`f(x)= {:{(x+asqrt2 sin x ,, 0 le x lt pi/4),( 2x cot x +b , , pi/4 le x le pi/2), (a cos 2 x - b sin x , , pi/2 lt x le pi):}`continuous function `AA x in [ 0,pi] " then " 5(a/b)^(2)` equals ....

Answer

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

\[ f(x) = \begin{cases} x + a\sqrt{2}\sin x & \text{for } 0 \leq x < \frac{\pi}{4} \\ 2x \cot x + b & \text{for } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{for } \frac{\pi}{2} < x \leq \pi \end{cases} \]

### Step 1: Ensure continuity at \( x = \frac{\pi}{4} \)

To ensure continuity at \( x = \frac{\pi}{4} \), we set the left-hand limit equal to the right-hand limit:

\[ f\left(\frac{\pi}{4}\right) \text{ from the first piece} = f\left(\frac{\pi}{4}\right) \text{ from the second piece} \]

Calculating \( f\left(\frac{\pi}{4}\right) \) from the first piece:

\[ f\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + a\sqrt{2}\sin\left(\frac{\pi}{4}\right) = \frac{\pi}{4} + a\sqrt{2} \cdot \frac{1}{\sqrt{2}} = \frac{\pi}{4} + a \]

Calculating \( f\left(\frac{\pi}{4}\right) \) from the second piece:

\[ f\left(\frac{\pi}{4}\right) = 2\left(\frac{\pi}{4}\right) \cot\left(\frac{\pi}{4}\right) + b = \frac{\pi}{2} + b \]

Setting these equal gives us:

\[ \frac{\pi}{4} + a = \frac{\pi}{2} + b \]

Rearranging this, we get:

\[ a - b = \frac{\pi}{4} \quad \text{(Equation 1)} \]

### Step 2: Ensure continuity at \( x = \frac{\pi}{2} \)

Next, we ensure continuity at \( x = \frac{\pi}{2} \):

\[ f\left(\frac{\pi}{2}\right) \text{ from the second piece} = f\left(\frac{\pi}{2}\right) \text{ from the third piece} \]

Calculating \( f\left(\frac{\pi}{2}\right) \) from the second piece:

\[ f\left(\frac{\pi}{2}\right) = 2\left(\frac{\pi}{2}\right) \cot\left(\frac{\pi}{2}\right) + b = 0 + b = b \]

Calculating \( f\left(\frac{\pi}{2}\right) \) from the third piece:

\[ f\left(\frac{\pi}{2}\right) = a \cos(2 \cdot \frac{\pi}{2}) - b \sin\left(\frac{\pi}{2}\right) = a \cdot (-1) - b = -a - b \]

Setting these equal gives us:

\[ b = -a - b \]

Rearranging this, we get:

\[ 2b = -a \quad \text{(Equation 2)} \]

### Step 3: Solve the equations

Now we have two equations:

1. \( a - b = \frac{\pi}{4} \)
2. \( 2b = -a \)

From Equation 2, we can express \( a \) in terms of \( b \):

\[ a = -2b \]

Substituting this into Equation 1 gives:

\[ -2b - b = \frac{\pi}{4} \]
\[ -3b = \frac{\pi}{4} \]
\[ b = -\frac{\pi}{12} \]

Now substituting \( b \) back into the expression for \( a \):

\[ a = -2\left(-\frac{\pi}{12}\right) = \frac{\pi}{6} \]

### Step 4: Calculate \( 5\left(\frac{a}{b}\right)^2 \)

Now we can calculate \( 5\left(\frac{a}{b}\right)^2 \):

\[ \frac{a}{b} = \frac{\frac{\pi}{6}}{-\frac{\pi}{12}} = -2 \]

Thus,

\[ 5\left(\frac{a}{b}\right)^2 = 5 \cdot (-2)^2 = 5 \cdot 4 = 20 \]

### Final Answer

The value of \( 5\left(\frac{a}{b}\right)^2 \) is \( \boxed{20} \).

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