On the interval [0, 1], the function x^ 25 (1 - x)^ 75 takes its … - Vidyadip

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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 6. Application of derivatives→STD 12 - 6. Application of derivatives — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Choose the correct answer from the given four options.Let $A$ and $B$ be two events such that $\text{P}(\text{A})=\frac{3}{8},\text{P}({\text{B}})=\frac{5}{8}$ and $\text{P}(\text{A}\cup\text{B})=\frac{3}{4}.$Then $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)\cdot\text{P}\Big(\frac{\text{A'}}{\text{B}}\Big)$ is equal to:

Area of the region $\left\{(x, y) \in \mathbb{R}^2: y \geq \sqrt{|x+3|}, 5 y \leq x+9 \leq 15\right\}$ is equal to

If $A = \int\limits_1^{\sin \theta } {\frac{t}{{1 + {t^2}}}} dt$ and $B = \int\limits_1^{\cos ec\theta } {\frac{dt}{{t\left( {1 + {t^2}} \right)}}} $ , (where $\theta \in \left( {0,\frac{\pi }{2}} \right))$, then the-value of $\left| {\begin{array}{*{20}{c}}
A&{{A^2}}&{ - B}\\
{{e^{A + B}}}&{{B^2}}&{ - 1}\\
1&{{A^2} + {B^2}}&{ - 1}
\end{array}} \right|$ is

Equations $x + y = 2,\,\,2x + 2y = 3$will have

The general solution $y(x)$ of the differential equation $\frac{{d\left( {\int\limits_x^y {dt} } \right)}}{{dy}} = x$ , is

The eqution of the plane $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\lambda(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})+\mu(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}})$ in scalar product from is:

A function $f$ satisfies the relation

$f(x) = f''(x) + f'''(x) + .......\infty $ where $f(x)$ is a differentiable function indefinitely. If $f(1) = 5$ , then the value of $f'(1) + f''(1)$ is equal to

The equation to the straight line passing through the points $(4, -5, -2)$ and $(-1, 5, 3)$ is

The solution of the differential equation $xy\frac{{dy}}{{dx}} = \frac{{(1 + {y^2})(1 + x + {x^2})}}{{(1 + {x^2})}}$ is

If $A = \left[ {\begin{array}{*{20}{c}}0&1 \\ 1&0\end{array}} \right],$then ${A^4}$=