CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Let $$f(x)$$ be a polynomial of degree four having extreme value at $$x=1$$ and $$x=2$$.If $$\displaystyle \lim_{x\rightarrow 0}(1+\dfrac{f(x)}{x^2})=3$$,then $$f(2)$$ is equal to ?


A
0
loader
B
4
loader
C
8
loader
D
49
loader

Solution

The correct option is D $$\dfrac{-4}{9}$$
Given $$f'(1)=f'(2)=0$$
     $$\displaystyle\lim_{x\rightarrow 0} \bigg(1+\dfrac{f(x)}{x^{2}}\bigg)=3$$      
     $$\displaystyle\lim _{x\rightarrow 0} \dfrac{f(x)}{x^{2}}=2\implies f(0)=0\ \ \ \ \ (\because $$ for applying L-hospital rule$$)$$
     Apply L-Hospital Rule
$$\implies \dfrac{f''(0)}{2}=2\implies f''(0)=4$$
Let $$f'(x)=(x-1)(x-2)(x-a)$$
   $$\implies f''(x)=(x-1)(x-2)+(x-2)(x-a)+(x-1)(x-a)$$
    $$f''(0)=2+2{a}+a=4\implies a=\dfrac{2}{3}$$
$$\implies f'(x)=(x-1)(x-2)(x-\dfrac{2}{3})=x^{3}-\dfrac{11}{3}x^{2}+4{x}-\dfrac{4}{3}$$
$$f(x)=\dfrac{x^{4}}{4}-\dfrac{11}{9}x^{3}+2{x^{2}}-\dfrac{4}{3}x+d$$
$$f(0)=0\implies d=0$$
$$f(2)=4-\dfrac{88}{9}+8-\dfrac{8}{3}=-\dfrac{4}{9}$$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image