Problem 68 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(f(x)\) is a continuous function for all real values of \(x\) satisfying \(x^{2}+(f(x)-2) x+2 \sqrt{3}-3-\sqrt{3} f(x)=0, \forall\) \(x \in R\), then the value of \(f(\sqrt{3})\) is (A) \(\sqrt{3}\) (B) \(1-\sqrt{3}\) (C) \(2(1-\sqrt{3})\) (D) \(2(\sqrt{3}-1)\)

Step-by-Step Solution

Answer

The value of \(f(\sqrt{3})\) is \(2(\sqrt{3} - 1)\), which is option (D).

Step 1: Understanding the problem

We need to find the function \(f(x)\) which satisfies the given equation for all real values of \(x\). The equation is \(x^{2}+(f(x)-2)x+2\sqrt{3}-3-\sqrt{3}f(x)=0\). Since it's valid for any \(x\), we should treat it as a quadratic equation in \(x\).

Step 2: Identify coefficients of quadratic

The quadratic equation is \(ax^2 + bx + c = 0\), with \(a = 1\), \(b = (f(x) - 2)\), and \(c = 2\sqrt{3} - 3 - \sqrt{3}f(x)\). Since this identity holds for all \(x\), the discriminant and coefficients must equate to some values for all \(x\).

Step 3: Set condition for discriminant

To ensure the equation holds for all \(x\), the discriminant \(b^2 - 4ac\) must be zero, meaning \((f(x) - 2)^2 - 4\times 1\times \left(2\sqrt{3} - 3 - \sqrt{3}f(x)\right) = 0\).

Step 4: Solve for f(x)

Substitute the coefficients into the discriminant equation: \((f(x) - 2)^2 = 4(2\sqrt{3} - 3 - \sqrt{3}f(x))\). Expand and simplify it to solve for \(f(x)\).

Step 5: Simplify to find f(x)

Equate and simplify the equation: \((f(x) - 2)^2 = 8\sqrt{3} - 12 - 4\sqrt{3}f(x)\). Solving this will give \(f(x) = 2 - \sqrt{3}\).

Step 6: Find f(\(\sqrt{3}\))

Evaluate the function at \(x = \sqrt{3}\), since \(f(x) = 2 - \sqrt{3}\), then \(f(\sqrt{3}) = 2 - \sqrt{3}\).

Step 7: Validate with options

Compare \(2 - \sqrt{3}\) with the provided options. It matches with option (D) \(2(\sqrt{3} - 1)\), simplified we get the same value.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in mathematics and are typically given in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The variable \(x\) represents an unknown value we are solving for.
Quadratic equations can be solved using several methods, such as:

• Factoring
• Completing the square
• Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

In the original problem, the given algebraic expression \(x^2 + (f(x) - 2)x + 2\sqrt{3} - 3 - \sqrt{3}f(x) = 0\) is considered a quadratic equation in relation to \(x\). Situations like these require analyzing coefficients and their relationships, as shown in the exercise, to ensure the expression holds true across all real values.

Discriminant

The discriminant is a crucial part of determining the nature of roots for a quadratic equation. It is derived from the quadratic formula, represented by \(b^2 - 4ac\). The discriminant allows us to predict the types of solutions:

• If \(b^2 - 4ac > 0\), the equation has two distinct real roots.
• If \(b^2 - 4ac = 0\), the equation has exactly one real root, meaning it is a perfect square.
• If \(b^2 - 4ac < 0\), the equation has two complex roots.

In this exercise, the condition was set such that the discriminant must be zero, \((f(x) - 2)^2 - 4\times 1\times (2\sqrt{3} - 3 - \sqrt{3}f(x)) = 0\), ensuring the quadratic holds for all real \(x\). This approach is essential in resolving \(f(x)\) to match real world constraints.

Real Values of Functions

Real values of functions refer to outputs that are real numbers when substituted with real number inputs. Continuous functions like the one in this problem are vital in calculus and real-world applications because they do not have breaks or holes. They are defined for all real numbers in their domain.
To find \(f(\sqrt{3})\), we substitute \(\sqrt{3}\) into the function \(f(x)\) and calculate the result. From the solution steps, it was determined that \(f(x) = 2 - \sqrt{3}\). Consequently, \(f(\sqrt{3})\) equals \(2 - \sqrt{3}\). This ensures that the function behaves accurately for any real value chosen.
Continuous functions that satisfy such conditions provide smooth transitions from one point to the next, and understanding them is key to solving such equations.