The function f is defined by f(x) = x^p (1 - | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the function $f(x) = x^p (1 - x)^q$. To find the critical points,we find the first derivative $f'(x)$ and set it to $0$. Using the product r

Vedclass · Accepted Answer

$\frac{p}{p+q}$

Vedclass · Answer

(D) Given the function $f(x) = x^p (1 - x)^q$. To find the critical points,we find the first derivative $f'(x)$ and set it to $0$. Using the product rule: $f'(x) = p x^{p-1} (1 - x)^q + x^p \cdot q (1 - x)^{q-1} (-1)$. $f'(x) = x^{p-1} (1 - x)^{q-1} [p(1 - x) - qx]$. $f'(x) = x^{p-1} (1 - x)^{q-1} [p - px - qx]$. $f'(x) = x^{p-1} (1 - x)^{q-1} [p - (p + q)x]$. Setting $f'(x) = 0$,we get critical points at $x = 0$,$x = 1$,and $x = \frac{p}{p+q}$. Since $p, q$ are positive integers,for $x$ in the interval $(0, 1)$,the sign of $f'(x)$ changes from positive to negative at $x = \frac{p}{p+q}$. Therefore,the function has a maximum value at $x = \frac{p}{p+q}$.

The function $f$ is defined by $f(x) = x^p (1 - x)^q$ for all $x \in R$,where $p, q$ are positive integers. The function has a maximum value for $x$ equal to:

Similar Questions

The maximum area of the rectangle that can be inscribed in a circle of radius $r$ is

If $y = a \ln x + bx^2 + x$ has its extremum value at $x = 1$ and $x = 2,$ then $(a, b) =$

If $g(x) = 2f(2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3)$,$\forall x \in R$ and $f''(x) > 0$,$\forall x \in R$,then $g'(x) > 0$ for $x$ belonging to

For all real values of $x,$ the minimum value of $\frac{1-x+x^{2}}{1+x+x^{2}}$ is

At $x=0$,$f(x)=\cos x-1+\frac{x^2}{2}-\frac{x^3}{3}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module