Question

Let in the given domain $f:\left[ 0,\dfrac{\pi }{2} \right]\to R:$ we have a function $f(x)=\sin x$ and in domain $g:\left[ 0,\dfrac{\pi }{2} \right]\to R:$ we have $g(x)=\cos x$. Show that each one of f and g is one-one but $\left( f+g \right)$ is not one-one.

Answer 1

Hint: To prove that the given function is one-one, assume two elements ${{x}_{1}}\text{ and }{{x}_{2}}$ in the set of the domain of the given function and show that, if $f({{x}_{1}})=f({{x}_{2}})$ then, ${{x}_{1}}={{x}_{2}}$. If we are getting any other relation between ${{x}_{1}}$ and ${{x}_{2}}$ then the function is not one-one.

Complete step-by-step solution -
For the function $f(x)=\sin x$.
Let us consider any two elements ${{x}_{1}}\text{ and }{{x}_{2}}$ in the domain of $f(x)$. Now, substituting $f({{x}_{1}})=f({{x}_{2}})$, we get,
$\sin {{x}_{1}}=\sin {{x}_{2}}$, we know that in the domain $\left[ 0,\dfrac{\pi }{2} \right]$ the value of sine ranges from 0 to 1, with a particular value for each angle. Therefore, for $\sin {{x}_{1}}=\sin {{x}_{2}}$, we must have ${{x}_{1}}={{x}_{2}}$.
Therefore, $f(x)=\sin x$ is a one-one function.
For the function $g(x)=\cos x$.
Let us consider any two elements ${{x}_{1}}\text{ and }{{x}_{2}}$ in the domain of $g(x)$ . Now, substituting $g({{x}_{1}})=g({{x}_{2}})$, we get,
$\cos {{x}_{1}}=\cos {{x}_{2}}$, we know that in the domain $\left[ 0,\dfrac{\pi }{2} \right]$ the value of cosine ranges from 1 to 0, with a particular value for each angle. Therefore, for $\cos {{x}_{1}}=\cos {{x}_{2}}$, we must have ${{x}_{1}}={{x}_{2}}$.
Therefore, $g(x)=\cos x$ is a one-one function.
Considering the function $\left( f+g \right)$.
$(f+g)=\sin x+\cos x$
Let us consider any two elements ${{x}_{1}}\text{ and }{{x}_{2}}$ in the domain of $\left( f+g \right)$. Now, substituting $(f+g)({{x}_{1}})=(f+g)({{x}_{2}})$, we get,
$\begin{aligned} & \sin {{x}_{1}}+\cos {{x}_{1}}=\sin {{x}_{2}}+\cos {{x}_{2}} \\\ & \Rightarrow \sin {{x}_{1}}-\sin {{x}_{2}}=\cos {{x}_{2}}-\cos {{x}_{1}} \\\ \end{aligned}$
Applying the formula: $\sin a-\sin b=2\sin \left( \dfrac{a-b}{2} \right)\cos \left( \dfrac{a+b}{2} \right)$ and$\cos a-\cos b=2\sin \left( \dfrac{b-a}{2} \right)\sin \left( \dfrac{a+b}{2} \right)$, we get,

& 2\sin \left( \dfrac{{{x}_{1}}-{{x}_{2}}}{2} \right)\cos \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)=2\sin \left( \dfrac{{{x}_{1}}-{{x}_{1}}}{2} \right)\sin \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right) \\\ & 2\sin \left( \dfrac{{{x}_{1}}-{{x}_{2}}}{2} \right)\cos \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)-2\sin \left( \dfrac{{{x}_{1}}-{{x}_{2}}}{2} \right)\sin \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)=0 \\\ & 2\sin \left( \dfrac{{{x}_{1}}-{{x}_{2}}}{2} \right)\left( \cos \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)-\sin \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right) \right)=0 \\\ & \Rightarrow \sin \left( \dfrac{{{x}_{1}}-{{x}_{2}}}{2} \right)\text{ or }\left( \cos \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)-\sin \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right) \right)=0 \\\ \end{aligned}$$ Now, if $$\sin \left( \dfrac{{{x}_{1}}-{{x}_{2}}}{2} \right)=0$$, then $$\dfrac{{{x}_{1}}-{{x}_{2}}}{2}=0$$ because $\sin {{0}^{\circ }}=0$. Therefore, $$\begin{aligned} & {{x}_{1}}-{{x}_{2}}=0 \\\ & \Rightarrow {{x}_{1}}={{x}_{2}} \\\ \end{aligned}$$ Also, if $$\left( \cos \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)-\sin \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right) \right)=0$$, then $$\cos \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)=\sin \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)$$ Dividing both sides by $$\cos \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)$$, we get, $1=\tan \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)$ We know that, $\tan \dfrac{\pi }{4}=1$. Therefore, $\begin{aligned} & \tan \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)=\tan \dfrac{\pi }{4} \\\ & \Rightarrow \left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2} \right)=\dfrac{\pi }{4} \\\ & \Rightarrow {{x}_{1}}+{{x}_{2}}=\dfrac{\pi }{2} \\\ & \Rightarrow {{x}_{1}}=\dfrac{\pi }{2}-{{x}_{2}} \\\ \end{aligned}$ Therefore, we can see that there is another relation between ${{x}_{1}}\text{ and }{{x}_{2}}$ other than ${{x}_{1}}={{x}_{2}}$, for the function $\left( f+g \right)$. Hence, $\left( f+g \right)$ is not one-one. Note: We can easily check that the functions f and g are one-one in the given domain by the help of graph. Also, we can easily say that at $0$ and $\dfrac{\pi }{2}$ the value of the function $\left( f+g \right)$ is 1. Therefore, the function $\left( f+g \right)$ cannot be one-one. But the above solution is the true way of solving the question.