A) \[a=-\frac{12}{13},b=\frac{15}{39}\]
B) \[a=-\frac{7}{13},b=\frac{6}{13}\]
C) \[a=\frac{12}{13},b=-\frac{15}{39}\]
D) \[a=-\frac{7}{13},b=-\frac{6}{13}\]
Correct Answer: C
Solution :
[c] Differentiating both sides, we get \[\frac{3\sin x+2\cos x}{3\cos x+2\sin x}=\]\[a+\frac{b(2\cos x-3\sin x)}{(2\sin x+3\cos x)}\] \[=\frac{\sin x(2a-3b)+\cos x\,(3a+2b)}{(3\cos x+2\sin x)}\] Comparing like terms on both sides, we get \[3=2a-3b,2=3a+2b\] Or \[a=\frac{12}{13},b=-\frac{15}{39}\]You need to login to perform this action.
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