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

Question

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

Accepted Answer

Answer: A. 0 \mathop {\lim }\limits_{x 	o 0} \left[ {1 + {{f\left( x ight)} \over {{x^2}}}} ight] = 3 \Rightarrow \mathop {Lim}\limits_{x 	o 0} {{f\left( x ight)} \over {{x^2}}} = 2 So, f(x) contains terms in x{}^2,{x^3} and {x^4} Let f\left( x ight) = {a_1}{x^2} + {a_2}{x^3} + {a_3}{x^4} Since \mathop {\lim }\limits_{x 	o 0} {{f\left( x ight)} \over {{x^2}}} = 2 \Rightarrow {a_1} = 2 Hence, f\left( x ight) = 2{x^2} + {a_2}{x^3} + {a_3}{x^4} f'\left( x ight) = 4x + 3{a_2}{x^2} + 4{a_3}{x^3} As given: f'\left( 1 ight) = 0 and f'\left( 2 ight) = 0 Hence, 4 + 3{a_2} + 4{a_3} = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 ight) and 8 + 12{a_2} + 32{a_3} = 0\,\,\,\,\,...\left( 1 ight) By 4x\left( {eq1} ight) - eq\left( 2 ight), we get 16 + 12{a_2} + 16{a_3} - \left( {8 + 12{a_2} + 32{a_3}} ight) = 0 \Rightarrow 8 - 16{a_3} = 0 \Rightarrow {a_3} = 1/2 and by eqn. \left( 1 ight),4 + 3{a_2} + 4/2 = 0 \Rightarrow {a_2} = - 2 \Rightarrow f\left( x ight) = 2{x^2} - 2{x^3} + {1 \over 2}{x^4} f\left( 2 ight) = 2 	imes 4 - 2 	imes 8 + {1 \over 2} 	imes 16 = 0

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

Answer

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