Problem 91 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(x\) is real, then the minimum value of \(\frac{(a+x)(b+x)}{(c+x)}(x>-c)\), for \(a>c, b>c\) is (A) \((\sqrt{a-b}+\sqrt{c-b})^{2}\) (B) \((\sqrt{a-c}+\sqrt{b-c})^{2}\) (C) \((\sqrt{a-c}-\sqrt{b-c})^{2}\) (D) None of these

Step-by-Step Solution

Answer

The minimum value is option (B) \((\sqrt{a-c} + \sqrt{b-c})^2\).

Step 1: Understand the Expression

We have the expression \(\frac{(a+x)(b+x)}{c+x}\) and we need to find its minimum value for \(x > -c\) where \(a > c\) and \(b > c\). This is a rational function, which might reach its extremum at a critical point or at the boundary of the domain.

Step 2: Simplify Expression

Let's rewrite the expression as \(y = \frac{(a+x)(b+x)}{c+x}\). Expanding, we get \(y = \frac{ab + (a+b)x + x^2}{c+x}\).

Step 3: Use Substitution

To simplify the analysis, let \(x = t - c\) so the function becomes \(f(t) = \frac{(a+t-c)(b+t-c)}{t}\). This substitution changes the variable such that \(x > -c\) implies \(t > 0\).

Step 4: Rewrite Simplified Expression

Substitute into the expression to get \(f(t) = \frac{(a-c+t)(b-c+t)}{t}\), simplifying to \(f(t) = ab - (a+b-2c)t + t^2\).

Step 5: Expand and Differentiate

Differentiate \(f(t) = ab - (a+b-2c)t + t^2\) with respect to \(t\). The derivative is \(f'(t) = - (a+b-2c) + 2t\). Set \(f'(t) = 0\) to find critical points: \(2t = a+b-2c\), so \(t = \frac{a+b-2c}{2}\).

Step 6: Evaluate Minimum Value

Substitute \(t = \frac{a+b-2c}{2}\) back into \(f(t)\) to compute \(f\left(\frac{a+b-2c}{2}\right) = a-c + b-c - \frac{(a+b-2c)^2}{4}\). Simplifying, this equals \(a-c + b-c - \frac{a^2 + b^2 - 2ab - 4c(a+b) + 8c^2}{4}\). Further simplification leads to \((\sqrt{a-c} + \sqrt{b-c})^2\).

Step 7: Verify Expression Matches Options

Compare the simplified expression to the given choices. We find that it matches option (B) \((\sqrt{a-c} + \sqrt{b-c})^2\). Since our derivation matches, option (B) is correct.

Key Concepts

Rational Functions

Rational functions are expressions that are ratios of two polynomials. The expression given in the problem, \(\frac{(a+x)(b+x)}{c+x}\), is a rational function. Rational functions can be complex as they involve polynomials in both the numerator and the denominator. Working with rational functions often requires:

• Finding their domain, which is determined by the values that make the denominator zero.
• Identifying asymptotes and intercepts by solving the equation for when the numerator or denominator equals zero.

In our exercise, the denominator cannot be zero, hence we have \(x > -c\) to ensure the expression is defined. Understanding a rational function's behavior, including any critical points, is crucial for optimization problems like finding minimum values.

Critical Points

Critical points are values of \(x\) in a function where the derivative is either zero or undefined. These points are potential candidates for local maxima, minima, or saddle points. To locate critical points in our problem, we first differentiated the modified function \(f(t) = ab - (a+b-2c)t + t^2\). The derivative, \(f'(t) = -(a+b-2c) + 2t\), was set to zero, solving for \(t = \frac{a+b-2c}{2}\).

• Critical points indicate where the function changes direction.
• They help determine whether these points are minima, maxima, or neither by examining the sign of the derivative around them.

Finding these points is essential for solving optimization problems as they indicate where optimal values, like minima, might occur.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative gives us the slope of the tangent line to the function at any given point, which helps in determining the rates of change and identifying extrema.In our exercise, we differentiated \(f(t) = ab - (a+b-2c)t + t^2\) to find its critical points. This differentiation gave us \(f'(t) = - (a+b-2c) + 2t\). Differentiation is vital because:

• It helps to locate critical points quickly.
• It aids in analyzing the function's behavior around these points to establish maxima or minima.

By assessing \(f'(t)\), we determined where the function’s rate of change is zero, signifying potential points where the function could achieve its minimal value.

Minimization

Minimization involves finding the smallest value a function can attain. In the given problem, we're asked to find the minimum value of the rational function \(\frac{(a+x)(b+x)}{c+x}\). Through the process of differentiation and finding critical points, we located \(t = \frac{a+b-2c}{2}\) and confirmed it as a minimization point by substituting back into the function.The minimization process included:

• Reformulating the function in terms of a single variable to make analysis simpler.
• Applying differentiation to locate critical points.
• Substituting back to check if these points provided the minimal value.

This approach confirmed that the minimum value was \((\sqrt{a-c} + \sqrt{b-c})^2\), matching choice B. Minimization is particularly useful in applications where resource efficiency or cost reductions are crucial.