Let f(x) be a polynomial of degree 5 such tha | vedclass

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(B) Given $\lim _{x \rightarrow 0}\left(2+\frac{f(x)}{x^{3}}\right)=4$,we have $\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=2$. Since $f(x)$ is a polyn

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$x=1$ is a point of minima and $x=-1$ is a point of maxima of $f$.

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(B) Given $\lim _{x \rightarrow 0}\left(2+\frac{f(x)}{x^{3}}\right)=4$,we have $\lim _{x \rightarrow 0} \frac{f(x)}{x^{3}}=2$. Since $f(x)$ is a polynomial of degree $5$,let $f(x) = ax^5 + bx^4 + cx^3 + dx^2 + ex + g$. For the limit to exist and be $2$,we must have $g=e=d=0$ and $c=2$. Thus,$f(x) = ax^5 + bx^4 + 2x^3$.$f'(x) = 5ax^4 + 4bx^3 + 6x^2$.Since $x=\pm 1$ are critical points,$f'(1) = 5a + 4b + 6 = 0$ and $f'(-1) = 5a - 4b + 6 = 0$.Adding these gives $10a + 12 = 0 \Rightarrow a = -6/5$. Subtracting gives $8b = 0 \Rightarrow b = 0$.So,$f(x) = 2x^3 - \frac{6}{5}x^5$.$f'(x) = 6x^2 - 6x^4 = 6x^2(1-x^2) = 6x^2(1-x)(1+x)$.For $x  0$. For $0  0$. For $x > 1$,$f'(x) At $x = -1$,$f'(x)$ changes from negative to positive,so it is a point of local minima.At $x = 1$,$f'(x)$ changes from positive to negative,so it is a point of local maxima.Option $B$ states $x=1$ is a point of minima and $x=-1$ is a point of maxima,which is false.

Let $f(x)$ be a polynomial of degree $5$ such that $x=\pm 1$ are its critical points. If $\mathop {\lim }\limits_{x \to 0} \left(2+\frac{f(x)}{x^{3}}\right)=4,$ then which one of the following is not true?

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