A) \[12{{x}^{2}}-7x+1=0\]
B) \[{{x}^{2}}+19x-120=0\]
C) \[{{x}^{2}}17x+60=0\]
D) \[{{x}^{2}}-7x+12=0\]
Correct Answer: A
Solution :
[A]\[\alpha =\underset{n\to \infty }{\mathop{\lim }}\,\frac{1+4+9+....+{{n}^{2}}}{{{n}^{3}}+1}\]\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n\,\,(n+1)\,\,(2n+1)}{6\,\,({{n}^{3}}+1)}\] |
\[\beta =\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{\sin 2x}{2x} \right)\,\,\left( \frac{8x}{\sin 8x} \right)\,\,\times \,\,\left[ \frac{2x}{8x} \right]\] |
\[\beta =\frac{1}{4}\] |
\[\alpha \beta =\frac{1}{12}\] |
\[\alpha +\beta =\frac{7}{12}\] |
\[12{{x}^{2}}-7x+1=0\] |
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