Q.Identify the function shown in the grap
(A) sin−1 𝑥
(B) sin−1(2𝑥)
(C) sin−1 ( 𝑥

We are solving

The “$\pm$” means we have two separate equations to consider. But before diving into algebra, we must respect the domains of $\sin^{-1}$.

1. Domain restrictions first

For $\sin^{-1}(2-x)$ to be defined:

For $\sin^{-1}x$ to be defined:

Intersecting these gives

So the only possible real value is $x = 1$.

2. Check $x = 1$ in the equation

Compute each term:

So the left side becomes

$\frac{\pi}{2} - 2\cdot\frac{\pi}{2} = \frac{\pi}{2} - \pi = -\frac{\pi}{2}.$ …

Let's break down this problem by first understanding the individual inverse hyperbolic functions involved, their domains, and their ranges. The notation $hx$ is slightly ambiguous; in the context of such problems, if $h$ is not defined as a function, it typically implies either $h=1$ (so the argument is $x$) or $h$ is a positive constant, in which case the range of the expression remains the same as if the argument were simply $x$. We will proceed assuming the argument is a variable, say $y$, and the function is $f(y) = \text{sech}^{-1} y + \text{csch}^{-1} y$.

Concept and Intuition

To find the range of a sum of functions, we first need to determine the domain where both functions are defined. Then, we analyze the behavior of each function over this common domain and, if necessary, the behavior of their sum. A common pitfall is to confuse inverse hyperbolic functions with inverse trigonometric functions, which have different properties and identities. For instance, there is no simple identity like $\sec^{-1} x + \csc^{-1} x = \pi/2$ for inverse hyperbolic functions.

We will use the definitions and properties of inverse hyperbolic functions:

• Inverse Hyperbolic Secant: $\text{sech}^{-1} x = \cosh^{-1} \left(\frac{1}{x}\right) = \log \left(\frac{1 + \sqrt{1-x^2}}{x}\right)$.
  • Domain: $(0, 1]$
  • Range: $[0, \infty)$
• Inverse Hyperbolic Cosecant: $\text{csch}^{-1} x = \sinh^{-1} \left(\frac{1}{x}\right) = \log \left(\frac{1}{x} + \sqrt{\frac{1}{x^2}+1}\right) = \log \left(\frac{1 + \sqrt{1+x^2}}{|x|}\right)$.
  • Domain: $(-\infty, \infty) \setminus \{0\}$
  • Range: $(-\infty, \infty) \setminus \{0\}$

Step-by-Step Solution

1. Determine the common domain:

   Let the given function be $f(y) = \text{sech}^{-1} y + \text{csch}^{-1} y$.

   The domain of $\text{sech}^{-1} y$ is $(0, 1]$.

   The domain of $\text{csch}^{-1} y$ is $(-\infty, \infty) \setminus \{0\}$.

   For $f(y)$ to be defined, $y$ must be in the intersection of these two domains:

   Domain of $f(y) = (0, 1] \cap ((-\infty, \infty) \setminus \{0\}) = (0, 1]$.

2. Analyze the behavior of each function over the common domain $(0, 1]$:

   • For $\text{sech}^{-1} y$ on $(0, 1]$:

     • As $y \to 0^+$, $\text{sech}^{-1} y = \log \left(\frac{1 + \sqrt{1-y^2}}{y}\right) \to \log \left(\frac{1 + \sqrt{1-0}}{0^+}\right) = \log \left(\frac{2}{0^+}\right) \to \infty$.
     • At $y = 1$, $\text{sech}^{-1} 1 = \log \left(\frac{1 + \sqrt{1-1^2}}{1}\right) = \log(1) = 0$.
     • So, the range of $\text{sech}^{-1} y$ for $y \in (0, 1]$ is $[0, \infty)$.

   • For $\text{csch}^{-1} y$ on $(0, 1]$:

     • As $y \to 0^+$, $\text{csch}^{-1} y = \log \left(\frac{1 + \sqrt{1+y^2}}{y}\right) \to \log \left(\frac{1 + \sqrt{1+0}}{0^+}\right) = \log \left(\frac{2}{0^+}\right) \to \infty$.
     • At $y = 1$, $\text{csch}^{-1} 1 = \log \left(\frac{1 + \sqrt{1+1^2}}{1}\right) = \log(1+\sqrt{2})$.
     • So, the range of $\text{csch}^{-1} y$ for $y \in (0, 1]$ is $[\log(1+\sqrt{2}), \infty)$.

3. Determine the monotonicity of $f(y)$ on $(0, 1]$:

   To find the range of the sum, it's helpful to know if the function is monotonic. Let's find the derivative of $f(y)$: …