Question

Let \( f(x)=(1-1/x)2,x>0\). Then

• A. \(f\) is increasing in \((0,2)\) and decreasing  in \((2,∞)\)
• B. \(f \) is decreasing in \((0,2)\) and increasing  in \((2,∞) \)
• C. \(f \) is increasing in \((0,1)\) and decreasing in \((1,∞)\)
• D. \( f\) is decreasing in \((0,1)\) and increasing  in \((1,∞) \)
• E. \(f\) is increasing in \((0,∞)\)

Answer 1

The Correct Option is D

Given that:

\(f(x)=(1-\dfrac{1}{x})^{2}\)

\(f'(x)=2(1-\dfrac{1}{x})(\dfrac{-1}{x^2})\)

now put, \(f'(x)=0\)

so ,\(f'(x)=2(1-\dfrac{1}{x})(\dfrac{-1}{x^2})=0\)

\((1-\dfrac{1}{x})(\dfrac{-1}{x^2})=0\)

but here for any positive value of \(x\), \((\dfrac{-1}{x^2})\) can not be zero 

hence , \((1-\dfrac{1}{x})=0\)

          \(⇒x=1\) ⇢one critical point

Now, on analyzing the intervals based on the behavior of the derivative we can state that,

For\(0<x<1,\) \(f′(x)<0\). Since \(f′(x)\) is negative in this interval, the function f(x) is decreasing in \((0,1)\).

For \(x>1\), \(f′(x)>0\). Since \(f′(x)\) is positive in this interval, the function f(x) is increasing in \((1,∞)\).

Hence , \(f\) is decreasing in \((0,1)\) and increasing in \((1,∞)\)(_Ans)