If `f(x)=sinx+cosx `and `g(x)=x^2-1`, then `g(f (x)) `is invertible in the domain .A. `[0, pi//2]`B. `[-pi//4, pi//4]`C. `[-pi//4, pi//2]`D. `[0, pi]`

If `f(x)=sinx+cosx `and `g(x)=x^2-1`, then `g(f (x)) `is invertible in

1.	If `f(x)=sinx+cosx `and `g(x)=x^2-1`, then `g(f (x)) `is invertible in the domain .A. `[0, pi//2]`B. `[-pi//4, pi//4]`C. `[-pi//4, pi//2]`D. `[0, pi]`
Answer» Correct Answer - B Clearly, range `(f) = [-1,1] sub` domina (g)= R `therefore" " gof: R to R` is given by `" "gof(x) = g(f(x))` `" "gof(x) = g(sin x + cosx) = (sin x + cos x)^(2) -1= sin x` For `gof` to be bijective, we must have `-(pi)/(2) le 2x le (pi)/(2)rArr x in [-pi//4,pi//4]` Hence, `gof:[-pi//4,pi//4] uu [-1,1]`

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