Problem 45 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If for all \(x, y\) the function \(f\) is defined by \(f(x)+f(y)+\) \(f(x) \cdot f(y)=1\) and \(f(x)>0\), then (A) \(f^{\prime}(x)\) does not exist (B) \(f^{\prime}(x)=0\) for all \(x\) (C) \(f^{\prime}(0)

Step-by-Step Solution

Answer

(B) \(f'(x) = 0\) for all \(x\).

Step 1: Understand the Function

We are given that \(f(x) + f(y) + f(x) \cdot f(y) = 1\) for all \(x\) and \(y\). This equation suggests that the function is independent of \(x\) and \(y\), because the left side must equal 1 regardless of the values of \(x\) and \(y\).

Step 2: Assume a Constant Function

Assume that \(f(x) = c\) for some constant \(c\). Substitute into the equation to verify: \(c + c + c \cdot c = 1\). This simplifies to \(2c + c^2 = 1\).

Step 3: Solve for c

Solve the equation \(c^2 + 2c - 1 = 0\) using the quadratic formula, \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=2\), and \(c=-1\). This yields \(c = \frac{-2 \pm \sqrt{2^2 + 4}}{2} = \frac{-2 \pm \sqrt{8}}{2}\).

Step 4: Simplify the Solutions for c

\(\sqrt{8} = 2\sqrt{2}\), hence \(c = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}\). Since \(f(x) > 0\), we choose \(c = -1 + \sqrt{2}\).

Step 5: Determine f'(x)

Since \(f(x) = c = -1 + \sqrt{2}\), it is a constant function. The derivative of a constant function is zero, so \(f'(x) = 0\) for all \(x\).

Step 6: Identify the Correct Option

Since \(f'(x) = 0\) for all \(x\), option (B) is correct: \(f'(x) = 0\) for all \(x\).

Key Concepts

Constant Function

A constant function is one of the simplest types of functions you'll encounter in mathematics. It assigns the same value to every point in its domain. For example, if you have a function defined as \( f(x) = c \), where \( c \) is a constant, this is a constant function. No matter what \( x \) is, the output remains \( c \).

In practical terms, consider a bank account that earns no interest, and has a constant balance. This balance does not change, no matter how many years pass. Mathematically, you could say the function \( f(x) \) describing the balance of the account is a constant function.

In relation to derivatives, if \( f(x) \) is a constant function, the slope of the curve at every point is zero. Therefore, the derivative of a constant function is always zero. This nature makes constant functions particularly useful in various mathematical and statistical modeling.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two. It has the general form \( ax^2 + bx + c = 0 \) where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic equations can model parabolic paths, such as the arc of a ball being thrown.

To find the solutions for a quadratic equation, you can use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

It's important to note that the expression under the square root sign, \( b^2 - 4ac \), is known as the discriminant. The value of the discriminant determines the nature of the roots:

• If it's positive, there are two distinct real roots.
• If zero, there's exactly one real root, also known as a double root or repeated root.
• If negative, there are no real roots, but two complex roots.

Understanding how to solve quadratic equations is crucial, as this knowledge extends further into complex algebraic problems and real-world applications.

Derivative of Constants

The derivative of a constant function is straightforward. When you have a constant function, like \( f(x) = c \), where \( c \) is any constant, its derivative is zero. This is because a constant function does not change; thus, it has a slope of zero everywhere on the graph.

To grasp why this is, remember that the derivative measures the rate at which a function changes. Since a constant remains the same across all values of \( x \), its rate of change is zero. This can be expressed mathematically as:

• \( f'(x) = 0 \)

Learning this concept is key in calculus, where many complex equations can include constant terms. This fundamental understanding simplifies how we handle derivatives, allowing more focus on parts of functions that exhibit change.

Positive Functions

A positive function is a function that yields positive outputs for all inputs from its domain. For example, if \( f(x) > 0 \) for all \( x \), then \( f \) is a positive function. These functions can be crucial in modeling scenarios where negative values do not make sense, such as calculating area or probability.

When dealing with specific problems, if a function is defined as positive like \( f(x) = -1 + \sqrt{2} \) which results from our previous example, it's important to ensure the solution meets this condition. Here, the challenge is to verify the chosen function is indeed positive according to the constraint.

Positive functions are frequently seen in real-world applications where only positive quantities are meaningful, such as determining the length, weight, or anything that cannot logically be negative. Understanding how to work with these functions is an important skill in both pure and applied mathematics.