Considering the principal values of inverse trigonometric functions, the positive real values of ‘x’ satisfying tan^-1(x) + tan^-1(2x) = $\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$ is :

The correct option is (-3 + √17)/4.

To find the positive real value of x that satisfies the given equation, we start with the original relation:

${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(2x\right)=\frac{\pi }{4}$

We apply the formula for the sum of two inverse tangent functions, ${\tan }^{-1}\left(A\right)+{\tan }^{-1}\left(B\right)={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$:

${\tan }^{-1}\left(\frac{x+2x}{1-x·2x}\right)=\frac{\pi }{4}$

Taking the tangent of both sides, we get:

$\frac{3x}{1-2x^2}=\tan \left(\frac{\pi }{4}\right)$

Since $\tan \left(\frac{\pi }{4}\right)=1$, the equation becomes:

$\frac{3x}{1-2x^2}=1$

Cross-multiplying yields:

$3x=1-2x^2$

Rearranging this expression into standard quadratic form:

$2x^2+3x-1=0$

Now, we solve for x using the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, with $a=2$, $b=3$, and $c=-1$:

$x=\frac{-3\pm \sqrt{3^2-4\left(2\right)(-1)}}{2\left(2\right)}$

Simplifying inside the radical:

$x=\frac{-3\pm \sqrt{9+8}}{4}$

$x=\frac{-3\pm \sqrt{17}}{4}$

Because the question specifies that x must be a positive real value, and $\sqrt{17}\approx 4.12>3$, we discard the negative root and choose the positive value:

$x=\frac{-3+\sqrt{17}}{4}$