Problem 106 - A Complete Resource Book in Mathematics for JEE Main

Question

Column-I I. Let \(f\) and \(g\) be differentiable functions satisfying \(g^{\prime}(a)=2\), \(g(a)=b\) and \(f o g=\mathrm{I}\) (identity function). Then, \(f^{\prime}(b)=\) II. Let \(f(x)=\left\\{\begin{array}{cc}(1+|\sin x|)^{a / \sin x \mid} & , \quad-\frac{\pi}{6}

Step-by-Step Solution

Answer

I: C, II: - , III: -, IV: B

Step 1: Understanding the Given Condition

Since the composition of functions \(f \circ g\) is the identity function, it implies \(f(g(x)) = x\). This leads to prime differentiation \(\frac{d}{dx}f(g(x)) = \frac{d}{dx}\,x \Rightarrow f'(g(x))g'(x) = 1\).

Step 2: Solve for \(f'(b)\)

Substituting \(x = a\), we get \(f'(g(a))g'(a) = 1\). We know \(g(a) = b\) and \(g'(a) = 2\), hence \(f'(b) \cdot 2 = 1\). So, \(f'(b) = \frac{1}{2}\).

Step 3: Find the Appropriate Value of \(a\) for Continuity

For \(-\frac{\pi}{6} < x < 0\), evaluate the limit \(\lim_{x \to 0^-} f(x) = (1 + 0)^0 = 1\). For \(0 < x < \frac{\pi}{6}\), evaluate \(\lim_{x \to 0^+} e^{\tan 0} = 1\). For continuity at \(x=0\), we need \(e^{2\sqrt{3}} = 1\). Since this isn't possible, find \(\lim_{x \to 0^-} (1 + |\sin x|)^{a/\sin x} = \lim_{x \to 0^+} e^{\tan 2x^3}\) to solve for \(a \).

Step 4: Solve for \(f(\pi/4)\) for Continuity

Consider the limit \(\lim_{x \to \pi/4} (\sin 2x)\tan^2 2x\). As \(x \to \pi/4\), \(\sin(2x) \to 1\) and \(\tan(2x) \to \infty\). Thus, for continuity at \(x=\pi/4\), evaluate \(\lim_{x \to \pi/4} (\sin 2x)(\tan 2x)^2\) and solve for \(f(\pi/4)\).

Step 5: Determine \(f(0)\) for Continuity

Evaluate \( \lim_{x \to 0} \left(\frac{\sqrt[3]{1+x} - \sqrt[4]{1+x}}{x}\right)\). Use L'Hôpital's Rule as the expression has an indeterminate form of \(\left[\frac{0}{0}\right]\). Taking derivatives leads to \(\frac{\frac{1}{3}(1 + x)^{-\frac{2}{3}} - \frac{1}{4}(1 + x)^{-\frac{3}{4}}}{1}\) at \(x = 0\), which simplifies to \(\frac{1}{3} - \frac{1}{4} = \frac{1}{12}\). This ensures \(f(x)\) is continuous at \(x=0\).

Key Concepts

Differentiable Functions

In calculus, a function is said to be differentiable if it has a derivative at every point in its domain. This is an important concept because differentiability implies continuity, which intuitively means that the function has no "jumps" or "holes". Here is a simple way to understand this: think about differentiability as the function having a well-behaved tangent line at every point you are interested in.

• A differentiable function allows you to compute its slope at any given point.
• A function is smooth if it is differentiable at all points in its range.
• While a function might be continuous, it is not necessarily differentiable. For instance, functions with sharp turns or cusps might not be differentiable there.

Differentiability is a key concept in solving problems involving rates of change, optimizations, and in finding tangent lines.

Composition of Functions

Composition of functions involves creating a new function by applying one function to the result of another function. If you have two functions, say \( f(x) \) and \( g(x) \), their composition is expressed as \( f(g(x)) \). This means you compute \( g(x) \) first, and then use the result to compute \( f \).

• For composition, the output of the first function needs to be valid as the input into the second.
• The order in which you apply the functions is crucial, as \( f(g(x)) \) is typically not the same as \( g(f(x)) \).
• A well-known property is that the composition of a function and its inverse, \( f \circ g = I \), yields the identity function, \( I(x) = x \). This was used in the exercise to find \( f'(b) \) for differentiable functions \( f \) and \( g \).

Understanding the composition of functions is imperative in solving problems involving nested functions and transformations, simplifying complexity, and establishing functional relationships.

Limits and L'Hôpital's Rule

Limits are used to define the value that a function approaches as the input approaches some value. They are foundational to calculus and are essential for precise definitions of concepts like continuity and differentiability.The limit notation \( \lim_{x \to a} f(x) = L \) means that as \( x \) gets arbitrarily close to \( a \), \( f(x) \) approaches \( L \).L'Hôpital's Rule is a powerful tool to solve limits involving indeterminate forms such as \([0/0]\) or \([\infty/\infty]\). When a limit has these forms, you can differentiate the numerator and the denominator separately to simplify the evaluation process:

• Applied correctly, L'Hôpital's rule can make tough limits solvable by transforming the original expression into a more tractable form.
• It requires the derivatives of the numerator and denominator to exist near the point of interest.
• Always check if directly substituting or simplifying first can avoid unnecessary application of the rule.

Understanding how to use limits and L'Hôpital's rule is critical for analyzing and interpreting function behavior near specific points, especially when direct substitution isn't possible.