MCQ
Maximum value of $x{(1 - x)^2}$ when $0 \le x \le 2$, is
- A${2 \over {27}}$
- ✓${4 \over {27}}$
- C$5$
- D$0$
✓
Answer
Correct option: B.
${4 \over {27}}$
b
(b) Given $f(x) = x{(1 - x)^2}$, $f(x) = {x^3} - 2{x^2} + x$
(b) Given $f(x) = x{(1 - x)^2}$, $f(x) = {x^3} - 2{x^2} + x$
Now $f'(x) = 3{x^2} - 4x + 1$
Put $f'(x) = 0$ i.e., $3{x^2} - 4x + 1 = 0$
$3{x^2} - 3x - x + 1 = 0$ ==> $x = 1,\,\,1/3$
$f''(x) = 6x - 4$
$\therefore f''\,(1) = 2 = $ positive and $f''(1/3) = - 2 = $ $-ve$
Hence maximum value will be at $x = \frac{1}{3}$
Maximum value $f\,\left( {\frac{1}{3}} \right) = \frac{4}{{27}}$.
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