Inverse Circular Functions 2 Question 11

11. $E _1=x \in R: x \neq 1$ and $\frac{x}{x-1}> 0 $ and $E _2=x \in E _1: \sin ^{-1} \log _e \frac{x}{x-1}$ is a real number

(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $-\frac{\pi}{2}, \frac{\pi}{2}$ ). Let $f: E _1 \rightarrow R$ be the function defined by $f(x)=\log _e \frac{x}{x-1}$ and $g: E _2 \rightarrow R$ be the function defined by $g(x)=\sin ^{-1} \log _e \frac{x}{x-1}$.

(2018 Adv.)

List I List II
P. The range of $f$ is 1. $-\infty, \frac{1}{1-e} \cup \frac{e}{e-1}, \infty$
Q. The range of $g$ contains 2. $(0,1)$
The domain of $f$
contains
3. $-\frac{1}{2}, \frac{1}{2}$
S. The domain of $g$ is 4. $(-\infty, 0) \cup(0, \infty)$
5. $-\infty, \frac{e}{e-1}$

The correct option is

(a) $P \rightarrow 4 ; Q \rightarrow 2 ; R \rightarrow 1 ; S \rightarrow 1$

(b) $P \rightarrow 3 ; Q \rightarrow 3 ; R \rightarrow 6 ; S \rightarrow 5$

(c) $P \rightarrow 4 ; Q \rightarrow 2 ; R \rightarrow 1 ; S \rightarrow 6$

(d) $P \rightarrow 4 ; Q \rightarrow 3 ; R \rightarrow 6 ; S \rightarrow 5$

Numerical Value Based

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Answer:

Correct Answer: 11. (a)

Solution:

  1. We have,

$ \begin{aligned} & E _1=x \in R: x \neq 1 \text { and } \frac{x}{x-1}>0 \\ \therefore & E _1=\frac{x}{x-1}>0 \\ & E _1=x \in(-\infty, 0) \cup(1, \infty) \end{aligned} $

and

$E _2=x \in E _1: \sin ^{-1} \log _e \frac{x}{x-1}$ is a real number

$ E _2=-1 \leq \log _e \frac{x}{x-1} \leq 1 \Rightarrow $

$ e^{-1} \leq \frac{x}{x-1} \leq e $

Now, $\quad \frac{x}{x-1} \geq e^{-1} \Rightarrow \frac{x}{x-1}-\frac{1}{e} \geq 0$

$\Rightarrow \frac{e x-x+1}{e(x-1)} \geq 0 \Rightarrow \frac{x(e-1)+1}{(x-1) e} \geq 0$

$\Rightarrow x \in-\infty, \frac{1}{1-e} \cup(1, \infty)$

Also,

$ \Rightarrow \quad \frac{(e-1) x-e}{x-1} \geq 0 $

$\Rightarrow x \in(-\infty, 1) \cup \frac{e}{e-1}, \infty$

So, $\quad E _2=-\infty, \frac{1}{1-e} \cup \frac{e}{e-1}, \infty$

$\therefore$ The domain of $f$ and $g$ are

$ -\infty, \frac{1}{1-e} \cup \frac{e}{e-1}, \infty $

and Range of $\frac{x}{x-1}$ is $R^{+}-{1}$

$\Rightarrow$ Range of $f$ is $R-{0}$ or $(-\infty, 0) \cup(0, \infty)$

Range of $g$ is $-\frac{\pi}{2}, \frac{\pi}{2}-{0}$ or $-\frac{\pi}{2}, 0 \cup 0, \frac{\pi}{2}$

Now, $P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 1$



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