Answer

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**Hint:**We here have been given the a function $y={{\sin }^{2}}x+4\sin x++5$ and we need to find its range and domain. For this we will first see what kind of functions are there combined in y and what values of x are valid for it. Hence, we will get the domain of y. Then, as we can see that y is a quadratic equation, we will assume sinx to be equal to t and then convert it into a complete square and then we will keep the value of t back to sinx. Then, we will get the range of y by putting the values of sinx in the completed square term. The lower limit of the range will be given at the point sinx is minimum and the upper limit here sinx is maximum. Hence, we will get the range of y. Then we will see which of the given options match the value of calculated domain and range and hence we will get our answer.

**Complete step by step answer:**

Here, we have been given the function $y={{\sin }^{2}}x+4\sin x++5$ and we need to find its domain and range.

Now, here we can see that ‘y’ is a quadratic equation in sinx, i.e. in a trigonometric function. We know that the domain of both a polynomial function and a trigonometric function is all real numbers, i.e. $\left( -\infty ,\infty \right)$.

Here also, if we keep any value of x, only the value of sinx will vary but since all values of sinx are real, x can take any real value. Hence, the domain of y is all real numbers. Thus we can say that:

$D:\left( -\infty ,\infty \right)$

Now, we have to find the range of y. We will do this by completing the square of y.

This is done as follows:

We have been given that:

$y={{\sin }^{2}}x+4\sin x+5$

Now, let us assume sinx to be equal to ‘t’. Thus, we get:

$y={{t}^{2}}+4t+5$

Now, we can write 5=4+1. Thus, we get:

$\begin{align}

& y={{t}^{2}}+4t+5 \\

& \Rightarrow y={{t}^{2}}+4t+4+1 \\

& \Rightarrow y={{\left( t \right)}^{2}}+2\left( 2 \right)\left( t \right)+{{\left( 2 \right)}^{2}}+1 \\

\end{align}$

Now, we can see that this is in the form of ${{a}^{2}}+2ab+{{b}^{2}}$ which is equal to ${{\left( a+b \right)}^{2}}$. Thus we get the value of y as:

$\begin{align}

& y={{\left( t \right)}^{2}}+2\left( 2 \right)\left( t \right)+{{\left( 2 \right)}^{2}}+1 \\

& \Rightarrow y={{\left( t+2 \right)}^{2}}+1 \\

\end{align}$

Now, putting t=sinx back in y, we get:

$\begin{align}

& y={{\left( t+2 \right)}^{2}}+1 \\

& \Rightarrow y={{\left( \sin x+2 \right)}^{2}}+1 \\

\end{align}$

Now, we know that the minimum value of sinx is -1 and the maximum value is 1.

Hence, the minimum value of y will also be at sinx=-1. Thus the minimum value of y is:

$\begin{align}

& y={{\left( \sin x+2 \right)}^{2}}+1 \\

& \Rightarrow y={{\left( -1+2 \right)}^{2}}+1 \\

& \Rightarrow y=1+1 \\

& \Rightarrow y=2 \\

\end{align}$

Hence, the minimum value of y is 2.

Now, the maximum value of y will be at sinx=1. Thus the maximum value of y is given as:

$\begin{align}

& y={{\left( \sin x+2 \right)}^{2}}+1 \\

& \Rightarrow y={{\left( 1+2 \right)}^{2}}+1 \\

& \Rightarrow y=9+1 \\

& \Rightarrow y=10 \\

\end{align}$

Hence, the maximum value of y is 10.

Thus, we get the range of y as [2,10]

Hence, we can say that:

R: [2,10]

Now, that we have our D and R, we can see that option (D) is the only correct option.

**So, the correct answer is “Option D”.**

**Note:**We here do not need to assume sinx as t to change y into a complete square. We just did it so that we can assume the complete sinx as a different variable as when there is sinx, people may confuse and take only x as the variable and not the complete sinx. We can do it either way according to our convenience just as long the complete sinx is considered as the variable.

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