Problem 112 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(f(x)=\int_{0}^{x}\left(1+t^{3}\right)^{-1 / 2} d t\) and \(g\) is the inverse of \(f\), then the value of \(\frac{g^{\prime \prime}}{g^{2}}\) is (A) \(\frac{1}{2}\) (B) \(\frac{3}{2}\) (C) 1 (D) Cannot be determined

Step-by-Step Solution

Answer

\(\frac{g''}{g^2} = \frac{3}{2}\) (Option B)

Step 1: Understanding the Relationship Between Inverses

We're given that \(g\) is the inverse of \(f\). Generally, if \(g = f^{-1}\), then \(f(g(x)) = x\) and by implicit differentiation \(f'(g(x)) \cdot g'(x) = 1\).

Step 2: Differentiating the Given Function

First, compute \(f'(x)\). We know \(f(x)=\int_{0}^{x}(1+t^3)^{-1/2} dt\). According to the Fundamental Theorem of Calculus, \(f'(x) = (1+x^3)^{-1/2}\).

Step 3: Using the Inverse Relationship

Now apply the inverse function relationship: \((1+g(x)^3)^{-1/2} \cdot g'(x) = 1\). Solving for \(g'(x)\), we get \(g'(x) = (1+g(x)^3)^{1/2}\).

Step 4: Second Derivative of the Inverse Function

Differentiate \(g'(x)\) to find \(g''(x)\):\[g''(x) = \frac{d}{dx}((1+g(x)^3)^{1/2})\]Use the chain rule here:\[g''(x) = \frac{1}{2}(1+g(x)^3)^{-1/2}\cdot 3g(x)^2 \cdot g'(x)\]Substituting \(g'(x) = (1+g(x)^3)^{1/2}\), we get:\[g''(x) = \frac{3}{2}g(x)^2\]

Step 5: Computing \(\frac{g''}{g^2}\)

Now compute \(\frac{g''}{g^2}\):\[\frac{g''}{g^2} = \frac{\frac{3}{2}g(x)^2}{g(x)^2} = \frac{3}{2}\]

Key Concepts

Inverse Function Theorem

The Inverse Function Theorem is a fundamental concept in differential calculus. It provides us with the conditions under which a function has an inverse that is also differentiable. This theorem is applicable when we have a function \( f \) that is continuously differentiable. For such a function, if the derivative \( f'(a) \) is not zero at a point \( a \), then the function is locally invertible around \( f(a) \). This means there exists an inverse function \( g \) near \( f(a) \), and that inverse is also differentiable.
The theorem further tells us how to compute the derivative of the inverse function. Specifically, if \( y = f(x) \) and \( x = g(y) \), the derivative of the inverse function \( g'(y) \) is given by:
\[ g'(y) = \frac{1}{f'(g(y))} \]
This relationship was used in the step-by-step solution when we differentiated implicitly to find \( g'(x) \). This establishes the core relationship between a function and its inverse, allowing us to solve problems involving inverse functions with confidence.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges differentiation and integration, showing that they are essentially inverse operations. This theorem has two main parts:

• The first part relates the integral of a function to its antiderivative: For a continuous function \( f \) over an interval \([a, b]\), the function \( F \), defined by \( F(x) = \int_{a}^{x} f(t) \, dt \), is an antiderivative of \( f \).
• The second part states that if \( F \) is an antiderivative of \( f \) on an interval, then: \[ f(x) = F'(x) \]

In the context of our problem, when we calculated \( f(x) = \int_{0}^{x} (1+t^3)^{-1/2} dt \), we applied the second part of this theorem. By differentiating the integral, we immediately found \( f'(x) \), which was needed to determine the derivative properties of its inverse \( g \).
This theorem is crucial, as it seamlessly connects integral calculus with differential calculus, allowing us to switch between the two as needed.

Chain Rule

The Chain Rule is an essential tool in differential calculus, allowing us to differentiate composite functions. Suppose we have two functions, \( u(x) \) and \( v(x) \). If \( y = u(v(x)) \), the Chain Rule helps find \( \frac{dy}{dx} \) as follows:

• First, compute the derivative of \( u \) with respect to \( v \), \( \frac{du}{dv} \).
• Then, compute the derivative of \( v \) with respect to \( x \), \( \frac{dv}{dx} \).
• The derivative of the composite function is then: \[ \frac{dy}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} \]

In our solution process, we used the Chain Rule when differentiating \( g'(x) \). The function \( g(x) \) was nested inside the expression \((1+g(x)^3)^{1/2}\). So, its derivative, \( g''(x) \), was calculated using:
\[ g''(x) = \frac{1}{2}(1+g(x)^3)^{-1/2} \, \cdot \, 3g(x)^2 \, \cdot \, g'(x) \].
Here, the Chain Rule allowed us to correctly handle the nested function, enabling us to solve for \( \frac{g''}{g^2} \) and reach the final solution efficiently.