Question

If a real valued function \( f(x) = \begin{cases} (1 + \sin x)^{\csc x} & , -\frac{\pi}{2} < x < 0 \\ a & , x = 0 \\ \frac{e^{2/x} + e^{3/x}}{ae^{2/x} + be^{3/x}} & , 0 < x < \frac{\pi}{2} \end{cases} \) is continuous at \(x = 0\), then \(ab = \)

• A. \(e \)
• B. \(e^2 \)
• C. \(1 \)
• D. \(-1 \)

Hint:

For piecewise functions to be continuous at a point \(x=c\), the left-hand limit, the right-hand limit, and the function value at \(x=c\) must all be equal. For indeterminate forms like \(1^\infty\), use the property \(\lim_{x \to c} [g(x)]^{h(x)} = e^{\lim_{x \to c} h(x)[g(x)-1]}\). For limits involving exponential terms like \(e^{k/x}\) as \(x \to 0^+\), divide by the term with the largest positive exponent in the denominator.

Answer 1

The Correct Option is C

For the function \(f(x)\) to be continuous at \(x = 0\), the following condition must be satisfied: \[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \] We are given \(f(0) = a\). 
Step 1: Evaluate the left-hand limit (\(\lim_{x \to 0^-} f(x)\)).
For \(-\pi/2<x<0\), \(f(x) = (1 + \sin x)^{\operatorname{cosec} x}\). This limit is of the indeterminate form \(1^\infty\). We can evaluate it using the formula: \(\lim_{x \to c} [g(x)]^{h(x)} = e^{\lim_{x \to c} h(x)[g(x)-1]}\).
Here, \(g(x) = 1 + \sin x\) and \(h(x) = \operatorname{cosec} x = \frac{1}{\sin x}\).
\[ \lim_{x \to 0^-} (1 + \sin x)^{\operatorname{cosec} x} = e^{\lim_{x \to 0^-} \operatorname{cosec} x ((1 + \sin x) - 1)} \] \[ = e^{\lim_{x \to 0^-} \frac{1}{\sin x} (\sin x)} \] \[ = e^{\lim_{x \to 0^-} 1} = e^1 = e \] So, from the continuity condition, we have \(a = e\). 
Step 2: Evaluate the right-hand limit (\(\lim_{x \to 0^+} f(x)\)).
For \(0<x<\pi/2\), \(f(x) = \frac{e^{2/x} + e^{3/x}}{ae^{2/x} + be^{3/x}}\).
As \(x \to 0^+\), \(1/x \to \infty\). In the expression, \(e^{3/x}\) grows faster than \(e^{2/x}\).
To evaluate the limit, divide the numerator and the denominator by \(e^{3/x}\): \[ \lim_{x \to 0^+} \frac{e^{2/x} + e^{3/x}}{ae^{2/x} + be^{3/x}} = \lim_{x \to 0^+} \frac{\frac{e^{2/x}}{e^{3/x}} + \frac{e^{3/x}}{e^{3/x}}}{a\frac{e^{2/x}}{e^{3/x}} + b\frac{e^{3/x}}{e^{3/x}}} \] \[ = \lim_{x \to 0^+} \frac{e^{2/x - 3/x} + 1}{ae^{2/x - 3/x} + b} \] \[ = \lim_{x \to 0^+} \frac{e^{-1/x} + 1}{ae^{-1/x} + b} \] As \(x \to 0^+\), \(1/x \to \infty\), which means \(-1/x \to -\infty\). Therefore, \(e^{-1/x} \to 0\).
Substituting this into the limit expression: \[ = \frac{0 + 1}{a(0) + b} = \frac{1}{b} \] So, the right-hand limit is \(\frac{1}{b}\). 
Step 3: Equate the limits and \(f(0)\) to find \(a\) and \(b\).
From the continuity condition: \[ e = a = \frac{1}{b} \] From this, we have two equations:
1. \(a = e\)
2. \(e = \frac{1}{b} \implies b = \frac{1}{e}\)
Step 4: Calculate the value of \(ab\). \[ ab = (e) \times \left(\frac{1}{e}\right) = 1 \] The final answer is $\boxed{1}$.