Discussion

Explanation:

Let us multiply and divide the given expression with √x 

The given expression becomes $\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}}}\times \sqrt{\mathrm{x}+\sqrt{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{\frac{\mathrm{x}+\sqrt{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}}{x}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{\frac{\mathrm{x}}{\mathrm{x}}+\frac{\sqrt{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}}{\mathrm{x}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\frac{\sqrt{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}}{\mathrm{x}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{\frac{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}{{\mathrm{x}}^2}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{\frac{{\mathrm{x}}^2}{{\mathrm{x}}^2}+\frac{\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}{{\mathrm{x}}^2}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{1+\sqrt{\frac{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}{{\mathrm{x}}^4}}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{1+\sqrt{\frac{{\mathrm{x}}^4}{{\mathrm{x}}^4}+\frac{\sqrt{{\mathrm{x}}^8+...\infty }}{{\mathrm{x}}^4}}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\frac{{\mathrm{x}}^8+...\infty }{x^8}}}}}$

Finally, we will get the following expression

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...\infty }}}}}$   ...(1)

Let y = $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...\infty }}}}}$

⇒ y = $\sqrt{1+y}$

⇒ y² = 1 + y

⇒ y² - y - 1 = 0

⇒ y = $\frac{-(-1)\pm \sqrt{{(-1)}^2-4\times 1\times -1}}{2\times 1}$ = $\frac{1+\sqrt{5}}{2}$ [we will accept only positive value]

From (1), the original expression becomes = $\sqrt{x}\times \left(\frac{1+\sqrt{5}}{2}\right)$

Hence, option (b).

» Your doubt will be displayed only after approval.