CBSE 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 $16- x ^2 \geq 0$
$=16 \geq x ^2$
$= x ^2 \leq 16$
$= x ^2-16 \leq 0$
$= x ^2-4^2 \leq 0$
$=( x +4)( x -4) \leq 0$
$= x \geq-4$ and $x \leq 4$
$\therefore x \in[-4,4]$
In addition, $f ( x )$ is also undefined when $16- x ^2=0$ because denominator will be zero and the result will be indeterminate.
$16-x^2=0$
$\Rightarrow x= \pm 4$
Hence, $x \in[-4,4]-\{-4,4\}$
$\therefore x \in(-4,4)$
Thus, domain of $f=(-4,4)$
Let $f ( x )= y$
$\Rightarrow \frac{1}{\sqrt{16-x^2}}=y$
$\Rightarrow\left(\frac{1}{\sqrt{16-x^2}}\right)^2=y^2$
$\Rightarrow \frac{1}{16-x^2}=y^2$
$=1=\left(16-x^2\right) y ^2$
$=1=16 y^2-x^2 y^2$
$=x^2 y^2+1-16 y^2=0$
$=\left(y^2\right) x^2+(0) x +\left(1-16 y^2\right)=0$
As $x \in R$, the discriminant of this quadratic equation in $x$ must be non$-$negative.
$=0^2-4\left(y^2\right)\left(1-16 y^2\right) \geq 0$
$=-4 y^2\left(1-16 y^2\right) \geq 0$
$=4 y^2\left(1-16 y^2\right) \leq 0$
$=1-16 y^2 \leq 0\left[\because y^2 \geq 0\right]$
$=16 y^2-1 \geq 0$
$\Rightarrow(4 y)^2-1^2 \geq 0$
$=(4 y+1)(4 y-1) \geq 0$
$=4 y \leq-1$ and $4 y \geq$
$\Rightarrow y \leq-\frac{1}{4}$ and $y \geq \frac{1}{4}$
$\Rightarrow y \in\left(-\infty,-\frac{1}{4}\right] \cup\left[\frac{1}{4}, \infty\right)$
$\Rightarrow f(x) \in\left(-\infty,-\frac{1}{4}\right] \cup\left[\frac{1}{4}, \infty\right)$
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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