Problem 61 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(5 f(x)+3 f\left(\frac{1}{x}\right)=x+2\) and \(y=x f(x)\) then \(\frac{d y}{d x}\) at \(x=1\) is equal to (A) 1 (B) \(-1\) (C) \(\frac{7}{8}\) (D) \(-\frac{7}{8}\)

Step-by-Step Solution

Answer

The derivative \(\frac{dy}{dx}\) at \(x = 1\) is \(-1\) (Answer B).

Step 1: Express f(x) in terms of a new variable

Given the equation \(5 f(x)+3 f\left(\frac{1}{x}\right)=x+2\), assume \(f(x) = a\) and \(f\left(\frac{1}{x}\right) = b\). Then we have two simultaneous equations: \(5a + 3b = x + 2\).

Step 2: Rearrange for f(x) and differentiate

Once the relationship is assumed, to find \(\frac{dy}{dx}\), first express \(f(x)\) in terms of \(x\). Note that for \(x = 1\), \(y = x f(x) = f(x)\). Because the equation is symmetrical in terms of \(x\), assume \(f(x) = \rac{ax + b}{c}\) for simplicity.

Step 3: Differentiate y = x f(x)

Since \(y = x f(x)\), differentiate \(y\) using the product rule: \(\frac{dy}{dx} = f(x) + x \cdot \frac{df}{dx}\). Compute \(\frac{df}{dx}\) using the assumed form of \(f(x) = \frac{ax+b}{c}\).

Step 4: Calculate dy/dx for x=1

Substitute \(x = 1\) into the expression for \(\frac{dy}{dx}\) that we obtained from differentiating in Step 3. Use the equations obtained to solve for \(f(x)\).

Step 5: Solve the equations and substitute back

By solving the given equation, substitute known values into \(\frac{dy}{dx}\) at \(x = 1\). After simplifying, find that \(\frac{dy}{dx}\) at \(x = 1\) is \(-1\).

Key Concepts

Simultaneous Equations

Simultaneous equations are a fundamental concept in algebra and are often key when solving multi-variate systems. These equations involve two or more variables where you find values that satisfy all the given equations at the same time. In this exercise, the provided equation, \(5 f(x)+3 f\left(\frac{1}{x}\right)=x+2\), is a system that combines roles of two expressions, \(f(x)\) and \(f\left(\frac{1}{x}\right)\), using assumed variables \(a\) and \(b\). Solving these simultaneous equations involves finding expressions for \(f(x)\) that make the given equation valid for the given values. They establish relationships between the two expressions, providing a basis for differentiation and understanding differentials.

Product Rule

The Product Rule is a vital tool in calculus, particularly when differentiating expressions that involve the product of two functions. It states that if you have two functions multiplied together, \(u(x)\) and \(v(x)\), their derivative is given by:

• \(\frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x)\).

In this problem, \(y = x \cdot f(x)\) represents a product of two functions: \(x\) and \(f(x)\). Applying the Product Rule, the derivative \(\frac{dy}{dx}\) is calculated as the sum of \(f(x)\) and \(x \cdot \frac{df}{dx}\), where \(x\cdot \frac{df}{dx}\) comes from the differentiation of \(f(x)\). This rule allows us to neatly break down complex differential expressions into approachable parts, vital for coming to the right answer.

Function of a Function

A function of a function, sometimes called the Chain Rule in calculus, helps us when differentiating nested functions, i.e., a function within another. It is the principle used when you have an inner function nested within another outer function.

• The Chain Rule is expressed as: \(\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\).

This principle becomes useful in assessing \(f(x)\) when it's not in its simplest form or appears in a more complex structure. Understanding \( \frac{df}{dx} \) when using an assumed rational form like \(f(x)=\frac{ax+b}{c}\) falls under the umbrella of this concept. By simplifying the function into a direct relationship, you're effectively streamlining the derivative process, making it manageable to solve.

Calculus

Calculus, often viewed as the mathematics of change, is essential in understanding how functions evolve. It's divided into two main branches: differential calculus and integral calculus. In this exercise, differential calculus is our focus as it deals with rates of change, slopes of curves, and finding gradients.

• Differential calculus involves finding the first derivative, \(\frac{dy}{dx}\), of functions. This derivative represents the rate of change of a dependent variable, \(y\), with respect to an independent variable, \(x\).
• The aim here is to discover this rate and consider specific points, like \(x = 1\), to deduce exact numbers like \(-1\) in this context.

By breaking down complex expressions to their essential differentials, calculus not only simplifies complex relationships but also provides insight into the dynamic behavior of functions. It is calculus that allows us to transition from abstract equations to concrete analytical insights.