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

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