Question:
On the interval $[0, 1]$, the function $x^{25}(1 - x)^{75}$ takes its maximum value at the point
On the interval $[0, 1]$, the function $x^{25}(1 - x)^{75}$ takes its maximum value at the point
Updated On: Apr 15, 2024
- $0$
- $ \frac{1}{4}$
- $ \frac{1}{2}$
- $ \frac{1}{3}$
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Let $f\left(x\right) = x^{25}\left(1 - x\right)^{75}, x \in \left[0, 1\right]$
$\Rightarrow f '\left(x\right) = 25x^{24} \left(1 - x\right)^{75} - 75x^{25} \left(1 - x\right)^{74}$
$= 25x^{24} \left(1 - x\right)^{74} \left(1 - x\right) - 3x$
$= 25 x^{24} \left(1 - x\right)^{74} \left(1 - 4x\right)$
We can see that $f '\left(x\right)$ is positive for $x \frac{1}{4}$.
Hence, $f \left(x\right)$ attains maximum at $x = \frac{1}{4}.$
$\Rightarrow f '\left(x\right) = 25x^{24} \left(1 - x\right)^{75} - 75x^{25} \left(1 - x\right)^{74}$
$= 25x^{24} \left(1 - x\right)^{74} \left(1 - x\right) - 3x$
$= 25 x^{24} \left(1 - x\right)^{74} \left(1 - 4x\right)$
We can see that $f '\left(x\right)$ is positive for $x \frac{1}{4}$.
Hence, $f \left(x\right)$ attains maximum at $x = \frac{1}{4}.$
Was this answer helpful?
0
0
Top Questions on Application of derivatives
- Let \(A\) be the region enclosed by the parabola \(y^2 = 2x\) and the line \(x = 24\). Then the maximum area of the rectangle inscribed in the region \(A\) is ________.
- JEE Main - 2024
- Mathematics
- Application of derivatives
- For the function \(f(x) = \sin x + 3x - \frac{2}{\pi}(x^2 + x), \quad x \in \left[0, \frac{\pi}{2}\right],\)consider the following two statements:
1. \( f \) is increasing in \( \left(0, \frac{\pi}{2}\right) \).
2. \( f' \) is decreasing in \( \left(0, \frac{\pi}{2}\right) \).
Between the above two statements- JEE Main - 2024
- Mathematics
- Application of derivatives
- Let the set of all values of \( p \), for which \[ f(x) = (p^2 - 6p + 8)(\sin^2 2x - \cos^2 2x) + 2(2 - p)x + 7 \] does not have any critical point, be the interval \( (a, b) \). Then \( 16ab \) is equal to _____ .
- JEE Main - 2024
- Mathematics
- Application of derivatives
Area of region enclosed by curve y=x3 and its tangent at (–1,–1)
- JEE Main - 2023
- Mathematics
- Application of derivatives
The minimum of \(f(x)=\sqrt{(10-x^2)}\) in the interval \([-3,2]\) is
- KEAM - 2023
- Mathematics
- Application of derivatives
View More Questions
Questions Asked in VITEEE exam
- Two identical blocks A and B, each of mass 'm' resting on a smooth floor are connected by a light spring of natural length L and spring constant K, with the spring at its natural length. A third identical block 'C' (mass m) moving with a speed v along the line joining A and B collides with A. the maximum compression in the spring is
- VITEEE - 2023
- work, energy and power
- Assertion :
The binding energy of the nucleus increases with the increase in atomic number.
Reason :
Heavier elements have a greater number of non-radioactive isotopes than radioactive isotopes. - The half-life of radioactive radon is 3.8 days. The time at the end of which \(\frac{1}{20}th\) of the Radon sample will remain undecayed is (given log10e=0.4343)
- Which element has more electron gain enthalpy among chalcogen group?
- VITEEE - 2022
- sulphur
- Coin is tossed till heads is obtained what is expectation of no. of coin tosses
- VITEEE - 2022
- Random Variables
View More Questions
Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives