Let f: R rightarrow R be a function defined b | vedclass

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

Question

(C) The derivative of the function is given by $f'(x) = n_{1}(x-3)^{n_{1}-1}(x-5)^{n_{2}} + n_{2}(x-3)^{n_{1}}(x-5)^{n_{2}-1}$.Simplifying this,we get

Vedclass · Accepted Answer

For $n_{1} = 3, n_{2} = 5$,there exists $\alpha \in (3, 5)$ where $f$ attains a local maximum.

Vedclass · Answer

(C) The derivative of the function is given by $f'(x) = n_{1}(x-3)^{n_{1}-1}(x-5)^{n_{2}} + n_{2}(x-3)^{n_{1}}(x-5)^{n_{2}-1}$.Simplifying this,we get $f'(x) = (x-3)^{n_{1}-1}(x-5)^{n_{2}-1} [n_{1}(x-5) + n_{2}(x-3)] = (x-3)^{n_{1}-1}(x-5)^{n_{2}-1} [(n_{1}+n_{2})x - (5n_{1}+3n_{2})]$.The critical point in $(3, 5)$ is $x = \frac{5n_{1}+3n_{2}}{n_{1}+n_{2}}$.For $n_{1}=3, n_{2}=5$,$f'(x) = (x-3)^{2}(x-5)^{4} [8x - 30] = 8(x-3)^{2}(x-5)^{4} (x - 3.75)$.Since $(x-3)^{2}$ and $(x-5)^{4}$ are always non-negative,the sign of $f'(x)$ changes from negative to positive at $x = 3.75$,which implies a local minimum,not a maximum. Thus,option $C$ is not true.

Let $f: R \rightarrow R$ be a function defined by $f(x) = (x - 3)^{n_{1}}(x - 5)^{n_{2}}$,where $n_{1}, n_{2} \in N$. Which of the following is $\text{NOT}$ true?

Similar Questions

The maximum value $M$ of $f(x) = 3^x + 5^x - 9^x + 15^x - 25^x$,as $x$ varies over all real numbers,satisfies:

In the interval $[-2, 4]$,the absolute maximum of $f(x) = 2x^3 - 3x^2 - 12x + 5$ occurs at $x =$

The shortest distance of the point $(0,0)$ from the curve $y = e^x + e^{-x}$ is-

Maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is

Slope of the tangent to the curve $y=2 e^x \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)$ where $0 \leq x \leq 2 \pi$ is minimum at $x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module