Question

Prove that the following function

$f(x) = logx - {tan}^{ - 1} \: x \: is \: always \: strictly \: increasing$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = logx - {tan}^{ - 1}x$

Let first define the domain. As logx is defined when x > 0

Now, On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx} f(x) = \dfrac{d}{dx}[logx - {tan}^{ - 1}x]$

$\rm \: f'(x) = \dfrac{1}{x} - \dfrac{1}{1 + {x}^{2} }$

$\rm \: f'(x) = \dfrac{ {x}^{2} + 1 - x}{x( {x}^{2} + 1)}$

can be further rewritten as

$\rm \: f'(x) = \dfrac{ {x}^{2}- x + 1}{x( {x}^{2} + 1)}$

We know,

If in a quadratic expression f(x) = ax² + bx + c such that, a > 0 and Discriminant, D = b² - 4ac < 0, then f(x) > 0

Now, here

x² - x + 1 is such that coefficient of x² = 1 > 0 and Discriminant, D = 1 - 4 × 1 × 1 = 1 - 4 = - 3 < 0

So,

$\rm\implies \:\rm \: {x}^{2} - x + 1 > 0$

and

$\rm \: {x}^{2} + 1 > 0$

$\rm \: As \: x \: \in \: (0, \infty ), \: so \: x > 0$

$\rm\implies \:\dfrac{ {x}^{2} - x + 1 }{x( {x}^{2} + 1) } > 0$

$\rm\implies \:f'(x) > 0$

$\rm\implies \:f(x) \: is \: strictly \: increasing.$

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$