Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2.

If \(\lim_{x \rightarrow 0} \left[1 + \frac {f(x)} {x^2} \right] = 3\) then f(2) is?

Concept:

To find extreme values of a function f , set f'(x)=0 and solve.

Calculation:

Consider a polynomial f(x) = \(\rm ax^4+bx^3+cx^2+dx+e\)

\(\rm \lim_{x \rightarrow 0} \left[1 + \frac {f(x)} {x^2} \right] = 3\\ ⇒ \lim_{x \rightarrow 0}\frac {f(x)} {x^2}=2\\ ⇒ \lim_{x \rightarrow 0}\frac {\rm ax^4+bx^3+cx^2+dx+e}{x^2}=2\\ ⇒ \lim_{x \rightarrow 0}\rm ax^2+bx+c+\frac d x+\frac{e}{x^2}=2\\\)

Now, let d = e = o

 \(So, \space \rm \lim_{x \rightarrow 0}\rm ax^2+bx+c=2\\ ⇒ c = 2\)

f(x) = \(\rm ax^4+bx^3+2x^2\)

f'(x) = \(\rm 4ax^3+3bx^2+4x\)

Here, extreme values are 1 and 2, so f'(1) = f'(2) = 0

f'(1) = 4a + 3b + 4= 0 

Multiply above equation by 4, we get 

16a + 12b + 16 = 0 ....(1)

And f'(2) = 32a + 12b + 8 =0 .....(2)

Subtract (1) from (2), we get 

16a - 8 = 0

⇒ a = 1/2

Put a = 1/2 in (1), we get 

8 + 12b + 16 = 0

⇒ b = -2 

f(x) = \(\rm \frac 1 2x^4-2x^3+2x^2\)

 \(\Rightarrow f(2) =\rm \frac 1 2(2)^4-2(2)^3+2(2)^2\\ \Rightarrow 8 - 16 +8\\ \Rightarrow 0 \)