Question

Determine the points of maxima and minima of the function $f\left( x \right) = \dfrac{1}{8}\ln x - bx + {x^2},x > 0$, where $b \geqslant 0$ is a constant.
$\left( a \right){\text{min}}{\text{.at }}x = \dfrac{1}{4}\left( {b + \sqrt {{b^2} - 1} } \right),{\text{ max at }}x = \dfrac{1}{4}\left( {b - \sqrt {{b^2} - 1} } \right)$
$\left( b \right){\text{min}}{\text{.at }}x = \dfrac{1}{4}\left( {b - \sqrt {{b^2} - 1} } \right),{\text{ max at }}x = \dfrac{1}{4}\left( {b + \sqrt {{b^2} - 1} } \right)$
$\left( c \right){\text{min}}{\text{.at }}x = \dfrac{1}{4}\left( {b + \sqrt {{b^2} + 1} } \right),{\text{ max at }}x = \dfrac{1}{4}\left( {b - \sqrt {{b^2} - 1} } \right)$
$\left( d \right){\text{min}}{\text{.at }}x = \dfrac{1}{4}\left( {b + \sqrt {{b^2} - 1} } \right),{\text{ max at }}x = \dfrac{1}{4}\left( {b - \sqrt {{b^2} + 1} } \right)$

Answer 1

In this particular question to find the maxima and minima differentiate the given function w.r.t x and equate to solve and solve for x, then again differentiate the given function and calculate its value on previous calculated x value if we got positive than it is a minima and if we got negative than it is a maxima so use these concepts to reach the solution of the question.

Complete step by step answer:
Given function.
$f\left( x \right) = \dfrac{1}{8}\ln x - bx + {x^2},x > 0$
Now differentiate this function w.r.t x and equate to zero we have,
$\Rightarrow \dfrac{d}{{dx}}f\left( x \right) = f'\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{8}\ln x - bx + {x^2}} \right) = 0$
Now as we know that $\dfrac{d}{{dx}}\ln x = \dfrac{1}{x},\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$, so use this property in the above equation we have,
$\Rightarrow \left( {\dfrac{1}{{8x}} - b + 2x} \right) = 0$
$\Rightarrow 1 - 8bx + 16{x^2} = 0$
$\Rightarrow 16{x^2} - 8bx + 1 = 0$
Now this is a quadratic equation so apply quadratic formula we have,
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where a = 16, b = -8b, c = 1
$\Rightarrow x = \dfrac{{8b \pm \sqrt {{{\left( { - 8b} \right)}^2} - 4\left( {16} \right)} }}{{2\left( {16} \right)}}$
$\Rightarrow x = \dfrac{{8b \pm \sqrt {64{b^2} - 64} }}{{32}} = \dfrac{{8b \pm 8\sqrt {{b^2} - 1} }}{{32}} = \dfrac{{b \pm \sqrt {{b^2} - 1} }}{4}$
$\Rightarrow x = \dfrac{{b + \sqrt {{b^2} - 1} }}{4},\dfrac{{b - \sqrt {{b^2} - 1} }}{4}$
So as we see that b should never be less than one other the above values of x will become complex.
Therefore b $\geqslant$ 1.
Now again differentiate the function w.r.t x we have,
$\Rightarrow \dfrac{d}{{dx}}f'\left( x \right) = f''\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{{8x}} - b + 2x} \right)$
$\Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{x^{ - 1}}}}{8} - b + 2x} \right)$
$\Rightarrow f''\left( x \right) = \left( {\dfrac{{ - 1{x^{ - 1 - 1}}}}{8} + 2} \right)$, as the differentiation of constant term is zero.
$\Rightarrow f''\left( x \right) = \left( {\dfrac{{ - 1}}{{8{x^2}}} + 2} \right)$
Now when, $x = \dfrac{{b + \sqrt {{b^2} - 1} }}{4}$ so the value of f” (x) is
$\Rightarrow f''\left( x \right) = \dfrac{{ - 1}}{{8{{\left( {\dfrac{{b + \sqrt {{b^2} - 1} }}{4}} \right)}^2}}} + 2$
Now simplify we have,
$\Rightarrow f''\left( x \right) = \dfrac{{ - 16}}{{8{{\left( {b + \sqrt {{b^2} - 1} } \right)}^2}}} + 2$
$\Rightarrow f''\left( x \right) = \dfrac{{ - 2}}{{{{\left( {b + \sqrt {{b^2} - 1} } \right)}^2}}} + 2$
As b $\geqslant$ 1 so the value of $\left( {b + \sqrt {{b^2} - 1} } \right)$ when b = 1 is 1
$\Rightarrow f''\left( x \right) = \dfrac{{ - 2}}{{{{\left( 1 \right)}^2}}} + 2 = 0$
So at b = 1 function neither minima nor maxima.
So when, $1 < b < 2$, the value of $\left( {b + \sqrt {{b^2} - 1} } \right) > 1$
$\Rightarrow \dfrac{2}{{b + \sqrt {{b^2} - 1} }} = \dfrac{2}{{ > 1}} < 2$
$\Rightarrow f''\left( x \right) = \dfrac{{ - 2}}{{{{\left( {b + \sqrt {{b^2} - 1} } \right)}^2}}} + 2$ = positive.
So when, $x = \dfrac{{b + \sqrt {{b^2} - 1} }}{4}$ the given function is at minimum position.
Now when, $x = \dfrac{{b - \sqrt {{b^2} - 1} }}{4}$ so the value of f” (x) is
$\Rightarrow f''\left( x \right) = \dfrac{{ - 1}}{{8{{\left( {\dfrac{{b - \sqrt {{b^2} - 1} }}{4}} \right)}^2}}} + 2$
Now simplify we have,
$\Rightarrow f''\left( x \right) = \dfrac{{ - 16}}{{8{{\left( {b - \sqrt {{b^2} - 1} } \right)}^2}}} + 2$
$\Rightarrow f''\left( x \right) = \dfrac{{ - 2}}{{{{\left( {b - \sqrt {{b^2} - 1} } \right)}^2}}} + 2$
As b $\geqslant$ 1 so the value of $\left( {b - \sqrt {{b^2} - 1} } \right)$ when b = 1 is 1
$\Rightarrow f''\left( x \right) = \dfrac{{ - 2}}{{{{\left( 1 \right)}^2}}} + 2 = 0$
So at b = 1 function neither minima nor maxima.
So when, $1 < b < 2$, the value of $\left( {b - \sqrt {{b^2} - 1} } \right) < 1$
$\Rightarrow \dfrac{2}{{b - \sqrt {{b^2} - 1} }} = \dfrac{2}{{ < 1}} > 2$
$\Rightarrow f''\left( x \right) = \dfrac{{ - 2}}{{{{\left( {b + \sqrt {{b^2} - 1} } \right)}^2}}} + 2$ = negative.
So when, $x = \dfrac{{b - \sqrt {{b^2} - 1} }}{4}$ the given function is at maximum position.
So this is the required answer.

So, the correct answer is “Option A”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation properties such as $\dfrac{d}{{dx}}\ln x = \dfrac{1}{x},\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ so differentiate the given function according to these properties as above.