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**Hint:**For checking whether a function is one-one or not, we need to check its graph. If we draw a horizontal line on the graph, which cuts it at more than one point, then we say that the function is not one—one. But if every possible horizontal line cuts the graph at most one point, then we say that the function is a one-to-one function.

**Complete step-by-step solution:**

According to the question, the function is

$f\left( x \right) = 3\sin 2x + 5$

Considering the graph of the above function, we have

Now, we know that a one-to-one function is a function in which every value of the co domain is the image of at most one value of the domain. This means that every horizontal line drawn on the graph of a one-to-one function to intersect it, will cut it at most on point. So we draw a horizontal line, say

$y = 5$, on the above graph so that we get

As we can see that the horizontal line $y = 5$ is cutting the graph at multiple points. So we can say that the given function is not a one-to-one function.

Now, we know that a function is invertible if and only if it is injective as well as subjective, that is, it is one-to-one as well as onto. But since we have shown above that the given function is not one-to-one, so it is not invertible.

**Hence, the inverse of the given function does not exist.**

**Note:**

We should keep in mind that the inverse of a function does not mean to simply express x in terms of y. The inverse of a function is itself a function and so it must follow all the properties of a function. Hence, for a function to be invertible, firstly it must be one-to-one as well as onto. The inverse of a function will only exist if it is a one-one function.

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