Q244 / 270 · NDA 2018 (Phase 2) · Mathematics
What is the minimum value of $[x(x-1)+1]^{\frac{1}{3}}$, where $0 \lt x \lt 1$ ?
- Correct answerA$\left(\frac{3}{4}\right)^{\frac{1}{3}}$
- B1
- C$\frac{1}{3}$
- D$\left(\frac{3}{8}\right)^{\frac{1}{3}}$
Solution
$[x(x-1)+1]^{1 / 3}$ is minimum when $x^2-x+1$ is minimum, $0 \leq x \leq 1$
$\begin{aligned}
& x^2-x+1=\left(x-\frac{1}{2}\right)^2+\left(1-\frac{1}{4}\right) \\
& =\left(x-\frac{1}{2}\right)^2+\frac{3}{4}
\end{aligned}$
minimum value of $x^2-x+1=\frac{3}{4}$
$\text { Required value }=\left(\frac{3}{4}\right)^{1 / 3}$
Hence, the correct answer is option: 'A'.