Application of Derivatives 4 Question 46

####48. 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: 48. $6: 6+\pi$

Solution:

  1. $f(x)$ 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 $

$\Rightarrow \quad(4 x-b)^{2}=b^{2}-1$

$\Rightarrow \quad(4 x-b)^{2}=(b-1)(b+1) \quad[b \geq 0$, 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^{2}}{4} \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 $x=\beta=\frac{1}{4}\left(b+\sqrt{b^{2}-1}\right)$



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