Question

Let \[f\left( x \right)\] be a polynomial of degree four 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\left( 2 \right) $ is equal to
(a) –4
(b) 0
(c) 4
(d) –8

Answer 1

Hint: We start solving the problem by assuming the polynomial as \[f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e\]. We then find the values of d and e using the fact that $ \displaystyle \lim_{x \to 0}\dfrac{1}{{{x}^{n}}}=\infty $ for $ n\ge 1 $ . We then apply the limit to find the value of c. We make use of the fact that if a function $ f\left( x \right) $ has extremum at $ x=a $ , then $ {{f}^{'}}\left( a \right)=0 $ at $ x=1 $ and $ x=2 $ to find the values of a and b. We then substitute these values in the assumed polynomial to get the required polynomial. We then substitute 2 in place of x to get the required answer.

Complete step by step answer:
According to the problem, we are given that \[f\left( x \right)\] is a polynomial of degree four having extreme values at $ x=1 $ and $ x=2 $ . We need to find the value of $ f\left( 2 \right) $ if $ \displaystyle \lim_{x \to 0}\left[ 1+\dfrac{f\left( x \right)}{{{x}^{2}}} \right]=3 $ .
Let us assume \[f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e\] ---(1) as the maximum power of x will be 4. So, we get \[\dfrac{f\left( x \right)}{{{x}^{2}}}=a{{x}^{2}}+bx+c+\dfrac{d}{x}+\dfrac{e}{{{x}^{2}}}\]. Let us substitute this in given limit.
So, we have $ \displaystyle \lim_{x \to 0}\left[ 1+a{{x}^{2}}+bx+c+\dfrac{d}{x}+\dfrac{e}{{{x}^{2}}} \right]=3 $ .
We know that $ \displaystyle \lim_{x \to 0}\dfrac{1}{{{x}^{n}}}=\infty $ for $ n\ge 1 $ , which means that the limit has to be $ \infty $ if d, e are non-zero real numbers. But the given limit is finite so, the values of d, e must be 0.
So, we get $ d=e=0 $ ---(2).
\[\Rightarrow \displaystyle \lim_{x \to 0}\left[ 1+a{{x}^{2}}+bx+c+\dfrac{d}{x}+\dfrac{e}{{{x}^{2}}} \right]=1+a{{\left( 0 \right)}^{2}}+b\left( 0 \right)+c+0+0=3\].
\[\Rightarrow 1+c=3\].
\[\Rightarrow c=2\] ---(3).
Let us substitute equations (2) and (3) in equation (1).
So, we get \[f\left( x \right)=a{{x}^{4}}+b{{x}^{3}}+2{{x}^{2}}\] ---(4).
According to the problem, we have given that the function \[f\left( x \right)\] has extreme values at $ x=1 $ and $ x=2 $ .
We know that if a function $ f\left( x \right) $ has extremum at $ x=a $ , then $ {{f}^{'}}\left( a \right)=0 $ .
Let us first find $ {{f}^{'}}\left( x \right) $ . So, let differentiate both sides of equation (4) w.r.t x.
So, we get \[{{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( a{{x}^{4}}+b{{x}^{3}}+2{{x}^{2}} \right)\].
 $ \Rightarrow {{f}^{'}}\left( x \right)=4a{{x}^{3}}+3b{{x}^{2}}+4x $ .
So, we have $ {{f}^{'}}\left( 1 \right)=0 $ .
 $ \Rightarrow {{f}^{'}}\left( 1 \right)=4a{{\left( 1 \right)}^{3}}+3b{{\left( 1 \right)}^{2}}+4\left( 1 \right) $ .
 $ \Rightarrow 0=4a+3b+4 $ .
 $ \Rightarrow 4a=-3b-4 $ ---(5).
Now, we have $ {{f}^{'}}\left( 2 \right)=0 $ .
 $ \Rightarrow {{f}^{'}}\left( 2 \right)=4a{{\left( 2 \right)}^{3}}+3b{{\left( 2 \right)}^{2}}+4\left( 2 \right) $ .
 $ \Rightarrow 0=4a\left( 8 \right)+12b+8 $ ---(6).
Let us substitute equation (5) in equation (6).
 $ \Rightarrow 0=\left( -3b-4 \right)\left( 8 \right)+12b+8 $ .
 $ \Rightarrow 0=-24b-32+12b+8 $ .
 $ \Rightarrow 0=-12b-24 $ .
 $ \Rightarrow 12b=-24 $ .
 $ \Rightarrow b=-2 $ ---(7).
Let us substitute equation (7) in equation (5).
 $ \Rightarrow 4a=-3\left( -2 \right)-4 $ .
 $ \Rightarrow 4a=6-4 $ .
 $ \Rightarrow 4a=2 $ .
 $ \Rightarrow a=\dfrac{1}{2} $ ---(8).
So, we get $ f\left( x \right)=\dfrac{{{x}^{4}}}{2}-2{{x}^{3}}+2{{x}^{2}} $ .
Now, we need to find the value of $ f\left( 2 \right) $ .
 $ \Rightarrow f\left( 2 \right)=\dfrac{{{\left( 2 \right)}^{4}}}{2}-2{{\left( 2 \right)}^{3}}+2{{\left( 2 \right)}^{2}} $ .
 $ \Rightarrow f\left( 2 \right)=\dfrac{16}{2}-2\left( 8 \right)+2\left( 4 \right) $ .
 $ \Rightarrow f\left( 2 \right)=8-16+8 $ .
 $ \Rightarrow f\left( 2 \right)=0 $ .
So, we have found the value of $ f\left( 2 \right) $ as 0.
$ \therefore $ The correct option for the given problem is (b).

Note:
We should not stop solving the problem after finding the polynomial as this is the most common mistake done by students. We can see that the given problems contain huge amount of calculation so, we need to perform each step carefully to avoid mistakes and confusion. We need to know that degree of the polynomial is the highest degree of the independent variable present in the polynomial. Similarly, we can expect the problems to find the zeroes of the obtained polynomial.