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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