\( 4a \)
We start with the given equation: \[ \frac{1}{x^4 + x^2 + 1} = \frac{Ax + B}{x^2 + ax + 1} + \frac{Cx + D}{x^2 - ax + 1} \]
Step 1:
Find a common denominator for the right-hand side: \[ \frac{Ax + B}{x^2 + ax + 1} + \frac{Cx + D}{x^2 - ax + 1} = \frac{(Ax + B)(x^2 - ax + 1) + (Cx + D)(x^2 + ax + 1)}{(x^2 + ax + 1)(x^2 - ax + 1)} \] Expanding the numerators: \[ (Ax + B)(x^2 - ax + 1) = Ax^3 - Aax^2 + Ax + Bx^2 - Bax + B \] \[ (Cx + D)(x^2 + ax + 1) = Cx^3 + Cax^2 + Cx + Dx^2 + Dax + D \] Adding these, \[ (Ax^3 - Aax^2 + Ax + Bx^2 - Bax + B) + (Cx^3 + Cax^2 + Cx + Dx^2 + Dax + D) \] Grouping like terms: \[ (A + C)x^3 + (-Aa + Ca + B + D)x^2 + (A + C)x + (-Ba + Da + B + D) \] Equating with the left-hand side: \[ 1 = (A + C)x^3 + (-Aa + Ca + B + D)x^2 + (A + C)x + (-Ba + Da + B + D) \] Since there is no \( x^3 \), \( x^2 \), and \( x \) terms on the left-hand side, we set: \[ A + C = 0, \quad -Aa + Ca + B + D = 0, \quad A + C = 0, \quad -Ba + Da + B + D = 1 \] Solving these equations, we get: \[ A = -C, \quad B + D = Aa - Ca, \quad B + D = 1 \] From \( B + D = 1 \) and substituting the values, we find: \[ A + B - C + D = 2a. \] Thus, the correct answer is \( \mathbf{2a} \).
If the coefficient of \( x^r \) in the expansion of \( (1 + x + x^2)^{100} \) is \( a_r \), and \( S = \sum\limits_{r=0}^{300} a_r \), then
\[ \sum\limits_{r=0}^{300} r a_r = \]
Given below are two statements, one is labelled as Assertion (A) and the other one labelled as Reason (R).
Assertion (A): \[ 1 + \frac{2.1}{3.2} + \frac{2.5.1}{3.6.4} + \frac{2.5.8.1}{3.6.9.8} + \dots \infty = \sqrt{4} \] Reason (R): \[ |x| <1, \quad (1 - x)^{-1} = 1 + nx + \frac{n(n+1)}{1.2} x^2 + \frac{n(n+1)(n+2)}{1.2.3} x^3 + \dots \]
\[ \text{The domain of the real-valued function } f(x) = \sin^{-1} \left( \log_2 \left( \frac{x^2}{2} \right) \right) \text{ is} \]
Let \( A = [a_{ij}] \) be a \( 3 \times 3 \) matrix with positive integers as its elements. The elements of \( A \) are such that the sum of all the elements of each row is equal to 6, and \( a_{22} = 2 \).
\[ \textbf{If } | \text{Adj} \ A | = x \text{ and } | \text{Adj} \ B | = y, \text{ then } \left( | \text{Adj}(AB) | \right)^{-1} \text{ is } \]