The point `P(a,b)` undergoes following transformation to a new co-ordinate `P'(-1/(sqrt2),7/sqrt2)`
(i) Reflection about `y=x`
(ii) Translation through 2 unit in the positive direction of x-axis
(iii)Rotation through an angle `pi/4` in anti-clockwise sense about the origin
Then the value of `2a-b` is

Options

A. 1
B. 7
C. 4
D. 9

Answer

To solve the problem, we will follow the transformations step by step and derive the values of \( a \) and \( b \).

### Step 1: Reflection about \( y = x \)

The point \( P(a, b) \) reflects about the line \( y = x \). This transformation interchanges the coordinates, resulting in the new point:
\[ P_1(b, a) \]

### Step 2: Translation through 2 units in the positive direction of the x-axis

Next, we translate the point \( P_1(b, a) \) by 2 units in the positive direction of the x-axis. This means we add 2 to the x-coordinate:
\[ P_2(b + 2, a) \]

### Step 3: Rotation through an angle \( \frac{\pi}{4} \) in the anti-clockwise sense about the origin

Now, we need to rotate the point \( P_2(b + 2, a) \) by an angle of \( \frac{\pi}{4} \). The rotation matrix for an angle \( \theta \) is given by:
\[ \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]
For \( \theta = \frac{\pi}{4} \), we have:
\[ \cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]
Thus, the rotation matrix becomes:
\[ \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \]

Now we apply this matrix to the point \( P_2(b + 2, a) \):
\[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} b + 2 \\ a \end{pmatrix} \]
Calculating this gives:
\[ x' = \frac{1}{\sqrt{2}}(b + 2) - \frac{1}{\sqrt{2}}(a) = \frac{b + 2 - a}{\sqrt{2}} \]
\[ y' = \frac{1}{\sqrt{2}}(b + 2) + \frac{1}{\sqrt{2}}(a) = \frac{b + 2 + a}{\sqrt{2}} \]

### Step 4: Setting the new coordinates equal to \( P'(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}) \)

We know the transformed coordinates equal \( P'(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}) \). Therefore, we set up the equations:
1. \(\frac{b + 2 - a}{\sqrt{2}} = -\frac{1}{\sqrt{2}}\)
2. \(\frac{b + 2 + a}{\sqrt{2}} = \frac{7}{\sqrt{2}}\)

### Step 5: Solving the equations

From the first equation:
\[ b + 2 - a = -1 \implies b - a = -3 \quad \text{(Equation 1)} \]

From the second equation:
\[ b + 2 + a = 7 \implies b + a = 5 \quad \text{(Equation 2)} \]

### Step 6: Adding the equations

Now we add Equation 1 and Equation 2:
\[ (b - a) + (b + a) = -3 + 5 \]
This simplifies to:
\[ 2b = 2 \implies b = 1 \]

### Step 7: Substituting to find \( a \)

Substituting \( b = 1 \) into Equation 2:
\[ 1 + a = 5 \implies a = 4 \]

### Step 8: Finding \( 2a - b \)

Now we calculate \( 2a - b \):
\[ 2a - b = 2(4) - 1 = 8 - 1 = 7 \]

### Final Answer

Thus, the value of \( 2a - b \) is:
\[ \boxed{7} \]