Question

If the sum of roots of equation $ {{x}_{1}}=-3\,\,\,and\,\,\,{{y}_{1}}=4 $ is equal to sum of squares of their reciprocals, then $ \underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{x+1}{x+2} \right)}^{2x+1}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1-\frac{1}{x+2} \right)}^{2x+1}} $ and $ =\underset{x\to \infty }{\mathop{\lim }}\,{{\left[ {{\left( 1-\frac{1}{x+2} \right)}^{x+2}} \right]}^{\frac{2x+1}{x+2}}} $ are in

• A. GP
• B. HP
• C. AP
• D. None of these

Answer 1

Answer: (C)

Let roots of equation $ {{Q}_{1}}{{R}_{2}}\ne {{Q}_{2}}{{R}_{1}} $ are $ {{Q}_{1}}{{R}_{2}}={{Q}_{2}}{{R}_{1}} $ and $ s=\frac{{{t}^{2}}}{4} $ Then, $ T\propto V $ and $ T\propto {{V}^{2}} $ According to the question, $ T\propto \frac{1}{{{V}^{2}}} $ $ T\propto \frac{1}{V} $ $ \text{6}\times \text{1}{{0}^{-\text{7}}}\text{A}-{{\text{m}}^{\text{2}}} $ $ \text{5 g}/\text{c}{{\text{m}}^{\text{3}}} $ $ \text{8}\text{.3}\times \text{1}{{0}^{\text{6}}} $ $ \text{1}.\text{2}\times \text{1}{{0}^{-\text{7}}} $ $ \text{3}\times \text{1}{{0}^{-\text{6}}} $ $ CaC{{l}_{2}} $ $ \text{MgS}{{\text{O}}_{\text{4}}} $ $ \text{MgS}{{\text{O}}_{\text{4}}} $ $ CaC{{l}_{2}} $ Hence, $ A\to B,B\to C $ and $ C\to A $ are in AP.