MCQ
અંતરાલ $[0, 1]$ માટે,વિધેય ${x^{25}}{(1 - x)^{75}}$ એ . . . . આગળ મહતમ મૂલ્ય મેળવે.
- A$0$
- B$1/2$
- C$1/3$
- ✓$1/4$
$f'(x) = {x^{25}}(75){(1 - x)^{74}}( - 1) + 25{x^{24}}{(1 - x)^{75}}$
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.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.