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 \( x = 0 \). The function is defined as:

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

### Step 1: Find the limit as \( x \) approaches 0

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

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

Calculating the limit:

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

### Step 2: Simplify the numerator

As \( x \to 0 \):

- \( \sin x \to 0 \)
- \( \cos x \to 1 \)

Thus, the numerator becomes:

\[ a(1 - 0) + b(1) + 5 = a + b + 5 \]

### Step 3: Evaluate the limit

Now we have:

\[ \lim_{x \to 0} \frac{a + b + 5}{x^2} \]

For this limit to exist and equal 3, the numerator must be 0:

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

### Step 4: Apply L'Hôpital's Rule

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:

Differentiate the numerator and denominator:

- The derivative of the numerator \( a(1 - x \sin x) + b \cos x + 5 \) is:

\[ -a \sin x + a x \cos x - b \sin x \]

- The derivative of the denominator \( x^2 \) is \( 2x \).

Thus, we have:

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

### Step 5: Evaluate the new limit

As \( x \to 0 \):

- \( \sin x \to 0 \)
- \( \cos x \to 1 \)

This gives:

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

Setting this equal to 3 (the value of \( f(0) \)) gives:

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

### Step 6: Substitute \( b \) into Equation 1

Now substitute \( b = -6 \) into Equation 1:

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

### Step 7: Calculate \( \frac{b^4 + a}{5 + a} \)

Now we have \( a = 1 \) and \( b = -6 \):

\[ b^4 = (-6)^4 = 1296 \]

Now calculate:

\[ \frac{b^4 + a}{5 + a} = \frac{1296 + 1}{5 + 1} = \frac{1297}{6} \]

Thus, the final answer is:

\[ \frac{1297}{6} \]