If gx- -sin xsin 2xsin -1t dt- then

Given, nbsp;g(x)= ∫_sin x^sin 2xsin ^ 1(t) dt, Differentiating w.r.t. x, we get g'(x)=sin^ 1 (sin 2x)· 2cos 2x sin^ 1 (sin x)·cos x ⇒ g'(π2)=0·(2)( 1) π2·0 ⇒ ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

If gx= ∫sin x...

Question

$If g (x) = sin 2 x \int sin x {sin}^{- 1} (t) d t, then$

g^{'} (\frac{π}{2}) = - 2 π

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g^{'} (- \frac{π}{2}) = 2 π

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g^{'} (\frac{π}{2}) = 2 π

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g^{'} (- \frac{π}{2}) = - 2 π

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None of the above

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Solution

The correct option is E None of the above
$g^{'} (x) = {sin}^{- 1} (sin 2 x) .2 cos 2 x - {sin}^{- 1} (sin x) cos x$
$∴ g^{'} (\frac{π}{2}) = 0, g^{'} (- \frac{π}{2}) = 0$

No option is matching.

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If g(x)=sin2x∫sinxsin−1(t) dt, then

The correct option is E None of the aboveg′(x)=sin−1(sin2x).2cos2x−sin−1(sinx)cosx ∴g′(π2)=0, g′(−π2)=0 No option is matching.

$If g (x) = sin 2 x \int sin x {sin}^{- 1} (t) d t, then$

The correct option is E None of the above
$g^{'} (x) = {sin}^{- 1} (sin 2 x) .2 cos 2 x - {sin}^{- 1} (sin x) cos x$
$∴ g^{'} (\frac{π}{2}) = 0, g^{'} (- \frac{π}{2}) = 0$

No option is matching.