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Question

Let $$f(x)$$ be a polynomial of degree 6, which satisfies $$\displaystyle \lim_{x\rightarrow 0}\left ( 1+\frac{f\left ( x \right )}{x^{3}} \right )^{1/x}={e^{2}}$$ and local maximum at $$x= 1$$ and local minimum at $$x= 0$$ and $$2$$, then $$5f(3)$$ is equal to


Solution

$$\displaystyle \lim_{x\rightarrow 0}\left ( 1+\dfrac{f\left ( x \right )}{x^{3}} \right )^{1/x}={e^{2}}$$
If the limit is to exist,$$f(x)$$ cannot have terms with degree lesser than 3. 
So, let $$f(x)=ax^4+bx^5+cx^6$$
Hence,  $$\displaystyle \lim _{ x\rightarrow 0 }{ \left[ 1+ax+bx^{ 2 }+cx^{ 3 } \right] ^{\displaystyle \dfrac { 1 }{ x }  } } =\quad e^{ 2 }$$
Applying log on both sides gives
$$\displaystyle\lim _{ x\rightarrow 0 }{ \left(\displaystyle \dfrac { \log { \left[ 1+ax+bx^{ 2 }+cx^{ 3 } \right]  }  }{ x }  \right)  } =\quad \log { e^{ 2 } }$$
using the series $$\displaystyle\log {(1+x)} = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dots$$ 
$$\displaystyle\lim _{ x\rightarrow 0 }{ \left(\displaystyle \dfrac { ax+bx^{ 2 }+cx^{ 3 } }{ x }  \right)  } =2$$        (higher order terms becomes $$0$$) 
$$\Rightarrow$$   $$a=2$$
Therefore,  $$f(x)=2x^4+bx^5+cx^6$$
$$\Rightarrow$$ $$f^{\prime}(x)=8x^3+5bx^4+6cx^5$$
but given that $$f(x)$$ has local maximum at $$x=1$$ and local minimum at $$x=0$$ and $$2$$
i.e $$f^{\prime}(x)=0$$ when $$x=0,1,2$$
$$\Rightarrow$$ $$f^{\prime}(1)=8+5b+6c=0$$ and $$f^{\prime}(2)=4+5b+12c=0$$
solving above two equations for $$b$$ and $$c$$ gives
$$b=\displaystyle\dfrac{-12}{5}$$ and $$c=\displaystyle\dfrac{2}{3}$$
$$\therefore$$ $$\displaystyle f(x)=2x^4-\dfrac{12}{5}x^5+\dfrac{2}{3}x^6$$
$$\displaystyle 5f(3) = 5(162-\dfrac{12}{5}3^5+\dfrac{2}{3}3^6) = 324$$
Hence, $$5f(3) = 324$$

Mathematics

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