Question

Let {{E}_{1}}=\left\\{ x\in R:x\ne 1,\dfrac{x}{x-1}=0 \right\\} and {{E}_{2}}=\left\\{ x\in {{E}_{1}}:{{\sin }^{-1}}\left( {{\log }_{e}}\left( \dfrac{x}{x-1} \right) \right)\text{ is a real number} \right\\} .(Here the inverse of sine function assumes the value in the interval $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$). Let the function $f:{{E}_{1}}\to R$ is defined by $f\left( x \right)={{\log }_{e}}\dfrac{x}{x-1}$ and $g:{{E}_{1}}\to R$ be defined as $g\left( x \right)={{\sin }^{-1}}\left( {{\log }_{e}}\dfrac{x}{x-1} \right)$ $LIST-1$
P. The range of $f$ is Q. The range of $g$ contains
R.The domain of $f$contains$S. The domain of g is$
LIST-II 1.$\left( -\infty ,\dfrac{1}{e} \right]\bigcup \left[ \dfrac{e}{e-1},\infty \right)$
2.\left( 0,1 \right)$$$$$ 3.\left[ \dfrac{-1}{2},\dfrac{1}{2} \right] 4.$\left( -\infty ,0 \right)\bigcup \left( 0,\infty \right)
5.\left( -\infty ,\dfrac{e}{e-1} \right]$$$$$ 6.\left( -\infty ,0 \right)\bigcup \left( \dfrac{1}{2},\dfrac{e}{e-1} \right) $$$$ Choose the correct option$$$$ A.P\to 4,Q\to 2,R\to 1,S\to 1 B. $P\to 3,Q\to 3,R\to 6,S\to 5
C. P\to 4,Q\to 2,R\to 1,S\to 6$$$$$ D. P\to 4,Q\to 3,R\to 6,S\to 5$$$$$

Answer 1

As ${{E}_{1}}$ and ${{E}_{2}}$ are the domains of the function $f$ and $g$ you can express them in intervals using the conditions inside them. Then you proceed to find ranges of $f$ and $g$ as given within their definition.$$$$

Complete step-by-step answer:
As given in the question ${{E}_{1}}$ is defined as
{{E}_{1}}=\left\\{ x\in R:x\ne 1,\dfrac{x}{x-1}>0 \right\\}
So we can proceed with given condition inside the bracket,

& \dfrac{x}{x-1}>0 \\\ & \Rightarrow x\in \left( -\infty ,0 \right)\left( 1,\infty \right) \\\ \end{aligned}$$ we obtain that $x$ is contained in the interval $\left( -\infty ,0 \right)\bigcup \left( 1,\infty \right)$ which is the domain of the function $f$. .....(1) The expression $\dfrac{x}{x-1}$ inside ${{E}_{1}}$ is not defined only at $x=1$ and also $\left| \dfrac{x}{x-1} \right|>0$ . Then $x$ lies in the interval ${{R}^{+}}-\\{0\\}$. So the range of $f\left( x \right)={{\log }_{e}}\dfrac{x}{x-1}$ is $\left( -\infty ,0 \right)\bigcup \left( 0,\infty \right)$....(2) $$$$ Also we get from the definition of ${{E}_{2}}$ $${{E}_{2}}=\left\\{ x\in {{E}_{1}}:{{\sin }^{-1}}\left( {{\log }_{e}}\left( \dfrac{x}{x-1} \right) \right)\text{ is a real number} \right\\}$$ Again we can proceed with given condition inside the bracket with data from the question that inverse of sine is well defined because sine takes the value from the interval$\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$. Using the fact that the domain of any sine inverse function is contained in between -1 to 1, $$\begin{aligned} & -1\le {{\log }_{e}}\left( \dfrac{x}{x-1} \right)\le 1 \\\ & \Rightarrow \dfrac{1}{e}\le \dfrac{x}{x-1}\le e \\\ & \Rightarrow \dfrac{x}{x-1}-\dfrac{1}{e}\ge 0\text{ and }\dfrac{x}{x-1}-e\le 0 \\\ & \Rightarrow \dfrac{\left( e-1 \right)x=1}{e\left( x-1 \right)}\ge 0\text{ and }\dfrac{\left( e-1 \right)x-e}{x-1}\ge 0 \\\ & \Rightarrow x\in \left( -\infty ,\dfrac{1}{e} \right]\bigcup \left( 1,\infty \right)\text{ and }x\in \left( -\infty ,1 \right]\bigcup \left[ \dfrac{e}{e-1},\infty \right) \\\ \end{aligned}$$ Combing above intervals we obtain that $x$ is contained in the interval $\left( -\infty ,\dfrac{1}{e} \right]\bigcup \left[ \dfrac{e}{e-1},\infty \right)$ which is the domain of the function $g$ ...(3)$$$$ The function $g$ is defined as $g\left( x \right)={{\sin }^{-1}}\left( {{\log }_{e}}\dfrac{x}{x-1} \right)$. The expression ${{\log }_{e}}\dfrac{x}{x-1}$ inside sine inverse function is not defined only at $x=1$ and also the range of the sine inverse function is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$. So the range of $g$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}$..(4) $$$$ From the conclusions (1),(2),(3),(4) and the list-I and list-II, we get that $P\to 4,S\to 1$. The range of $g$ contains only $2.\left( 0,1 \right)$ from list-II and domain of $f$contains only $1.\left( -\infty ,\dfrac{1}{e} \right]\bigcup \left[ \dfrac{e}{e-1},\infty \right)$ Hence the correct choice is $P\to 4,Q\to 2,R\to 1,S\to 1$.$$$$ **So, the correct answer is “Option A”.** **Note:** We need to be careful of inequalities while determining domains and ranges of functions. It is important to not to confuse between the domains of inverse trigonometric functions and note that the base of logarithmic function here is $e$.