Question

Let $\lim_{\varepsilon \to 0^+} \int_\varepsilon^x \frac{b t \cos 4t - a \sin 4t}{t^2} \, dt = \frac{a \sin 4x}{x} - 1,\quad (0<x<\frac{\pi}{4})$. Then $a$ and $b$ are given by

• A. a=2, b=2
• B. a=1/4, b=1
• C. a=-1, b=4
• D. a=2, b=4

Answer 1

The Correct Option is B

Let us define the function inside the integral: \[ I(x) = \int_\varepsilon^x \frac{b t \cos 4t - a \sin 4t}{t^2} \, dt \] We are given: \[ \lim_{\varepsilon \to 0^+} I(x) = \frac{a \sin 4x}{x} - 1 \]

Step 1: Differentiate both sides with respect to \( x \). Since the integral has variable upper limit \( x \), we can apply the Fundamental Theorem of Calculus:

\[ \frac{d}{dx} \left( \int_\varepsilon^x \frac{b t \cos 4t - a \sin 4t}{t^2} \, dt \right) = \frac{b x \cos 4x - a \sin 4x}{x^2} \]

Differentiate the right-hand side: \[ \frac{d}{dx} \left( \frac{a \sin 4x}{x} - 1 \right) \] Use quotient rule: \[ \frac{d}{dx} \left( \frac{a \sin 4x}{x} \right) = \frac{x \cdot a \cdot 4 \cos 4x - a \sin 4x}{x^2} = \frac{4a x \cos 4x - a \sin 4x}{x^2} \]

So, equating both derivatives: \[ \frac{b x \cos 4x - a \sin 4x}{x^2} = \frac{4a x \cos 4x - a \sin 4x}{x^2} \]

Multiply both sides by \( x^2 \): \[ b x \cos 4x - a \sin 4x = 4a x \cos 4x - a \sin 4x \]

Cancel \( -a \sin 4x \) from both sides: \[ b x \cos 4x = 4a x \cos 4x \]

Divide both sides by \( x \cos 4x \) (which is nonzero in given interval \( (0, \frac{\pi}{4}) \)): \[ b = 4a \]

Step 2: Now use the original equation and evaluate at \( x \to 0^+ \). Use the given: \[ \lim_{\varepsilon \to 0^+} \int_\varepsilon^x \frac{b t \cos 4t - a \sin 4t}{t^2} dt = \frac{a \sin 4x}{x} - 1 \] As \( x \to 0 \), use expansion: \[ \sin 4x \approx 4x - \frac{(4x)^3}{6} + \dots \Rightarrow \frac{\sin 4x}{x} \to 4 \Rightarrow \frac{a \sin 4x}{x} \to 4a \]

So: \[ \lim_{x \to 0^+} \left( \int_\varepsilon^x \ldots \right) \to 4a - 1 \] But this must match the limit of the integral from \( \varepsilon \to 0 \), so check the limit of the integral.

We observe that: \[ \lim_{\varepsilon \to 0^+} \int_\varepsilon^x \frac{b t \cos 4t - a \sin 4t}{t^2} \, dt = \frac{a \sin 4x}{x} - 1 \Rightarrow \text{LHS} \to \text{RHS as } \varepsilon \to 0 \]

From earlier, we concluded: \[ \frac{a \sin 4x}{x} - 1 \to 4a - 1 \text{ as } x \to 0 \] But observe: \[ \text{If the integrand is not improper at } t = 0, \text{ the limit must exist and be finite.} \] So: \[ \lim_{x \to 0} \left( \frac{a \sin 4x}{x} - 1 \right) = 4a - 1 \Rightarrow \text{To avoid divergence, we must have } 4a - 1 = 0 \Rightarrow a = \frac{1}{4} \]

Now substitute back: \[ b = 4a = 4 \cdot \frac{1}{4} = 1 \]

Answer:

\[ \boxed{a = \frac{1}{4}, \quad b = 1} \]