Question 64

If $$4x=\sqrt{5}+2$$, then $$x-\frac{1}{16x}$$ ?

Solution

Given : $$4x=\sqrt{5}+2$$

=> $$x=\frac{\sqrt{5}+2}{4}$$

To find : $$x-\frac{1}{16x}$$

= $$\frac{\sqrt{5}+2}{4} - \frac{4}{16(\sqrt{5}+2)}$$

= $$\frac{\sqrt{5}+2}{4}-\frac{1}{4(\sqrt{5}+2)}$$

= $$\frac{(\sqrt{5}+2)^2-1}{4(\sqrt{5}+2)}$$

= $$\frac{5+4+4\sqrt{5}-1}{4(\sqrt{5}+2)}$$

= $$\frac{8+4\sqrt{5}}{8+4\sqrt{5}}=1$$

=> Ans - (A)

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