If g(x)=∫sinxsin(2x)sin−1(t)dt,then: (a)gprime(2π)=−2π (b) gprime(−2π)=−2π (c)gprime(−2π)=2π (d) gprime(2π)=2π

Name: Video solution 1: If g(x)=∫sinxsin(2x)sin−1(t)dt,then: (a)gprime(2π)=−2π (b) gprime(−2π)=−2π (c)gprime(−2π)=2π (d) gprime(2π)=2π
Uploaded: 2021-01-01T18:41:02.000Z
Duration: 2 min
Description: Video solution 1: If g(x)=∫sinxsin(2x)sin−1(t)dt,then: (a)gprime(2π)=−2π (b) gprime(−2π)=−2π (c)gprime(−2π)=2π (d) gprime(2π)=2π

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If $g (x) = \int_{s i n x}^{sin (2 x)} sin^{- 1} (t) d t, t h e n :$ (a) $g^{p r i m e} (\frac{π}{2}) = - 2 π$ (b) $g^{p r i m e} (- \frac{π}{2}) = - 2 π$ (c) $g^{p r i m e} (- \frac{π}{2}) = 2 π$ (d) $g^{p r i m e} (\frac{π}{2}) = 2 π$

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g (x) = \int_{s i n x}^{sin (2 x)} sin^{- 1} (t) d t, t h e n :

(a)

g^{p r i m e} (\frac{π}{2}) = - 2 π

(b)

g^{p r i m e} (- \frac{π}{2}) = - 2 π

(c)

g^{p r i m e} (- \frac{π}{2}) = 2 π

(d)

g^{p r i m e} (\frac{π}{2}) = 2 π

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Question Text	If $g (x) = \int_{s i n x}^{sin (2 x)} sin^{- 1} (t) d t, t h e n :$ (a) $g^{p r i m e} (\frac{π}{2}) = - 2 π$ (b) $g^{p r i m e} (- \frac{π}{2}) = - 2 π$ (c) $g^{p r i m e} (- \frac{π}{2}) = 2 π$ (d) $g^{p r i m e} (\frac{π}{2}) = 2 π$
Answer Type	Video solution: 1
Upvotes	150

If g(x)=∫sinxsin(2x)​sin−1(t)dt,then: (a)gprime(2π​)=−2π (b) gprime(−2π​)=−2π (c)gprime(−2π​)=2π (d) gprime(2π​)=2π

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If $g (x) = \int_{s i n x}^{sin (2 x)} sin^{- 1} (t) d t, t h e n :$ (a) $g^{p r i m e} (\frac{π}{2}) = - 2 π$ (b) $g^{p r i m e} (- \frac{π}{2}) = - 2 π$ (c) $g^{p r i m e} (- \frac{π}{2}) = 2 π$ (d) $g^{p r i m e} (\frac{π}{2}) = 2 π$