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 values of $x$ satisfying the equation:
${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(2x\right)=\frac{\pi }{4}$
we can proceed step-by-step.

Step 1: Rearrange the equation
Subtract ${\tan }^{-1}\left(x\right)$ from both sides of the equation:
${\tan }^{-1}\left(2x\right)=\frac{\pi }{4}-{\tan }^{-1}\left(x\right)$

Step 2: Apply the tangent function to both sides
Taking the tangent on both sides:
$\tan \left({\tan }^{-1}\left(2x\right)\right)=\tan \left(\frac{\pi }{4}-{\tan }^{-1}\left(x\right)\right)$

Using the definition of the inverse function, the left-hand side simplifies to:
$\tan \left({\tan }^{-1}\left(2x\right)\right)=2x$

For the right-hand side, we apply the trigonometric subtraction identity $\tan (A-B)=\frac{\tan \left(A\right)-\tan \left(B\right)}{1+\tan \left(A\right)\tan \left(B\right)}$:
$2x=\frac{\tan \left(\frac{\pi }{4}\right)-\tan \left({\tan }^{-1}\left(x\right)\right)}{1+\tan \left(\frac{\pi }{4}\right)\tan \left({\tan }^{-1}\left(x\right)\right)}$

Substituting the known values $\tan \left(\frac{\pi }{4}\right)=1$ and $\tan \left({\tan }^{-1}\left(x\right)\right)=x$:
$2x=\frac{1-x}{1+x}$

Step 3: Solve the algebraic equation
Multiply both sides by $(1+x)$ to eliminate the denominator:
$2x(1+x)=1-x$
$2x+2x^2=1-x$
Rearranging the terms into standard quadratic form $a{x}^2+bx+c=0$:
$2x^2+3x-1=0$

Step 4: Apply the quadratic formula
Using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a=2$, $b=3$, and $c=-1$:
$x=\frac{-3\pm \sqrt{3^2-4\left(2\right)(-1)}}{2\left(2\right)}$
$x=\frac{-3\pm \sqrt{9+8}}{4}$
$x=\frac{-3\pm \sqrt{17}}{4}$

Step 5: Select the positive root
Since the question specifies finding the positive real value of $x$ (so $x>0$), we discard the negative root and choose the positive one:
$x=\frac{-3+\sqrt{17}}{4}$