Question

Total number of values of x, satisfying $(\sqrt{3}+1{)}^{2x}+(\sqrt{3}-1{)}^{2x}=2^{3x}$ , is equal to :

• A. 2
• B. 3
• C. 1
• D. None

Answer 1

The correct option is C

1

$(\sqrt{3}+1{)}^{2x}+(\sqrt{3}-1{)}^{2x}=2^{3x}=(2\sqrt{2}{)}^{2x}$

$\Rightarrow$ ${\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}^{2x}+{\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}^{2x}=1$

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

$\Rightarrow$ x = 1 is a solution.

$Also0<\frac{2+\sqrt{3}}{4}<1$
$,0<$$\frac{2-\sqrt{3}}{4}<1$ $\therefore$$ifx>1$$,{\left(\frac{2+\sqrt{3}}{4}\right)}^x$$<\frac{2+\sqrt{3}}{4}$$,{\left(\frac{2-\sqrt{3}}{4}\right)}^x$$<\frac{2-\sqrt{3}}{4}$

$\Rightarrow$ ${\left(\frac{2+\sqrt{3}}{4}\right)}^x$$+{\left(\frac{2-\sqrt{3}}{4}\right)}^x<1.$

$\therefore$ if x $<$ 1 , ${\left(\frac{2+\sqrt{3}}{4}\right)}^x+{\left(\frac{2-\sqrt{3}}{4}\right)}^x>1.$ Hence x = 1 is the only solution