Application of Derivatives 4 Question 15
####16. Let $f(x)$ be a polynomial of degree four having extreme values at $x=1$ and $x=2$. If $\lim _{x \rightarrow 0} 1+\frac{f(x)}{x^{2}}=3$, then $f(2)$ is equal to
(a) -8
(b) -4
(c) 0
(d) 4
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Answer:
Correct Answer: 16. (c)
Solution:
- PLAN Any function have extreme values (maximum or minimum) at its critical points, where $f^{\prime}(x)=2$.
Since, the function have extreme values at $x=1$ and $x=2$.
$ \begin{array}{ll} \therefore & f^{\prime}(x)=0 \text { at } x=1 \text { and } x=2 \\ \Rightarrow & f^{\prime}(1)=0 \text { and } f^{\prime}(2)=0 \end{array} $
Also, it is given that,
$ \begin{array}{rlrl} \Rightarrow & & \lim _{x \rightarrow 0} 1+\frac{f(x)}{x^{2}} & =3 \\ \Rightarrow & 1+\lim _{x \rightarrow 0} \frac{f(x)}{x^{2}} & =3 \\ \Rightarrow & \lim _{x \rightarrow 0} \frac{f(x)}{x^{2}} & =2 \end{array} $
$\Rightarrow \quad f(x)$ will be of the form $a x^{4}+b x^{3}+2 x^{2}$.
$[\because f(x)$ is of four degree polynomial]
$ \begin{aligned} & \text { Let } \quad f(x)=a x^{4}+b x^{3}+2 x^{2} \\ & \Rightarrow \quad f^{\prime}(x)=4 a x^{3}+3 b x^{2}+4 x \\ & \Rightarrow \quad f^{\prime}(1)=4 a+3 b+4=0 \\ & \text { and } \quad f^{\prime}(2)=32 a+12 b+8=0 \\ & \Rightarrow \quad 8 a+3 b+2=0 \end{aligned} $
On solving Eqs. (i) and (ii),
$ \begin{array}{ll} \text { we get } & a=\frac{1}{2}, b=-2 \\ \therefore & f(x)=\frac{x^{4}}{2}-2 x^{3}+2 x^{2} \\ \Rightarrow & f(2)=8-16+8=0 \end{array} $
