Question

Let $f:X\to \left[ 1,27 \right]$ be a function defined by $f\left( x \right)=5\sin x+12\cos x+14$. Find the set X so that f is both one – one and onto?

Answer 1

Assume the given expression $f\left( x \right)=5\sin x+12\cos x+14$ as $f\left( x \right)=a\cos x+b\sin x+c$ and convert it into a function containing the single trigonometric function by using the conversion$a\cos x+b\sin x=\sqrt{{{a}^{2}}+{{b}^{2}}}\cos \left( x-{{\tan }^{-1}}\left( \dfrac{b}{a} \right) \right)$. Now, for the function to be one – one substitutes the argument of the cosine function in the range $\left[ 0,\pi \right]$ and finds the value of x. Now, find the range of $f\left( x \right)$ and check if it is equal to the co – domain $\left[ 1,27 \right]$ or not. If they are equal then it is onto for the above obtained range of x. this range of x will be our answer.

Complete step by step answer:
Here we have been provided with the function $f\left( x \right)=5\sin x+12\cos x+14$ which is defined for $f:X\to \left[ 1,27 \right]$. We are asked to find the set X such that the given function is both one – one and onto. First we need to simplify the function such that it contains a single trigonometric function.
Now, an expression of the form $a\cos x+b\sin x$ can be written as $\sqrt{{{a}^{2}}+{{b}^{2}}}\cos \left( x-{{\tan }^{-1}}\left( \dfrac{b}{a} \right) \right)$ which contains single trigonometric function, i.e. the cosine function. So we have,
$\begin{aligned} & \Rightarrow f\left( x \right)=\sqrt{{{5}^{2}}+{{12}^{2}}}\cos \left( x-{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right)+14 \\\ & \Rightarrow f\left( x \right)=13\cos \left( x-{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right)+14 \\\ \end{aligned}$
(1) We know that one – one function is a type of function where for each value of x we have only one value of $f\left( x \right)$. We know that the value of the cosine function oscillates in the range $\left[ -1,1 \right]$. As we travel from 0 to $\pi$ it covers all the values in the range $\left[ -1,1 \right]$. After that it will again start to take the same values for different values of x, so for the above function to be one – one the argument of the cosine function must lie in the range $\left[ 0,\pi \right]$. Therefore we have,
$\begin{aligned} & \Rightarrow 0\le x-{{\tan }^{-1}}\left( \dfrac{5}{12} \right)\le \pi \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{5}{12} \right)\le x\le \pi +{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \\\ & \therefore x\in \left[ {{\tan }^{-1}}\left( \dfrac{5}{12} \right),\pi +{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right] \\\ \end{aligned}$
So, for the above range of x the function is one – one.
(2) Now, a function is onto if and only if its range is equal to the co – domain. The co – domain for the given function $f\left( x \right)$ is provided $\left[ 1,27 \right]$. We need to determine its range. We know that the value of cosine function ranges from -1 to 1, so we have,
$\Rightarrow -1\le \cos \left( x-{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right)\le 1$
Multiplying both the sides with 13 we get,
$\Rightarrow -13\le 13\cos \left( x-{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right)\le 13$
Adding 14 in each term we get,

& \Rightarrow -13+14\le 13\cos \left( x-{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right)+14\le 13+14 \\\ & \Rightarrow 1\le f\left( x \right)\le 27 \\\ & \therefore f\left( x \right)\in \left[ 1,27 \right] \\\ \end{aligned}$$ Clearly we can see that the range of $f\left( x \right)$ is equal to its co – domain, therefore $f\left( x \right)$ is onto for the above obtained range of x. **Hence, the set X for which the given function is both one – one and onto is given by $X\in \left[ {{\tan }^{-1}}\left( \dfrac{5}{12} \right),\pi +{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right]$** **Note:** One important point you may note is that the function $f\left( x \right)$ is not onto for the above obtained range of X only but it is for all real values of x. however, it is one – one for the above obtained range of X only so we have to consider the intersection of the two sets which gives $X\in \left[ {{\tan }^{-1}}\left( \dfrac{5}{12} \right),\pi +{{\tan }^{-1}}\left( \dfrac{5}{12} \right) \right]$ as the answer. Remember the conversion formula $a\cos x+b\sin x=\sqrt{{{a}^{2}}+{{b}^{2}}}\cos \left( x-{{\tan }^{-1}}\left( \dfrac{b}{a} \right) \right)$ because it is very helpful in these types of question where you have to determine the maxima of minima values.