Question
Let $$f\left( x \right) = \sin x$$ and $$g\left( x \right) = \ln \left| x \right|.$$ If the ranges of the composition functions $$fog$$ and $$gof$$ are $${R_1}$$ and $${R_2}$$ respectively, then
A.
$${R_1} = \left\{ {u: - 1 \leqslant u < 1} \right\},{R_2} = \left\{ {v: - \infty < v < 0} \right\}$$
B.
$${R_1} = \left\{ {u: - \infty < u < 0} \right\},{R_2} = \left\{ {v: - 1 \leqslant v \leqslant 0} \right\}$$
C.
$${R_1} = \left\{ {u: - 1 < u < 1} \right\},{R_2} = \left\{ {v: - \infty < v < 0} \right\}$$
D.
$${R_1} = \left\{ {u: - 1 \leqslant u \leqslant 1} \right\},{R_2} = \left\{ {v: - \infty < v \leqslant 0} \right\}$$
Answer :
$${R_1} = \left\{ {u: - 1 \leqslant u \leqslant 1} \right\},{R_2} = \left\{ {v: - \infty < v \leqslant 0} \right\}$$
Solution :
$$\eqalign{
& {\text{We have }}fog\left( x \right) = f\left( {g\left( x \right)} \right) = \sin \left( {\ln \left| x \right|} \right) \cr
& \therefore {R_1} = \left\{ {u: - 1 \leqslant u \leqslant 1} \right\}\,\left( {\because - 1 \leqslant \sin \theta \leqslant 1,\forall \theta } \right) \cr
& {\text{Also }}gof\,\left( x \right) = g\left( {f\left( x \right)} \right) = \ln \left| {\sin x} \right| \cr
& \because 0 \leqslant \left| {\sin x} \right| \leqslant 1 \cr
& \therefore - \infty < \ln \left| {\sin x} \right| \leqslant 0 \cr
& \therefore {R_2} = \left\{ {v: - \infty < v \leqslant 0} \right\} \cr} $$