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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
If $\int_{}^{} {(\sin 2x - \cos 2x)} \;dx = \frac{1}{{\sqrt 2 }}\sin (2x - a) + b$, then
  • A
    $a = \frac{\pi }{4},\;b = 0$
  • B
    $a = - \frac{\pi }{4},\;b = 0$
  • C
    $a = \frac{{5\pi }}{4},\;b = $any constant
  • ✓
    $a = - \frac{{5\pi }}{4},\;b = $any constant
✓

Answer

Correct option: D.
$a = - \frac{{5\pi }}{4},\;b = $any constant
d
(d)$\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 - \left[ {\frac{1}{{\sqrt 2 }}\sin 2x + \frac{1}{{\sqrt 2 }}\cos 2x} \right] = \sin (2x - a) + b\sqrt 2 $
$ \Rightarrow \sin \left( {2x + \frac{{5\pi }}{4}} \right) = \sin (2x - a) + b\sqrt 2 $
$ \Rightarrow b$ is any constant and $a = \frac{{ - 5\pi }}{4}$.

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