Let `f(x)={(e^x-1+ax)/x^2, x>0 and b, x=0 and sin(x/2)/x , x<0` , then

Options

A. f(X) is continuous at x=0 if a=- 1, ` b=(1)/(2)`
B. f(x) is discontinuous at x=0 if b`bne (1) /(2)`
C. f(x) has irremovable diisontinuity at x-0, if `a ne -1 `
D. f(x) has removable discontinuity at `x=0 if a-=-1, bne (1)/(2)`

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} \frac{e^x - 1 + ax}{x^2} & \text{if } x > 0 \\ b & \text{if } x = 0 \\ \frac{\sin(x/2)}{x} & \text{if } x < 0 \end{cases} \]

We need to determine the conditions under which the function is continuous at \( x = 0 \).

### Step 1: Find the left-hand limit as \( x \) approaches 0

For \( x < 0 \), we have:

\[ f(x) = \frac{\sin(x/2)}{x} \]

We need to calculate:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(x/2)}{x} \]

Using the substitution \( y = x/2 \) (thus \( x = 2y \)), we rewrite the limit:

\[ \lim_{y \to 0} \frac{\sin(y)}{2y} = \frac{1}{2} \cdot \lim_{y \to 0} \frac{\sin(y)}{y} = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

So,

\[ \lim_{x \to 0^-} f(x) = \frac{1}{2} \]

### Step 2: Find the right-hand limit as \( x \) approaches 0

For \( x > 0 \), we have:

\[ f(x) = \frac{e^x - 1 + ax}{x^2} \]

We need to calculate:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^x - 1 + ax}{x^2} \]

Using the Taylor series expansion for \( e^x \) around \( x = 0 \):

\[ e^x = 1 + x + \frac{x^2}{2} + O(x^3) \]

Substituting this into our limit gives:

\[ e^x - 1 = x + \frac{x^2}{2} + O(x^3) \]

Thus,

\[ e^x - 1 + ax = x + \frac{x^2}{2} + ax = (1 + a)x + \frac{x^2}{2} \]

Now substituting this back into our limit:

\[ \lim_{x \to 0^+} \frac{(1 + a)x + \frac{x^2}{2}}{x^2} = \lim_{x \to 0^+} \left( \frac{(1 + a)}{x} + \frac{1}{2} \right) \]

As \( x \to 0^+ \), the term \( \frac{(1 + a)}{x} \) will approach infinity unless \( 1 + a = 0 \). Therefore, for the limit to exist and be finite, we must have:

\[ 1 + a = 0 \implies a = -1 \]

If \( a = -1 \):

\[ \lim_{x \to 0^+} f(x) = \frac{1}{2} \]

### Step 3: Set the function value at \( x = 0 \)

For \( f(x) \) to be continuous at \( x = 0 \), we need:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]

Thus, we have:

\[ \frac{1}{2} = b \]

### Conclusion

For \( f(x) \) to be continuous at \( x = 0 \):

1. \( a = -1 \)
2. \( b = \frac{1}{2} \)

### Final Answer

The function \( f(x) \) is continuous at \( x = 0 \) if \( a = -1 \) and \( b = \frac{1}{2} \).

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