Problem 6 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int x\left[f\left(x^{2}\right) g^{\prime \prime}\left(x^{2}\right)-f^{\prime \prime}\left(x^{2}\right) g\left(x^{2}\right)\right] d x\) (A) \(f\left(x^{2}\right) g^{\prime}\left(x^{2}\right)-g\left(x^{2}\right) f^{\prime}\left(x^{2}\right)+c\) (B) \(\frac{1}{2}\left[f\left(x^{2}\right) g\left(x^{2}\right) f^{\prime}\left(x^{2}\right)\right]+c\) (C) \(\frac{1}{2}\left[f\left(x^{2}\right) g^{\prime}\left(x^{2}\right)-g\left(x^{2}\right) f^{\prime}\left(x^{2}\right)\right]+c\) (D) none of the above

Step-by-Step Solution

Answer

The answer is (C).

Step 1: Recognize the Structure of the Integral

The given integral expression is \( \int x \left[ f(x^2) g''(x^2) - f''(x^2) g(x^2) \right] \, dx \). This expression suggests that it may be related to the derivative of a product involving functions of \( x^2 \).

Step 2: Consider the Product Rule for Derivatives

The derivative of a product gives a structure \( u(x) v''(x) - u''(x) v(x) \). We assume a function form \( f(x^2) g(x^2) \). If we differentiate this once with respect to \( x \), we apply the chain rule: \( 2x[f'(x^2) g(x^2) + f(x^2) g'(x^2)] \). However, we need to focus on the second derivative terms.

Step 3: Apply the Second Derivative in Product Rule Form

For the product \( u(x^2) = f(x^2) \) and \( v(x^2) = g(x^2) \), their second derivatives are considered. The expression \( f(x^2) g''(x^2) - f''(x^2) g(x^2) \) resembles the result from differentiating the determinant form of functions: \( u(x^2) \cdot v'(x^2) - v(x^2) \cdot u'(x^2) \) in a similar manner, related to the Wronskian.

Step 4: Simplify the Integral Expression

The term \( x \) outside the bracket resembles multiplication by \( 2x \), the factor coming out of differentiating \( x^2 \). Therefore, by simplifying and recognizing the reduced factor in derivation, it matches the form which gives \( \frac{1}{2} \left[ f(x^2) g'(x^2) - g(x^2) f'(x^2) \right] \).

Step 5: Match with Given Options

By comparing the simplified expression to the given options, it directly matches option (C): \( \frac{1}{2} \left[ f(x^2) g'(x^2) - g(x^2) f'(x^2) \right] + c \). Thus, this integral is equivalent to option (C).

Key Concepts

Integration

Integration is a fundamental concept in calculus, focusing on finding the total, or accumulated, value across a specific range. This process is essentially the reverse of differentiation. When dealing with integrals, the goal is to find a function, known as the antiderivative, that describes the accumulated area under a curve represented by a mathematical function.

In the context of indefinite integrals, like in the provided problem, we aim to find a general function whose derivative matches the integrand (the function being integrated). While the exercise includes specific expressions, the principle remains that we seek an antiderivative form, potentially complemented by a constant of integration, denoted as 'c'.

To find such an antiderivative, we often employ rules and methods similar to differentiation. These include recognizing patterns and applying integration techniques like substitution. In our particular exercise, recognizing the structure of the integrand suggests the usefulness of other calculus rules, such as:

• Product Rule – helpful when dealing with integrands that appear as products of functions.
• Chain Rule – assists in managing complex functions nested within others.
  

Product Rule

The Product Rule is a technique often discussed alongside differentiation but can offer insights when tackling integration challenges. This rule provides a method to determine the derivative of the product of two functions, say, if you have functions \( u(x) \) and \( v(x) \).

The Product Rule states that the derivative of their product is given by:
\[ (u(x) \, v(x))' = u'(x) v(x) + u(x) v'(x) \]
Now, though primarily for differentiation, understanding this rule aids in recognizing and simplifying the integration of certain products. When reversing the process (finding the integral), we think conceptually of what combination of derivatives could yield the given integrand.

In practice, if you encounter integrands that seem convoluted as products of derivatives, consideration of the Product Rule simplifies the task by leveraging relationships between these derived components. In the given exercise, these observations, paired with the structure of the integral, inform us about the transformation needed, leading us toward the solution that matches with option (C).

Chain Rule

The Chain Rule is another backbone principle of calculus that helps manage composite functions. This rule is vital when dealing with integrals that involve compositions, as it explains how derivatives of these layers affect the derivative as a whole.

Mathematically, if you have a composite function \( y = f(g(x)) \), the Chain Rule tells you that the derivative \( y' \) is the derivative of \( f \) with respect to \( g \), multiplied by the derivative of \( g \) with respect to \( x \). Formally, it is expressed as:
\[ \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \]
In integration, understanding the Chain Rule becomes instrumental when evaluating integrals of composite functions, allowing you to perform substitutions that simplify the integral into a recognizable form.

• This concept ties back to our exercise where we examined functions of \( x^2 \), necessitating application of the Chain Rule logic when considering derivatives of these substituted forms.
• The structure informed a recognition similar to applying inverse chain derivatives, thus guiding us toward identifying the appropriate integral outcome, ultimately aligned with option (C).