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 two pieces of the function defined 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 1: Find the limit of \( f(x) \) as \( x \) approaches 0 from the left (\( x \to 0^- \))

For \( x < 0 \):

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

As \( x \to 0^- \):
- \( \sin x \to 0 \)
- \( \cos x \to 1 \)

Thus, we can substitute these limits into the function:

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

This limit approaches infinity unless \( a + b + 5 = 0 \).

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

For \( x > 0 \):

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

As \( x \to 0^+ \):

Using the limit \( \lim_{x \to 0} (2 + x^2)^{\frac{1}{x}} \):

This can be rewritten using the exponential function:

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

Using L'Hôpital's Rule:

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

Thus:

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

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

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

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

This gives us:

\[ \frac{a + b + 5}{0} = 1 \]

Since the left-hand limit approaches infinity, for \( f(x) \) to be continuous, we require:

\[ a + b + 5 = 0 \]

### Conclusion

The correct statement for \( f(x) \) to be continuous at \( x = 0 \) is:

\[ a + b + 5 = 0 \]