Problem :

Consider a function $f: [0, \frac{\pi}{2}] \to \mathbb R$ given by $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \to \mathbb R$ given by $g(x) =\cos x$. Then which of the following statement(s) are correct

(a) $f$ and $g$ are one-one

(b) $f$ is one one but $g$ is not one-one

(c) $f +g$ is one-one

(d) $f+g$ is not one-one

My working :

Taking horizontal line test for one-one function (which states, if a horizontal line cuts a graph at only one point, then the function will be a one one function) Therefore, $\sin x$ in the restricted domain which is $[0,\frac{\pi}{2}]$ is a one-one function.

Also $\cos x$ function in similar manner is one-one function in the given domain, therefore, option (a) is true.

I am confused with the graph of $\sin x + \cos x$ in the given domain (I am unable to draw a graph here, as I don't know how to draw this graph using latex, so please excuse. or guide how to draw this graph) . The graph of $\sin x + \cos x$ doesn't satisfy the horizontal line test in my opinion...please suggest whether I am right or wrong...

Also suggest how to draw a graph of $f(x) + g(x)$ here in LaTeX.. Thanks a lot...


You are correct. Both $f$ and $g$ are one-to-one on the given intervals.

Note that: $$ (f+g)(0)=\sin0+\cos0=0+1=1+0=\sin\pi/2+\cos\pi/2=(f+g)(\pi/2) $$ Hence, $f+g$ is not one-to-one. Here's a graph that verifies that $f+g$ indeed fails the horizontal line test. Note that by using trig identites we can rewrite $f+g$ as: $$ \sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right) $$ which suggests that $f+g$ has an axis of symmetry at $x=\pi/4$.


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