A) \[\frac{4\sqrt{2}}{3a}\]
B) 2
C) \[\frac{1}{12a}\]
D) None of these
Correct Answer: A
Solution :
Let \[u={{x}^{2}}+{{y}^{2}},\] and \[{{y}'}'=a{{e}^{x}}+b{{e}^{-x}}\], then \[y'=\frac{-\,a\sin (\log x)}{x}+\frac{b\cos (\log x)}{x}\]. Again diff. w.r.t. x, we get \[y={{\sin }^{2}}\alpha +{{\cos }^{2}}(\alpha +\beta )+2\sin \alpha \sin \beta \cos (\alpha +\beta )\]= \[\frac{1}{3a}\,\left( \frac{{{\sec }^{4}}t}{\sin t} \right)\] \[\therefore \,\,{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{t=\pi /4}}=\frac{1}{3a}.\frac{4}{1/\sqrt{2}}=\frac{4\sqrt{2}}{3a}\].
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