Question

If $g\left(x\right)={\int }_{\sin x}^{\sin \left(2x\right)}{\sin }^{-1}\left(t\right)dt$, then

• A. $g^{{}^{\prime }}\left(\frac{\pi }{2}\right)=-2\pi$
• B. $g^{{}^{\prime }}(-\frac{\pi }{2})=-2\pi$
• C. $g^{{}^{\prime }}(-\frac{\pi }{2})=2\pi$
• D. $g^{{}^{\prime }}\left(\frac{\pi }{2}\right)=2\pi$

Answer 1

Answer: (A), (C)

The correct options are
A $g^{{}^{\prime }}\left(\frac{\pi }{2}\right)=-2\pi$
C $g^{{}^{\prime }}(-\frac{\pi }{2})=2\pi$

Remember $si{n}^{-1}\left(sinx\right)=x\forall x\in [-\pi ,\pi ]$

$g\left(x\right)={\int }_{sinx}^{sin\left(2x\right)}si{n}^{-1}\left(t\right)dt$

$\therefore g^{\prime }\left(x\right)=si{n}^{-1}\left(sin2x\right)\times cos2x\times 2-si{n}^{-1}\left(sinx\right)\times cosx$

$⟹g^{\prime }\left(\frac{\pi }{2}\right)=si{n}^{-1}\left(sin\right(2\frac{\pi }{2}\left)\right)\times cos\left(2\frac{\pi }{2}\right)\times 2-si{n}^{-1}\left(sin\frac{\pi }{2}\right)\times cos\frac{\pi }{2}$

$g^{\prime }\left(\frac{\pi }{2}\right)=si{n}^{-1}\left(sin\pi \right)\times cos\pi \times 2-si{n}^{-1}\left(sin\frac{\pi }{2}\right)\times cos\frac{\pi }{2}$

$⟹g^{\prime }\left(\frac{\pi }{2}\right)=-2\pi$

Also, $g^{\prime }(-\frac{\pi }{2})=si{n}^{-1}\left(sin\right(2\frac{-\pi }{2}\left)\right)\times cos\left(2\frac{-\pi }{2}\right)\times 2-si{n}^{-1}\left(sin\frac{-\pi }{2}\right)\times cos\frac{-\pi }{2}$

$g^{\prime }(-\frac{\pi }{2})=si{n}^{-1}\left(sin\right(-\pi \left)\right)\times cos(-\pi )\times 2-si{n}^{-1}\left(sin\frac{-\pi }{2}\right)\times cos\frac{-\pi }{2}$

$⟹g^{\prime }(-\frac{\pi }{2})=2\pi$