Question

Let\[f(x) = \sin x\]and \[g(x) = {\log _e}\left| x \right|\], If the ranges of the composition function \[fog\]and\[gof\]are \[{R_1}\]and\[{R_2}\]respectively, then
A. \[{R_1}\]=\[\left\{ {u: - 1 \le u < 1} \right\}\], \[{R_2}\]=\[\left\{ {v: - \infty < v < 0} \right\}\]
B. \[{R_1}\]=\[\left\{ {u: - \infty < u < 0} \right\}\],\[{R_2} = \left\{ {v: - \infty < v < 0} \right\}\]
C. \[{R_1}\]=\[\left\{ {u: - 1 < u < 1} \right\}\], \[{R_2} = \left\{ {v: - \infty < v < 0} \right\}\]
D. \[{R_1}\]=\[\left\{ {u: - 1 \le u \le 1} \right\}\],\[{R_2} = \left\{ {v: - \infty < v \le 0} \right\}\]

Answer 1

Hint: When we put the value of \[g(x)\]in \[f(x)\]we get \[fog\]and when we put the value of\[f(x)\]in \[g(x)\]we get \[gof\]


Complete step by step solution: Given that \[f(x) = \sin x\]and \[g(x) = {\log _e}\left| x \right|\], Now find \[fog\]and\[gof\]with domain and range
1st \[fog\]=\[f(gx) = \sin (gx) = \sin ({\log _e}\left| x \right|)\]
We know \[\left| x \right| \in R\], So \[{\log _e}\left| x \right| \in R\]
And \[\sin x\]vary from – 1 to 1, So for \[fog\]range \[{R_1}\]=\[\left[ { - 1,1} \right]\] --------(1)
Now, 2nd \[gof\]=\[g(fx) = g(\sin x) = {\log _e}\left| {\sin x} \right|\]
We know the value of \[\left| {\sin x} \right|\]vary from 0 to 1, So the range of \[{\log _e}\left| {\sin x} \right| = - \infty < {\log _e}\left| {\sin x} \right| \le 0\]
Hence range of \[{R_2}\]=\[{\log _e}\left| {\sin x} \right| = \left[ { - \infty ,0} \right]\] ------(2)
Taking both the equation we get \[{R_1}\]=\[\left\{ {u: - 1 \le u \le 1} \right\}\],\[{R_2} = \left\{ {v: - \infty < v \le 0} \right\}\]


Thus, Option (D) is correct.

Note:most of the student Students put wrong values to find composition function like andbecause of misunderstanding of the functions. So student must know the ranges of basic functions like modulus, linear, trigonometric functions, and understanding of composition function.