If the function f(x), defined as `f(x)={{:((a(1-xsinx)+b cosx+5)/(x^(2)),":", xne0),(3,":",x=0):}` is continuous at `x=0`, then the value of `(b^(4)+a)/(5+a)` is equal to

Answer

To solve the problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This means that the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) \).

### Step 1: Set up the limit
The function is defined as:
\[ f(x) = \frac{a(1 - x \sin x) + b \cos x + 5}{x^2}, \quad x \neq 0 \]
and \( f(0) = 3 \).

We need to find:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{a(1 - x \sin x) + b \cos x + 5}{x^2} \]
This limit must equal 3 for continuity.

### Step 2: Expand the terms using Taylor series
Using Taylor series expansions around \( x = 0 \):
- \( \sin x \approx x - \frac{x^3}{6} + O(x^5) \)
- \( \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \)

Substituting these into the function:
\[ 1 - x \sin x \approx 1 - x\left(x - \frac{x^3}{6}\right) = 1 - x^2 + O(x^4) \]
\[ \cos x \approx 1 - \frac{x^2}{2} \]

Now substituting these approximations into \( f(x) \):
\[ f(x) \approx \frac{a(1 - x^2) + b(1 - \frac{x^2}{2}) + 5}{x^2} \]
\[ = \frac{a + b + 5 - ax^2 - \frac{bx^2}{2}}{x^2} \]

### Step 3: Simplify the limit expression
Now, simplifying the numerator:
\[ = \frac{(a + b + 5) + \left(-a - \frac{b}{2}\right)x^2}{x^2} \]
\[ = \frac{a + b + 5}{x^2} - a - \frac{b}{2} \]

Taking the limit as \( x \to 0 \):
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left(\frac{a + b + 5}{x^2} - a - \frac{b}{2}\right) \]

For this limit to equal 3, the term \( a + b + 5 \) must be 0 (to avoid infinity):
\[ a + b + 5 = 0 \quad \Rightarrow \quad a + b = -5 \quad \text{(Equation 1)} \]

### Step 4: Set the remaining limit equal to 3
Now we need the remaining part to equal 3:
\[ -a - \frac{b}{2} = 3 \quad \text{(Equation 2)} \]

### Step 5: Solve the system of equations
From Equation 1:
\[ b = -5 - a \]

Substituting this into Equation 2:
\[ -a - \frac{-5 - a}{2} = 3 \]
\[ -a + \frac{5 + a}{2} = 3 \]
Multiplying through by 2 to eliminate the fraction:
\[ -2a + 5 + a = 6 \]
\[ -a + 5 = 6 \]
\[ -a = 1 \quad \Rightarrow \quad a = -1 \]

Now substituting \( a = -1 \) back into Equation 1:
\[ -1 + b = -5 \quad \Rightarrow \quad b = -4 \]

### Step 6: Calculate the required expression
Now we need to find the value of \( \frac{b^4 + a}{5 + a} \):
\[ b^4 = (-4)^4 = 256 \]
\[ a = -1 \]
So:
\[ \frac{256 + (-1)}{5 + (-1)} = \frac{255}{4} = 63.75 \]

### Final Answer
The value of \( \frac{b^4 + a}{5 + a} \) is \( \frac{255}{4} \) or 63.75.