MHT CET · Maths · Application of Derivatives
The function \(\mathrm{f}(x)=x^3-6 x^2+9 x+2\) has maximum value when \(x\) is
- A 1
- B 2
- C 3
- D 6
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{f}(x)=x^3-6 x^2+9 x+2 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2-12 x+9
\end{aligned}\)
For maximum or minimum,
\(\begin{aligned}
& \mathrm{f}^{\prime}(x)=0 \\
& \Rightarrow 3 x^2-12 x+9=0 \\
& \Rightarrow 3(x-1)(x-3)=0 \\
& \Rightarrow x=1,3
\end{aligned}\)
Now, \(\mathrm{f}^{\prime \prime}(x)=6 x-12\)
\(\therefore \quad \mathrm{f}^{\prime \prime}(1)=-6 < 0\)
\(\therefore \quad \mathrm{f}(x)\) is maximum at \(x=1\).
& \mathrm{f}(x)=x^3-6 x^2+9 x+2 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2-12 x+9
\end{aligned}\)
For maximum or minimum,
\(\begin{aligned}
& \mathrm{f}^{\prime}(x)=0 \\
& \Rightarrow 3 x^2-12 x+9=0 \\
& \Rightarrow 3(x-1)(x-3)=0 \\
& \Rightarrow x=1,3
\end{aligned}\)
Now, \(\mathrm{f}^{\prime \prime}(x)=6 x-12\)
\(\therefore \quad \mathrm{f}^{\prime \prime}(1)=-6 < 0\)
\(\therefore \quad \mathrm{f}(x)\) is maximum at \(x=1\).
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