Problem 100 - Calculus for JEE ADVANCED

Question

\(\frac{1}{2}(\) gof \()(x)=2 x^{2}-5 x+2\) or \(\quad \frac{1}{2} g[f(x)]=2 x^{2}-5 x+2\) \(\therefore \quad\left[\\{f(x)\\}^{2}+\\{f(x)\\}-2\right]=2\left[2 x^{2}-5 x+2\right]\) or \(f(x)^{2}+f(x)-\left(4 x^{2}-10 x+6\right)=0\) \(\therefore \quad f(x)=\frac{-1 \pm \sqrt{1+4\left(4 x^{2}-10 x+6\right)}}{2}\) \(=\frac{-1 \pm \sqrt{\left(16 x^{2}-40 x+25\right)}}{2}=\frac{-1 \pm(4 x-5)}{2}\) \(=2 x-3\) or \(-2 x+2\)

Step-by-Step Solution

Answer

The solutions for \( f(x) \) are \( 2x - 3 \) and \( -2x + 2 \).

Step 1: Identify the given equation

We are given that \( \frac{1}{2} g[f(x)] = 2x^2 - 5x + 2 \). We need to find \( f(x) \) such that the equation holds.

Step 2: Eliminate the coefficient

Multiply both sides of the equation \( \frac{1}{2} g[f(x)] = 2x^2 - 5x + 2 \) by 2 to eliminate the fraction: \( g[f(x)] = 4x^2 - 10x + 4 \).

Step 3: Re-arrange to standard polynomial form

Re-write the expression using \( f(x) \) as a variable in a polynomial using the substitution form \( f(x)^2 + f(x) = g[f(x)] \). Thus we have \( f(x)^2 + f(x) - (4x^2 - 10x + 4) = 0 \).

Step 4: Solve the quadratic equation

The quadratic equation is \( f(x)^2 + f(x) - (4x^2 - 10x + 4) = 0 \). Use the quadratic formula \( f(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here \( a = 1 \), \( b = 1 \), and \( c = -(4x^2 - 10x + 4) \).

Step 5: Calculate the discriminant

The discriminant \( \Delta = b^2 - 4ac \) is calculated as follows: \( \Delta = 1^2 - 4(1)(-(4x^2 - 10x + 4)) = 1 + 16x^2 - 40x + 16 \). Simplify to \( 16x^2 - 40x + 25 \).

Step 6: Find solutions for f(x)

Substitute the discriminant back into the quadratic formula: \( f(x) = \frac{-1 \pm \sqrt{16x^2 - 40x + 25}}{2} \). Simplify: \( f(x) = \frac{-1 \pm (4x - 5)}{2} \).

Step 7: Separate into two cases

Consider both cases from the simplified equation: \( f(x) = \frac{-1 + 4x - 5}{2} = 2x - 3 \) or \( f(x) = \frac{-1 - 4x + 5}{2} = -2x + 2 \).

Key Concepts

Function Composition

Function composition is a method of combining two functions to form a new function. This involves applying one function to the results of another. Consider functions \( f(x) \) and \( g(x) \). The notation \( (g \circ f)(x) \) represents the function \( g \) applied to the result of \( f(x) \).
In the exercise, we see \( \frac{1}{2} g[f(x)] = 2x^2 - 5x + 2 \). Here, \( g[f(x)] \) indicates that \( f(x) \) is used first, and its result is the input for \( g \). This structure is key to solving complex problems.

• Function composition allows breaking down complex expressions into simpler pieces.
• It is crucial in scenarios where nested functions are involved.
• Understanding the order of operations in function composition is important.

Discriminant

The discriminant is a part of the quadratic formula, used to determine the nature of the roots of a quadratic equation. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant \( \Delta \) is given by \( b^2 - 4ac \).
The discriminant value can classify the roots of the quadratic equation:

• If \( \Delta > 0 \), there are two distinct real roots.
• If \( \Delta = 0 \), there is exactly one real root (repeated root).
• If \( \Delta < 0 \), there are no real roots, but two complex roots.

In the problem, the discriminant was calculated as \( 16x^2 - 40x + 25 \), helping identify the roots of the given equation.

Polynomial Equation

A polynomial equation is an expression made up of terms, which are variables raised to a power and multiplied by coefficients. The general form of a polynomial of degree \( n \) is \( a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \). Each term involves the variable \( x \) raised to a non-negative integer power.
In the exercise, we dealt with the quadratic polynomial equation \( f(x)^2 + f(x) - (4x^2 - 10x + 4) = 0 \), where \( f(x) \) was treated as a variable.
Understanding polynomial equations is essential as they appear in many mathematical contexts:

• Polynomials with degree 2, like our example, are called quadratics.
• They can model various real-world scenarios, such as physics and economics.
• The solutions of polynomial equations are the roots of the equation.

Quadratic Formula

The quadratic formula is a classic method to find the solutions of a quadratic equation \( ax^2 + bx + c = 0 \). The formula is written as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]This formula provides a straightforward way to find the roots of any quadratic equation, assuming the coefficients \( a \), \( b \), and \( c \) are known.
Using the quadratic formula in the exercise involved substituting \( a = 1 \), \( b = 1 \), and \( c = -(4x^2 - 10x + 4) \) to solve for \( f(x) \).

• The quadratic formula requires computing the discriminant, which influences the type of roots obtained.
• It is versatile and applicable when factoring is difficult.
• This formula is a critical tool in algebra for solving quadratic equations thoroughly.