The maximum value of the function $f(x)=x+\sqrt{1-x}$ in the interval [0, 1] is,

$\frac{5}{4}$

The correct answer is Option (2) → $\frac{5}{4}$

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

for max. value $f'(c)=0$ and $f''(c)<0$

Now,

$f'(c)=1-\frac{1}{2\sqrt{1-x}}=0$

$⇒2\sqrt{1-x}=1$

$⇒4(1-x)=1$

$⇒x=\frac{3}{4}$

$∴f(\frac{3}{4})=\frac{3}{4}+\sqrt{1-\frac{3}{4}}=\frac{3}{4}+\frac{1}{2}$

$=\frac{5}{4}$