Arts (English Medium) इयत्ता ११ - CBSE Question Bank Solutions

If the A.M. of two positive numbers a and b (a > b) is twice their geometric mean. Prove that:

\[a : b = (2 + \sqrt{3}) : (2 - \sqrt{3}) .\]

If one A.M., A and two geometric means G₁ and G₂ inserted between any two positive numbers, show that \[\frac{G_1^2}{G_2} + \frac{G_2^2}{G_1} = 2A\]

\[\lim_{x \to 0} \frac{\sqrt{1 + x + x^2} - 1}{x}\]

\[\lim_{x \to 0} \frac{2x}{\sqrt{a + x} - \sqrt{a - x}}\] 

\[\lim_{x \to 0} \frac{\sqrt{a^2 + x^2} - a}{x^2}\] 

\[\lim_{x \to 0} \frac{\sqrt{1 + x} - \sqrt{1 - x}}{2x}\]

\[\lim_{x \to 2} \frac{\sqrt{3 - x} - 1}{2 - x}\] 

\[\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\] 

\[\lim_{x \to 0} \frac{x}{\sqrt{1 + x} - \sqrt{1 - x}}\] 

\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x - 1}\] 

\[\lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 + 3 - 2}}\] 

\[\lim_{x \to 3} \frac{\sqrt{x + 3} - \sqrt{6}}{x^2 - 9}\] 

\[\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^2 - 1}\] 

\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] 

\[\lim_{x \to 2} \frac{\sqrt{x^2 + 1} - \sqrt{5}}{x - 2}\] 

\[\lim_{x \to 2} \frac{x - 2}{\sqrt{x} - \sqrt{2}}\] 

\[\lim_{x \to 7} \frac{4 - \sqrt{9 + x}}{1 - \sqrt{8 - x}}\] 

\[\lim_{x \to 0} \frac{\sqrt{a + x} - \sqrt{a}}{x\sqrt{a^2 + ax}}\]

\[\lim_{x \to 5} \frac{x - 5}{\sqrt{6x - 5} - \sqrt{4x + 5}}\]