Question

$f\left(x\right)=(x-1)|(x-2)(x-3)|$ then $f$ decreases in the interval?

• A. $(2-\frac{1}{\sqrt{3}},2)$
• B. $(2,2+\frac{1}{\sqrt{3}})$
• C. $(3,\infty )$
• D. $(2+\frac{1}{\sqrt{3}},4)$

Answer 1

Answer: (A), (B)

The correct options are
A $(2-\frac{1}{\sqrt{3}},2)$
B $(2,2+\frac{1}{\sqrt{3}})$

$f\left(x\right)=x^3-6x^2+11x-6f^{\prime }\left(x\right)=3x^2-12x+11$

For $f\left(x\right)$ to decrease $\to f^{\prime }\left(x\right)<0$

$\Rightarrow 3x^2-12x+11<0\Rightarrow 3x^2-12x+12-1<0\Rightarrow 3(x^2-4x+4)-1<0\Rightarrow 3{(x-2)}^2-1<0\Rightarrow {(x-2)}^2-\frac{1}{3}<1\Rightarrow (x-(2+\frac{1}{\sqrt{3}}))(x-(2-\frac{1}{\sqrt{3}}))<0\Rightarrow x\in (2-\frac{1}{\sqrt{3}},2+\frac{1}{\sqrt{3}})$

Hence, correct answer is

$x\in (2-\frac{1}{\sqrt{3}},2+\frac{1}{\sqrt{3}})$