Question

Let $ f(x) = \begin{cases} (1+ax)^{1/x} & , x<0 \\1+b & , x = 0 \\\frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} & , x>0 \end{cases} $ be continuous at x = 0. Then $ e^a bc $ is equal to

• A. 64
• B. 72
• C. 48
• D. 36

Hint:

For continuity, \( \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a) \). Use L'Hopital's rule when dealing with indeterminate forms.

Answer 1

The Correct Option is C

To establish continuity of the function \( f(x) \) at \( x = 0 \), the left-hand limit, the function's value at \( x = 0 \), and the right-hand limit must be equivalent.

1. Left-hand limit as \( x \to 0^- \):

For \( x<0 \), \( f(x) = (1 + ax)^{1/x} \). The limit as \( x \) approaches 0 from the left is calculated using logarithms:

\[ \begin{align*} \lim_{x \to 0^-} (1 + ax)^{1/x} &= \exp\left(\lim_{x \to 0^-} \frac{\ln(1 + ax)}{x}\right) \\ &= \exp(a \cdot \lim_{x \to 0^-} \frac{x}{x}) = \exp(a \cdot 1) = e^a. \end{align*} \]

The left-hand limit is \( e^a \).

1. Function value at \( x = 0 \):

The function's value at \( x = 0 \) is \( f(0) = 1 + b \).

1. Right-hand limit as \( x \to 0^+ \):

For \( x>0 \), \( f(x) = \frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} \). The limit as \( x \) approaches 0 from the right is determined using L'Hospital's Rule due to indeterminate forms:

\[ \begin{align*} \lim_{x \to 0^+} \frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} &= \lim_{x \to 0^+} \frac{\frac{1}{2}(x+4)^{-1/2}}{\frac{1}{3}(x+c)^{-2/3}} \\ &= \lim_{x \to 0^+} \frac{3(x+4)^{-1/2}}{2(x+c)^{-2/3}} \\ &= \frac{3}{2} \cdot \frac{4^{-1/2}}{c^{-2/3}} = \frac{3}{2} \cdot \frac{1}{2} \cdot c^{2/3} = \frac{3}{4} c^{2/3}. \end{align*} \]

The right-hand limit is \(\frac{3}{4} c^{2/3}\).

1. Continuity condition:

For continuity at \( x = 0 \), the following equality must hold:

\[ e^a = 1 + b = \frac{3}{4} c^{2/3}. \]

Solving these equations yields:

From \( e^a = 1 + b \), \( b = e^a - 1 \).

From \( e^a = \frac{3}{4} c^{2/3} \), solving for \( c \) gives \( c = \left(\frac{4}{3}\right)^{3/2} e^{3a/2} \).

Substituting these into \( e^a bc \) results in:

\[ \begin{align*} e^a bc &= e^a \cdot (e^a - 1) \cdot \left(\frac{4}{3}\right)^{3/2} e^{3a/2} \\ &= e^{5a/2} \cdot (e^a - 1) \cdot \left(\frac{4}{3}\right)^{3/2}. \end{align*} \]

Given the condition \( e^a bc = 48 \), the final result is 48.

The answer is 48.