Question:

If the function f(x)=sin3x+αsinxβcos3xx3,f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}, xRx \in \mathbb{R} , is continuous at x=0 x = 0 , then f(0) f(0) is equal to:

Updated On: Mar 20, 2025
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The Correct Option is D

Solution and Explanation

f(x)=sin3x+αsinxβcos3xx3 f(x) = \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3} is continuous at x=0 x = 0 .

limx03x(3x33)++α(xx333)β(1(3x)22)x3=f(0) \lim_{x \to 0} \frac{3x - \left(\frac{3x^3}{3}\right) + \dots + \alpha \left(\frac{x - \frac{x^3}{3}}{3}\right) - \beta \left(1 - \frac{(3x)^2}{2} \dots \right)}{x^3} = f(0)

Continuing with the limit:

limx0β+x(3+α)+9βx22+(273+α3)x3x3=f(0) \lim_{x \to 0} \frac{-\beta + x(3 + \alpha) + \frac{9 \beta x^2}{2} + \left(-\frac{27}{3} + \frac{\alpha}{3}\right)x^3 \dots}{x^3} = f(0)

For existence:

β=0,3+α=0,273+α3=f(0) \beta = 0, \quad 3 + \alpha = 0, \quad -\frac{27}{3} + \frac{\alpha}{3} = f(0)

Calculating:

α=3,276=36=f(0) \alpha = -3, \quad -\frac{27}{6} = -\frac{3}{6} = f(0)  
f(0)=27+36=4 f(0) = \frac{-27 + 3}{6} = -4

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