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Model Paper 3 — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 32 Marks
Question
Find the domain and range of the real valued function $f(x)=\frac{1}{\sqrt{16-x^2}}$.
✓

Answer

We know the square of a real number is never negative. 
Clearly, $f ( x )$ takes real values only when $16- x ^2 \geq 0$
$\begin{array}{l}=16 \geq x ^2 \\ = x ^2 \leq 16 \\ = x ^2-16 \leq 0 \\ = x ^2-4^2 \leq 0 \\ =( x +4)( x -4) \leq 0 \\ = x \geq-4 \text { and } x \leq 4 \\ \therefore x \in[-4,4]\end{array}$
In addition, $f ( x )$ is also undefined when $16- x ^2=0$ because denominator will be zero and the result will be indeterminate.
$\begin{array}{l}16-x^2=0 \Rightarrow x= \pm 4 \\
\text { Hence, } x \in[-4,4]-\{-4,4\} \\
\therefore x \in(-4,4)\end{array}$
Thus, domain of $f=(-4,4)$
$\begin{array}{l}\text { Let } f ( x )= y \\ \Rightarrow \frac{1}{\sqrt{16-x^2}}=y \\ \Rightarrow\left(\frac{1}{\sqrt{16-x^2}}\right)^2=y^2 \\ \Rightarrow \frac{1}{16-x^2}=y^2 \\ =1=\left(16-x^2\right) y ^2 \\ =1=16 y^2-x^2 y^2 \\ =x^2 y^2+1-16 y^2=0 \\ =\left(y^2\right) x^2+(0) x +\left(1-16 y^2\right)=0\end{array}$
As $x \in R$, the discriminant of this quadratic equation in $x$ must be non-negative.
$\begin{array}{l}=0^2-4\left(y^2\right)\left(1-16 y^2\right) \geq 0 \\ =-4 y^2\left(1-16 y^2\right) \geq 0 \\ =4 y^2\left(1-16 y^2\right) \leq 0 \\ =1-16 y^2 \leq 0\left[\because y^2 \geq 0\right] \\ =16 y^2-1 \geq 0 \\ \Rightarrow(4 y)^2-1^2 \geq 0 \\ =(4 y+1)(4 y-1) \geq 0 \\ =4 y \leq-1 \text { and } 4 y \geq \\ \Rightarrow y \leq-\frac{1}{4} \text { and } y \geq \frac{1}{4} \\ \Rightarrow y \in\left(-\infty,-\frac{1}{4}\right] \cup\left[\frac{1}{4}, \infty\right) \\ \Rightarrow f(x) \in\left(-\infty,-\frac{1}{4}\right] \cup\left[\frac{1}{4}, \infty\right)\end{array}$
However, y is always positive because it is the reciprocal of a non-zero square root. 
$\therefore f(x) \in\left[\frac{1}{4}, \infty\right)$
Thus, range of $f =\left[\frac{1}{4}, \infty\right)$
Thus , is the required domain and range of the function

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