Question

Where does $f\left(x\right)=x+\sqrt{1-x};0<x<1$ decrease?

• A. $(\frac{3}{4},1)$
• B. $(0,1)$
• C. $(0,\frac{3}{4})$
• D. $(\frac{3}{4},\infty )$

Answer 1

The correct option is A $(\frac{3}{4},1)$

$f\left(x\right)=x+\sqrt{1-x}$

$f^{{}^{\prime }}\left(x\right)=1-\frac{1}{2}\frac{1}{\sqrt{1-x}}$

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

$⟹1-\frac{1}{2}\frac{1}{\sqrt{1-x}}<0$

$⟹2\sqrt{1-x}<1$

$⟹4(1-x)<1$

$⟹x>\frac{3}{4}$

Also, $(1-x)>0\Rightarrow x<1$

Hence, the answer is option (A).