Problem 11 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)\) be a function satisfying the condition \(f(-x)=f(x)\), for all real \(x\). If \(f^{\prime}(0)\) exists, then its value is (A) 0 (B) 1 (C) \(-1\) (D) None of these

Step-by-Step Solution

Answer

The value of \( f'(0) \) is 0, so the correct answer is (A) 0.

Step 1: Understanding Even Functions

The first step is to recognize the condition given: \( f(-x) = f(x) \) for all real \( x \). This condition implies that the function \( f(x) \) is even. An even function is symmetric about the y-axis.

Step 2: Deriving the Property of Derivatives at Zero

For an even function, we can infer certain properties related to its derivative. Specifically, if an even function is differentiable at \( x = 0 \), its derivative at that point, \( f'(0) \), must be zero. This is because the slope of the tangent line at \( x=0 \) is horizontal due to symmetry about the y-axis.

Step 3: Applying the Property

Given that \( f'(0) \) exists, and \( f(x) \) is an even function, we apply the property that \( f'(0) = 0 \). Therefore, the value of \( f'(0) \) is calculated to be 0.

Key Concepts

Symmetry about the y-axis

An even function is one where the condition \( f(-x) = f(x) \) holds true for all real numbers \( x \). This property means that the function is symmetric about the y-axis.
Imagine folding the graph along the y-axis. For an even function, every point on the left of the y-axis has a matching point at the same height on the right side.
Some common examples of even functions include \( f(x) = x^2 \) and \( f(x) = \cos(x) \). These graphs exemplify this mirror-like symmetry perfectly.
Symmetrical functions often have unique properties, such as making certain calculations and visualizations more straightforward, due to their balanced structure.

Derivative at a point

The derivative of a function at a point, like \( f'(0) \), gives us valuable information about the slope of the tangent line at that specific point.
When considering an even function, like the one in the exercise, we have a condition which affects the derivative at the point \( x = 0 \).

• For even functions symmetric about the y-axis, the derivative at the origin \( f'(0) \) is always 0.
• This happens because at \( x = 0 \), the symmetry implies no net change in direction, making the tangent line horizontal.

Understanding this helps us make quick assessments about function behavior without needing explicit computation at every point. It simplifies prediction and analysis precisely at the origin.

Differentiability

Differentiability refers to a function's ability to have a derivative, which essentially represents the function's rate of change at any given point.
For a function to be differentiable at a point, it must be both continuous and smooth (no sharp corners or cusps) around that area.
In the context of our exercise, when we say \( f(x) \) is differentiable at 0, it confirms that we can calculate \( f'(0) \). For even functions, if differentiable at the origin:

• The derivative \( f'(0) \) will be 0.
• This follows from the symmetry about the y-axis, where the slope of the tangent at the origin (0,0) aligns horizontally.

This reinforces the importance of understanding the nature of the function and its behavior at specific points, helping students to grasp broader calculus concepts effectively.