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Consider a function f:[0,pi/2]->R given ...

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.

Text Solution

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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`
...
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