Question

The range of the function $f(x) = \tan \sqrt {\dfrac{{{\pi ^2}}}{9} - {x^2}}$ is
$\left( 1 \right){\text{ }}\left[ {0,3} \right]$
$\left( 2 \right){\text{ }}\left[ {0,\sqrt 3 } \right]$
$\left( 3 \right){\text{ }}\left( { - \infty ,\infty } \right)$
$\left( 4 \right)$ None of these

Answer 1

Take the value under the root greater than equal to zero $\left( { \geqslant 0} \right)$ . Then make the coefficient of ${x^2}$ positive . Then find the domain of the function. By using the domain of the function, find the value of range of the function.

Complete step by step answer:
The given function is $f(x) = \tan \sqrt {\dfrac{{{\pi ^2}}}{9} - {x^2}}$ .
Because the value under the root is always greater than equal to zero $\left( { \geqslant 0} \right)$ due to which the value under the root should be positive . So , we can say that
$\dfrac{{{\pi ^2}}}{9} - {x^2} \geqslant 0$
On multiplying the left side by negative sign $' - '$ the coefficient of ${x^2}$ becomes positive and the inequality sign changes as shown below
${x^2} - \dfrac{{{\pi ^2}}}{9} \leqslant 0$
Now the left hand side term is of the form $\left( {{a^2} - {b^2}} \right)$ . Therefore , by applying the formula $\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$ we get
$\left( {x - \dfrac{\pi }{3}} \right)\left( {x + \dfrac{\pi }{3}} \right) \leqslant 0$
Therefore , $\left( {x - \dfrac{\pi }{3}} \right) \leqslant 0$ and $\left( {x + \dfrac{\pi }{3}} \right) \leqslant 0$
$\Rightarrow {\text{ x}} \leqslant \dfrac{\pi }{3}$ and ${\text{x}} \leqslant - \dfrac{\pi }{3}$
From this we can say that $x \in \left[ { - \dfrac{\pi }{3},\dfrac{\pi }{3}} \right]$ . This is the domain of the function .
Now , $x$ lies from $- \dfrac{\pi }{3}$ to $\dfrac{\pi }{3}$ . Therefore $- \dfrac{\pi }{3} \leqslant x \leqslant \dfrac{\pi }{3}$ . On squaring we get ,
$0 \leqslant {x^2} \leqslant \dfrac{{{\pi ^2}}}{9}$
Where $0$ is the minimum value and $\dfrac{{{\pi ^2}}}{9}$ is the maximum value . Now multiply by negative sign by the which inequality sign changes as shown below
$0 \geqslant - {x^2} \geqslant - \dfrac{{{\pi ^2}}}{9}$
Again on adding $\dfrac{{{\pi ^2}}}{9}$ we get
$\dfrac{{{\pi ^2}}}{9} \geqslant - {x^2} + \dfrac{{{\pi ^2}}}{9} \geqslant 0$
On applying square root we get
$\dfrac{\pi }{3} \geqslant \sqrt {\dfrac{{{\pi ^2}}}{9} - {x^2}} \geqslant 0$
We need the value of $\tan$ and $\tan$ is an increasing function . Therefore inequalities will not be changed. $\therefore$ on multiplying the above equation by $\tan$ we get
$\tan \dfrac{\pi }{3} \geqslant \tan \sqrt {\dfrac{{{\pi ^2}}}{9} - {x^2}} \geqslant \tan 0$
The value of $\tan \dfrac{\pi }{3}$ is $\sqrt 3$ and that of $\tan 0$ is $0$
$\therefore$ $\sqrt 3 \geqslant \tan \sqrt {\dfrac{{{\pi ^2}}}{9} - {x^2}} \geqslant 0$
Hence, the Range of the function is $\left[ {0,\sqrt 3 } \right]$ .
Thus , the correct option is $\left( 2 \right){\text{ }}\left[ {0,\sqrt 3 } \right]$.

Note:
When the derivative of a function is always positive then that function is increasing in its domain. $f(x) = \tan x$ is an increasing function in $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ . The range is the resulting values that the dependent variable can have as $x$ varies throughout the domain. Whereas the domain of a function is the specific set of values that the independent variable in a function can take on .