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Question Number 170155 by cortano1 last updated on 17/May/22

G i v e n f (x) = x \sqrt{1 - x + \sqrt{1 - x}}

w h e r e 0 ⩽ x ⩽ 1

f i n d m a x f (x)

Answered by Mathspace last updated on 18/May/22

x \in [0, 1] \Rightarrow x = c o s θ \Rightarrow

f (x) = c o s θ \sqrt{1 - c o s θ + \sqrt{1 - c o s θ}}

= c o s θ \sqrt{2 {s i n}^{2} (\frac{θ}{2}) + \sqrt{2 {s i n}^{2} \frac{θ}{2}}}

= c o s θ \sqrt{2 {s i n}^{2} (\frac{θ}{2}) + \sqrt{2} s i n (\frac{θ}{2})}

= c o s θ \sqrt{\sqrt{2} s i n (\frac{θ}{2})} \sqrt{\sqrt{2} s i n (\frac{θ}{2}) + 1}

⩽ (^{4} \sqrt{2}) \sqrt{1 + \sqrt{2}}

\dots .

Commented by cortano1 last updated on 18/May/22

w r o n g