Problem 37 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the function \(f(x)=\left\\{\begin{array}{cl}(1-|\tan x|)^{\frac{a}{\left.\right|^{\tan x \mid}}} & , \frac{-\pi}{4}

Step-by-Step Solution

Answer

Option (C) is correct: \( a = \frac{-3}{2}, b = e^{3/2} \).

Step 1: Understanding Continuity at a Point

A function \( f(x) \) is continuous at a point \( x = c \) if the following conditions are met:1. \( f(c) \) is defined.2. \( \lim_{x \to c} f(x) = f(c) \).For this problem, the function \( f(x) \) is defined at \( x = 0 \) as \( f(0) = b \). We need to ensure the limit of \( f(x) \) as \( x \to 0 \) from both sides equals \( b \).

Step 2: Calculate Left-Hand Limit (LHL)

For \( x \) approaching 0 from the left \((x \to 0^-)\), use the function definition \( (1-| an x|)^{\frac{a}{| an x|}} \).As \( x \to 0^- \), \( \tan x \to 0 \), so \( | an x| \to 0 \). Apply the limit:\[ \lim_{| an x| \to 0} (1-| an x|)^{\frac{a}{| an x|}} = e^{-a} \] Thus, \( \lim_{x \to 0^-} f(x) = e^{-a} \).

Step 3: Calculate Right-Hand Limit (RHL)

For \( x \to 0^+ \), the function is \( e^{\frac{\sin 3x}{\sin 2x}} \).As \( x \to 0 \), \( \sin 3x \approx 3x \) and \( \sin 2x \approx 2x \) by small angle approximations.\[ \lim_{x \to 0^+} e^{\frac{\sin 3x}{\sin 2x}} = \lim_{x \to 0^+} e^{\frac{3}{2}} = e^{3/2} \]Thus, \( \lim_{x \to 0^+} f(x) = e^{3/2} \).

Step 4: Applying Continuity Condition

For \( f(x) \) to be continuous at \( x = 0 \), the left-hand limit and right-hand limit must both equal \( f(0) = b \).From Step 2: \( e^{-a} = b \).From Step 3: \( e^{3/2} = b \).Therefore, equate the two limits: 1. \( e^{-a} = e^{3/2} \)Solving gives \( -a = \frac{3}{2} \), so \( a = \frac{-3}{2} \). Finally, \( b = e^{3/2} \).

Step 5: Identify Correct Option

From the calculations, we found \( a = \frac{-3}{2} \) and \( b = e^{3/2} \).Comparing this with the options given:- Option (C) is \( a = \frac{-3}{2}, b = e^{3/2} \), which is correct.

Key Concepts

Limit of a Function

The limit of a function is a fundamental concept in calculus. It describes the behavior of a function as the input approaches a particular point. For a function \( f(x) \) to be continuous at a point \( x = c \), the limit as \( x \) approaches \( c \) must equal the function's value at \( c \): \( \lim_{x \to c} f(x) = f(c) \). This means there should be no sudden jumps or breaks in the graph of the function at that point.

Understanding limits is essential for solving continuity problems because it determines whether the two-sided limits match the function's value at that point.

• Left-Hand Limit (LHL): This checks the behavior of the function as \( x \to c^- \) (from values less than \( c \)).
• Right-Hand Limit (RHL): This checks the behavior as \( x \to c^+ \) (from values greater than \( c \)).

For our exercise, continuity at \( x = 0 \) required both these limits to equal the defined value of \( b \). We calculated \( e^{-a} \) for the LHL and \( e^{3/2} \) for the RHL, and since both need to equal \( b \) for continuity, we set \( b = e^{3/2} \).

Trigonometric Functions

Trigonometric functions like \( \tan x \), \( \sin x \), and \( \cos x \) play a crucial role in calculus due to their periodic nature and unique properties. In this problem, we see \( \tan x \) in the expression \((1-|\tan x|)^{\frac{a}{|\tan x|}} \) for the interval \( -\pi/4 < x < 0 \). As \( x \) approaches 0 from the left, we need to evaluate the limit as \( \tan x \) approaches 0.

Key properties include:

• \( \tan x \) can be expressed as \( \sin x / \cos x \).
• As \( x \to 0 \), \( \tan x \to 0 \) since both \( \sin x \) and \( \cos x \) approach 0 and 1 respectively.
• The function \((1-|\tan x|)^{\frac{a}{|\tan x|}}\) is a form of \((1 - u)^{1/u} \to e^{-1}\) as \(u \to 0\), crucial for calculating limits involving trigonometric functions.

These properties allow us to simplify problems involving limits, as seen with the LHL calculation in the solution.

Exponential Functions

Exponential functions, noted by the form \( e^x \), serve as one of the most important classes of functions in mathematics due to their unique growth properties and derivatives. In the exercise, we analyze the function \( e^{\frac{\sin 3x}{\sin 2x}} \) for values of \( x \) approaching 0 from the right (\(0 < x < \pi/4\)). Understanding the behavior of exponential functions helps in dealing with expressions involving exponential growth or decay.

Essential characteristics include:

• Exponential functions have the property that \( e^{0} = 1 \).
• The derivative of \( e^x \) is \( e^x \), showing exponential growth.
• For small \( x \), \( \sin x \approx x \), a useful approximation for limits.

In the problem, as \( x \to 0 \), \( \sin 3x \approx 3x \) and \( \sin 2x \approx 2x \) lead to \( e^{\frac{3}{2}} \). This matches the RHL, establishing part of the function’s continuity at \( x=0 \). Understanding these concepts allowed us to find \( b = e^{3/2} \) effectively.