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JEE Advanced 2009Paper 2
Exam
JEE Advanced
Year
2009
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2009 · Paper 2 · Mathematics · Application of Derivatives

The maximum value of the function f(x)=2 x^3-15 x^2+36 x-48 on the...

The maximum value of the function f(x)=2x3−15x2+36x−48f(x)=2 x^3-15 x^2+36 x-48 on the set A={x∣x2+20≤9x}A=\left\{x \mid x^2+20 \leq 9 x\right\} isThe maximum value of the function f(x)=2 x^3-15 x^2+36 x-48 on the set A=\ x x^2+20 ≤ 9 x\ is

Correct answer

5

Step-by-step solution

Given, A={x∣x2+20≤9x}\quad A=\left\{x \mid x^2+20 \leq 9 x\right\} ={x∣x∈[4,5]}=\{x \mid x \in[4,5]\} Now, f′(x)=6(x2−5x+6)f^{\prime}(x)=6\left(x^2-5 x+6\right) Put f′(x)=0⇒x=2,3\quad f^{\prime}(x)=0 \Rightarrow x=2,3 f(2)=−20,f(3)=−21f(2)=-20, f(3)=-21, f(4)=−16,f(5)=7f(4)=-16, f(5)=7 The maximum value of the function f(x)=2 x^3-15 x^2+36 x-48 on the set A=\ x x^2+20 ≤ 9 x\ is
From graph, maximum of f(x)f(x) on set AA is f(5)=7f(5)=7.
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