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Find the range of the following functions given by (i) f(x)=\frac{3}{2-x^{2}} (ii) f(x)=1-|x-2|
CBSE | Class 11 | Maths | NCERT Exemplar | Relations and Functions
Find the range of the following functions given by (i) f(x)=\frac{3}{2-x^{2}} (ii) f(x)=1-|x-2|

Solution:

(i)  f(x)=\frac{3}{2-x^{2}}

As per the question,

Let \mathrm{f}(\mathrm{x})=\mathrm{y},

\begin{array}{l} y=\frac{3}{2-x^{2}} \\ \Rightarrow 2-x^{2}=\frac{3}{y} \\ \Rightarrow x^{2}=2-\frac{3}{y} \end{array}

But, it is known that, x^{2} \geq 0

\begin{array}{r} 2-\frac{3}{y} \geq 0 \\ \Rightarrow \frac{2 y-3}{y} \geq 0 \end{array}

\Rightarrow y>0 and 2 y-3 \geq 0

\Rightarrow y>0 and 2 y \geq 3

\Rightarrow y>0 and y \geq 3 / 2

Or f(x)>0 and f(x) \geq 3 / 2

\begin{array}{l} f(x) \in(-\infty, 0) \cup[3 / 2, \infty) \\ \Rightarrow f(x) \in(-\infty, 0) \cup\left[\frac{3}{2}, \infty\right) \end{array}

As a result, the range of f=(-\infty, 0) \cup[3 / 2, \infty)

(ii) f(x)=1-|x-2|

As per the question,

For real value of f,

|x-2| \geq 0

When we add negative sign, we have

Or -|x-2| \leq 0

When we add 1 we have

\Rightarrow 1-|x-2| \leq 1

Or f(x) \leq 1

\Rightarrow f(x) \in(-\infty, 1]

As a result, the range of f=(-\infty, 1]

i

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