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Let  \mathrm{f: \mathbb{R} \rightarrow \mathbb{R}} be a function defined by \mathrm{f(x)=(x-3)^{n_{1}}(x-5)^{n_{2}}, n_{1}, n_{2} \in \mathrm{N}} . Then, which of the folllowing is NOT true?

Option: 1

For \mathrm{n_{1}=3, n_{2}=4} , there exists  \mathrm{\alpha \in(3,5)} where \mathrm{f} attains local maxima.


Option: 2

For \mathrm{n_{1}=4, n_{2}=3} , there exists  \mathrm{\alpha \in(3,5)} where \mathrm{f} attains local minima.


Option: 3

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


Option: 4

For \mathrm{n_{1}=4, n_{2}=6} , there exists  \mathrm{\alpha \in(3,5)} where \mathrm{f} attains local maxima.


Answers (1)

best_answer

\mathrm{f(x) =(x-3)^{n_{1}}(x-5)^{n_{2}}} \\

\mathrm{f^{\prime}(x) =n_{1}(x-3)^{n_{1}-1}(x-5)^{n_{2}}+ n_{2}(x-3)^{n_{1}}(x-5)^{n_{2}-1}} \\

\mathrm{=(x-3)^{n_{1}-1} \cdot(x-5)^{n_{2}-1}\left[\begin{array}{l} n_{1}(x-5)+n_{2}(x-3) \\ \end{array}\right]}

\mathrm{=(x-3)^{n_{1}-1} \cdot(x-5)^{n_{2}-1}\left[\begin{array}{l} (n_{1}+n_{2})x-(5n_{1}+3n_{2}) \\ \end{array}\right]}

For option 3

\mathrm{f^{\prime}(x)=(x-3)^{2}(x-5)^{4}(8 x-30)}

x = 15/4 is a local minima

( Using first order derivative test )

Therefore option 3 is wrong.

We can similarly check other options.

Hence answer is option 3

Posted by

Divya Prakash Singh

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