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I have attached the pic of the questiona along with my solution written below.. the correct answer is zero which doesn't match with my ans, requesting you to please check where am I going wrong..

I have attached the pic of the  questiona along with my solution written below.. the correct answer is zero which doesn't match with my ans, requesting you to please check where am I going wrong..

Question Image
Grade:12

1 Answers

Aditya Gupta
2075 Points
one year ago
this is the definition of a conceptual doubt.
lets take an eg. if x belongs to (pi/2, pi), then what is the value of sin-1(sinx) ?? clearly it cant be x since the range of sin-1 is [ – pi/2, pi/2] while x lies in (pi/2, pi) acc to our assumption. hence we need to write sin-1(sinx)= sin-1(sin(pi – x))= pi – x, clearly pi – x lies in (0, pi/2) which is a subset of [ – pi/2, pi/2].
now, instead of letting x= cos^2y, we first define cos-1(sqrtx)= y.....(1)
given that x lies in (0, 1/2) or sqrtx lies in (0, 1/sqrt2) or y=cos-1(sqrtx) lies in (pi/4, pi/2).
from (1), we get x= cos^2y.
so that f(x)= 2sin-1(|siny|) + sin-1(2|sinycosy|) (using sqrt(t^2)= |t| for all real t)
since y lies in (pi/4, pi/2), both siny and cosy are positive (bcoz ½ is less than pi/2). however, 2y lies in (pi/2, pi). or pi – 2y lies in (0, pi/2).
so, f(x)= 2sin-1(siny) + sin-1(sin2y)
= 2y + sin-1(sin(pi – 2y))
= 2y + pi – 2y
f(x) pi
so f’(x)= 0
KINDLY APPROVE :))

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