Question

Let $f\left( x \right)$ be a polynomial of degree 4 having extreme values at x = 1 and x = 2 if $\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x \right)}{{{x}^{2}}}+1 \right]=3$, then $f\left( -1 \right)$ is equal to
$\begin{aligned} & a.\dfrac{1}{2} \\\ & b.\dfrac{3}{2} \\\ & c.\dfrac{5}{2} \\\ & d.\dfrac{9}{2} \\\ \end{aligned}$

Answer 1

First we need to write the four degree polynomial, $f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0$ and use the condition of any function which has extreme values at the critical points has $f'\left( x \right)=0$. Also, use the limit rules and subtraction operation between the obtained equations in order to find the values of a and b, and finally substitute x = -1 in the obtained $f\left( x \right)$ to get the required result.

Complete step by step answer:
We know that any function which has extreme values, that are maximum and minimum at the critical points, then we can say that $f'\left( x \right)=0$. From the question, we can say that the function has extreme values at x = and x = 2.
Therefore, for $f'\left( x \right)=0$ at x = 1 and at x = 2, we can say that,
$f'\left( 1 \right)=0$ and $f'\left( 2 \right)=0$
It is given that the polynomial $f\left( x \right)$ has a degree of four.
Therefore, $f\left( x \right)$ will be $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0$.
Hence, $f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0.........(i)$
We also have,
$\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{f\left( x \right)}{{{x}^{2}}}+1 \right]=3$
We know that $\underset{x\to 0}{\mathop{\lim }}\,k=k$, therefore, we can write,
$\begin{aligned} & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{{{x}^{2}}}+\underset{x\to 0}{\mathop{\lim }}\,1=3 \\\ & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{{{x}^{2}}}+1=3 \\\ & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{{{x}^{2}}}=3-1 \\\ & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{f\left( x \right)}{{{x}^{2}}}=2 \\\ \end{aligned}$
If we take the entire $f\left( x \right)$ and divide it by ${{x}^{2}}$, we get,
$f\left( x \right)=a{{x}^{2}}+bx+c+\dfrac{d}{x}+\dfrac{e}{{{x}^{2}}}=0$
In the above expression, if we apply the limits, we will get a not defined result. Hence, we will only consider,
$f\left( x \right)=a{{x}^{2}}+bx+c$
Therefore, we can say that, d = 0 and e = 0 for the limit value to be finite.
Now,
$\begin{aligned} & \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\left[ a{{x}^{2}}+bx+c \right] \\\ & =a\left( 0 \right)+b\left( 0 \right)+c \\\ \end{aligned}$
We know that,
$\begin{aligned} & \underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=2 \\\ & c=2 \\\ \end{aligned}$
Let us substitute the values d = 0, e = 0 and c = 2 in equation (i), so we get,
$f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+2{{x}^{2}}$
Differentiating the above expression with respect to x, we get,
$f'\left( x \right)=4a{{x}^{3}}+3b{{x}^{2}}+4x$
We have,
$f'\left( 1 \right)=0$
So, we can write,
$\begin{aligned} & f'\left( 1 \right)=4a{{\left( 1 \right)}^{3}}+3b{{\left( 1 \right)}^{2}}+4\left( 1 \right)=0 \\\ & \Rightarrow 4a+3b+4=0.........(ii) \\\ \end{aligned}$
Also, we know that,
$f'\left( 2 \right)=0$
So, we can write,
$\begin{aligned} & f'\left( 2 \right)=4a{{\left( 2 \right)}^{3}}+3b{{\left( 2 \right)}^{2}}+4\left( 2 \right)=0 \\\ & \Rightarrow 32a+12b+8=0.........(iii) \\\ \end{aligned}$
Let us solve equations (ii) and (iii). So, we have,
$\begin{aligned} & 4a+3b+4=0 \\\ & 32a+12b+8=0 \\\ \end{aligned}$
We will subtract the above equations, to get the value of a and b.
So, we will first divide equation (iii) by 4 throughout, so we get,
$8a+3b+2=0.........(iv)$
Now, we will subtract equation (2) from equation (iv), so we get,
$\begin{aligned} & \left( 8a+3b+2 \right)-\left( 4a+3b+4 \right)=0 \\\ & 8a+3b+2-4a-3b-4=0 \\\ & 4a-2=0 \\\ & 4a=2 \\\ & a=\dfrac{2}{4} \\\ & a=\dfrac{1}{2} \\\ \end{aligned}$
On substituting $a=\dfrac{1}{2}$ in equation (ii), we get,
$\begin{aligned} & 4\left( \dfrac{1}{2} \right)+3b+4=0 \\\ & 2+3b+4=0 \\\ & 3b+6=0 \\\ & 3b=-6 \\\ & b=-\dfrac{6}{3} \\\ & b=-2 \\\ \end{aligned}$
Now, let us substitute $a=\dfrac{1}{2},b=-2$ in $f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+2{{x}^{2}}$, so we will get,
$\begin{aligned} & f\left( x \right)=\dfrac{1}{2}{{x}^{4}}+\left( -2 \right)b{{x}^{3}}+2{{x}^{2}} \\\ & =\dfrac{{{x}^{4}}}{2}-2b{{x}^{3}}+2{{x}^{2}} \\\ \end{aligned}$
We need the value of $f\left( -1 \right)$, therefore, substitute the value x = -1 in the above equation. So, we get,
$\begin{aligned} & f\left( -1 \right)=\dfrac{{{\left( -1 \right)}^{4}}}{2}-2{{\left( -1 \right)}^{3}}+2{{\left( -1 \right)}^{2}} \\\ & =\dfrac{1}{2}-2\left( -1 \right)+2\left( 1 \right) \\\ & =\dfrac{1}{2}+2+2 \\\ & =\dfrac{1}{2}+4 \\\ & =\dfrac{1+8}{2} \\\ & =\dfrac{9}{2} \\\ \end{aligned}$
Therefore, the value of $f\left( -1 \right)=\dfrac{9}{2}$.

So, the correct answer is “Option D”.

Note: In mathematics, a limit is the value that a function approaches as the input approaches some value. Limits are essential in calculus and are used to define continuity, derivatives and integrals.