Q64 / 150 · MHT CET 2024 (11 May Shift 1) · Mathematics
If $\mathrm{f}(x)=x^3-6 x^2+9 x+3$ is monotonically decreasing function, then $x$ lies in
- A$(3, \infty)$
- Correct answerB$(1,3)$
- C$[3, \infty)$
- D$[0,3]$
Solution
$\begin{array}{ll}
& \mathrm{f}(x)=x^3-6 x^2+9 x+3 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=3 x^2-12 x+9
\end{array}$
Since $\mathrm{f}(x)$ is monotonically decreasing, $\begin{aligned} & \mathrm{f}^{\prime}(x) \lt 0 \\ & \Rightarrow 3 x^2-12 x+9 \lt 0 \\ & \Rightarrow x^2-4 x+3 \lt 0 \\ & \Rightarrow(x-3)(x-1) \lt 0 \\ & \Rightarrow x \in(1,3) \end{aligned}$
Since $\mathrm{f}(x)$ is monotonically decreasing, $\begin{aligned} & \mathrm{f}^{\prime}(x) \lt 0 \\ & \Rightarrow 3 x^2-12 x+9 \lt 0 \\ & \Rightarrow x^2-4 x+3 \lt 0 \\ & \Rightarrow(x-3)(x-1) \lt 0 \\ & \Rightarrow x \in(1,3) \end{aligned}$
Hence, the correct answer is option: 'B'.