MHT CET · Maths · Application of Derivatives
The function \(\mathrm{f}(x)=2 x^3-9 x^2+12 x+2\) is decreasing in
- A \(1 \lt x \lt 2\)
- B \(x \lt 1\) or \(x \gt 2\)
- C \(x \lt -1\) or \(x \gt -2\)
- D \(-2 \lt x \lt -1\)
Answer & Solution
Correct Answer
(A) \(1 \lt x \lt 2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \mathrm{f}(x)=2 x^3-9 x^2+12 x+2 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=6 x^2-18 x+12
\end{aligned}\)
For decreasing function, \(\mathrm{f}^{\prime}(x) \lt 0\)
\(\begin{aligned}
& \Rightarrow 6 x^2-18 x+12 \lt 0 \\
& \Rightarrow x^2-3 x+2 \lt 0 \\
& \Rightarrow(x-1)(x-2) \lt 0 \\
& \Rightarrow x \in(1,2)
\end{aligned}\)
& \mathrm{f}(x)=2 x^3-9 x^2+12 x+2 \\
\therefore \quad & \mathrm{f}^{\prime}(x)=6 x^2-18 x+12
\end{aligned}\)
For decreasing function, \(\mathrm{f}^{\prime}(x) \lt 0\)
\(\begin{aligned}
& \Rightarrow 6 x^2-18 x+12 \lt 0 \\
& \Rightarrow x^2-3 x+2 \lt 0 \\
& \Rightarrow(x-1)(x-2) \lt 0 \\
& \Rightarrow x \in(1,2)
\end{aligned}\)
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