Question

If \(f(x) = \begin{cases} \frac{a\sin x - bx + cx^2 + x^3}{2\log(1+x) - 2x^3 + x^4} & , x \neq 0 \\ 0 & , x = 0 \end{cases}\) is continuous at \(x = 0\), then

• A. \(a = 2b\)
• B. \(a = b\)
• C. \(a = b = c\)
• D. \(b = c\)

Hint:

In limit problems involving parameters where the limit is finite (or zero), the lowest degree terms in the numerator and denominator usually determine the leading behavior. Equate the coefficient of the lowest power to satisfy the limit condition.

Answer 1

The Correct Option is B

Step 1: Condition for Continuity:
For \(f(x)\) to be continuous at \(x=0\), \(\lim_{x \to 0} f(x) = f(0) = 0\).
Step 2: Expand Numerator and Denominator:
Using Maclaurin series for small \(x\): Numerator \(N(x) = a(x - \frac{x^3}{6} + \dots) - bx + cx^2 + x^3\) \(N(x) = (a - b)x + cx^2 + (1 - \frac{a}{6})x^3 + \dots\) Denominator \(D(x)\): Looking at the expression \(2\log(1+x) - 2x^3 + x^4\). Note: There seems to be a slight ambiguity in the image between \(2x\) and \(2x^3\), but usually, the denominator is dominated by the lowest power term. Expansion of \(2\log(1+x) = 2(x - \frac{x^2}{2} + \frac{x^3}{3} - \dots) = 2x - x^2 + \dots\). Since the other terms in the denominator are higher powers (\(-2x^3, x^4\)), the leading term of the denominator is \(2x\).
Step 3: Evaluate Limit:
\[ \lim_{x \to 0} \frac{(a - b)x + cx^2 + \dots}{2x - x^2 + \dots} = \lim_{x \to 0} \frac{(a - b) + cx + \dots}{2 - x + \dots} = \frac{a - b}{2} \] For the limit to be 0, we must have: \[ \frac{a - b}{2} = 0 \implies a = b \] The coefficient \(c\) corresponds to the \(x^2\) term, which vanishes as \(x \to 0\) compared to the constant leading term after division by \(x\), so no specific condition on \(c\) is required for the limit to be 0. Final Answer:
\(a = b\).