Answer
Verified
393k+ views
Hint: We can find the domain by finding the value of x that makes the function undefined and subtracting it from the set of real numbers. For that we equate the denominator to 0 and solve for x. For finding the range, we can write the interval of ${x^2}$ . Then we can transform it to the function by addition, multiplication and taking reciprocals, to get the range of the function.
Complete step-by-step answer:
We know that domain of a function is the set of all the possible values that the variable in that function takes. We know that a real valued function is not defined when the denominator becomes zero and the term inside the root is negative.
Here, we have the function $f\left( x \right) = \dfrac{3}{{2 - {x^2}}}$ .
The function has a denominator. So, the range must not have the value which makes the denominator 0.
We can equate the denominator to 0 and solve it to get that value of x.
$ \Rightarrow 2 - {x^2} = 0$
On rearranging, we get,
$ \Rightarrow {x^2} = 2$
On taking the square root, we get,
$ \Rightarrow x = \pm \sqrt 2 $
So, x cannot take the values $ - \sqrt 2 $ and $\sqrt 2 $ .
Therefore, the domain of the function is set of all the real numbers except the numbers $ - \sqrt 2 $ and $\sqrt 2 $ .
$ \Rightarrow D = R - \left\{ {\; - \sqrt 2 ,\sqrt 2 } \right\}$
Range of a function is the set of all the values that the function gives for the values of variables in the domain.
$f\left( x \right) = \dfrac{3}{{2 - {x^2}}}$
We know that, $\infty > {x^2} \geqslant 0$
On multiplying with -1, the equality reverses,
\[ \Rightarrow \] $ - \infty < - {x^2} \leqslant 0$
On adding 2, we get,
\[ \Rightarrow \] $ - \infty < 2 - {x^2} \leqslant 2$
According to the domain of the function, ${x^2} \ne 2$ . So $2 - {x^2}$ cannot take value 0.
$ \Rightarrow - \infty < 2 - {x^2} < 0$ and $0 < 2 - {x^2} < 2$
On taking the reciprocal, we get,
$ \Rightarrow \dfrac{1}{{ - \infty }} > \dfrac{1}{{2 - {x^2}}} > \dfrac{1}{0},\,\,\,\,\dfrac{1}{0} > \dfrac{1}{{2 - {x^2}}} \geqslant \dfrac{1}{2}$
On multiply throughout with 3, we get,
$ \Rightarrow \dfrac{3}{{ - \infty }} > \dfrac{3}{{2 - {x^2}}} > \dfrac{3}{0},\,\,\,\dfrac{3}{0} > \dfrac{3}{{2 - {x^2}}} \geqslant \dfrac{3}{2}$
We know that $\dfrac{1}{\infty } = 0$ and $\dfrac{1}{0} = \infty $
$ \Rightarrow 0 > \dfrac{3}{{2 - {x^2}}} > - \infty ,\,\,\infty > \dfrac{3}{{2 - {x^2}}} \geqslant \dfrac{3}{2}$
So, y can take values in the interval $\left( { - \infty ,0} \right) \cup \left[ {\dfrac{3}{2},\infty } \right)$
Therefore, the domain of the function is $R - \left\{ {\; - \sqrt 2 ,\sqrt 2 } \right\}$ and range is $\left( { - \infty ,0} \right) \cup \left[ {\dfrac{3}{2},\infty } \right)$.
Note: Domain of a function is the set of all the values that the variable can take. Range of a function is the set of all the values that the function gives for the values of variables in the domain. While calculating the range we must consider that in the range of the function ${x^2} \ne 2$ and change the interval accordingly. To verify our answer, we can check what are the values of y when ${x^2}$ tends to infinity, 2, zero and negative infinity.
Complete step-by-step answer:
We know that domain of a function is the set of all the possible values that the variable in that function takes. We know that a real valued function is not defined when the denominator becomes zero and the term inside the root is negative.
Here, we have the function $f\left( x \right) = \dfrac{3}{{2 - {x^2}}}$ .
The function has a denominator. So, the range must not have the value which makes the denominator 0.
We can equate the denominator to 0 and solve it to get that value of x.
$ \Rightarrow 2 - {x^2} = 0$
On rearranging, we get,
$ \Rightarrow {x^2} = 2$
On taking the square root, we get,
$ \Rightarrow x = \pm \sqrt 2 $
So, x cannot take the values $ - \sqrt 2 $ and $\sqrt 2 $ .
Therefore, the domain of the function is set of all the real numbers except the numbers $ - \sqrt 2 $ and $\sqrt 2 $ .
$ \Rightarrow D = R - \left\{ {\; - \sqrt 2 ,\sqrt 2 } \right\}$
Range of a function is the set of all the values that the function gives for the values of variables in the domain.
$f\left( x \right) = \dfrac{3}{{2 - {x^2}}}$
We know that, $\infty > {x^2} \geqslant 0$
On multiplying with -1, the equality reverses,
\[ \Rightarrow \] $ - \infty < - {x^2} \leqslant 0$
On adding 2, we get,
\[ \Rightarrow \] $ - \infty < 2 - {x^2} \leqslant 2$
According to the domain of the function, ${x^2} \ne 2$ . So $2 - {x^2}$ cannot take value 0.
$ \Rightarrow - \infty < 2 - {x^2} < 0$ and $0 < 2 - {x^2} < 2$
On taking the reciprocal, we get,
$ \Rightarrow \dfrac{1}{{ - \infty }} > \dfrac{1}{{2 - {x^2}}} > \dfrac{1}{0},\,\,\,\,\dfrac{1}{0} > \dfrac{1}{{2 - {x^2}}} \geqslant \dfrac{1}{2}$
On multiply throughout with 3, we get,
$ \Rightarrow \dfrac{3}{{ - \infty }} > \dfrac{3}{{2 - {x^2}}} > \dfrac{3}{0},\,\,\,\dfrac{3}{0} > \dfrac{3}{{2 - {x^2}}} \geqslant \dfrac{3}{2}$
We know that $\dfrac{1}{\infty } = 0$ and $\dfrac{1}{0} = \infty $
$ \Rightarrow 0 > \dfrac{3}{{2 - {x^2}}} > - \infty ,\,\,\infty > \dfrac{3}{{2 - {x^2}}} \geqslant \dfrac{3}{2}$
So, y can take values in the interval $\left( { - \infty ,0} \right) \cup \left[ {\dfrac{3}{2},\infty } \right)$
Therefore, the domain of the function is $R - \left\{ {\; - \sqrt 2 ,\sqrt 2 } \right\}$ and range is $\left( { - \infty ,0} \right) \cup \left[ {\dfrac{3}{2},\infty } \right)$.
Note: Domain of a function is the set of all the values that the variable can take. Range of a function is the set of all the values that the function gives for the values of variables in the domain. While calculating the range we must consider that in the range of the function ${x^2} \ne 2$ and change the interval accordingly. To verify our answer, we can check what are the values of y when ${x^2}$ tends to infinity, 2, zero and negative infinity.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE