$$ \because $$    f(x) has extremum values at x = 1, and x = 2

$$ \because $$  f'(1) = 0 and f'(2) = 0

As, f(x) is a polynomial of degree 4.

Suppose f(x) = Ax⁴ + Bx³ + cx² + Dx + E

$$ \because $$   $$\mathop {\lim }\limits_{x \to 0} $$ $$\left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right)$$ = 3

$$ \Rightarrow $$$$\,\,\,$$$$\mathop {\lim }\limits_{x \to 0} \left( {{{A{x^4} + B{x^3} + C{x^2} + Dx + E} \over {{x^2}}} + 1} \right)$$ = 3

$$ \Rightarrow $$   $$\mathop {\lim }\limits_{x \to 0} $$ $$\left( {A{x^2} + Bx + C + {D \over x} + {E \over {{x^2}}} + 1} \right)$$ = 3

As limit has finite value, so D = 0 and E = 0

Now A(0)² + B(0) + C + 0 + 0 + 1 = 3

$$ \Rightarrow $$   c + 1 = 3 $$ \Rightarrow $$ c = 2

f'(x) = 4Ax³ + 3Bx² + 2Cx + D

f'(1) = 0 $$ \Rightarrow $$ 4A(1) + 3B(1) + 2C(1) + D = 0

$$ \Rightarrow $$$$\,\,\,$$ 4A + 3B = $$-$$ 4     . . . .(1)

f'(2) = 0 $$ \Rightarrow $$ 4A(8) + 3B(4) + 2C(2) + D = 0

$$ \Rightarrow $$ 8A + 3B = $$-$$ 2      . . . . .(2)

From equations (1) and (2), we get

A = $${1 \over 2}$$ and B = $$-$$ 2

So, f(x) = $${{{x_4}} \over 2} - 2{x^3} + 2x{}^2$$

Therefore, f($$-$$ 1) = $${{{{( - 1)}^4}} \over 2} - 2{( - 1)^3} + 2{( - 1)^2}$$

= $${1 \over 2} + 2 + 2 = {9 \over 2}$$

Hence f($$-$$1) = $${9 \over 2}$$