SATHEE: Application of Derivatives 4 Question 46

Application of Derivatives 4 Question 46

####48. Determine the points of maxima and minima of the function $f (x) = \frac{1}{8} In x - b x + x^{2}, x > 0$ , where $b \geq 0$ is a constant.

(1996, 5M)

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Answer:

Correct Answer: 48. $6 : 6 + π$

Solution:

$f (x)$ is a differentiable function for $x > 0$ .

Therefore, for maxima or minima, $f^{'} (x) = 0$ must satisfy.

Given,

$f (x) = \frac{1}{8} \ln x - b x + x^{2}, x > 0$

$\Rightarrow f^{'} (x) = \frac{1}{8} \cdot \frac{1}{x} - b + 2 x$

$For f^{'} (x) = 0$

$\Rightarrow \frac{1}{8 x} - b + 2 x = 0$

$\Rightarrow 16 x^{2} - 8 b x + 1 = 0$

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

$\Rightarrow (4 x - b)^{2} = (b - 1) (b + 1) [b \geq 0$ , 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 $b = 1$ ,

$f^{'} (x) = \frac{1}{8 x} - 1 + 2 x = \frac{2}{x} x^{2} - \frac{1}{2} x + \frac{1}{16} = \frac{2}{x} x - {\frac{1}{4}}^{2}$

We have to check the sign of $f^{'} (x)$ at $x = 1 / 4$ .

Interval	Sign of $f^{'} (x)$	Nature of $f (x)$
$- \infty, 0$	$- v e$	$↓$
$0, \frac{1}{4}$	$+ ve$	$↑$
$\frac{1}{4}, \infty$	$+ ve$	$↑$

From sign chart, it is clear that $f^{'} (x)$ has no change of sign in left and right of $x = 1 / 4$ .

Case III When $b > 1$ , then

$\begin{aligned} f^{'} (x) & = \frac{1}{8 x} - b + 2 x = \frac{2}{x} x^{2} - \frac{1}{2} b x + \frac{1}{16} \\ = \frac{2}{x} x - \frac{b^{2}}{4} - \frac{1}{16} (b^{2} - 1) \\ = \frac{2}{x} x - \frac{b}{4} - \frac{1}{4} \sqrt{b^{2} - 1} x - \frac{b}{4} + \frac{1}{4} \sqrt{b^{2} - 1} \\ = \frac{2}{x} (x - α) (x - β) \end{aligned}$

where, $α < β$ and $α = \frac{1}{4} (b - \sqrt{b^{2} - 1})$ and

$β = \frac{1}{4} (b + \sqrt{b^{2} - 1})$ . 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 $x = α$

$= \frac{1}{4} (b - \sqrt{b^{2} - 1})$

and $f (x)$ has a minima at $x = β = \frac{1}{4} (b + \sqrt{b^{2} - 1})$

Application of Derivatives 4 Question 46

Answer:

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Application of Derivatives 4 Question 46

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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