ & x=2 a \& x=3 a \\ & x_1=2 a \quad x_2=3 a \\ & x_1 x_2=54 \\ & 6 a^2=54 \\ & a=3 \\ & a+x_1+x_2 \\ & 3+2 \times 3+3 \times 3=18 \\ & \text { option }(2)\end{aligned}\)
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