#### Question

If(x) = x+\[\frac{1}{x}\],x > 0, then its greatest value is

(a) \[-\] 2

(b) 0

(c) 3

(d) none of these

#### Solution

(d) none of these

\[\text { Given }: f\left( x \right) = x + \frac{1}{x}\]

\[ \Rightarrow f'\left( x \right) = 1 - \frac{1}{x^2}\]

\[\text { For a local maxima or a local minima, we must have } \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow 1 - \frac{1}{x^2} = 0\]

\[ \Rightarrow x^2 - 1 = 0\]

\[ \Rightarrow x^2 = 1\]

\[ \Rightarrow x = \pm 1\]

\[ \Rightarrow x = 1 \left( \text { Given }: x>0 \right)\]

\[\text { Now,} \]

\[f''\left( x \right) = \frac{2}{x^3}\]

\[ \Rightarrow f''\left( 1 \right) = 2 > 0\]

\[\text { So, x = 1 is a local minima } .\]

Is there an error in this question or solution?

Solution for question: If(X) = X+ 1 X ,X > 0, Then Its Greatest Value is (A) − 2 (B) 0 (C) 3 (D) None of These concept: Graph of Maxima and Minima. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science