Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 52 Marks
Question
Find the domain and the range of the real function: $f(x)=\frac{1}{\sqrt{x^2-1}}$
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Answer
Here we have, $f(x)=\frac{1}{\sqrt{x^2-1}}$ we need to find where the function is defined The condition for the function to be defined $\begin{array}{l}x^2-1>0 \\ \Rightarrow x^2>1 \\ \Rightarrow x>1\end{array}$ So, the domain of the function is the set of all the real numbers greater than 1 The domain of the function, $D _{\{ f ( x )\}}=(1, \infty)$ Now put any value of x within the domain set we get the value of the function always a fraction whose denominator is not equalled to 0 The range of the function, $R _{ f ( x )}=(0,1)$.
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