MHT CET · Maths · Application of Derivatives
If \(\mathrm{f}(x)=x^3-6 x^2+9 x+3\) is monotonically decreasing function, then \(x\) lies in
- A \((3, \infty)\)
- B \((1,3)\)
- C \([3, \infty)\)
- D \([0,3]\)
Answer & Solution
Correct Answer
(B) \((1,3)\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& \mathrm{f}(x)=x^3-6 x^2+9 x+3 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2-12 x+9
\end{array}\)
Since \(\mathrm{f}(x)\) is monotonically decreasing,
\(\begin{aligned}
& \mathrm{f}^{\prime}(x) \lt 0 \\
& \Rightarrow 3 x^2-12 x+9 \lt 0 \\
& \Rightarrow x^2-4 x+3 \lt 0 \\
& \Rightarrow(x-3)(x-1) \lt 0 \\
& \Rightarrow x \in(1,3)
\end{aligned}\)
& \mathrm{f}(x)=x^3-6 x^2+9 x+3 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2-12 x+9
\end{array}\)
Since \(\mathrm{f}(x)\) is monotonically decreasing,
\(\begin{aligned}
& \mathrm{f}^{\prime}(x) \lt 0 \\
& \Rightarrow 3 x^2-12 x+9 \lt 0 \\
& \Rightarrow x^2-4 x+3 \lt 0 \\
& \Rightarrow(x-3)(x-1) \lt 0 \\
& \Rightarrow x \in(1,3)
\end{aligned}\)
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