If `g(x)=(1-a^x+x a^xloga)/(x^2*a^x), x<0` `((2a)^x-xlog(2a)-1)/(x^2), x >0` (where a > 0) then find a and `g(0)` so that `g(x)` is continuous at `x=0.`

Options

A. (a)`(-1)/(sqrt(2))`
B. (b)`(1)/(sqrt(2))`
C. (c)2
D. (d)-2

Answer

To find the value of \( a \) and \( g(0) \) such that the function \( g(x) \) is continuous at \( x = 0 \), we need to analyze the limits of \( g(x) \) as \( x \) approaches 0 from both sides.

### Step 1: Define the function for \( x < 0 \) and \( x > 0 \)

For \( x < 0 \):
\[ g(x) = \frac{1 - a^x + x a^x \log a}{x^2 a^x} \]

For \( x > 0 \):
\[ g(x) = \frac{(2a)^x - x \log(2a) - 1}{x^2} \]

### Step 2: Find the limit as \( x \to 0 \) from the left

We need to compute:
\[ \lim_{x \to 0^-} g(x) = \lim_{x \to 0} \frac{1 - a^x + x a^x \log a}{x^2 a^x} \]

As \( x \to 0 \):
- \( a^x \to 1 \)
- \( 1 - a^x \to 1 - 1 = 0 \)
- \( x a^x \log a \to 0 \)

This gives us a \( \frac{0}{0} \) form, so we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

Differentiate the numerator and denominator:
- Numerator: \( \frac{d}{dx}(1 - a^x + x a^x \log a) = -a^x \log a + a^x \log a + x a^x \frac{d}{dx}(\log a) = -a^x \log a + a^x \log a + x a^x \log a \)
- Denominator: \( \frac{d}{dx}(x^2 a^x) = 2x a^x + x^2 a^x \log a \)

So we have:
\[ \lim_{x \to 0^-} g(x) = \lim_{x \to 0} \frac{-a^x \log a + a^x \log a + x a^x \log a}{2x a^x + x^2 a^x \log a} \]

### Step 4: Simplify the limit

This simplifies to:
\[ \lim_{x \to 0} \frac{x a^x \log a}{2x a^x + x^2 a^x \log a} \]

Factoring out \( x a^x \):
\[ = \lim_{x \to 0} \frac{\log a}{2 + x \log a} \]
As \( x \to 0 \), this becomes:
\[ \frac{\log a}{2} \]

### Step 5: Find the limit as \( x \to 0 \) from the right

Now compute:
\[ \lim_{x \to 0^+} g(x) = \lim_{x \to 0} \frac{(2a)^x - x \log(2a) - 1}{x^2} \]

As \( x \to 0 \):
- \( (2a)^x \to 1 \)
- \( - x \log(2a) \to 0 \)

This also gives a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 6: Apply L'Hôpital's Rule to the right limit

Differentiate the numerator and denominator:
- Numerator: \( \frac{d}{dx}((2a)^x - x \log(2a) - 1) = (2a)^x \log(2a) - \log(2a) \)
- Denominator: \( \frac{d}{dx}(x^2) = 2x \)

Thus:
\[ \lim_{x \to 0^+} g(x) = \lim_{x \to 0} \frac{(2a)^x \log(2a) - \log(2a)}{2x} \]

This simplifies to:
\[ = \lim_{x \to 0} \frac{\log(2a)}{2} \]

### Step 7: Set the limits equal for continuity

For \( g(x) \) to be continuous at \( x = 0 \):
\[ \frac{\log a}{2} = \frac{\log(2a)}{2} \]

This implies:
\[ \log a = \log(2a) \]
\[ \log a = \log 2 + \log a \]
Thus:
\[ 0 = \log 2 \quad \Rightarrow \quad a = 2 \]

### Step 8: Find \( g(0) \)

Now substitute \( a = 2 \) into either limit:
\[ g(0) = \frac{\log 2}{2} = \frac{\log(4)}{2} = 1 \]

### Final Answer
Thus, \( a = 2 \) and \( g(0) = 1 \).