Let `f(x) = {{:((a(1-x sin x)+b cos x + 5)/(x^(2))",",x lt 0),(3",",x = 0),([1 + ((cx + dx^(3))/(x^(2)))]^(1//x)",",x gt 0):}` If f is continuous at x = 0, then (a + b + c + d) is

Options

A. (a)5
B. (b)-5
C. (c)log 3 - 5
D. (d)5 - log 3

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). The function is defined piecewise as follows:

\[ f(x) = \begin{cases} \frac{a(1 - x \sin x) + b \cos x + 5}{x^2} & \text{for } x < 0 \\ 3 & \text{for } x = 0 \\ \left(1 + \frac{cx + dx^3}{x^2}\right)^{\frac{1}{x}} & \text{for } x > 0 \end{cases} \]

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0 from the left (\( x \to 0^- \))

We need to compute:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{a(1 - x \sin x) + b \cos x + 5}{x^2} \]

As \( x \to 0 \), we know that \( \sin x \to 0 \) and \( \cos x \to 1 \). Thus, we can simplify:

\[ 1 - x \sin x \to 1 \quad \text{and} \quad \cos x \to 1 \]

So,

\[ \lim_{x \to 0^-} f(x) = \frac{a(1) + b(1) + 5}{0} = \frac{a + b + 5}{0} \]

This limit tends to infinity unless \( a + b + 5 = 0 \). Therefore, we set:

\[ a + b + 5 = 0 \quad \text{(Equation 1)} \]

### Step 2: Find the limit of \( f(x) \) as \( x \) approaches 0 from the right (\( x \to 0^+ \))

Next, we compute:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \left(1 + \frac{cx + dx^3}{x^2}\right)^{\frac{1}{x}} \]

This can be rewritten as:

\[ \lim_{x \to 0^+} \left(1 + \frac{c}{x} + d x\right)^{\frac{1}{x}} \]

For the limit to exist and be finite, \( c \) must equal 0. Thus, we have:

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

### Step 3: Set the limits equal for continuity at \( x = 0 \)

Since \( f(0) = 3 \), we set the limits equal:

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

From the left, we have:

\[ \frac{a + b + 5}{0} \to \infty \quad \text{(if } a + b + 5 \neq 0\text{)} \]

From the right, we have:

\[ 1 = 3 \quad \text{(not possible)} \]

Thus, we need to ensure that both limits equal 3. This leads us to:

\[ \lim_{x \to 0^-} f(x) = 3 \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 3 \]

### Step 4: Solve the equations

From Equation 1:

\[ a + b + 5 = 0 \implies a + b = -5 \quad \text{(Equation 2)} \]

Now, we need to find \( d \) such that:

\[ \lim_{x \to 0^+} f(x) = 3 \]

From our earlier analysis, we know:

\[ e^{d \cdot 0} = 1 \implies d = 0 \]

### Step 5: Substitute \( c \) and \( d \)

Since \( c = 0 \) and \( d = 0 \), we substitute back into our equations:

1. From Equation 2: \( a + b = -5 \)
2. We need to find \( a + b + c + d = -5 + 0 + 0 = -5 \)

### Final Answer

Thus, \( a + b + c + d = -5 \).