Consider a function `f:[0,pi/2]->R` given by `f(x)=sin x and g:[0,pi/2 – InterviewSolution

1.	Consider a function `f:[0,pi/2]->R` given by `f(x)=sin x and g:[0,pi/2]->R` given by `g(x)=cos x.` Show that f and g are one-one, but `f+g` is not one-one.
Answer» one-one: `a_1!=a_2`, then `f(a_1)!xf(a_2)` `f(a_1)=f(a_2)`,then `(a_1=a_2)` `sqrt2(sinx/sqrt2+cosx/sqrt2)` `=sqrt2(sinxcos45+cosxsin45)` `=sqrt2(sin(A+B)` `=sqrt2sin(x+45)` `a_1!=a_2` `f(g)(a_1)=(f+g)(a_2)` this function is one-one.

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