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Let $f(x)$ be a polynomial of degree four having extreme values at $x=1$ and $x=2$. If $\displaystyle \lim_{x\to0}\left[1+\frac{f(x)}{x^2}\right]=3$, then $f(2)$ is equal to:

(1) $-8$

(2) $-4$

(3) $0$

(4) $4$

The answer is $0$. Can anyone explain the answer with steps?

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    what did you try? – 2017-01-11
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    Maybe $\dfrac{1+f}{x^2}$ – 2017-01-11
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    The solution given in my book is as follows but i am not able to understand it:- – 2017-01-11
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    limx→0 (x^2+f(x)/x^2)=3,since , limit exits hence x^2+f(x)=ax^4+bx^3+3x^2 ⇒ f(x)=ax^4+bx^3+2x^2⇒ ⇒ f(x)=4ax^3+3bx^2+4x⇒ also f(x)=0 at x=1,2 ⇒ a=1/2,b=−2 ⇒ f(x)=x^4/2−2x^3+2x^2⇒ ⇒ f(x)=8−16+8=0 – 2017-01-11
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    i am not able to understand this part:- – 2017-01-11
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    limit exits hence x^2+f(x)=ax^4+bx^3+3x^2 – 2017-01-11

2 Answers 2

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Hint

$$\lim_{x\rightarrow 0}\frac{f(x)}{x^2}=\lim_{x\rightarrow 0}\left(ax^2+bx+c+\frac{d}{x}+\frac{e}{x^2}\right)=2$$

that give us,

$$ d=e=0 \quad \text{and}\quad c=2$$

so,

$$f(x)=ax^4+bx^3+2x^2 \rightarrow f'(x)=4ax^3+3bx^2+4x$$

and $f(x)$ has $x=1$ and $x=2$ as extreme values, so

$$f'(1)=f'(2)=0$$

Can you finish?

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    i am not able to understand why d and e have to be zero – 2017-01-11
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    if not, the limit goes to $\pm \infty$ – 2017-01-11
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    should c always be 2 as in my textbook they take c as 2 – 2017-01-11
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    $c=2$ for sure, take a look at the answer. If you make $x \rightarrow 0$ at $ax^2+bx+c$ will remain just $c$. – 2017-01-11
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    Is the answer clear? – 2017-01-11
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I can give you some hints. As far as I can tell, the real question is to find the polynomial, and then the answer for the multiple choice question will be found by plugging in $x=2$.

First, write your polynomial as a generic degree four polynomial $f(x)=ax^4+bx^3+cx^2+dx+e$. Then the limit condition says that $$\lim_{x\to0}\left[1+ax^2bx+c+\frac{d}{x}+\frac{e}{x^2}\right]=3,$$ which immediately implies that $d=0$ and $e=0$ (do you see why?), and then you can also find the value of the coefficient $c$ from this.

Now you have your polynomial as $f(x)=ax^4+bx^3+cx^2$, with a known value of $c$ here. Find the derivative of $f(x)$: $$f'(x)=4ax^3+3bx^2+2cx=4ax\left(x^2+\frac{3b}{4a}x+\frac{2c}{4a}\right).$$ But since $x=1$ and $x=2$ are extreme points, they are roots of the derivative, and so the quadratic part in the parentheses above is equal to $(x-1)(x-2)=x^2-3x+2$. By equating the coefficients (remember that $c$ already has a numerical value), we'll find the values of $a$ and $b$.