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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
MCQ
Choose the correct answer The maximum value of $[\text{x(x}-1)+1]^\frac{1}{3},0\leq\text{x}\leq1]\text{is:}$
  • A
    $\Big(\frac{1}{3}\Big)^{\frac{1}{3}}$
  • B
    $\frac{1}{2}$
  • ✓
    $1$
  • D
    $0$
✓

Answer

Correct option: C.
$1$
(C) 1
Let $\text{f}\text{(x)}=[\text{x(x}-1)+1]^{\frac{1}{3}}$ $=(\text{x}^2-\text{x}+1)^{\frac{1}{3}},0\leq\text{x}\leq1\ \dots\text{(i)}$
$\therefore\ \ \text{f}'\text{(x)}=\frac{1}{3}(\text{x}^2-\text{x}+1)^{\frac{-2}{3}}\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}+1)$ $=\frac{(2\text{x}-1)}{3(\text{x}^2-\text{x}+1)^{\frac{2}{3}}}$
Now $\text{f}'\text{(x)}=0\ \Rightarrow\ \frac{(2\text{x}-1)}{3(\text{x}^2-\text{x}+1)^{\frac{2}{3}}}=0$ $\Rightarrow\ 2\text{x}-1=0$
$\Rightarrow\ \text{x}=\frac{1}{2}\ $[Turning point] and it belongs to the given enclosed interval $0\leq\text{x}\leq1$ i.e.,[0, 1].
$\text{At}\text{ x}=\frac{1}{2},\ \text{from eq.(i)},\ \text{f}\Big(\frac{1}{2}\Big)$ $=\Big(\frac{1}{4}-\frac{1}{2}+1\Big)^{\frac{1}{3}}$ $=\Big(\frac{1-2+4}{4}\Big)^{\frac{1}{3}}=\Big(\frac{3}{4}\Big)^{\frac{1}{3}}<1$
$\text{At}\text{ x}=0,\ \text{form eq.(i)},\ \text{f}(0)=(1)^{\frac{1}{3}}=1$
$\text{At}\text{ x}=1,\ \text{from eq.(i)},\ \text{f}(1)=(1-1+1)^{\frac{1}{3}}=(1)^{\frac{1}{3}}=1$
$\therefore\ $ Maximum value of f(x) is 1.

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