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Let $f(x)=\log|x|$ and $g(x) =\sin x$. If $A$ is the range of $f(g(x))$ and $B$ is the range of $g(f(x))$ then $A\cap B$ is

  1. $[-1,0]$
  2. $[-1,0)$
  3. $[-\infty ,0]$
  4. $[-\infty ,1]$
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2 Answers

Best answer
10 votes
10 votes

f(g(x)) = log|g(x)| =log |sin x|

  0\leq |sin x| \leq1

which is given to input for log function

and we know in domain of [0,1] log will give range (- \infty ,0]

so here range of f(g(x)) = (- \infty ,0]

g(f(x)) = sin(f(x))  = sin(log|x|)

log|x| will gives values in between - \infty to + \infty

and which is given the input for sin function

-1\leq sinx \leq 1

so range of g(f(x)) is [-1,1]

common of both is [-1,0]

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1 votes
1 votes
I think it should be A as the log|sinx| { |sinx| vary from 1 to 0 } so range of log|sinx| is -inf to 0
for sin(log|x|) we have log|x| range over Real number so sin(x) range over -1 to 1 and hence intersection is
[-1,0]
Answer:

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