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 $\left[ {0,\dfrac{3}{{\sqrt 2 }}} \right]$ .