Question

Let f(x) be a polynomial of degree 4 having extreme values at x=1 and x=2.
If $\displaystyle \lim_{x \to 0}\left( 1+\dfrac{f\left( x \right)}{{{x}^{2}}} \right)=3$ , then f(2) is equal to
(a) -8
(b) -4
(c) 0
(d) 4

Answer 1

Let the polynomial be $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0$ . Use the point that the extremums of the polynomial are at x=2 and x=1, which implies $f'\left( 1 \right)=\text{0 and }f'\left( 2 \right)=0$ . Also, use the condition $\displaystyle \lim_{x \to 0} \left( 1+\dfrac{f\left( x \right)}{{{x}^{2}}} \right)=3$ to get the value of a, b, c, d and e. Remember, that a number divided by 0 is finite and defined if and only if the number is also zero.

Complete step-by-step answer:
Let us start the solution to the above question by letting the four degree polynomial to be $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0$ .
It is given that $\displaystyle \lim_{x \to 0}\left( 1+\dfrac{f\left( x \right)}{{{x}^{2}}} \right)=3$ . So, if we substitute the polynomial, we get
$\displaystyle \lim_{x \to 0}\left( 1+\dfrac{a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e}{{{x}^{2}}} \right)=3$
$\Rightarrow \displaystyle \lim_{x \to 0}\left( 1+a{{x}^{2}}+bx+c+\dfrac{d}{x}+\dfrac{e}{{{x}^{2}}} \right)=3$
Now, we know that a number divided by 0 is finite and defined if and only if the number is also zero. So, d=e=0. Therefore, our equation is:
$\displaystyle \lim_{x \to 0}\left( 1+a{{x}^{2}}+bx+c \right)=3$
Now, if we put the limit, i.e., x=0, we get
$1+a\times {{0}^{2}}+b\times 0+c=3$
$\Rightarrow 1+c=3$
$\Rightarrow c=2$
So, we can say $f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+2{{x}^{2}}$ . If we differentiate f(x) using the formula $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ , we get
$f'\left( x \right)=4a{{x}^{3}}+3b{{x}^{2}}+4x$
As it is given that that the extremums of the polynomial are at x=2 and x=1, we can say that $f'\left( 1 \right)=\text{0 and }f'\left( 2 \right)=0$ .
$f'\left( 1 \right)=0$
$\Rightarrow 4a\times {{1}^{3}}+3b\times {{1}^{2}}+4\times 1=0$
$\Rightarrow 4a+3b+4=0.............(i)$
$f'\left( 2 \right)=0$
$\Rightarrow 4a\times {{2}^{3}}+3b\times {{2}^{2}}+4\times 2=0$
$\Rightarrow 32a+12b+8=0...........(ii)$
Now, we will multiply equation (i) by 4 and subtract it from equation (ii). On doing so, we get
$32a+12b+8-4\left( 4a+3b+4 \right)=0$
$\Rightarrow 32a+12b+8-16a-12b-16=0$
$\Rightarrow 16a-8=0$
$\Rightarrow 16a=8$
$\Rightarrow a=\dfrac{1}{2}$
Now, if we put this value of a in equation (i), we get
$4\times \dfrac{1}{2}+3b+4=0$
$\Rightarrow 3b+6=0$
$\Rightarrow b=-2$
So, we can say that the polynomial is $f\left( x \right)=\dfrac{1}{2}\times {{x}^{4}}-2{{x}^{3}}+2{{x}^{2}}$ . So, if we put x=2 in this function, we get
$f\left( 2 \right)=\dfrac{1}{2}\times {{\left( 2 \right)}^{4}}-2{{\left( 2 \right)}^{3}}+2{{\left( 2 \right)}^{2}}=8-16+8=0$

So, the correct answer is “Option C”.

Note: It is not always necessary that you can find out the polynomial using the method but it is a sure thing that all the data required will be given by using the above interpretations only. Also, it is not a necessary thing that you should know the polynomial, you might reach the answer without knowing the polynomial as well, depending on what is asked.