If $\mathrm{f}(x)=x^3-6 x^2+9 x+3$ is monotonically decreasing function, then $x$ lies in
- $(3, \infty)$
- $(1,3)$
- $[3, \infty)$
- $[0,3]$
Solution
Since $\mathrm{f}(x)$ is monotonically decreasing, $\begin{aligned} & \mathrm{f}^{\prime}(x) \lt 0 \\ & \Rightarrow 3 x^2-12 x+9 \lt 0 \\ & \Rightarrow x^2-4 x+3 \lt 0 \\ & \Rightarrow(x-3)(x-1) \lt 0 \\ & \Rightarrow x \in(1,3) \end{aligned}$
Asked in: MHT CET 2024 (11 May Shift 1)
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