Question

The range of the function \[y = 3\sin \left( {\sqrt {\dfrac{{{\pi ^2}}}{{16}} - {x^2}} } \right)\] is
1) \[\left[ {0,\sqrt {\dfrac{3}{2}} } \right]\]
2) [0, 1]
3) \[\left[ {0,\dfrac{3}{{\sqrt 2 }}} \right]\]
4) \[[0,\infty ]\]

Answer 1

Hint: A function relates an input to an output. Domain is defined as the entire set of values possible for independent variables. The Range is found after substituting the possible x- values to find the y-values. Here, we are given a function and we need to find the range of the given function. For this, we will find the domain of the function. And then, substitute that value in place of x to get the final output.

Complete step-by-step answer:
Given that,
\[y = 3\sin \left( {\sqrt {\dfrac{{{\pi ^2}}}{{16}} - {x^2}} } \right)\]
Here, we will find the range of this function, as below
\[\dfrac{{{\pi ^2}}}{{16}} - {x^2} \geqslant 0\]
\[ \Rightarrow \dfrac{{{\pi ^2}}}{{16}} \geqslant {x^2}\]
\[ \Rightarrow {x^2} \leqslant \dfrac{{{\pi ^2}}}{{16}}\]
\[ \Rightarrow 0 \leqslant {x^2} \leqslant \dfrac{{{\pi ^2}}}{{16}}\]
Thus, here the range of the domain is \[\left[ {0,\dfrac{\pi }{4}} \right]\] .
Let \[y = f(x) = 3\sin \left( {\sqrt {\dfrac{{{\pi ^2}}}{{16}} - {x^2}} } \right)\] -------- (1)
First, we will find y-min as below:
\[{y_{\min }}\] will be min when\[\dfrac{{{\pi ^2}}}{{16}}\] should be min.
This means that \[{x^2}\] should be maximum.
\[ \Rightarrow {x^2}\] will be max at \[\dfrac{{{\pi ^2}}}{{16}}\] .
\[\therefore {x^2} = \dfrac{{{\pi ^2}}}{{16}}\] (i.e. we are using \[x = \dfrac{\pi }{4}\])
Thus, substituting this value in equation (1), we will get,
\[\therefore {y_{\min }} = 3\sin \left( {\sqrt {\dfrac{{{\pi ^2}}}{{16}} - \dfrac{{{\pi ^2}}}{{16}}} } \right)\]
\[ \Rightarrow {y_{\min }} = 3\sin (0)\]
\[ \Rightarrow {y_{\min }} = 3(0)\] \[(\because \sin (0) = 0)\]
\[ \Rightarrow {y_{\min }} = 0\]
Next, to find y-max as below:
\[{y_{\max }}\] will be max, when \[\dfrac{{{\pi ^2}}}{{16}}\] will be max.
This means that x2 will be mine.
\[ \Rightarrow {x^2} = 0\] (i.e. we are using x = 0 here)
Thus, substituting this value in equation (1), we will get,
\[\therefore {y_{\max }} = 3\sin \left( {\sqrt {\dfrac{{{\pi ^2}}}{{16}} - 0} } \right)\]
\[ \Rightarrow {y_{\max }} = 3\sin \left( {\dfrac{\pi }{4}} \right)\]
\[ \Rightarrow {y_{\max }} = 3 \times \dfrac{1}{{\sqrt 2 }}\] \[\left( {\because \sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}} \right)\]
\[ \Rightarrow {y_{\max }} = \dfrac{3}{{\sqrt 2 }}\]
Thus, \[y \in \left[ {0,\dfrac{3}{{\sqrt 2 }}} \right]\]
So, the correct answer is “Option 3”.

Note: Let, \[f(x) = y = 3\sin \left( {\sqrt {\dfrac{{{\pi ^2}}}{{16}} - {x^2}} } \right)\]
Thus, here the range of the domain is \[\left[ {0,\dfrac{\pi }{4}} \right]\] .
First, we will substitute x = 0 in the given equation:
\[\therefore f(0) = 3\sin \left( {\sqrt {\dfrac{{{\pi ^2}}}{{16}} - {{(0)}^2}} } \right)\]
\[ \Rightarrow f(0) = 3\sin \left( {\sqrt {\dfrac{{{\pi ^2}}}{{16}}} } \right)\]
\[ \Rightarrow f(0) = 3\sin \left( {\dfrac{\pi }{4}} \right)\]
\[ \Rightarrow f(0) = 3 \times \dfrac{1}{{\sqrt 2 }}\] \[\left( {\because \sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}} \right)\]
\[ \Rightarrow f(0) = \dfrac{3}{{\sqrt 2 }}\]
Next, we will substitute \[x = \dfrac{\pi }{4}\] in the given equation:
Similarly, we will get, \[f(\dfrac{\pi }{4}) = \dfrac{3}{{\sqrt 2 }}\] .
Hence, the range of the given function is \[\left[ {0,\dfrac{3}{{\sqrt 2 }}} \right]\] .