Question 89

If $$\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=194$$, then what is the value of x?

Solution

Given : $$\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=194$$

=> $$\frac{(x+\sqrt{x^2-1})^2+(x-\sqrt{x^2-1})^2}{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}=194$$

=> $$\frac{(x^2+x^2-1+2x\sqrt{x^2-1})+(x^2+x^2-1-2x\sqrt{x^2-1})}{x^2-(x^2-1)}=194$$

=> $$\frac{4x^2-2}{1}=194$$

=> $$4x^2=194+2=196$$

=> $$x^2=\frac{196}{4}=49$$

=> $$x=\sqrt{49}=7$$

=> Ans - (C)

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: