Integral Calculus Question 247
Question: If $ \int_{{}}^{{}}{(\sin 2x-\cos 2x)}\ dx=\frac{1}{\sqrt{2}}\sin (2x-a)+b $ , then
[Roorkee 1978; MP PET 2001]
Options:
A) $ a=\frac{\pi }{4},\ b=0 $
B) $ a=-\frac{\pi }{4},\ b=0 $
C) $ a=\frac{5\pi }{4},\ b= $ any constant
D) $ a=-\frac{5\pi }{4},\ b= $ any constant
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Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{(\sin 2x-\cos 2x),dx=\frac{1}{\sqrt{2}}\sin (2x-a)+b} $
$ \Rightarrow -\frac{1}{2}(\sin 2x+\cos 2x)=\frac{1}{\sqrt{2}}\sin (2x-a)+b $
$ \Rightarrow -[ \frac{1}{\sqrt{2}}\sin 2x+\frac{1}{\sqrt{2}}\cos 2x ]=\sin (2x-a)+b\sqrt{2} $
$ \Rightarrow \sin ( 2x+\frac{5\pi }{4} )=\sin (2x-a)+b\sqrt{2} $
$ \Rightarrow b $ is any constant and $ a=\frac{-5\pi }{4} $ .
BETA
