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 solve the problem, we need to ensure that the given function \( f(x) \) is continuous at \( x = 0 \). Continuity at this point means that the left-hand limit, the value of the function at \( x = 0 \), and the right-hand limit must all be equal.

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

We have the function \( f(x) = (1 + ax)^{1/x} \) for \( x < 0 \). To find the limit as \( x \) approaches zero from the left, we use the logarithm:

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

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

1. Value at \( x = 0 \):

The value of the function at \( x = 0 \) is given by \( f(0) = 1 + b \).

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

For \( x > 0 \), \( f(x) = \frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} \). We need to find the limit as \( x \) approaches zero from the right:

Applying L'Hospital's Rule (since both numerator and denominator approach 0), we get:

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

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

1. Ensure continuity:

For continuity at \( x = 0 \), set the left-hand limit equal to the value at zero and the right-hand limit:

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

Solve these equations simultaneously.

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

From \( e^a = \frac{3}{4} c^{2/3} \), we get \((e^a)^3 = \left(\frac{3}{4}\right)^3 c^2\), solve to find \( c \).

\[ c = \left(\frac{4}{3}\right)^{3/2} e^{3a/2}. \]

Combine all to find \( e^a bc \):

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

Given all conditions satisfy \( e^a bc = 48 \).

Thus, the answer is 48.

Answer 2

For continuity at x = 0, we need \( f(0^-) = f(0) = f(0^+) \). 

\( f(0^-) = \lim_{x \to 0^-} (1+ax)^{1/x} = e^{\lim_{x \to 0^-} \frac{1}{x} \ln(1+ax)} = e^{\lim_{x \to 0^-} \frac{ax}{x}} = e^a \) \( f(0) = 1+b \) \( f(0^+) = \lim_{x \to 0^+} \frac{(x+4)^{1/2} - 2}{(x+c)^{1/3} - 2} \) 

Using L'Hopital's rule: \( f(0^+) = \lim_{x \to 0^+} \frac{\frac{1}{2}(x+4)^{-1/2}}{\frac{1}{3}(x+c)^{-2/3}} \) \( f(0^+) = \frac{\frac{1}{2}(4)^{-1/2}}{\frac{1}{3}(c)^{-2/3}} = \frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{3} c^{-2/3}} = \frac{\frac{1}{4}}{\frac{1}{3} c^{-2/3}} = \frac{3}{4} c^{2/3} \) 

From \( f(0^-) = f(0) \), we have \( e^a = 1+b \). From \( f(0) = f(0^+) \), we have \( 1+b = \frac{3}{4} c^{2/3} \). 
Also, we know that if \( (x+c)^{1/3} - 2 \) is in the denominator, then \( (x+c)^{1/3} - 2 = 0 \) at x = 0. \( c^{1/3} - 2 = 0 \) \( c^{1/3} = 2 \) \( c = 8 \) 

Now, \( 1+b = \frac{3}{4} (8)^{2/3} = \frac{3}{4} (2^3)^{2/3} = \frac{3}{4} \cdot 4 = 3 \) \( b = 2 \) 

Also, \( e^a = 1+b = 3 \) \( a = \ln 3 \) 

Therefore, \( e^a bc = 3 \cdot 2 \cdot 8 = 48 \)