MCQ
On the interval $[0, 1],$  the function ${x^{25}}{(1 - x)^{75}}$ takes its maximum value at the point
  • A
    $0$
  • B
    $1/2$
  • C
    $1/3$
  • $1/4$

Answer

Correct option: D.
$1/4$
d
(d) $f(x) = {x^{25}}{(1 - x)^{75}}$

$f'(x) = {x^{25}}(75){(1 - x)^{74}}( - 1) + 25{x^{24}}{(1 - x)^{75}}$

For maxima and minima,

$ - 75{x^{25}}{(1 - x)^{74}} + 25{x^{24}}{(1 - x)^{75}} = 0$

==> $25{x^{24}}{(1 - x)^{74}}[(1 - x) - 3x] = 0$

==> Either $x = 0$or $x = 1$ or $x = \frac{1}{4}$

At $x = \frac{1}{4},\;\;f'\,\left( {\frac{1}{4} - h} \right) > 0$ and $f'\left( {\frac{1}{4} + h} \right) < 0$

$\therefore f(x)$ is maximum at $x = \frac{1}{4}$.

Trick: Check with the options.

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