Question

Let \[f:R\to R\] be a function such that $f(x)=ax+3\sin x+4\cos x$. Then $f(x)$ is invertible if,
A. \[a\in (-5,5)\]
B. \[a\in (-\infty ,-5)\]
C. \[a\in (0,+\infty )\]
D. None of these.

Answer 1

Hint: We know that a function will be invertible ${{f}^{-1}}$ only if it is a one to one function and a function is one to one if ${{f}^{'}}(x)\ge 0$ for all $x$or ${{f}^{'}}(x)\le 0$ for all $x$. We will first differentiate the function and find ${{f}^{'}}(x)$ and then calculate ${{f}^{'}}(x)\ge 0$ and ${{f}^{'}}(x)\le 0$ to determine $a$.

Complete step by step solution: We are given a function $f(x)=ax+3\sin x+4\cos x$ when \[f:R\to R\] and we have to determine when the function $f(x)$ will be invertible.
We know that the inverse ${{f}^{-1}}$ will exist only when a function is one to one and a function is one-to one if ${{f}^{'}}(x)\ge 0$ for all $x$ or ${{f}^{'}}(x)\le 0$ for all $x$.
So we will now calculate ${{f}^{'}}(x)$ by differentiating function $f(x)=ax+3\sin x+4\cos x$,
${{f}^{'}}(x)=a+3\cos x-4\sin x$
We will now calculate ${{f}^{'}}(x)\ge 0$ and ${{f}^{'}}(x)\le 0$ and determine $a$.
First we will take ${{f}^{'}}(x)\ge 0$.
$\begin{align}
  & a+3\cos x-4\sin x\ge 0 \\
 & a\ge 4\sin x-3\cos x
\end{align}$
From the above expression we can say that the value of $a$ should always be greater than $4\sin x-3\cos x$ .
Now we know that the maximum value of $a\sin x+b\cos x$ is $a\sin x+b\cos x=\sqrt{{{a}^{2}}+{{b}^{2}}}$. So,
$\begin{align}
  & a\ge 4\sin x-3\cos x \\
 & a\ge \sqrt{{{4}^{2}}+{{3}^{2}}} \\
 & a\ge \sqrt{25} \\
 & a\ge 5 \\
\end{align}$
We will now take${{f}^{'}}(x)\le 0$.
\[\begin{align}
  & a+3\cos x-4\sin x\le 0 \\
 & a\le 4\sin x-3\cos x
\end{align}\]
From the above expression we can say that the value of $a$ should always be less than $4\sin x-3\cos x$ .
And we know that the minimum value of $a\sin x+b\cos x$ is $a\sin x+b\cos x=-\sqrt{{{a}^{2}}+{{b}^{2}}}$. So,
$\begin{align}
  & a\le 4\sin x-3\cos x \\
 & a\le -\sqrt{{{4}^{2}}+{{3}^{2}}} \\
 & a\le -\sqrt{25} \\
 & a\le -5 \\
\end{align}$
Now the value of $a$ will be either $a\ge 5$or $a\le -5$ that is $a\in (-\infty ,-5)\cup (5,\infty )$.
The function $f(x)=ax+3\sin x+4\cos x$ is invertible if $a\in (-\infty ,-5)\cup (5,\infty )$. Hence the correct option is (B).

Note: A one to one function can be defined as a type of function in which each of the input from the domain has a unique output or exactly one output.
A function can be easily determined one to one by the horizontal line test. If there is no horizontal line intersecting the graph of function in more than one point, then that function is one to one.