Let $${f\left( x \right)}$$  be a polynomial of degree four having extreme values at $$x = 1$$  and $$x = 2.$$  If $$\mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{{f\left( x \right)}}{{{x^2}}}} \right] = 3,$$     then $$f\left( 2 \right)$$  is equal to:

Solution :
$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{{f\left( x \right)}}{{{x^2}}}} \right] = 3, \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right)}}{{{x^2}}} = 2 \cr & {\text{So,}}\,f\left( x \right)\,{\text{contains terms in }}{x^2},{x^3}\,{\text{and }}{x^4}{\text{.}} \cr & {\text{Let }}f\left( x \right) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4} \cr & {\text{since }}\mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right)}}{{{x^2}}} = 2 \Rightarrow {a_1} = 2 \cr & {\text{Hence, }}f\left( x \right) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4} \cr & f'\left( x \right) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3} \cr & {\text{As given }}:f'\left( 1 \right) = 0{\text{ and }}f'\left( 2 \right) = 0 \cr & {\text{Hence, }}4 + 3{a_2} + 4{a_3} = 0\,......\left( 1 \right) \cr & {\text{and }}8 + 12{a_2} + 32{a_3} = 0\,......\left( 2 \right) \cr & {\text{By}}\,4x\left( {{\text{eq}}1} \right) - {\text{eq}}\left( 2 \right),\,{\text{we get}} \cr & 16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} \right) = 0 \cr & \Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = \frac{1}{2} \cr & {\text{and by eqn}}{\text{. }}\left( 1 \right),4 + 3{a_2} + \frac{4}{2} = 0 \Rightarrow {a_2} = - 2 \cr & \Rightarrow f\left( x \right) = 2{x^2} - 2{x^3} + \frac{1}{2}{x^4} \cr & f\left( 2 \right) = 2 \times 4 - 2 \times 8 + \frac{1}{2} \times 16 = 0 \cr} $$