Problem 7 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Show that the function \(f(x)=\sin x, \quad x\) is a rationalno. \(=x, \quad x\) is an irrationalno. has positive derivative at \(x=0\) but \(f(x)\) is not increasing at \(x=0\).

Step-by-Step Solution

Answer

In conclusion, the function \(f(x)\) has a positive derivative at \(x = 0\), with \(f'(0) = 1\). However, by providing a counterexample with \(x_{1}=\frac{\sqrt{2}}{2}\) (irrational) and \(x_{2}=\frac{1}{2}\) (rational), we demonstrated that \(f(x)\) is not increasing at \(x = 0\), as \(f(x_{1}) > f(x_{2})\) even though \(0 < x_{1} < x_{2}\).

Step 1: Compute the left-hand side (LHS) derivative at x = 0

Recall that the left-hand derivative at x is given by: \[ f'(x^-) = \lim_{h \to 0^-} \frac{f(x+h) - f(x)}{h} \] As x approaches 0 from the left, these x values are irrational numbers. Thus, we can compute the LHS derivative at x = 0 as: \[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{h - 0}{h} = \lim_{h \to 0^-} 1 \]

Step 2: Compute the right-hand side (RHS) derivative at x = 0

Now, compute the right-hand derivative at x = 0 as follows: \[ f'(x^+) = \lim_{h \to 0^+} \frac{f(x+h) - f(x)}{h} \] As x approaches 0 from the right, these x values are rational numbers. Thus, we can compute the RHS derivative at x = 0 as: \[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\sin h - 0}{h} \] Now, using L'Hôpital's Rule: \[ \lim_{h \to 0^+} \frac{\sin h}{h} = \lim_{h \to 0^+} \frac{\cos h}{1} = \cos 0 = 1 \]

Step 3: Determine the derivative at x = 0

Since both the LHS and RHS derivatives are equal and exist at x = 0, the derivative of the function at x = 0 exists and is given by: \[ f'(0) = f'(0^-) = f'(0^+) = 1 \] So, the function has a positive derivative at x = 0.

Step 4: Provide a counterexample to show f(x) is not increasing at x = 0

To show that the function is not increasing at x = 0, we need to find an irrational number x1 and a rational number x2 such that \(0 < x_{1} < x_{2}\) and \(f(x_{1}) > f(x_{2})\). Let's consider \(x_{1} = \frac{\sqrt{2}}{2}\) (which is irrational) and \(x_{2} = \frac{1}{2}\) (which is rational). Then, we have: \[ f(x_{1}) = f\left(\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2} \quad \text{and} \quad f(x_{2}) = f\left(\frac{1}{2}\right) = \sin\left(\frac{1}{2}\right) \] Since \(\sin\left(\frac{1}{2}\right) < \frac{\sqrt{2}}{2}\), we have \(f(x_{1}) > f(x_{2})\) even though \(0 < x_{1} < x_{2}\). This proves that f(x) is not increasing at x = 0. In conclusion, we have shown that the function f(x) has a positive derivative at x = 0 but is not increasing at x = 0.

Key Concepts

Derivative

In calculus, the derivative of a function measures how the function's value changes as the input changes. Imagine it as the slope of the tangent line to the function at a particular point. This makes it a crucial tool for understanding how functions behave.
To compute a derivative, we often use the limit definition:

• For a function \( f(x) \), its derivative at a point \( x \) is \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).

This method is applied in the step-by-step solution to evaluate the derivative \( f'(0) \) of the function \( f(x) \) at \( x = 0 \). The derivative tells us the function is positive at this point, yet it does not guarantee the function is increasing. It's essential to distinguish between having a positive derivative and the function being strictly increasing.

L'Hôpital's Rule

L'Hôpital's Rule is a handy technique for finding limits of indeterminate forms, typically \( 0/0 \) or \( \infty/\infty \). It works by taking the derivative of the numerator and the derivative of the denominator, and then evaluating the limit again.
For example, if the limit of a function \( \frac{f(x)}{g(x)} \) yields an indeterminate form, we can write:

• \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \)

This rule is applied when computing the right-hand side derivative at \( x = 0 \) in the solution. The application of L'Hôpital's Rule simplifies the limit \( \lim_{h \to 0^+} \frac{\sin h}{h} \) to \( \cos 0 \), which equals 1. This result, together with the left derivative, confirms the existence and positivity of the derivative at \( x = 0 \).

Rational and Irrational Numbers

Rational and irrational numbers form the two main classes of real numbers in mathematics. Understanding them is essential, especially when analyzing functions that react differently based on the type of number.

• Rational numbers: These can be expressed as a ratio of two integers, such as \( \frac{1}{2} \) or \( -3 \).
• Irrational numbers: These cannot be expressed as a simple fraction. Common examples are \( \sqrt{2} \), \( \pi \), and \( e \).

In the exercise, the property of rational and irrational numbers is used to analyze the behavior of the function \( f(x) \) near \( x = 0 \). Specifically, it distinguishes the derivative evaluation on rational and irrational approaches to zero and demonstrates that the function is not increasing at \( x = 0 \). This understanding underpins why such calculations are valid across different number types.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in calculus and appear frequently in limits, derivatives, and integrals. They describe the relationships of angles and sides in right-angled triangles.

• \( \sin(x) \) represents the sine function, which plots a wave-like curve known as a sinusoid.
• \( \cos(x) \) represents the cosine function, which is a leading hand in calculus due to its derivatives and integrals.

In the solution, the function \( f(x) = \sin x \) is analyzed. Specifically, understanding the limit approach \( \lim_{h \to 0^+} \frac{\sin h}{h} = 1 \) is derived from the behavior of trigonometric functions near zero. This behavior is crucial for evaluating the derivative and demonstrating the function's non-increasing property at \( x = 0 \). Recognizing these functions' periodicity and continuity helps students tackle similar problems in calculus effectively.