Question

The domain of the function $ \sqrt{{{\log }_{e}}({{x}^{2}}-6x+6)} $ is

• A. $ (-\infty ,3-\sqrt{3}]\cup [3+\sqrt{3},\infty ] $
• B. $ (-\infty ,3-\sqrt{3}]\cup (3+\sqrt{3},\infty ] $
• C. $ (-\infty ,1]\cup [5,\infty ) $
• D. $ (-\infty ,1)\cup (5,\infty ) $

Answer 1

Answer: (B)

Let $ ({{x}_{1}},{{y}_{1}}) $ be the required point. We have the given curve $ y={{x}^{3}}-2{{x}^{2}}-x $ ...(i) $ \frac{dy}{dx}=3{{x}^{2}}-4x-1 $ $ {{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},{{y}_{1}})}}=3x_{1}^{2}-4{{x}_{1}}-1 $ This is the slope of the tangent to the curve but the tangent is parallel to the line $ y=3x-2 $ $ \therefore $ $ 3x_{1}^{2}-4{{x}_{1}}-1=3 $ $ \Rightarrow $ $ 3x_{1}^{2}-4{{x}_{1}}-4=0 $ $ \Rightarrow $ $ ({{x}_{1}}-2)(3{{x}_{1}}+2)=0 $ $ \Rightarrow $ $ {{x}_{1}}=2,-\frac{2}{3} $ Since, the point $ ({{x}_{1}},{{y}_{1}}) $ lies on the curve (i), $ \therefore $ At $ {{x}_{1}}=2{{y}_{1}}={{2}^{3}}-2{{(2)}^{2}}-2=-2 $ at $ {{x}_{1}}=-\frac{2}{3},{{y}_{1}}={{\left( -\frac{2}{3} \right)}^{3}}-2{{\left( -\frac{2}{3} \right)}^{2}}+\frac{2}{3}=-\frac{14}{27} $ Hence, the required points are $ (2,-2) $ and $ \left( -\frac{2}{3},-\frac{14}{27} \right) $