Question

The range of the function \[f\left( x \right) = \frac{x + 2}{\left| x + 2 \right|}\],x ≠ −2 is

 

पर्याय

• (a) {−1, 1}

• (b) {−1, 0, 1}

• (c) {1}

• (d) (0, ∞)

   

Answer 1

(a) {−1, 1}

\[f\left( x \right) = \frac{x + 2}{\left| x + 2 \right|}\]  , x ≠ −2

\[\text{ Let } y = \frac{x + 2}{\left| x + 2 \right|}\]
\[\text{ For }  \left| x + 2 \right| > 0, \]
\[\text{ or } x > - 2 , \]
\[y = \frac{x + 2}{x + 2} = 1\]
\[\text{ For }  \left| x + 2 \right| < 0, \]
\[\text{ or }  x < - 2, \]
\[y = \frac{x + 2}{- (x + 2)} = - 1\]
\[\text{ Thus } , y = { - 1, 1}\]
\[\text{ or range }  f(x) = { - 1, 1} .\]