Animated Solution for Mathematics - Differentiation: Let $f(x)={\log }_e(\sin x),(0<x<\pi )$ and $g(x)={\sin }^{-1}(e^{-x}),(x\ge 0)$. If $\alpha$ is a positive real number such that $a=(fog{)}^{\prime }(\alpha )$ and $b=(fog)(\alpha )$, then :

We are given two functions: $f(x)={\log }_e(\sin x)$ and $g(x)={\sin }^{-1}(e^{-x})$. We are tasked with finding the relationship between $a=(f\circ g{)}^{\prime }(\alpha )$ and $b=(f\circ g)(\alpha )$.

Imagine you are a signal passing through a machine. The input $x$ first enters the 'g-machine', which transforms it into ${\sin }^{-1}(e^{-x})$. Then, that output is fed into the 'f-machine', which takes the sine of that value and then finds its natural logarithm.

Mathematically, we write this composition as:

Many students immediately reach for the chain rule, attempting to differentiate ${\log }_e(\sin x)$ and ${\sin }^{-1}(e^{-x})$ separately. While that is technically correct, it is a path filled with potential for error. Instead, look at the inner term: $\sin ({\sin }^{-1}(e^{-x}))$.

Recall the fundamental definition of inverse trigonometric functions. The function ${\sin }^{-1}(\theta )$ returns an angle whose sine is $\theta$. Therefore, $\sin ({\sin }^{-1}(\theta ))$ is simply $\theta$, provided $\theta$ is within the valid domain.

Since $x\ge 0$, we know that $e^{-x}$ is always between $0$ and $1$. This is well within the domain $[-1,1]$. Thus, the entire trigonometric mess collapses into a single, beautiful term:

Now, substitute this back into our expression for the composition. The function becomes:

Using the fundamental property of logarithms, ${\log }_e(e^k)=k$, we see that the logarithm and the exponential function effectively cancel each other out. We are left with the simplest possible function:

Isn't that breathtaking? All that initial complexity was just a mask for a simple linear relationship.

Now that we have $(f\circ g)(x)=-x$, finding $a$ and $b$ becomes trivial. For $b$, we simply evaluate the function at $x=\alpha$:

For $a$, we find the derivative of the composition with respect to $x$ and evaluate it at $\alpha$:

Since the derivative of a constant $-1$ is always $-1$, the value of $a$ is independent of $\alpha$. We have $a=-1$ and $b=-\alpha$.

We have successfully distilled the problem down to $a=-1$ and $b=-\alpha$. When you look at the options provided, you are now equipped to test them with absolute confidence.

Remember, the JEE Advanced is not just about calculation; it is about the courage to simplify. You have navigated the composition, identified the identity, and arrived at the solution. Keep this mindset—look for the simplification, trust the math, and never let the complexity of the notation distract you from the underlying beauty of the function.