A) 2.5
B) \[\frac{x}{2}-\frac{y}{3}=1\]
C) \[\frac{x}{-2}+\frac{y}{1}=1\]
D) \[\frac{x}{2}-\frac{y}{3}=-1\]
Correct Answer: B
Solution :
Clearly, \[{{\sin }^{2}}(A+B)\] is continuous on [2, 3] an differentiable on (2, 3). So, by Lagrange's mean value theorem, ther exists a e (2, 3) such that\[\frac{{{a}^{2}}}{{{a}^{2}}+{{(1-b)}^{2}}}\] \[\frac{{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]\[\frac{{{a}^{2}}}{{{(a+b)}^{2}}}\] \[\frac{{{a}^{2}}}{{{b}^{2}}+{{(1-a)}^{2}}}\] \[\Delta ABC,\]\[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=ac+\sqrt{3}ab,\]\[(a\times b)\times c=\frac{1}{3}|b|\,\,|c|\]\[\theta \]\[\theta \]\[\frac{2\sqrt{2}}{3}\]
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