Problem 88 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the function \(f(x)=\frac{\sin 3 x+a \sin 2 x+b \sin x}{x^{5}}, x \neq 0\) is continuous at \(x=0\), then (A) \(a=-4\) (B) \(b=5\) (C) \(a=4\) (D) \(f(0)=1\)

Step-by-Step Solution

Answer

(A) \(a = -4\), (B) \(b = 5\), (D) \(f(0) = 1\).

Step 1: Understanding Continuity

For the function \( f(x) \) to be continuous at \( x = 0 \), the limit of \( f(x) \) as \( x \) approaches 0 must exist and equal \( f(0) \). This implies \( \lim_{x \to 0} \frac{\sin 3x + a \sin 2x + b \sin x}{x^5} = f(0) \).

Step 2: Applying L'Hôpital's Rule

The expression \( \lim_{x \to 0} \frac{\sin 3x + a \sin 2x + b \sin x}{x^5} \) is an indeterminate form \( \frac{0}{0} \) at \( x = 0 \). We will use L'Hôpital's Rule, which states that for such limits, \( \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} \), provided the latter limit exists.

Step 3: Differentiating the Numerator and Denominator

Differentiate the numerator \( N(x) = \sin 3x + a \sin 2x + b \sin x \). The derivative is \( N'(x) = 3 \cos 3x + 2a \cos 2x + b \cos x \). Differentiate the denominator \( D(x) = x^5 \) to get \( D'(x) = 5x^4 \).

Step 4: First Application of L'Hôpital's Rule

Now apply L'Hôpital's Rule: \( \lim_{x \to 0} \frac{N'(x)}{D'(x)} = \lim_{x \to 0} \frac{3 \cos 3x + 2a \cos 2x + b \cos x}{5x^4} \). Evaluating this still gives \( \frac{0}{0} \), so apply L'Hôpital's rule again.

Step 5: Second Derivative Calculation

Differentiate the numerator: \( (N'(x))' = -9 \sin 3x - 4a \sin 2x - b \sin x \). Differentiate the denominator again: \( (D'(x))' = 20x^3 \).

Step 6: Second Application of L'Hôpital's Rule

Apply L'Hôpital's Rule again: \( \lim_{x \to 0} \frac{(N'(x))'}{(D'(x))'} = \lim_{x \to 0} \frac{-9 \sin 3x - 4a \sin 2x - b \sin x}{20x^3} \). This still results in \( \frac{0}{0} \), requiring another differentiation.

Step 7: Continue Derivating

Proceed with differentiating the numerator and denominator two more times until the first non-zero constant part is left to ensure the continuity condition.

Step 8: Solve for Coefficients

After calculating and simplifying to the non-zero constant term, equate the limit with \( f(0) = 1 \). The remaining expressions will help you solve for \( a \) and \( b \). After organizing terms and variable powers, solve the equations: \( \frac{6b - 8a + 54}{120} = 1 \). Solve for \( a \) and \( b \).

Step 9: Determine Values

From the equation \( 6b - 8a + 54 = 120 \), solve to find \( a = -4 \) and \( b = 5 \). Substituting these values, you find \( f(0) = 1 \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a wonderful tool for calculus students, used to evaluate limits that produce the indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It helps us find the limit of a ratio of two functions when direct substitution doesn't work.

Here's how it works:

• Identify that the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) yields \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Differentiation comes into play. Differentiate the numerator \( f(x) \) to get \( f'(x) \).
• Also, differentiate the denominator \( g(x) \) to get \( g'(x) \).
• Evaluate the limit of the new quotient: \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \).

This process can be repeated if the new limit also presents an indeterminate form, as seen in the provided problem where it was applied multiple times. Each differentiation step simplifies the limits further until the indeterminate form is resolved.

Indeterminate form

The term "indeterminate form" refers to expressions in calculus where substitution into a function's limit results in undefined values, typically \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms indicate that the limit cannot be directly determined without further manipulation.

Why does this happen?

• When both the numerator and denominator of a fraction approach zero, it becomes unclear what the fraction tends towards: \( \frac{0}{0} \).
• Likewise, if both tend towards infinity: \( \frac{\infty}{\infty} \).

These forms necessitate techniques like L'Hôpital's Rule to resolve them. By differentiating the numerator and the denominator separately, we aim to find a new limit that doesn't produce an indeterminate form. This peeling away of complexity helps uncover the true nature of the limit, as was needed in the function given in the exercise.

Differentiation

Differentiation is a fundamental concept in calculus, which allows us to find the rate of change or the slope of a function at any given point. When dealing with limits, especially when L'Hôpital's Rule is applied, differentiation becomes our best friend.

Why is it so crucial here?

• For L'Hôpital's Rule to be applicable, we need derivatives of both the numerator and denominator.
• Repeated differentiation may be necessary to break down complex expressions into simpler parts.
• Each differentiation step brings clarity to limits where indeterminate forms initially cloud outcomes.

In our example, differentiation was used multiple times. Each round produced new limits that were easier to handle. Derivatives of trigonometric functions and power functions both illustrated how every differentiation step provided the necessary tools, honing in on the exact value needed to ensure continuity at \( x = 0 \).