1) If g(x)=∫sin(2x)sinxsin−1(t)dt , then A) g′(−π2)=2π B) g′(−π2)=−2π C) g′(π2)=2π D) g′(π2)=−2π Answer: Option *Explanation:g(x)=∫sin(2x)sinxsin−1(t)dt g′(x)=2cos2xsin−1(sin2x)−cosxsin−1(sinx) g′(π2)=−2sin−1(0)=0 g′(−π2)=−2sin−1(0)=0 no option is matching
