Application Of Derivatives Ques 101
101. Determine the points of maxima and minima of the function $f(x)=\frac{1}{8} \operatorname{In} x-b x+x^{2}, x>0$, where $b \geq 0$ is a constant.
(1996, 5M)
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Answer:
Correct Answer: 101.Maxima at $x=\frac{\left(b-\sqrt{b^{2}-1}\right)}{4}$ and minima at $x=\frac{1}{4}\left(b+\sqrt{b^{2}-1}\right)$
Solution:
Formula:
Maxima and Minima of functions of one variable :
- $f(x)$ is a differentiable function for $x>0$.
Therefore, for maxima or minima, $f^{\prime}(x)=0$ must satisfy.
Given,
$ f(x)=\frac{1}{8} \ln x-b x+x^{2}, x>0 $
$\Rightarrow \quad f^{\prime}(x)=\frac{1}{8} \cdot \frac{1}{x}-b+2 x$
$ \text { For } \quad f^{\prime}(x)=0 $
$\Rightarrow \quad \frac{1}{8 x}-b+2 x=0$
$ \Rightarrow \quad 16 x^{2}-8 b x+1=0 $
$\Rightarrow \quad(4 x-b)^{2}=b^{2}-1 ……(i)$
$\Rightarrow \quad(4 x-b)^{2}=(b-1)(b+1) \quad[b \geq 0$, given $]$
Case I $0 \leq b<1$, has no solution. Since, RHS is negative in this domain and LHS is positive.
Case II When $b=1$, then $x=\frac{1}{4}$ is the only solution. When $b=1$,
$ f^{\prime}(x)=\frac{1}{8 x}-1+2 x=\frac{2}{x} (x^{2}-\frac{1}{2} x+\frac{1}{16})=\frac{2}{x} \quad (x-\frac{1}{4})^{2} $
We have to check the sign of $f^{\prime}(x)$ at $x=1 / 4$.
| Interval | Sign of $f^{\prime}(x)$ | Nature of $f(x)$ |
|---|---|---|
| $-\infty, 0$ | $-v e$ | $\downarrow$ |
| $(0, \frac{1}{4})$ | $+\mathrm{ve}$ | $\uparrow$ |
| $(\frac{1}{4}, \infty)$ | $+\mathrm{ve}$ | $\uparrow$ |
From sign chart, it is clear that $f^{\prime}(x)$ has no change of sign in left and right of $x=1 / 4$.
Case III When $b>1$, then
$ \begin{aligned} f^{\prime}(x) & =\frac{1}{8 x}-b+2 x=\frac{2}{x} (x^{2}-\frac{1}{2} b x+\frac{1}{16}) \\ & =\frac{2}{x} \quad [(x-\frac{b}{4})^{2} \quad-\frac{1}{16}\left(b^{2}-1\right)] \\ & =\frac{2}{x} \quad [(x-\frac{b}{4}-\frac{1}{4} \sqrt{b^{2}-1}) \quad (x-\frac{b}{4}+\frac{1}{4} \sqrt{b^{2}-1})] \\ & =\frac{2}{x}(x-\alpha)(x-\beta) \end{aligned} $
where, $\alpha<\beta$ and $\alpha=\frac{1}{4}\left(b-\sqrt{b^{2}-1}\right)$ and
$\beta=\frac{1}{4}\left(b+\sqrt{b^{2}-1}\right)$. From sign scheme, it is clear that
$
0 \text {, for } 0<x<\alpha $
$ \begin{aligned} f^{\prime}(x)<0, & \text { for } \alpha<x<\beta \\
0, & \text { for } x>\beta \end{aligned} $
By the first derivative test, $f(x)$ has a maxima at $x=\alpha$
$ =\frac{1}{4}\left(b-\sqrt{b^{2}-1}\right) $
and $f(x)$ has a minima at $x=\beta=\frac{1}{4}\left(b+\sqrt{b^{2}-1}\right)$
BETA
