• # question_answer Let f(x) be a polynomial of degree 5 such that $x=\pm 1$ are its critical points. If $\underset{x\to 0}{\mathop{\lim }}\,\left( 2+\frac{f(x)}{{{x}^{3}}} \right)=4$, then which one of the following is not true? [JEE MAIN Held on 07-01-2020 Evening] A) f is an odd function B) x =1 is a point of minima and $x=1$ is a point of maxima of f. C) $f\left( 1 \right)4f\left( 1 \right)=4$ D) x =1 is a point of maxima and $x=1$ is a point of minimum of f.

 [b] $\because$ f(x) is a five degree polynomial such that $\underset{x\to 0}{\mathop{\lim }}\,\left( 2+\frac{f(x)}{{{x}^{3}}} \right)=4$ let $f(x)=a{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}$ $\underset{x\to 0}{\mathop{\lim }}\,\left( 2+\frac{a\,{{x}^{5}}+b{{x}^{4}}+c{{x}^{3}}}{{{x}^{3}}} \right)=4$ $\Rightarrow \,\,2+c=4\Rightarrow \,c=2$ Now, $f'(x)=5a{{x}^{4}}+4b{{x}^{3}}+3c{{x}^{2}}$ $={{x}^{2}}(5a{{x}^{2}}+4bx+3c)$ $\because \,\,f'(1)\,=0\Rightarrow \,5a+4b+6=0$ and $f'(-1)=0\Rightarrow \,5a-4b+6=0$ $\therefore \,\,b=0,\,a=-\frac{6}{5}$ $\therefore \,f(x)=-\frac{6}{5}{{x}^{5}}+2{{x}^{3}}$ $f'(x)=-6{{x}^{4}}+6{{x}^{2}}=-\,6{{x}^{2}}(x+1)(x-1)$ It is clear that maxima at x = 1 and minima at$x=1$. and $f\left( 1 \right)4f\left( 1 \right)=4$