Question

Let $E_1 = \begin{Bmatrix} x \in R : x \neq 1 \text{ and } \frac{x}{x-1} > 0 \end{Bmatrix}$ and $E_2 = \begin{Bmatrix} x \in E_1 : \sin^{-1} \left( \log_e \left( \frac{x}{x-1} \right) \right) \text{ is a real number} \end{Bmatrix}$; (Here, the inverse trigonometric function $\sin^{-1} x$ assumes values in $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$).

Let $f: E_1 \rightarrow R$ be the function defined by $f(x) = \log_e \left( \frac{x}{x-1} \right)$ and

$g; E_2 \rightarrow R$ be the function defined by $g(x) = \sin^{-1} \left( \log_e \left( \frac{x}{x-1} \right) \right)$. [JEE Advanced 2018]

List - I List - II

(P) The range of $f$ is (1) $\left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right)$

(Q) The range of $g$ contains (2) $(0,1)$

(R) The domain of $f$ contains (3) $\left[ -\frac{1}{2}, \frac{1}{2} \right]$

(S) The domain of $g$ is (4) $(-\infty, 0) \cup (0, \infty)$

(5) $\left( -\infty, \frac{e}{e-1} \right]$

(6) $(-\infty, 0) \cup \left[ \frac{1}{2}, \frac{e}{e-1} \right]$

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

Answer 1

Answer: (A)

P $\rightarrow$ 4; Q $\rightarrow$ 2; R $\rightarrow$ 1; S $\rightarrow$ 1

Answer 2

The problem requires us to determine the domain and range of two functions, $f$ and $g$, which are defined based on sets $E_1$ and $E_2$.

Step 1: Determine the set $E_1$. $E_1 = \begin{Bmatrix} x \in R : x \neq 1 \text{ and } \frac{x}{x-1} > 0 \end{Bmatrix}$

For $\frac{x}{x-1} > 0$, we analyze the signs of the numerator and denominator:

Case 1: $x > 0$ and $x-1 > 0 \implies x > 1$. Case 2: $x < 0$ and $x-1 < 0 \implies x < 0$.

So, $\frac{x}{x-1} > 0$ when $x \in (-\infty, 0) \cup (1, \infty)$. The condition $x \neq 1$ is already satisfied by this interval.

Thus, $E_1 = (-\infty, 0) \cup (1, \infty)$.

Step 2: Determine the set $E_2$. $E_2 = \begin{Bmatrix} x \in E_1 : \sin^{-1} \left( \log_e \left( \frac{x}{x-1} \right) \right) \text{ is a real number} \end{Bmatrix}$

For $\sin^{-1}(y)$ to be a real number, its argument $y$ must satisfy $-1 \le y \le 1$.

Here, $y = \log_e \left( \frac{x}{x-1} \right)$.

So, we need $-1 \le \log_e \left( \frac{x}{x-1} \right) \le 1$.

Since $\log_e$ is an increasing function, we can exponentiate with base $e$:

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

$\frac{1}{e} \le \frac{x}{x-1} \le e$.

We solve two inequalities:

(a) $\frac{x}{x-1} \ge \frac{1}{e}$

$\frac{x}{x-1} - \frac{1}{e} \ge 0 \implies \frac{ex - (x-1)}{e(x-1)} \ge 0 \implies \frac{(e-1)x + 1}{e(x-1)} \ge 0$.

Since $e > 0$ and $e-1 > 0$, this simplifies to $\frac{(e-1)x + 1}{x-1} \ge 0$.

The critical points are $x = 1$ and $x = -\frac{1}{e-1}$.

Using a sign analysis, the solution is $x \in (-\infty, -\frac{1}{e-1}] \cup (1, \infty)$.

(b) $\frac{x}{x-1} \le e$

$\frac{x}{x-1} - e \le 0 \implies \frac{x - e(x-1)}{x-1} \le 0 \implies \frac{(1-e)x + e}{x-1} \le 0$.

Since $1-e < 0$, we can multiply by $-1$ and reverse the inequality: $\frac{(e-1)x - e}{x-1} \ge 0$.

The critical points are $x = 1$ and $x = \frac{e}{e-1}$.

Using a sign analysis, the solution is $x \in (-\infty, 1) \cup [\frac{e}{e-1}, \infty)$.

The values of $x$ for which $-1 \le \log_e \left( \frac{x}{x-1} \right) \le 1$ are the intersection of the solutions for (a) and (b):

$\left( (-\infty, -\frac{1}{e-1}] \cup (1, \infty) \right) \cap \left( (-\infty, 1) \cup [\frac{e}{e-1}, \infty) \right)$

This intersection is $x \in (-\infty, -\frac{1}{e-1}] \cup [\frac{e}{e-1}, \infty)$.

Finally, $E_2$ is the intersection of this result with $E_1 = (-\infty, 0) \cup (1, \infty)$.

Since $-\frac{1}{e-1} < 0$, $(-\infty, -\frac{1}{e-1}] \subset (-\infty, 0)$.

Since $\frac{e}{e-1} > 1$, $[\frac{e}{e-1}, \infty) \subset (1, \infty)$.

Therefore, $E_2 = (-\infty, -\frac{1}{e-1}] \cup [\frac{e}{e-1}, \infty)$.

Note that $\frac{1}{1-e} = -\frac{1}{e-1}$.

So, $E_2 = \left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right)$.

Matching the options:

(S) The domain of $g$ is:

The domain of $g$ is $E_2$.

$E_2 = \left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right)$, which matches option (1).

So, S $\rightarrow$ 1.

(R) The domain of $f$ contains:

The domain of $f$ is $E_1 = (-\infty, 0) \cup (1, \infty)$.

We need to find which option is a subset of $E_1$.

Option (1) is $E_2$. Since $E_2 \subset E_1$, $E_1$ contains $E_2$. So R $\rightarrow$ 1 is possible.

Option (6) is $(-\infty, 0) \cup \left[ \frac{1}{2}, \frac{e}{e-1} \right]$. This set is not contained in $E_1$ because $\frac{1}{2}$ is not in $(1, \infty)$.

Therefore, R $\rightarrow$ 1.

(P) The range of $f$ is:

$f(x) = \log_e \left( \frac{x}{x-1} \right)$. The domain of $f$ is $E_1 = (-\infty, 0) \cup (1, \infty)$.

Let $t = \frac{x}{x-1}$.

If $x \in (-\infty, 0)$: As $x \to -\infty$, $t = \frac{1}{1 - 1/x} \to 1$. As $x \to 0^-$, $t \to 0^+$. So $t \in (0, 1)$.

If $x \in (1, \infty)$: As $x \to 1^+$, $t = \frac{1}{1 - 1/x} \to \infty$. As $x \to \infty$, $t \to 1$. So $t \in (1, \infty)$.

Thus, for $x \in E_1$, the range of $\frac{x}{x-1}$ is $(0, 1) \cup (1, \infty)$.

Now we find the range of $\log_e(t)$ for $t \in (0, 1) \cup (1, \infty)$.

If $t \in (0, 1)$, $\log_e(t) \in (-\infty, 0)$.

If $t \in (1, \infty)$, $\log_e(t) \in (0, \infty)$.

Combining these, the range of $f(x)$ is $(-\infty, 0) \cup (0, \infty) = R \setminus \{0\}$.

This matches option (4).

So, P $\rightarrow$ 4.

(Q) The range of $g$ contains:

$g(x) = \sin^{-1} \left( \log_e \left( \frac{x}{x-1} \right) \right)$. The domain of $g$ is $E_2$.

For $x \in E_2$, we know that $\frac{1}{e} \le \frac{x}{x-1} \le e$.

Let $Y = \log_e \left( \frac{x}{x-1} \right)$.

Since $\frac{1}{e} \le \frac{x}{x-1} \le e$, taking $\log_e$ (which is increasing), we get:

$\log_e \left( \frac{1}{e} \right) \le \log_e \left( \frac{x}{x-1} \right) \le \log_e(e)$

$-1 \le Y \le 1$.

The function $g(x) = \sin^{-1}(Y)$.

The range of $\sin^{-1}(Y)$ for $Y \in [-1, 1]$ is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

So, the range of $g$ is exactly $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

We need to find which of the given options is contained in $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Note that $\frac{\pi}{2} \approx 1.57$.

Option (2) is $(0,1)$. This is contained in $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Option (3) is $\left[ -\frac{1}{2}, \frac{1}{2} \right]$. This is also contained in $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Since the question asks "The range of $g$ contains", both (2) and (3) are possible.

However, in JEE Advanced, typically for "contains", the most specific (largest possible) set is preferred if multiple options are subsets. Both are valid.

Let's look at the multiple choice options provided for the final answer.

(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

From our analysis: P $\rightarrow$ 4 R $\rightarrow$ 1 S $\rightarrow$ 1

Let's check the options (A), (C), (D) for consistency with P, R, S.

Option (A): P $\rightarrow$ 4; R $\rightarrow$ 1; S $\rightarrow$ 1. This is consistent. For Q, it says Q $\rightarrow$ 2. Option (C): P $\rightarrow$ 4; R $\rightarrow$ 1. But S $\rightarrow$ 6, which is incorrect. So (C) is out. Option (D): P $\rightarrow$ 4. But R $\rightarrow$ 6, which is incorrect. S $\rightarrow$ 5 is also incorrect. So (D) is out. Option (B): P $\rightarrow$ 3 (incorrect); R $\rightarrow$ 6 (incorrect); S $\rightarrow$ 5 (incorrect). So (B) is out.

This implies that option (A) is the correct choice.

Let's confirm Q $\rightarrow$ 2. The range of $g$ is $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.

Option (2) is $(0,1)$. Since $(0,1) \subset \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$, Q $\rightarrow$ 2 is correct.

Final mapping: P $\rightarrow$ 4 Q $\rightarrow$ 2 R $\rightarrow$ 1 S $\rightarrow$ 1

This matches option (A).

The final answer is $\boxed{A}$