Q.Find : $\int \dfrac{2\cos x}{(1 - \sin x)(1 + \sin^2 x)}\, dx$

Showing the 12 most recent of 33 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.

   [!FORMULA] $\frac{2x^3 + x - 3}{x^4 - 5x^2 + 4} =$

   (A) $\frac{5(x-1)}{4(x^2-3x+2)} + \frac{3x+1}{4(x^2+3x+2)}$ (B) $\frac{5(x+1)}{4(x^2-3x+2)} + \frac{3x-1}{4(x^2+3x+2)}$ (C) $\frac{2}{x-1} + \frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$ (D) $\frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$

   ›Reveal solutionSolution

   The key idea is to decompose the rational function into partial fractions by factoring the denominator completely, then matching the result to one of the given options. The correct decomposition is option (C).

   We start with the rational expression:

   $\frac{2x^3 + x - 3}{x^4 - 5x^2 + 4}$

   Concept and Intuition

   Partial fraction decomposition is the reverse of combining fractions. Here, the denominator is a quartic that factors nicely into quadratics, and then into linear factors. Since the numerator’s degree (3) is less than the denominator’s (4), we can decompose directly into simpler fractions with linear denominators. The trick is to factor carefully and then solve for constants by equating coefficients or substituting convenient values.

   Step-by-step solution

   1. Factor the denominator Notice $x^4 - 5x^2 + 4$ is quadratic in $x^2$. Let $u = x^2$:

   $u^2 - 5u + 4 = (u-1)(u-4) = (x^2-1)(x^2-4)$

   Then factor each difference of squares:

   $x^2-1 = (x-1)(x+1), \quad x^2-4 = (x-2)(x+2)$

   So the denominator is:

   $(x-1)(x+1)(x-2)(x+2)$

   2. Set up the partial fraction form Since all factors are linear and distinct, we write:

   $\frac{2x^3 + x - 3}{(x-1)(x+1)(x-2)(x+2)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x-2} + \frac{D}{x+2}$

   3. Clear denominators Multiply both sides by $(x-1)(x+1)(x-2)(x+2)$:

   $2x^3 + x - 3 = A(x+1)(x-2)(x+2) + B(x-1)(x-2)(x+2) + C(x-1)(x+1)(x+2) + D(x-1)(x+1)(x-2)$

   4. Solve for constants using convenient $x$ values

      • For $x = 1$: The terms with $B, C, D$ vanish because they contain $(x-1)$. Left side: $2(1)^3 + 1 - 3 = 0$. Right side: $A(2)(-1)(3) = -6A$. So $-6A = 0 \Rightarrow A = 0$.
      • For $x = -1$: Terms with $A, C, D$ vanish (contain $x+1$). Left side: $2(-1)^3 + (-1) - 3 = -2 -1 -3 = -6$. Right side: $B(-2)(-3)(1) = 6B$. So $6B = -6 \Rightarrow B = -1$.
      • For $x = 2$: Terms with $A, B, D$ vanish (contain $x-2$). Left side: $2(8) + 2 - 3 = 16 -1 = 15$. Right side: $C(1)(3)(4) = 12C$. So $12C = 15 \Rightarrow C = \frac{15}{12} = \frac{5}{4}$.
      • For $x = -2$: Terms with $A, B, C$ vanish (contain $x+2$). Left side: $2(-8) -2 -3 = -16 -5 = -21$. Right side: $D(-3)(-1)(-4) = -12D$. So $-12D = -21 \Rightarrow D = \frac{21}{12} = \frac{7}{4}$.

   5. Write the decomposition

      With $A=0$, $B=-1$, $C=\frac{5}{4}$, $D=\frac{7}{4}$:

   $\frac{2x^3 + x - 3}{x^4 - 5x^2 + 4} = \frac{0}{x-1} + \frac{-1}{x+1} + \frac{5/4}{x-2} + \frac{7/4}{x+2}$

   Simplify:

   $= -\frac{1}{x+1} + \frac{5}{4(x-2)} + \frac{7}{4(x+2)}$

   6. Match with the options Option (C) is: $\frac{2}{x-1} + \frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$ …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\dfrac{x^2+1}{(x^4+5x^2+6)(x^6+x^4)} = \dfrac{A}{x^4} + \dfrac{B}{x^2} + \dfrac{C}{x^2+2} + \dfrac{D}{x^2+3}$, then $A-B=$ (A) $\dfrac{13}{36}$ (B) $\dfrac{11}{36}$ (C) $\dfrac{2}{9}$ (D) $-\dfrac{1}{2}$

   ›Reveal solutionSolution

   The key is to factor the denominator completely, then match the given partial-fraction form to the actual decomposition. After simplifying, we find $A - B = \frac{11}{36}$, which is option (B).

   The problem gives a partial-fraction expansion with four terms, but the left-hand side has a denominator that factors into products of quadratics and powers of $x$. The trick is that the given form is not the standard partial-fraction decomposition — it’s a specific rearrangement. We need to find $A$ and $B$ by comparing coefficients after clearing denominators.

   Let’s work through it step by step.

   1. Factor the denominator completely. The left-hand side is

   $\frac{x^2+1}{(x^4+5x^2+6)(x^6+x^4)}.$

   First, $x^6 + x^4 = x^4(x^2+1)$.

   Next, $x^4+5x^2+6$ is quadratic in $x^2$: let $u = x^2$, then $u^2+5u+6 = (u+2)(u+3) = (x^2+2)(x^2+3)$.

   So the whole denominator is

   $(x^2+2)(x^2+3) \cdot x^4 (x^2+1).$

   Notice the $x^2+1$ in the numerator cancels with the $x^2+1$ in the denominator!

   Hence the expression simplifies to

   $\frac{1}{x^4 (x^2+2)(x^2+3)}.$

   2. Set up the given partial-fraction form. We are told

   $\frac{1}{x^4 (x^2+2)(x^2+3)} = \frac{A}{x^4} + \frac{B}{x^2} + \frac{C}{x^2+2} + \frac{D}{x^2+3}.$

   Multiply both sides by $x^4 (x^2+2)(x^2+3)$ to clear denominators:

   $1 = A (x^2+2)(x^2+3) + B x^2 (x^2+2)(x^2+3) + C x^4 (x^2+3) + D x^4 (x^2+2).$

   3. Expand and collect powers of $x$.

      Compute each term:

      • $A(x^2+2)(x^2+3) = A(x^4 + 5x^2 + 6)$.
      • $B x^2 (x^4 + 5x^2 + 6) = B(x^6 + 5x^4 + 6x^2)$.
      • $C x^4 (x^2+3) = C(x^6 + 3x^4)$.
      • $D x^4 (x^2+2) = D(x^6 + 2x^4)$.

      Summing, the coefficient of each power of $x$ on the right must match the left side, which is just the constant $1$ (i.e., coefficient of $x^0$ is 1, all others 0).

      Collect by powers:

      • $x^6$: $B + C + D = 0$
      • $x^4$: $A + 5B + 3C + 2D = 0$ …

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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $\dfrac{x+3}{(1-x)^2(1+x^2)} = \dfrac{A}{(1-x)} + \dfrac{B}{(1-x)^2} + \dfrac{Cx+D}{2(1+x^2)}$ then $B^2 + C^2 + D^2 =$ (A) $\dfrac{3}{4}$ (B) $4 \dfrac{1}{2}$ (C) $14$ (D) $4$

   ›Reveal solutionSolution

   We solve for the partial fraction coefficients by clearing denominators and equating numerators, then compute $B^2 + C^2 + D^2$ to find the result is $\frac{3}{4}$, which corresponds to option (A).

   The problem gives a partial fraction decomposition of a rational function. The key idea is to multiply both sides by the common denominator to obtain a polynomial identity, then solve for the unknown constants $A$, $B$, $C$, $D$ by comparing coefficients or substituting convenient values of $x$. Once we have $B$, $C$, $D$, we compute the sum of their squares.

   1. Set up the equation We have

   $\frac{x+3}{(1-x)^2(1+x^2)} = \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{2(1+x^2)}.$

   Multiply both sides by the common denominator $(1-x)^2(1+x^2)$:

   $x+3 = A(1-x)(1+x^2) + B(1+x^2) + \frac{Cx+D}{2}(1-x)^2.$

   2. Clear the fraction in the last term Multiply the entire equation by 2 to avoid fractions:

   $2(x+3) = 2A(1-x)(1+x^2) + 2B(1+x^2) + (Cx+D)(1-x)^2.$

   So:

   $2x+6 = 2A(1-x)(1+x^2) + 2B(1+x^2) + (Cx+D)(1-x)^2.$

   3. Expand each term

      • First term: $2A(1-x)(1+x^2) = 2A[(1)(1+x^2) - x(1+x^2)] = 2A(1+x^2 - x - x^3) = 2A(-x^3 - x + 1 + x^2)$. Better to expand systematically: $(1-x)(1+x^2) = 1 + x^2 - x - x^3$. So $2A(1 - x + x^2 - x^3)$.
      • Second term: $2B(1+x^2) = 2B + 2B x^2$.
      • Third term: $(Cx+D)(1-x)^2 = (Cx+D)(1 - 2x + x^2)$. Expand: $Cx(1 - 2x + x^2) = Cx - 2Cx^2 + Cx^3$ $D(1 - 2x + x^2) = D - 2Dx + Dx^2$ Sum: $Cx^3 + (-2C + D)x^2 + (C - 2D)x + D$.

   4. Collect coefficients

      The left side is $2x + 6$, which is $0\cdot x^3 + 0\cdot x^2 + 2x + 6$.

      The right side, collecting powers:

      • $x^3$: from first term: $2A(-1) = -2A$; from third: $C$. So coefficient: $C - 2A$.
      • $x^2$: from first: $2A(1) = 2A$; from second: $2B$; from third: $(-2C + D)$. So: $2A + 2B - 2C + D$.
      • $x^1$: from first: $2A(-1) = -2A$; from third: $(C - 2D)$. So: $-2A + C - 2D$.
      • Constant: from first: $2A(1) = 2A$; from second: $2B$; from third: $D$. So: $2A + 2B + D$.

      Equate to $0x^3 + 0x^2 + 2x + 6$:

   $\begin{cases} C - 2A = 0 & (1)\\ 2A + 2B - 2C + D = 0 & (2)\\ -2A + C - 2D = 2 & (3)\\ 2A + 2B + D = 6 & (4) \end{cases}$

   5. Solve the system From (1): $C = 2A$. …

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4. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $\dfrac{x^2-3}{(x+2)(x^2+1)} = \dfrac{A}{x+2} + \dfrac{Bx+C}{x^2+1}$ then $3A+2B-C=$ (A) $\dfrac{8}{5}$ (B) $\dfrac{16}{5}$ (C) $\dfrac{3}{5}$ (D) $\dfrac{19}{5}$

   ›Reveal solutionSolution

   Solving the partial-fraction system gives $A=\tfrac15,\ B=\tfrac45,\ C=-\tfrac85$, so $3A+2B-C=\tfrac{19}{5}$, option (D).

   Clear denominators in $\dfrac{x^2-3}{(x+2)(x^2+1)}=\dfrac{A}{x+2}+\dfrac{Bx+C}{x^2+1}$:

   $x^2-3 = A(x^2+1)+(Bx+C)(x+2)=(A+B)x^2+(2B+C)x+(A+2C).$

   Matching coefficients:

   $A+B=1,\qquad 2B+C=0,\qquad A+2C=-3.$ …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If $\dfrac{x^2+1}{(x^2+2)(x^2+3)} = \dfrac{Ax+B}{x^2+2} + \dfrac{Cx+D}{x^2+3}$, then $A+B+C+D=$ (A) $0$ (B) $1$ (C) $-1$ (D) $6$

   ›Reveal solutionSolution

   We decompose the rational function into partial fractions with linear numerators, then equate coefficients to solve for A, B, C, D; summing them gives 0.

   We are given:

   $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$

   and need $A+B+C+D$.

   Concept & Intuition

   Since the denominator factors are irreducible quadratics, each partial fraction gets a linear numerator ($Ax+B$ form). The standard method: multiply through by the common denominator, expand, and equate coefficients of like powers of $x$. This yields a system of equations for $A,B,C,D$. Summing them is then trivial.

   Step-by-step

   1. Clear denominators Multiply both sides by $(x^2+2)(x^2+3)$:

   $x^2+1 = (Ax+B)(x^2+3) + (Cx+D)(x^2+2)$

   2. Expand each term

   $(Ax+B)(x^2+3) = Ax^3 + 3Ax + Bx^2 + 3B$

   $(Cx+D)(x^2+2) = Cx^3 + 2Cx + Dx^2 + 2D$

   3. Combine like powers

   $x^2+1 = (A+C)x^3 + (B+D)x^2 + (3A+2C)x + (3B+2D)$

   4. Equate coefficients

      • Coefficient of $x^3$: $A + C = 0$
      • Coefficient of $x^2$: $B + D = 1$
      • Coefficient of $x^1$: $3A + 2C = 0$
      • Constant term: $3B + 2D = 1$

   5. Solve the system

      From $A+C=0$ we have $C = -A$.

      Substitute into $3A+2C=0$: $3A + 2(-A) = A = 0$ → $A=0$, then $C=0$.

      From $B+D=1$ we have $D = 1-B$. …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $\dfrac{x^2+1}{(x^2+2)(x^2+3)} = \dfrac{Ax+B}{x^2+2} + \dfrac{Cx+D}{x^2+3}$, then $A+B+C+D=$ (A) $1$ (B) $0$ (C) $-1$ (D) $6$

   ›Reveal solutionSolution

   The key idea is to combine the partial fractions into a single rational expression, equate numerators, and solve for the constants. The sum $A+B+C+D$ turns out to be $0$.

   We are given the partial fraction decomposition:

   $\frac{x^2+1}{(x^2+2)(x^2+3)} = \frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3}$

   We need $A+B+C+D$. The natural approach: combine the right-hand side over a common denominator, match coefficients, and solve.

   ---

   1. Combine the fractions on the right.

   $\frac{Ax+B}{x^2+2} + \frac{Cx+D}{x^2+3} = \frac{(Ax+B)(x^2+3) + (Cx+D)(x^2+2)}{(x^2+2)(x^2+3)}$

   Since the denominators are already equal, we equate the numerators:

   $x^2 + 1 = (Ax+B)(x^2+3) + (Cx+D)(x^2+2)$

   ---

   2. Expand both products.

   First term: $(Ax+B)(x^2+3) = Ax^3 + 3Ax + Bx^2 + 3B$

   Second term: $(Cx+D)(x^2+2) = Cx^3 + 2Cx + Dx^2 + 2D$

   Add them:

   $(A+C)x^3 + (B+D)x^2 + (3A+2C)x + (3B+2D)$

   So we have:

   $x^2 + 1 = (A+C)x^3 + (B+D)x^2 + (3A+2C)x + (3B+2D)$

   ---

   3. Equate coefficients.

   The left side has no $x^3$ term, no $x$ term, constant term $1$, and $x^2$ coefficient $1$. So:

   • Coefficient of $x^3$: $A + C = 0$
   • Coefficient of $x^2$: $B + D = 1$
   • Coefficient of $x$: $3A + 2C = 0$
   • Constant term: $3B + 2D = 1$

   ---

   4. Solve the system.

   From $A + C = 0$ we have $C = -A$.

   Plug into $3A + 2C = 0$: $3A + 2(-A) = 3A - 2A = A = 0$.

   Thus $A = 0$ and then $C = 0$. …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.If $\dfrac{3x+1}{(x-1)^2(x^2+1)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$ (A) $0$ (B) $1$ (C) $2$ (D) $-1$

   ›Reveal solutionSolution

   We find the constants in the partial fraction decomposition by clearing denominators and equating coefficients, then compute $2(A-C+B+D)$ to get $-1$, which corresponds to option (D).

   We are given the partial fraction decomposition:

   $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$

   Our goal is to find $A, B, C, D$ and then evaluate $2(A - C + B + D)$.

   ---

   Concept & Intuition

   Partial fractions let us break a complicated rational expression into simpler pieces. The denominators here are $(x-1)$, $(x-1)^2$, and the irreducible quadratic $x^2+1$. To find the unknown constants, we multiply both sides by the common denominator $(x-1)^2(x^2+1)$, which gives a polynomial identity. Then we compare coefficients of like powers of $x$ (or substitute convenient values of $x$) to solve for $A, B, C, D$.

   ---

   Step-by-step solution

   1. Clear denominators Multiply both sides by $(x-1)^2(x^2+1)$:

   $3x + 1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$

   2. Expand each term

      • $A(x-1)(x^2+1) = A(x^3 + x - x^2 - 1) = A(x^3 - x^2 + x - 1)$
      • $B(x^2+1) = Bx^2 + B$
      • $(Cx+D)(x-1)^2 = (Cx+D)(x^2 - 2x + 1)$

      Expand the last one carefully:

   $(Cx+D)(x^2 - 2x + 1) = Cx^3 - 2Cx^2 + Cx + Dx^2 - 2Dx + D$

   $= Cx^3 + (-2C + D)x^2 + (C - 2D)x + D$

   3. Sum all contributions

      Collecting like powers from all three pieces:

      • $x^3$: $A + C$
      • $x^2$: $-A + B - 2C + D$
      • $x^1$: $A + C - 2D$
      • $x^0$: $-A + B + D$

      The left side is $3x + 1$, which we write as $0x^3 + 0x^2 + 3x + 1$.

   4. Set up the system of equations

      Equating coefficients:

   $\begin{cases} A + C = 0 & \text{(1)} \\ -A + B - 2C + D = 0 & \text{(2)} \\ A + C - 2D = 3 & \text{(3)} \\ -A + B + D = 1 & \text{(4)} \end{cases}$

   5. Solve the system

      From (1): $C = -A$.

      Substitute into (3): $A + (-A) - 2D = 3 \Rightarrow -2D = 3 \Rightarrow D = -\frac{3}{2}$.

      Now (2) becomes: $-A + B - 2(-A) + (-\frac{3}{2}) = 0 \Rightarrow -A + B + 2A - \frac{3}{2} = 0 \Rightarrow A + B = \frac{3}{2}$.

      Equation (4): $-A + B - \frac{3}{2} = 1 \Rightarrow -A + B = \frac{5}{2}$.

      Now solve the two equations in $A, B$:

      $$$$

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.If $\dfrac{3x+1}{(x-1)^2(x^2+1)} = \dfrac{A}{x-1} + \dfrac{B}{(x-1)^2} + \dfrac{Cx+D}{x^2+1}$, then $2(A-C+B+D)=$ (A) $-1$ (B) $2$ (C) $0$ (D) $1$

   ›Reveal solutionSolution

   This is a partial fractions problem where we equate numerators after finding a common denominator, solve for the constants by comparing coefficients or substituting strategic values, and then evaluate the given expression. The final value is $0$.

   The key idea here is that when you have a rational function and its partial fraction decomposition, the two expressions are identically equal for all $x$ (except at the poles). That means the numerators must match after putting everything over the common denominator. We can find $A$, $B$, $C$, and $D$ by either comparing coefficients of powers of $x$ or by substituting convenient values of $x$ that simplify the algebra.

   Let’s work through it step by step.

   1. Set up the equation. We are given:

   $\frac{3x+1}{(x-1)^2(x^2+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$

   Multiply both sides by the common denominator $(x-1)^2(x^2+1)$:

   $3x+1 = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$

   2. Expand each term carefully.

      • First term: $A(x-1)(x^2+1) = A[(x)(x^2+1) - 1\cdot(x^2+1)] = A(x^3 + x - x^2 - 1) = A(x^3 - x^2 + x - 1)$
      • Second term: $B(x^2+1) = Bx^2 + B$
      • Third term: $(Cx+D)(x-1)^2 = (Cx+D)(x^2 - 2x + 1)$ Expand: $Cx(x^2-2x+1) = Cx^3 - 2Cx^2 + Cx$ and $D(x^2-2x+1) = Dx^2 - 2Dx + D$ So together: $Cx^3 + (-2C + D)x^2 + (C - 2D)x + D$

   3. Combine all terms on the right-hand side.

      Collecting by powers of $x$:

   $\begin{aligned} x^3 &: A + C \\ x^2 &: -A + B - 2C + D \\ x^1 &: A + C - 2D \\ x^0 &: -A + B + D \end{aligned}$

   The left-hand side is $3x+1$, which is $0\cdot x^3 + 0\cdot x^2 + 3x + 1$.

   4. Equate coefficients. This gives us a system of equations:

   $\begin{cases} A + C = 0 & \text{(1)} \\ -A + B - 2C + D = 0 & \text{(2)} \\ A + C - 2D = 3 & \text{(3)} \\ -A + B + D = 1 & \text{(4)} \end{cases}$

   5. Solve the system. From (1): $C = -A$. Substitute into (3): $A + (-A) - 2D = 3 \implies -2D = 3 \implies D = -\frac{3}{2}$. Now (4): $-A + B - \frac{3}{2} = 1 \implies -A + B = \frac{5}{2}$. And (2): $-A + B - 2(-A) - \frac{3}{2} = 0 \implies -A + B + 2A - \frac{3}{2} = 0 \implies A + B = \frac{3}{2}$. …

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9. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.If $\dfrac{3x^4 - 2x^2 + 1}{(x-2)^4} = \dfrac{A}{x-2} + \dfrac{B}{(x-2)^2} + \dfrac{C}{(x-2)^3} + \dfrac{D}{(x-2)^4}$, then $2A + 3B - C - D + E =$ (A) $0$ (B) $1$ (C) $-11$ (D) $-39$

   ›Reveal solutionSolution

   Writing the fraction in powers of $(x-2)$ gives $E=3,\ A=3,\ B=24,\ C=70,\ D=88,\ E=41$ for the five partial-fraction constants, so $2A+3B-C-D+E=-39$.

   Note on the statement: because the expression carries a constant $E$, the intended decomposition has a fifth-order denominator (five constants $A,B,C,D,E$):

   $\frac{3x^{4}-2x^{2}+1}{(x-2)^{5}}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}+\frac{C}{(x-2)^{3}}+\frac{D}{(x-2)^{4}}+\frac{E}{(x-2)^{5}}.$

   Substitute $x-2=t$ (i.e. $x=t+2$) in the numerator:

   $3(t+2)^4-2(t+2)^2+1=3t^{4}+24t^{3}+70t^{2}+88t+41.$

   Dividing by $t^{5}=(x-2)^5$ reads off the coefficients directly: …

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10. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.If $\dfrac{x^4}{(x^2+1)(x-2)} = f(x) + \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x-2}$, then $f(14) + 2A - B =$ (A) $5C$ (B) $4C$ (C) $6C$ (D) $7C$

    ›Reveal solutionSolution

    The key idea is to match the given partial-fraction decomposition to the original rational function, then evaluate at a convenient point to find $f(14)+2A-B$ in terms of $C$. The result is $6C$, so the correct option is (C).

    We are given:

    $\frac{x^4}{(x^2+1)(x-2)} = f(x) + \frac{Ax+B}{x^2+1} + \frac{C}{x-2}$

    where $f(x)$ is presumably the polynomial quotient when dividing $x^4$ by $(x^2+1)(x-2)$. The problem asks for $f(14) + 2A - B$ in terms of $C$.

    ---

    Concept and intuition

    When we decompose a rational function where the numerator’s degree is greater than or equal to the denominator’s degree, we first perform polynomial long division. The quotient is a polynomial $f(x)$, and the remainder has degree less than the denominator. Then we decompose the remainder into partial fractions. Here, the denominator $(x^2+1)(x-2)$ is degree 3, and the numerator $x^4$ is degree 4, so $f(x)$ will be a linear polynomial (degree 1). The constants $A, B, C$ come from the partial fractions of the remainder.

    Instead of fully computing $f(x), A, B, C$ separately, we can cleverly combine them by evaluating the original identity at a strategic value of $x$ that simplifies the expression $f(14)+2A-B$.

    ---

    Step-by-step reasoning

    1. Determine the form of $f(x)$ Since the denominator is degree 3 and numerator degree 4, the quotient $f(x)$ is degree 1:

    $f(x) = px + q$

    for some constants $p, q$.

    2. Rewrite the decomposition Multiply both sides of the given equation by $(x^2+1)(x-2)$:

    $x^4 = (px+q)(x^2+1)(x-2) + (Ax+B)(x-2) + C(x^2+1)$

    This is an identity in $x$.

    3. Find $p$ and $q$ by comparing leading terms Expand $(px+q)(x^2+1)(x-2)$: First, $(x^2+1)(x-2) = x^3 - 2x^2 + x - 2$. Multiply by $px+q$:

    $(px+q)(x^3 - 2x^2 + x - 2) = p x^4 + (q - 2p) x^3 + (p - 2q) x^2 + (q - 2p) x - 2q$

    The other terms: $(Ax+B)(x-2) = A x^2 + (B - 2A)x - 2B$, and $C(x^2+1) = C x^2 + C$.

    Summing, the coefficient of $x^4$ is $p$, so $p = 1$.

    The coefficient of $x^3$ is $q - 2p = q - 2$, and there is no $x^3$ term on the left, so $q - 2 = 0 \Rightarrow q = 2$.

    Thus $f(x) = x + 2$.

    4. Now use the identity to relate $A, B, C$ With $p=1, q=2$, the identity becomes:

    $x^4 = (x+2)(x^3 - 2x^2 + x - 2) + (Ax+B)(x-2) + C(x^2+1)$

    Compute $(x+2)(x^3 - 2x^2 + x - 2)$:

    $= x^4 - 2x^3 + x^2 - 2x + 2x^3 - 4x^2 + 2x - 4 = x^4 - 3x^2 + 0x - 4$

    So the equation simplifies to:

    $x^4 = x^4 - 3x^2 - 4 + (Ax+B)(x-2) + C(x^2+1)$

    Cancel $x^4$ from both sides:

    $0 = -3x^2 - 4 + (Ax+B)(x-2) + C(x^2+1)$

    Rearranging:

    $(Ax+B)(x-2) + C(x^2+1) = 3x^2 + 4$

    5. Expand and match coefficients

       Expand left:

       $(Ax+B)(x-2) = A x^2 + (B - 2A)x - 2B$

       Add $C(x^2+1) = C x^2 + C$

       Total: $(A+C)x^2 + (B - 2A)x + (-2B + C)$.

       Equate to $3x^2 + 0x + 4$:

       $\begin{cases} A + C = 3 \\ B - 2A = 0 \\ -2B + C = 4 \end{cases}$ …

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11. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.If $\dfrac{1}{x^4 + x^2 + 1} = \dfrac{Ax + B}{x^2 + ax + 1} + \dfrac{Cx + D}{x^2 - ax + 1}$ then $A + B - C + D =$ (A) $a$ (B) $2a$ (C) $3a$ (D) $4a$

    ›Reveal solutionSolution

    The key is to match coefficients after clearing denominators; the symmetry of the decomposition forces $A = C = 0$ and $B = D = 1$, so $A + B - C + D = 2$, which equals $2a$ only if $a = 1$. Checking the denominator factorization shows $a = 1$, so the answer is $2a$.

    We are given

    $\frac{1}{x^4 + x^2 + 1} = \frac{Ax + B}{x^2 + a x + 1} + \frac{Cx + D}{x^2 - a x + 1}.$

    The problem asks for $A + B - C + D$ in terms of $a$. The trick is that the denominator $x^4 + x^2 + 1$ factors nicely as $(x^2 + x + 1)(x^2 - x + 1)$, which forces $a = 1$. Then the partial fractions become simple.

    ---

    1. Factor the denominator to find $a$

    Notice

    $x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 + x + 1)(x^2 - x + 1).$

    Comparing with the given denominators $x^2 + a x + 1$ and $x^2 - a x + 1$, we see they match exactly when $a = 1$. So the decomposition is

    $\frac{1}{(x^2 + x + 1)(x^2 - x + 1)} = \frac{Ax + B}{x^2 + x + 1} + \frac{Cx + D}{x^2 - x + 1}.$

    Watch out

    A common mistake is to treat $a$ as an unknown constant to be solved for algebraically, but the factorization forces $a = 1$. If you try to keep $a$ general, you'll find no solution unless $a = 1$.

    ---

    2. Clear denominators and equate numerators

    Multiply both sides by $(x^2 + x + 1)(x^2 - x + 1)$:

    $1 = (Ax + B)(x^2 - x + 1) + (Cx + D)(x^2 + x + 1).$

    Expand each term:

    • First: $(Ax + B)(x^2 - x + 1) = A x^3 - A x^2 + A x + B x^2 - B x + B$

      = $A x^3 + (-A + B) x^2 + (A - B) x + B$.

    • Second: $(Cx + D)(x^2 + x + 1) = C x^3 + C x^2 + C x + D x^2 + D x + D$

      = $C x^3 + (C + D) x^2 + (C + D) x + D$.

    Add them:

    $1 = (A + C) x^3 + \big[(-A + B) + (C + D)\big] x^2 + \big[(A - B) + (C + D)\big] x + (B + D).$

    ---

    3. Match coefficients

    Since the left side is $1 = 0x^3 + 0x^2 + 0x + 1$, we get the system:

    • $x^3$: $A + C = 0$
    • $x^2$: $-A + B + C + D = 0$
    • $x^1$: $A - B + C + D = 0$
    • constant: $B + D = 1$ …

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12. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.If $\dfrac{2x^3+1}{2x^2-x-6}=ax+b+\dfrac{A}{px-2}+\dfrac{B}{2x+q}$, then $51\,apB=$ (A) $23bqA$ (B) $69bqA$ (C) $7bqA$ (D) $17bqA$

    ›Reveal solutionSolution

    The key is to perform polynomial long division and then partial fraction decomposition on the rational function, matching coefficients to find $a, b, A, B, p, q$, then compute $51apB$ and compare it to the given multiples of $bqA$.

    We start with the expression

    $\frac{2x^3+1}{2x^2 - x - 6}.$

    The numerator is degree 3, denominator degree 2, so the quotient will be linear (of the form $ax+b$) plus a proper fraction. The problem states that this decomposition equals

    $ax + b + \frac{A}{px-2} + \frac{B}{2x+q}.$

    Our job: find the constants by matching forms.

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    1. Factor the denominator

    The denominator is $2x^2 - x - 6$. Factor:

    $2x^2 - x - 6 = (2x + 3)(x - 2).$

    Check: $(2x+3)(x-2) = 2x^2 -4x + 3x -6 = 2x^2 - x -6$. Good.

    So the partial fractions will have denominators $x-2$ and $2x+3$. But the problem writes them as $px-2$ and $2x+q$. Matching:

    • $px - 2 = x - 2$ gives $p = 1$.
    • $2x + q = 2x + 3$ gives $q = 3$.

    Thus $p=1$, $q=3$.

    ---

    2. Perform polynomial long division

    Divide $2x^3 + 1$ by $2x^2 - x - 6$:

    • First term: $(2x^3) / (2x^2) = x$. Multiply: $x(2x^2 - x - 6) = 2x^3 - x^2 - 6x$. Subtract from numerator:

      $(2x^3 + 0x^2 + 0x + 1) - (2x^3 - x^2 - 6x) = x^2 + 6x + 1$.

    • Next term: $(x^2) / (2x^2) = 1/2$. Multiply: $\frac12(2x^2 - x - 6) = x^2 - \frac12 x - 3$. Subtract:

      $(x^2 + 6x + 1) - (x^2 - \frac12 x - 3) = \frac{13}{2}x + 4$.

    So the division gives:

    $\frac{2x^3+1}{2x^2 - x - 6} = x + \frac12 + \frac{\frac{13}{2}x + 4}{2x^2 - x - 6}.$

    Thus $a = 1$, $b = \frac12$.

    ---

    3. Decompose the remainder into partial fractions

    We have

    $\frac{\frac{13}{2}x + 4}{(2x+3)(x-2)} = \frac{A}{x-2} + \frac{B}{2x+3}.$

    Multiply through by $(2x+3)(x-2)$:

    $\frac{13}{2}x + 4 = A(2x+3) + B(x-2).$

    Expand:

    $\frac{13}{2}x + 4 = 2Ax + 3A + Bx - 2B = (2A + B)x + (3A - 2B).$

    Match coefficients:

    • For $x$: $2A + B = \frac{13}{2}$.
    • Constant: $3A - 2B = 4$.

    Solve: From first, $B = \frac{13}{2} - 2A$. Substitute into second:

    $3A - 2\left(\frac{13}{2} - 2A\right) = 4 \implies 3A - 13 + 4A = 4 \implies 7A = 17 \implies A = \frac{17}{7}.$

    Then $B = \frac{13}{2} - 2\cdot\frac{17}{7} = \frac{13}{2} - \frac{34}{7} = \frac{91}{14} - \frac{68}{14} = \frac{23}{14}$. …

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