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If $g(x) = 1 - x$ and $h(x) = \frac{x}{x-1}$, then $\frac{g(h(x))}{h(g(x))}$ is:

  1. $\frac{h(x)}{g(x)}$
  2. $\frac{-1}{x}$
  3. $\frac{g(x)}{h(x)}$
  4. $\frac{x}{(1-x)^{2}}$
in Set Theory & Algebra by Boss (30.8k points) | 1.6k views
+4

Please edit the question! Correct question is given below:

+2
Thanks. Edited :)

3 Answers

+27 votes
Best answer
option a) is correct.

$g\left(h\left(x\right)\right) =g\left(\frac{x}{x-1}\right)$
    $=1-\frac{x}{x-1}$
    $= \frac{-1}{x-1}$

$h\left(g\left(x\right)\right) =h(1-x)$
  $=\frac{1-x}{-x}$

$\frac{g(h(x))}{h(g(x))} = \frac{x}{(1-x)(x-1)} = \frac{h(x)}{g(x)}$
       

option A)
by Active (1.5k points)
edited by
0
sir how iitk declare ans as a?
+1
Answer is A only. I guess the answer was put for a different question- corrected now :)
0 votes
Correct answer is A because of

(x/x-1)*(1/1-x)

g(x) = 1-x and h(x)=x/x-1

so h(x)/g(x)
by Boss (11.7k points)
–2 votes

The correct answer is (A) h(x) / g(x)

by Loyal (8k points)
Answer:

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