Let $$f(x)$$ be a polynomial of degree $$5$$ such that $$x = \pm 1$$ are its critical points. If $$\underset{x \rightarrow 0}{\lim} \left(2 + \dfrac{f(x)}{x^3} \right) = 4$$, then which one of the following is not true?
As $$\underset{x \rightarrow 0}{\lim} \left(2 + \dfrac{7(x)}{x^3} \right) $$ exist so $$f(x)$$ should be $$ax^5 + bx^4 + cx^3$$ $$\therefore \underset{x \rightarrow 0}{\lim} \left(2 + \dfrac{ax^5 + bx^4 + cx^3}{x^3} \right) = 4$$ $$2 + c = 4 \Rightarrow c = 2$$ $$f'(x) = 5 ax^4 + 4bx^3 + 6x^2$$ As at $$x = I_1$$ f has critical point $$\therefore f'(1) = 0 \Rightarrow 5a + 4b + b = 0$$ ___(I) $$f'(-1) = 0 \Rightarrow 5a - 4b + b = 0$$ ___(II) From equation (I) & (II) $$b = 0$$ $$a = -6/5$$ $$\therefore f(x) = \dfrac{-6}{5} x^5 + 2x^3$$ $$f'(x) = -6x^4 + 6x^2$$ $$= 6x^2 (-x^2 - 1)$$ $$= -6x^2 (x - 1)(x +1)$$ ref. image minima at $$x = -1$$ maxima at $$x = +1$$
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