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The maximum value of $[x(x-1)+1]^{\frac{1}{3}}, 0 \leq x \leq 1$ is
(a) $\left(\frac{1}{3}\right)^{\frac{1}{3}}$
(b) $\frac{1}{2}$
(c) 1
(d) 0
The maximum value of $[x(x-1)+1]^{\frac{1}{3}}, 0 \leq x \leq 1$ is
(a) $\left(\frac{1}{3}\right)^{\frac{1}{3}}$
(b) $\frac{1}{2}$
(c) 1
(d) 0
Solution:
1874 Upvotes
Verified Answer
(c) Let $\mathrm{y}=[\mathrm{x}(\mathrm{x}-1)+1]^{1 / 3}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(2 \mathrm{x}-1)}{3[\mathrm{x}(\mathrm{x}-1)+1]^{2 / 3}}, \frac{\mathrm{dy}}{\mathrm{dx}}=0$ at $\mathrm{x}=\frac{1}{2}$
$\frac{d y}{d x}$ Changes sign from $-$ ve to $+$ ve at $x=\frac{1}{2}$
$\therefore \quad \mathrm{y}$ is minimum at $\mathrm{x}=\frac{1}{2}$
Value of $y$ at $x=0,(0+1)^{1 / 3}=1^{1 / 3}=1$
Value of $\mathrm{y}$ at $\mathrm{x}=1,(0+1)^{1 / 3}=1^{1 / 3}=1$
$\therefore$ The maximum value of $y$ is 1.
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{(2 \mathrm{x}-1)}{3[\mathrm{x}(\mathrm{x}-1)+1]^{2 / 3}}, \frac{\mathrm{dy}}{\mathrm{dx}}=0$ at $\mathrm{x}=\frac{1}{2}$
$\frac{d y}{d x}$ Changes sign from $-$ ve to $+$ ve at $x=\frac{1}{2}$
$\therefore \quad \mathrm{y}$ is minimum at $\mathrm{x}=\frac{1}{2}$
Value of $y$ at $x=0,(0+1)^{1 / 3}=1^{1 / 3}=1$
Value of $\mathrm{y}$ at $\mathrm{x}=1,(0+1)^{1 / 3}=1^{1 / 3}=1$
$\therefore$ The maximum value of $y$ is 1.
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