Problem 34 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(f\) and \(g\) be differentiable functions on \(\mathbf{R}\) such that fog is the identity function. If for some \(a, b \in \mathbf{R}, g^{\prime}(a)=5\) and \(g(a)=b\), then \(f^{\prime}(b)\) is equal to: \(\quad\) [Jan. 9,2020 (II)] (a) \(\frac{1}{5}\) (b) 1 (c) 5 (d) \(\frac{2}{5}\)

Step-by-Step Solution

Answer

(a) \( \frac{1}{5} \)

Step 1: Understand the identity function composition

If \( f \circ g \) is the identity function, then it means that \( (f \circ g)(a) = f(g(a)) = a \) for any \( a \in \mathbb{R} \). In this problem, since \( g(a) = b \), it follows that \( f(b) = a \).

Step 2: Chain Rule for derivatives

To find \( f'(b) \), we apply the chain rule to \( f(g(a)) = a \). Differentiating both sides with respect to \( a \), we get:\[\frac{d}{da}f(g(a)) = \frac{d}{da}a\]Which simplifies to:\[f'(g(a)) \cdot g'(a) = 1\]Substituting \( g(a) = b \), we have:\( f'(b) \cdot g'(a) = 1 \).

Step 3: Solve for f'(b)

We know from the problem statement that \( g'(a) = 5 \). Plugging \( g'(a) = 5 \) into the equation from Step 2:\[f'(b) \cdot 5 = 1\]Solving for \( f'(b) \) gives us:\[f'(b) = \frac{1}{5}\]

Step 4: Select the correct answer

The possible answer choices are (a) \( \frac{1}{5} \), (b) 1, (c) 5, and (d) \( \frac{2}{5} \). From our calculation in Step 3, \( f'(b) = \frac{1}{5} \), therefore the correct answer is (a).

Key Concepts

Chain Rule

The chain rule is a fundamental tool in differential calculus. It allows us to differentiate composite functions, which are functions nested within other functions. The rule states that if you have two functions, say \( u(x) \) and \( v(x) \), and you want to find the derivative of their composition, then the derivative of \( v(u(x)) \) is given by:

• \( v'(u(x)) \cdot u'(x) \).

This rule is crucial when dealing with problems where functions are composed, like in the exercise you encountered. By systematically pulling apart the layers of the functions, you can calculate derivatives that might otherwise seem complicated.

Derivatives

In calculus, derivatives represent the rate of change of a function concerning its variable. The derivative of a function is denoted by \( f'(x) \) or sometimes \( \frac{df}{dx} \). It tells us how the function's output changes as the input changes.

• For example, if you drive a car and want to know your speed at a certain time, you are essentially looking for the derivative of your distance with respect to time.

In the given exercise, understanding derivatives is key to solving for \( f'(b) \). By correctly applying the chain rule and the given derivative \( g'(a) = 5 \), we find that \( f'(b) = \frac{1}{5} \). This shows how powerful derivatives are in analyzing and solving calculus problems.

Function Composition

Function composition refers to the process of applying one function to the results of another. If you have two functions, say \( f \) and \( g \), then the composition of \( f \, \text{and}\, g \), denoted by \( f(g(x)) \), means that \( g \) takes \( x \) as an input, and \( f \) takes the output of \( g \) and gives a final result.
The exercise illustrates this concept where \( f \circ g \) produces the identity function. By understanding how these functions interact, you gain deeper insight into how changes in \( g(x) \) will affect the outcome of \( f(g(x)) \). This is why knowing both the individual functions and their interactions through composition is critical in calculus.

Identity Function

An identity function is a mainstay in mathematics, defined as a function that returns its input unchanged. Mathematically, it is represented as \( f(x) = x \). It serves a special role because it acts as a neutral element in function composition.
In the context of the given exercise, the composition \( f \circ g \) being the identity function means that applying \( g \) followed by \( f \) gets you back to your starting point, which gives \( f(g(a)) = a \). This property plays a vital role in simplifying and understanding the relationship between the functions \( f \) and \( g \), and efficiently finding the derivative \( f'(b) \). Recognizing and using identity functions can significantly simplify complex problems involving multiple function compositions.