Problem 113 - A Complete Resource Book in Mathematics for JEE Main

Question

In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: If \(f(x)=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\), then \(f(x)\) is differentiable everywhere Reason: \(f^{\prime}(x)=\left\\{\begin{array}{l}\frac{-2}{1+x^{2}}, \text { if }|x|<1 \\\ \frac{2}{1+x^{2}}, \text { if }|x|>1\end{array}\right.\)

Step-by-Step Solution

Answer

D: Assertion(A) is False, Reason(R) is True

Step 1: Evaluate the Differentiability of f(x)

First, let's consider the function \( f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right) \). We need to evaluate if it is differentiable everywhere. The function inside \( \cos^{-1} \) is defined and continuous for all \( x \), and the range of \( \frac{2x}{1+x^2} \) for all real \( x \) is between -1 and 1, inclusive. Therefore, \( f(x) \) is differentiable for all x because the composition of a differentiable function (\( \cos^{-1} \)) with the inner function is differentiable.

Step 2: Check the Reason

Now, let's consider the derivative \( f'(x) \). According to the provided reason, the derivative is piecewise: \( f'(x) = \frac{-2}{1+x^2} \) for \(|x|<1\) and \( f'(x) = \frac{2}{1+x^2} \) for \(|x| > 1\). To determine the correctness, calculate the derivative of \( \cos^{-1}\left( \frac{2x}{1+x^2} \right) \) using the chain rule and verify if the piecewise derivative holds true at all points.

Step 3: Verify Derivative at Critical Points

Check continuity of \( f'(x) \) at \( x = 1 \) and \( x = -1 \). Substituting these points into the derivative expressions, we find that \( f'(x) \) is not defined at \( x = 1 \) and \( x = -1 \) in the given piecewise form, revealing discontinuity or non-differentiability at these points. This means the reason is incorrect as \( f(x) \) cannot be differentiable everywhere with a discontinuous derivative.

Step 4: Comparison between Assertion and Reason

Since the assertion claims \( f(x) \) is differentiable everywhere but the reason provides a piecewise derivative that fails at boundary points, we can conclude that the assertion is false. The reason provided itself is true in its piecewise expression, but irrelevant to a true differentiability proof.

Key Concepts

Chain Rule

When you are dealing with composite functions, like our function \( f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right) \), you will often use the chain rule to find the derivative. The chain rule is a fundamental concept in calculus that helps you differentiate compositions of functions. It states that if you have two functions \( g(x) \) and \( h(x) \), where \( f(x) = g(h(x)) \), the derivative \( f'(x) \) is given by the formula:

• \( f'(x) = g'(h(x)) \cdot h'(x) \)

Breaking it down:

• Find the derivative of the outer function \( g'(h(x)) \) and leave the inner function untouched in its argument.
• Then, multiply it with the derivative of the inner function, \( h'(x) \).

In practice, for our case, you'd determine the derivative of the \( \cos^{-1} \) part (the outer function) and multiply it by the derivative of \( \frac{2x}{1+x^2} \) (the inner function). This approach is crucial in ensuring you get the correct derivative for a complex, nested function.

Cosine Inverse Function

The cosine inverse function, also known as \( \cos^{-1} \) or arc-cosine, is defined to provide you with an angle whose cosine is a specific value. It is restricted to the range from 0 to \( \pi \) radians to maintain one-to-one correspondence and be invertible.

• The domain refers to the values that can be plugged into the function, ranging from -1 to 1.
• The range indicates the output, which is all angles from 0 to \( \pi \) radians.

Understanding these constraints will help you grasp why the inverse cosine function is differentiable for inputs within its defined domain. Its derivative is given by:

• \( \frac{d}{dx} \left[\cos^{-1}(x)\right] = -\frac{1}{\sqrt{1-x^2}} \)

This shows that the derivative is negative since it measures the rate of change in radian measure for changes in input. This is particularly useful when you're applying the chain rule where the cosine inverse appears as part of the composite function.

Piecewise Functions

A piecewise function is a function composed of multiple sub-functions, each valid over certain intervals of the domain. It is like having different formulas depending on different conditions or parts of the number line. For the derivative provided in the original problem, the function splits based on whether \( |x| < 1 \) or \( |x| > 1 \).

• In piecewise functions, continuity and differentiability can vary between segments. Each segment might have its own rule for differentiation or calculation, demanding a close evaluation of boundary points.
• It's crucial to check these boundaries, like \( x = 1 \) or \( x = -1 \), to ensure there are no jumps or undefined behaviors.

In the exercise, the piecewise derivative \( f'(x) \) suggested there's an issue at these boundaries, leading to discontinuities. This illustrates that piecewise definitions can be tricky: a function may be non-differentiable even when seemingly each piece is continuous and differentiable separately. Always check transitions smoothly link up without any holes or breaks.