TG EAMCET 2 May 2025 Shift 2 Paper

We are given:$\frac{x^2-3}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1}$First, get a common denominator to combine the right side:$\frac{A}{x+2} + \frac{Bx+C}{x^2+1} = \frac{A(x^2+1) + (Bx+C)(x+2)}{(x+2)(x^2+1)}$Set the numerators equal to each other:$x^2 - 3 = A(x^2+1) + (Bx+C)(x+2)$Expand the right side:$A(x^2+1) = A x^2 + A$$(Bx+C)(x+2) = Bx^2 + 2Bx + Cx + 2C = Bx^2 + (2B+C)x + 2C$Combine all terms on the right: $A x^2 + B x^2 = (A+B)x^2$$(2B+C)x$$A + 2C$So we get:$x^2 - 3 = (A+B)x^2 + (2B+C)x + (A+2C)$Now, match the coefficients for $x^2, x$, and the constant term:For $x^2: A+B=1$For $x: 2B+C=0$For constant: $A+2C = -3$Write the equations:1. $A+B=1$2. $2B+C=0$3. $A+2C = -3$Now solve these equations step by step:From equation 2: $2B+C=0 \longrightarrow C=-2B$Substitute $C = -2B$ into equation 3: $A+2(-2B) = -3A - 4B = -3$$A - 4B = -3$Now use $A+B=1$. Substitute $A = 1-B$ into the last equation:$1 - B - 4B = -3$$1 - 5B = -3$$-5B = -4$$B = \frac{4}{5}$Now $C = -2B = -2 \times \frac{4}{5} = \frac{-8}{5}$$A = 1-B = 1 - \frac{4}{5} = \frac{1}{5}$Now find $3A + 2B - C$ :$3\left(\frac{1}{5}\right) + 2\left(\frac{4}{5}\right) - \left(\frac{-8}{5}\right)$$= \frac{3}{5} + \frac{8}{5} + \frac{8}{5}$$= \frac{3+8+8}{5}$$= \frac{19}{5}$