Let `f (x)=[{:((a (1-x sin x)+ b cos x +5)/(x ^(2)),,, x lt 0), ((1+ ((dx + dx ^(3))/(dx ^(2))))^(1/x),,, x gt 0):}`
If f is continous at `x=0` then correct statement (s) is/are:

Options

A. `a+c=-1`
B. `b+x=-4`
C. `a+b=-5`
D. `c+d=an` irrational number

Answer

To determine the conditions under which the function \( f(x) \) is continuous at \( x = 0 \), we need to analyze the function piecewise, as given in the problem statement.

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

The function is defined as follows:

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

### Step 2: Find the limit of \( f(x) \) as \( x \) approaches 0 from the left

For \( x < 0 \), we need to evaluate:

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

As \( x \to 0 \), \( \sin x \to 0 \) and \( \cos x \to 1 \). Thus, we have:

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

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

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

### Step 3: Find the limit of \( f(x) \) as \( x \) approaches 0 from the right

For \( x > 0 \), we need to evaluate:

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

This simplifies to:

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

As \( x \to 0 \), this approaches \( 2^{\frac{1}{x}} \), which approaches \( e^{\ln(2)/x} \) and diverges unless \( d = 0 \). Hence, we set:

\[ d = 0 \quad \text{(Equation 2)} \]

### Step 4: Apply L'Hôpital's Rule for \( x < 0 \)

We need to evaluate the limit for \( x < 0 \) again using L'Hôpital's Rule since it is of the form \( \frac{0}{0} \):

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

Differentiating the numerator and denominator:

\[ \text{Numerator: } -a \sin x + a x \cos x - b \sin x \quad \text{Denominator: } 2x \]

Thus, we evaluate:

\[ \lim_{x \to 0^-} \frac{-a \sin x + ax \cos x - b \sin x}{2x} \]

Substituting \( x = 0 \):

\[ \frac{-0 + 0 - 0}{0} = \text{still } 0 \]

We apply L'Hôpital's Rule again:

\[ \lim_{x \to 0^-} \frac{-a \cos x + a \cos x - b \cos x}{2} = \frac{-b}{2} \]

Setting this equal to the limit from the right (which approaches \( 3 \)) gives us:

\[ -\frac{b}{2} = 3 \quad \Rightarrow \quad b = -6 \quad \text{(Equation 3)} \]

### Step 5: Solve the equations

Now we have three equations:

1. \( a + b + 5 = 0 \)
2. \( d = 0 \)
3. \( b = -6 \)

Substituting \( b = -6 \) into Equation 1:

\[ a - 6 + 5 = 0 \quad \Rightarrow \quad a - 1 = 0 \quad \Rightarrow \quad a = 1 \]

### Step 6: Summary of values

Now we have:

- \( a = 1 \)
- \( b = -6 \)
- \( c \) (not determined yet)
- \( d = 0 \)

### Step 7: Determine \( c \)

To find \( c \), we analyze the limit for \( x > 0 \):

\[ \lim_{x \to 0^+} \left(1 + \frac{cx + 0}{0}\right)^{\frac{1}{x}} = \text{needs to equal } 3 \]

This leads to:

\[ c = 0 \quad \text{(Equation 4)} \]

### Final Values

Thus, we have:

- \( a = 1 \)
- \( b = -6 \)
- \( c = 0 \)
- \( d = 0 \)

### Conclusion

The correct statements are:

- \( a + c = 1 + 0 = 1 \)
- \( b + c = -6 + 0 = -6 \)
- \( a + b = 1 - 6 = -5 \)
- \( c + d = 0 + 0 = 0 \)