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

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

The value of \(\frac{g''}{g^2}\) is \(\frac{3}{2}\). (Option B)

Step 1: Differentiate the integral form of f(x)

We start by differentiating the function \(f(x) = \int_{0}^{x}(1+t^3)^{-1/2} \, dt\) using the fundamental theorem of calculus. This gives us \(f'(x) = (1+x^3)^{-1/2}\).

Step 2: Express g(x) in terms of f(x)

Since \(g\) is the inverse of \(f\), we have \(f(g(x)) = x\). Differentiating both sides with respect to \(x\), we get \(f'(g(x)) \cdot g'(x) = 1\). Therefore, \(g'(x) = \frac{1}{f'(g(x))}\).

Step 3: Simplify g'(x) expression

From Step 1, we know \(f'(x) = (1+x^3)^{-1/2}\), so substituting \(g(x)\) for \(x\), we get \(f'(g(x)) = (1+g(x)^3)^{-1/2}\). The expression for \(g'(x)\) becomes \(g'(x) = (1+g(x)^3)^{1/2}\).

Step 4: Differentiate g'(x) to find g''(x)

Differentiate \(g'(x) = (1+g(x)^3)^{1/2}\) using the chain rule: \[g''(x) = \frac{1}{2}(1+g(x)^3)^{-1/2} \cdot 3g(x)^2 \cdot g'(x)\]This simplifies to: \[g''(x) = \frac{3}{2}g(x)^2(1+g(x)^3)^{-1/2} \cdot (1+g(x)^3)^{1/2}\]So, \(g''(x) = \frac{3}{2}g(x)^2\).

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

Using the result from Step 4, we calculate \(\frac{g''}{g^2}\):\[\frac{g''}{g^2} = \frac{\frac{3}{2}g(x)^2}{g(x)^2} = \frac{3}{2}\].

Key Concepts

differential_calculus

Differential calculus is all about understanding the rates at which things change. This branch of mathematics allows us to find the derivative of a function, which measures how the function's output value changes as its input value changes. In our exercise, differentiation plays a crucial role in determining the rate of change of the integral, \(f(x) = \int_{0}^{x}(1+t^3)^{-1/2} \, dt\).

Using the fundamental theorem of calculus, differentiating this integral form helps us to find \(f'(x)\), the first derivative of \(f(x)\). Here, \(f'(x) = (1+x^3)^{-1/2}\), which tells us how \(f(x)\) changes with \(x\). This derivative will later be essential when we explore the inverse function, \(g(x)\), and its derivatives.

fundamental_theorem_of_calculus

The fundamental theorem of calculus bridges the world of antiderivatives and integrals. It tells us that differentiation and integration are inverse processes. In simple terms, this theorem guarantees that if you have a function defined as an integral, like \(f(x) = \int_{0}^{x}(1+t^3)^{-1/2} \, dt\), differentiating it will give back the integrand as a function of \(x\).

This crucial principle allows us to differentiate \(f(x)\) directly to get \(f'(x) = (1+x^3)^{-1/2}\). This step is foundational in finding the properties of the inverse function \(g(x)\), helping to relate \(g'(x)\) and further derivatives like \(g''(x)\). Ultimately, it simplifies the process of working with complex integral and derivative relationships in calculus.

derivative_of_inverse_function

When dealing with inverse functions, such as \(g(x)\) being the inverse of \(f(x)\), understanding derivatives is crucial. The relationship \(f(g(x)) = x\) is key here. Differentiating this equation using the chain rule gives us \(f'(g(x)) \cdot g'(x) = 1\).

This simplifies to the formula \(g'(x) = \frac{1}{f'(g(x))}\). Utilizing the derivative \(f'(x) = (1+x^3)^{-1/2}\) from our earlier steps, we substitute to find \(g'(x) = (1+g(x)^3)^{1/2}\).

• This setup is crucial as it sets the stage for finding the second derivative \(g''(x)\).
• Using the chain rule again, \(g''(x)\) can be calculated, leading to \(g''(x) = \frac{3}{2}g(x)^2\).
• Finally, determining \(\frac{g''}{g^2}\) yields the result \(\frac{3}{2}\), providing a direct insight into the behavior of the inverse function's rate of change.

This exploration showcases the beauty of calculus in connecting derivatives and their inverses in a seamless mathematical dance.