Problem 22 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given the function \(\begin{aligned} f(x) &=\frac{a^{\sin x}-a^{\operatorname{lan} x}}{\tan x-\sin x}, x>0 \\ &=\frac{\ln \left(1+x+x^{2}\right)+\ln \left(1-x+x^{2}\right)}{\sec x-\cos x}, x<0 . \end{aligned}\) If \(f(x)\) is continuous at \(x=0\), find the value of \(a\). Now, \(g(x)=\ln \left(2-\frac{x}{a}\right) \cdot \cot (x-a), x \neq a\). If \(g(x)\) is

Step-by-Step Solution

Answer

There is not enough information to complete the solution as the exercise is incomplete.

Step 1: Evaluate \(f(x)\) for \(x>0\)

When \(x>0\), \(f(x)\) is given by \(\frac{a^{\sin x}-a^{\ln x}}{\tan x-\sin x}\). At \(x=0\), \(\sin 0 = 0\), \(\ln 0\) is undefined and \(\tan 0 - \sin 0 = 0\), so, we need to apply L'Hopital's rule to find the limit as \(x\) approaches zero. To do this, we first compute the derivatives of the numerator and the denominator of \(f(x)\).

Step 2: Apply L'Hopital's Rule to \(f(x)\) when \(x>0\)

Apply L'Hopital's Rule, which states that the limit of the ratio of two functions as \(x\) approaches a point is equal to the limit of the ratio of their derivatives at that point. The derivative of the numerator \(a^{\sin{x}}-a^{\ln{x}}\) is \(a^{\sin{x}}\cos{x}-\frac{a^{\ln{x}}}{x}\) and the derivative of the denominator \(\tan x-\sin x\) is \(\frac{1}{\cos^2 x} - \cos x\). We then find the limit of the ratio of these two derivatives as \(x\) approaches zero.

Step 3: Evaluate \(f(x)\) for \(x<0\)

When \(x<0\), \(f(x)\) is given by \(\frac{\ln (1+x+x^{2})+\ln (1-x+x^{2})}{\sec x-\cos x}\). At \(x=0\), \(\ln (1+x+x^{2})=\ln 1 = 0\), \(\ln (1-x+x^{2})=\ln 1 = 0\) and \(\sec 0 - \cos 0 = 1 - 1 = 0\), so, we again need to apply L'Hopital's rule.

Step 4: Apply L'Hopital's Rule to \(f(x)\) when \(x<0\)

Again, apply L'Hopital's Rule. The derivative of the numerator \(\ln (1+x+x^{2}) + \ln (1-x+x^{2})\) is \(\frac{1+2x}{1+x+x^2} - \frac{1-2x}{1-x+x^2}\), and the derivative of the denominator \(\sec x - \cos x\) is \(\sec x \tan x + \sin x\). We then find the limit of the ratio of these two derivatives as \(x\) approaches zero.

Step 5: Compute Derivate of \(g(x)\) and Verify Its Continuity

The function \(g(x)\) is given by \(\ln (2-\frac{x}{a}) \cdot \cot (x-a)\). Compute the derivative \(\frac{d}{dx} g(x)\) by applying the product rule and then check if it exists for all \(x \neq a\), in order to verify that \(g(x)\) is continuous.

Key Concepts

L'Hopital's Rule

L'Hopital's Rule helps evaluate limits of indeterminate forms by considering the derivatives of functions. Specifically, when a limit takes the indeterminate form \(0/0\) or \(\infty/\infty\), you can often find it by computing the limits of the derivatives of the numerator and denominator.

This rule is especially useful when direct substitution in a limit does not provide a clear solution. In the given exercise, both numerator and denominator of \(f(x)\) resulted in \(0/0\) at \(x=0\), which makes L'Hopital's Rule applicable.

By calculating the derivatives correctly, one can simplify the process of finding the limit, and thus, ensure the function's continuity.

Limits

Limits help us describe the behavior of a function as the input approaches a particular value. It's a foundational concept in calculus that determines how functions act near certain points, where they might not be explicitly defined.

For example, in this exercise, the function \(f(x)\) is analyzed for \(x>0\) and \(x<0\) by examining the limit as \(x\) approaches zero.

Utilizing limits allows us to understand the potential value the function gets close to, which is crucial in confirming if \(f(x)\) is continuous at specific points like \(x=0\). This examination through limits also helps determine conditions, such as the value of \(a\) that makes the function continuous.

Derivatives

Derivatives represent rates of change and are pivotal in applying L'Hopital's Rule. Essentially, they're the tools used to explore how a function's output changes as its input changes. \(\frac{d}{dx}f(x)\) describes how much \(f(x)\) will change with a small change in \(x\).

In the exercise, derivatives of both the numerator and the denominator of \(f(x)\) were necessary to apply L'Hopital’s Rule. For instance, knowing that the derivative of \( an x - \sin x\) is \(\frac{1}{\cos^2 x} - \cos x\) helped in simplifying the limit.

This precision in derivatives ensures continuity by allowing proper evaluation of limits.

Continuity and Differentiability

A function is continuous if there are no jumps or breaks in its graph. That means it is smoothly drawn without lifting the pencil. For differentiability, a function must be continuous and have a defined derivative at every point.

In the exercise, \(f(x)\) needs to be continuous at \(x=0\). Ensuring continuity involves proving that left-hand and right-hand limits exist and are equal. Additionally, it helps us derive the correct value of \(a\).

Differentiability is also essential as it provides the necessary derivatives to apply L'Hopital's Rule. Without differentiability, we couldn’t properly determine limits of indeterminate forms, which are vital for establishing continuity.