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

Hint: 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 .