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Applications of Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsApplications of Derivatives2 Marks
MCQ
The function $f(x)=x \sqrt{1-x},$ where $x \in(0,1)$ has local maximum at $x=$
  • ✓
    $\frac{2}{3}$
  • B
    $\frac{1}{4}$
  • C
    $\frac{3}{4}$
  • D
    $\frac{1}{3}$
✓

Answer

Correct option: A.
$\frac{2}{3}$
Given $,f(x)=x \sqrt{1-x}$
$f^{\prime}(x)=x \cdot \frac{-1}{2 \cdot \sqrt{1-x}}+\sqrt{1-x}$
$=\frac{-x+2(1-x)}{2 \sqrt{1-x}}$
$=\frac{2-3 x}{2 \sqrt{1-x}}$
Put $f^{\prime}(x)=0$ so, $x=\frac{2}{3}$ is the critical point
Now, $f^{\prime \prime}(x)=\frac{2 \sqrt{1-x}(-3)-(2-3 x) \cdot 2 \cdot \frac{(-1) \cdot 2}{\sqrt{1-x}}}{4(1-x)}$
$=\frac{-6(1-x)+(2-3 x)}{4(1-x)^{3 / 2}}$
$=\frac{-4+3 x}{4(1-x)^{3 / 2}}$
Now at $x=\frac{2}{3} f^{\prime \prime}(x)=- ve$
Hence, at $x=\frac{2}{3} f(x)$ has local maximum.

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