a : If `1/((x-2)(x^(2)+1))=A/(x-2)+(Bx+C)/(x^(2)+1) " then "A=1/5, B=-1/5, C=-2/5`.
R : `1/((x-a)(x^(2)+b))=1/(a^(2)+b)[1/(x-a)-(x+a)/(x^(2)+b)]`

Options

A. Both A & R are true and R is correct explanation of A
B. Both A & R are true and R is not correct explanation of A
C. A is true but R is false
D. A is false but R is true

Answer

To solve the equation

\[ \frac{1}{(x-2)(x^2+1)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+1} \]

we will find the values of \(A\), \(B\), and \(C\) using the method of partial fractions.

### Step 1: Combine the right-hand side
First, we need to express the right-hand side with a common denominator:

\[ \frac{A}{x-2} + \frac{Bx+C}{x^2+1} = \frac{A(x^2+1) + (Bx+C)(x-2)}{(x-2)(x^2+1)} \]

### Step 2: Expand the numerator
Now, we will expand the numerator:

\[ A(x^2 + 1) + (Bx + C)(x - 2) = Ax^2 + A + Bx^2 - 2Bx + Cx - 2C \]

Combining like terms gives us:

\[ (A + B)x^2 + (-2B + C)x + (A - 2C) \]

### Step 3: Set the numerators equal
Now, we set the numerators equal to each other since the denominators are the same:

\[ 1 = (A + B)x^2 + (-2B + C)x + (A - 2C) \]

### Step 4: Compare coefficients
Since the left-hand side has no \(x^2\) or \(x\) terms, we can compare coefficients:

1. Coefficient of \(x^2\): \(A + B = 0\) (Equation 1)
2. Coefficient of \(x\): \(-2B + C = 0\) (Equation 2)
3. Constant term: \(A - 2C = 1\) (Equation 3)

### Step 5: Solve the equations
From Equation 1, we can express \(B\) in terms of \(A\):

\[ B = -A \]

Substituting \(B = -A\) into Equation 2:

\[ -2(-A) + C = 0 \implies 2A + C = 0 \implies C = -2A \]

Now substituting \(C = -2A\) into Equation 3:

\[ A - 2(-2A) = 1 \implies A + 4A = 1 \implies 5A = 1 \implies A = \frac{1}{5} \]

Now substituting \(A\) back to find \(B\) and \(C\):

\[ B = -A = -\frac{1}{5} \]
\[ C = -2A = -2 \cdot \frac{1}{5} = -\frac{2}{5} \]

### Final Values
Thus, we have:

\[ A = \frac{1}{5}, \quad B = -\frac{1}{5}, \quad C = -\frac{2}{5} \]

### Conclusion
The values of \(A\), \(B\), and \(C\) are indeed:

\[ A = \frac{1}{5}, \quad B = -\frac{1}{5}, \quad C = -\frac{2}{5} \]