Q.Integrate the following function: $\frac{x}{(x + 1)(x + 2)}$

Showing the 12 most recent of 63 on this concept.

1. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.$\int \dfrac{x-1}{(x-2)(x-3)}\,dx=$ (A) $2\log|x-3|+\log|x-2|+c$ (B) $\log|x-3|-\log|x-2|+c$ (C) $\log|x-3|^2-\log|x+2|+c$ (D) $\log\left|\dfrac{(x-3)^2}{x-2}\right|+c$

   ›Reveal solutionSolution

   A rational function with distinct linear factors in the denominator — resolve into partial fractions, integrate each term as a log, then combine using log rules.

   Concept and Intuition

   Any proper rational function with distinct linear denominator factors can be split into simple fractions $\frac{A}{x-a}+\frac{B}{x-b}$, each of which integrates to $A\log|x-a|$. Combining the two logs at the end into a single log of a ratio/power lets you match against answer choices written as one combined logarithm.

   Step-by-Step Solution

   1. Write $\dfrac{x-1}{(x-2)(x-3)} = \dfrac{A}{x-2}+\dfrac{B}{x-3}$, so $x-1 = A(x-3)+B(x-2)$.
   2. Put $x=2$: $2-1 = A(2-3) \Rightarrow 1 = -A \Rightarrow A=-1$.
   3. Put $x=3$: $3-1 = B(3-2) \Rightarrow 2=B$.
   4. So $\displaystyle\int\frac{x-1}{(x-2)(x-3)}dx = -\log|x-2| + 2\log|x-3| + c$.
   5. Combine: $2\log|x-3|-\log|x-2| = \log\dfrac{|x-3|^2}{|x-2|} = \log\left|\dfrac{(x-3)^2}{x-2}\right|$.

   Common Mistakes

   • Getting the sign of $A$ wrong (it is $-1$, not $+1$), which flips the final combined-log form and matches the wrong option.
   • Combining logs incorrectly, e.g. writing $\log|x-3|^2+\log|x-2|$ instead of the correct difference.

   ✓Final answer

   The correct option is (D) — $\log\left|\dfrac{(x-3)^2}{x-2}\right|+c$.

   ANSWER: D

2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int \frac{dx}{\sin x + \sin 2x} =$ (A) $\frac{1}{4}\log|1-\cos x| + \frac{1}{3}\log|1+\cos x| - \frac{2}{3}\log|1+\cos 2x| + c$ (B) $\frac{1}{3}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{1}{3}\log|1+\cos 2x| + c$ (C) $\frac{1}{6}\log|1-\cos x| + \frac{1}{2}\log|1+\cos x| - \frac{2}{3}\log|1+2\cos x| + c$ (D) $\frac{1}{6}\log|1-\cos x| + \frac{1}{4}\log|1+\cos x| + \frac{2}{3}\log|1+2\cos x| + c$

   ›Reveal solutionSolution

   Factoring $\sin x+\sin2x=\sin x(1+2\cos x)$ and substituting $t=\cos x$ reduces this to a rational-function partial-fractions integral, giving $\frac16\log|1-\cos x|+\frac12\log|1+\cos x|-\frac23\log|1+2\cos x|+c$.

   Concept and Intuition

   Whenever an integral has $\sin x$ (an odd power effectively) times other cosine factors in the denominator, multiplying numerator and denominator by $\sin x$ turns $\sin^2x$ into $1-\cos^2x$, which lets us substitute $t=\cos x$ and reduce the whole problem to partial fractions of a rational function in $t$ — a completely mechanical final step.

   Step-by-Step Solution

   1. Factor the denominator: $\sin x+\sin2x = \sin x + 2\sin x\cos x = \sin x(1+2\cos x)$.
   2. So the integral is $\displaystyle\int\frac{dx}{\sin x(1+2\cos x)}$.
   3. Multiply top and bottom by $\sin x$: $\displaystyle\int\frac{\sin x\,dx}{\sin^2x(1+2\cos x)}=\int\frac{\sin x\,dx}{(1-\cos^2x)(1+2\cos x)}=\int\frac{\sin x\,dx}{(1-\cos x)(1+\cos x)(1+2\cos x)}$.
   4. Substitute $t=\cos x$, $dt=-\sin x\,dx$: integral $=-\displaystyle\int\frac{dt}{(1-t)(1+t)(1+2t)}$.
   5. Partial fractions: $\dfrac{1}{(1-t)(1+t)(1+2t)}=\dfrac{A}{1-t}+\dfrac{B}{1+t}+\dfrac{C}{1+2t}$. Evaluating at $t=1$: $A=\tfrac16$. At $t=-1$: $B=\tfrac12$. At $t=-\tfrac12$: $C=-\tfrac23$.
   6. So $-\displaystyle\int\left[\frac{1/6}{1-t}+\frac{1/2}{1+t}-\frac{2/3}{1+2t}\right]dt = \frac16\log|1-t|+\frac12\log|1+t|-\frac23\log|1+2t|+c$ (the sign flips on the $1/(1-t)$ term because $\int\frac{dt}{1-t}=-\log|1-t|$, and the overall integral carries a leading minus sign).
   7. Substitute back $t=\cos x$: $\dfrac16\log|1-\cos x|+\dfrac12\log|1+\cos x|-\dfrac23\log|1+2\cos x|+c$.

   Common Mistakes

   • Missing the overall minus sign introduced by $dt=-\sin x\,dx$, which flips the sign of every term if mishandled.
   • Misassigning the partial-fraction coefficients (double-check by plugging $t=0$: $A+B+C$ should equal $1$).

   ✓Final answer

   The correct option is (C) — $\frac{1}{6}\log|1-\cos x| + \frac{1}{2}\log|1+\cos x| - \frac{2}{3}\log|1+2\cos x| + c$.

   ANSWER: C

3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{x^3 - 1}{x^3 + x^2}\, dx = f(x) + \log(g(x)) + c$, $f(1) = 2$ and $g(-3) = \dfrac{3}{4}$, then $f(-2) + g(-2) =$ (A) $-\dfrac{9}{2}$ (B) $-\dfrac{1}{2}$ (C) $\dfrac{9}{4}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   A rational-function integral splits by partial fractions into a polynomial/rational part $f(x)$ plus a logarithmic part $\log(g(x))$; the two given data points pin down $f$ and $g$ exactly, letting us evaluate $f(-2)+g(-2)$. Answer: $-\tfrac12$.

   Concept and Intuition

   When an improper rational integrand is written $\frac{x^3-1}{x^3+x^2}$, polynomial division peels off the constant part, and partial fractions turn the remaining proper fraction into simple terms of the form $\frac{A}{x}, \frac{B}{x^2}, \frac{C}{x+1}$ whose antiderivatives are $A\log|x|$, $-B/x$, $C\log|x+1|$. Collecting all log terms into a single $\log(g(x))$ and all algebraic terms into $f(x)$ matches the form the question gives; the two numeric conditions are just there to confirm the constants (and resolve the sign inside the absolute value at negative $x$).

   Step-by-Step Solution

   1. Divide: $\dfrac{x^3-1}{x^3+x^2}=1-\dfrac{x^2+1}{x^3+x^2}=1-\dfrac{x^2+1}{x^2(x+1)}$.
   2. Partial fractions: $\dfrac{x^2+1}{x^2(x+1)}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x+1}$. Clearing denominators: $x^2+1=Ax(x+1)+B(x+1)+Cx^2$. Setting $x=0$: $B=1$. Setting $x=-1$: $2=C$. Matching $x^2$ coefficients: $1=A+C\Rightarrow A=-1$.
   3. So $\dfrac{x^2+1}{x^2(x+1)}=-\dfrac1x+\dfrac1{x^2}+\dfrac2{x+1}$, and the integrand is $1+\dfrac1x-\dfrac1{x^2}-\dfrac2{x+1}$.
   4. Integrate term by term: $\displaystyle\int\left(1+\frac1x-\frac1{x^2}-\frac{2}{x+1}\right)dx=x+\log|x|+\frac1x-2\log|x+1|+c=\left(x+\frac1x\right)+\log\left|\frac{x}{(x+1)^2}\right|+c$.
   5. So $f(x)=x+\dfrac1x$; check $f(1)=1+1=2$ ✓ (matches the given condition exactly, with no extra constant needed).
   6. $g(x)$ must equal $\left|\dfrac{x}{(x+1)^2}\right|$ so that $\log(g(x))=\log\left|\dfrac{x}{(x+1)^2}\right|$; check $g(-3)=\dfrac{3}{4}$ ✓ (matches exactly).
   7. So $g(x)=\dfrac{|x|}{(x+1)^2}$ for $x<0$ (i.e. $g(x)=\dfrac{-x}{(x+1)^2}$ there).
   8. $f(-2)=-2+\dfrac{1}{-2}=-2-\dfrac12=-\dfrac52$. $g(-2)=\dfrac{|-2|}{(-1)^2}=\dfrac{2}{1}=2$.
   9. $f(-2)+g(-2)=-\dfrac52+2=-\dfrac12$.

   Common Mistakes

   • Forgetting the $|x|$/sign subtlety in $g(x)$ for negative $x$, which would give the wrong sign in $g(-2)$.
   • Arithmetic slip in the partial-fraction coefficients.

   ✓Final answer

   The correct option is (B) — $-\dfrac{1}{2}$.

   ANSWER: B

4. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $\dfrac{A}{(x+2)^2}$ is one of the partial fractions of $\dfrac{x^2+3x+5}{(2x+1)(x+2)^2}$, then $A =$ (A) 2 (B) 1 (C) -2 (D) -1

   ›Reveal solutionSolution

   Using the cover-up (Heaviside) method at the repeated root $x=-2$ gives $A=-1$ directly, no full partial-fraction expansion needed.

   Concept and Intuition

   For a repeated linear factor $(x+2)^2$ in the denominator, the partial fraction decomposition has a term $\dfrac{A}{(x+2)^2}$ whose coefficient can be found by the cover-up method: multiply both sides by $(x+2)^2$, which cancels that factor from the denominator entirely, then substitute $x=-2$ into what remains (the other terms of the decomposition vanish or become finite/zero at this special substitution for the squared-factor coefficient specifically).

   Step-by-Step Solution

   1. Write $\dfrac{x^2+3x+5}{(2x+1)(x+2)^2}=\dfrac{P}{2x+1}+\dfrac{Q}{x+2}+\dfrac{A}{(x+2)^2}$.
   2. Multiply both sides by $(x+2)^2$: $\dfrac{x^2+3x+5}{2x+1} = P\cdot\dfrac{(x+2)^2}{2x+1}+Q(x+2)+A$.
   3. Substitute $x=-2$ (this kills the $P$ and $Q$ terms since they still carry a factor of $(x+2)$): $A=\dfrac{x^2+3x+5}{2x+1}\bigg|_{x=-2}$.
   4. Numerator at $x=-2$: $(-2)^2+3(-2)+5=4-6+5=3$. Denominator: $2(-2)+1=-3$.
   5. $A=\dfrac{3}{-3}=-1$.

   Common Mistakes

   • Forgetting to substitute into $(2x+1)$ as well and only evaluating the numerator, or making an arithmetic sign slip at $x=-2$.
   • Trying to solve for $A$ via a full system of equations (equating coefficients) when the direct cover-up substitution is much faster for this specific coefficient.

   ✓Final answer

   The correct option is (D) — $-1$.

   ANSWER: D

5. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $\dfrac{3x^2+x+2}{(3x^2+x+4)(3x^2+x+1)} = \dfrac{Ax+B}{3x^2+x+4} + \dfrac{Cx+D}{3x^2+x+1}$, then $(A+B) + (C+D) =$ (A) $\dfrac{1}{3}$ (B) $\dfrac{2}{3}$ (C) $1$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   Substituting $u=3x^2+x$ collapses the problem to an ordinary constant partial fraction in $u$, forcing $A=C=0$ and giving $(A+B)+(C+D)=1$.

   Concept and Intuition

   When a rational expression's numerator and both denominator factors are built from the same quadratic block $3x^2+x$ shifted by constants, it's really a partial-fraction problem in the single variable $u=3x^2+x$, not in $x$ directly. Recognizing this shortcut avoids a messy 4-unknown system in $x$.

   Step-by-Step Solution

   1. Let $u=3x^2+x$. The equation becomes $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{Ax+B}{u+4}+\dfrac{Cx+D}{u+1}$.
   2. Do ordinary partial fractions in $u$: $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{P}{u+4}+\dfrac{Q}{u+1}$.
   3. Cover-up at $u=-4$: $P = \dfrac{-4+2}{-4+1} = \dfrac{-2}{-3}=\dfrac23$. At $u=-1$: $Q=\dfrac{-1+2}{-1+4}=\dfrac{1}{3}$.
   4. So $\dfrac{u+2}{(u+4)(u+1)} = \dfrac{2/3}{u+4}+\dfrac{1/3}{u+1}$, which is entirely $x$-independent in its numerators.
   5. Matching with $\dfrac{Ax+B}{u+4}+\dfrac{Cx+D}{u+1}$ forces $Ax+B = \tfrac23$ (a constant) for all $x$, so $A=0,\ B=\tfrac23$; similarly $Cx+D=\tfrac13$ gives $C=0,\ D=\tfrac13$.
   6. Therefore $(A+B)+(C+D) = \left(0+\tfrac23\right)+\left(0+\tfrac13\right) = 1$.

   Common Mistakes

   • Trying to cross-multiply and match powers of $x$ directly in the original (degree-4-denominator) equation — much longer than the $u$-substitution shortcut.
   • Missing that $A$ and $C$ must be zero (since the true partial fractions have no $x$-dependence in the numerator at all).

   ✓Final answer

   The correct option is (C) — $1$.

   ANSWER: C

6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\dfrac{x^4}{(x-1)(x-2)}=f(x)+\dfrac{A}{x-1}+\dfrac{B}{x-2}$, then $f(-2)+A+B=$ (A) 32 (B) 28 (C) 22 (D) 20

   ›Reveal solutionSolution

   This tests polynomial long division combined with partial fractions — split $\dfrac{x^4}{(x-1)(x-2)}$ into a polynomial part $f(x)$ plus proper fractions. Answer: $f(-2)+A+B=20$.

   Concept and Intuition

   When the numerator's degree ($4$) is greater than or equal to the denominator's degree ($2$), a rational function isn't purely a sum of partial fractions — you first must divide out a polynomial quotient $f(x)$, leaving a proper-fraction remainder that partial-fractions cleanly. Here $f(x)$ is exactly that quotient (degree $4-2=2$).

   Step-by-Step Solution

   1. Divide $x^4$ by $x^2-3x+2$ (long division): $x^4 = (x^2-3x+2)(x^2+3x+7) + (15x-14)$ Check: $(x^2-3x+2)(x^2+3x+7) = x^4 - 15x+14$, so adding $15x-14$ recovers $x^4$. ✓
   2. So $f(x)=x^2+3x+7$, and the remainder gives $\dfrac{15x-14}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$.
   3. Clear denominators: $15x-14=A(x-2)+B(x-1)$.
   4. Put $x=1$: $15-14=A(-1)\Rightarrow 1=-A\Rightarrow A=-1$.
   5. Put $x=2$: $30-14=B(1)\Rightarrow B=16$.
   6. Compute $f(-2)=(-2)^2+3(-2)+7=4-6+7=5$.
   7. $f(-2)+A+B = 5+(-1)+16=20$.

   Common Mistakes

   • Forgetting $f(x)$ altogether and trying to partial-fraction $x^4/((x-1)(x-2))$ directly without first dividing out the polynomial part.
   • Sign slips when solving $A(x-2)+B(x-1)$ by substitution — always plug the root that kills the other term.

   ✓Final answer

   The correct option is (D) — 20.

   ANSWER: D

7. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\dfrac{4x}{(x^2-1)^2} = \dfrac{A_1}{x-1} + \dfrac{A_2}{(x-1)^2} + \dfrac{A_3}{x+1} + \dfrac{A_4}{(x+1)^2}$, then $A_1 + A_2 + A_3 + A_4 =$ (A) $-2$ (B) $1$ (C) $0$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   Clearing denominators and substituting convenient values of $x$ (the repeated roots plus two extra points) pins down all four constants; they add up to $0$.

   Concept and Intuition

   For a partial fraction decomposition with repeated linear factors, substituting the roots directly isolates the "squared-term" coefficients instantly, while substituting a couple of extra convenient values (like $x=0$ and $x=2$) gives enough equations to solve for the remaining linear-term coefficients.

   Step-by-Step Solution

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

   $4x=A_1(x-1)(x+1)^2+A_2(x+1)^2+A_3(x+1)(x-1)^2+A_4(x-1)^2.$

   2. Set $x=1$: $4=A_2(2)^2=4A_2 \Rightarrow A_2=1$.
   3. Set $x=-1$: $-4=A_4(-2)^2=4A_4 \Rightarrow A_4=-1$.
   4. Set $x=0$: $0=A_1(-1)(1)+A_2(1)+A_3(1)(1)+A_4(1)=-A_1+A_2+A_3+A_4$. Using $A_2=1,A_4=-1$: $0=-A_1+A_3 \Rightarrow A_1=A_3$.
   5. Set $x=2$: $8=A_1(1)(9)+A_2(9)+A_3(3)(1)+A_4(1)=9A_1+9A_2+3A_3+A_4$. Substitute $A_2=1,A_4=-1,A_3=A_1$: $8=9A_1+9+3A_1-1=12A_1+8 \Rightarrow A_1=0$, so $A_3=0$.
   6. Sum: $A_1+A_2+A_3+A_4=0+1+0-1=0$.

   Common Mistakes

   • Only using the two root substitutions ($x=1,-1$) and forgetting extra points are needed to separate $A_1$ from $A_3$.
   • Sign errors when expanding $(x-1)(x+1)^2$ and $(x+1)(x-1)^2$ at the chosen test points.

   ✓Final answer

   The correct option is (C) — $0$.

   ANSWER: C

8. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $\dfrac{-x^2+6x+1}{(x-1)^2(x^2+2)} = \dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{Cx-3}{x^2+2}$, then $A+B+C=$ (A) $7$ (B) $5$ (C) $3$ (D) $2$

   ›Reveal solutionSolution

   Clearing denominators and matching coefficients (using $x=1$ to isolate $B$ first) gives $A=0,\ B=2,\ C=0$, so $A+B+C=2$.

   Concept and Intuition

   This is a standard partial-fractions decomposition. The repeated linear factor $(x-1)^2$ lets us find $B$ instantly by substituting $x=1$ directly into the cleared equation (this kills every term except the one multiplying $B$). The remaining coefficients $A,C$ are then found by matching powers of $x$.

   Step-by-Step Solution

   1. Clear denominators: $-x^2+6x+1=A(x-1)(x^2+2)+B(x^2+2)+(Cx-3)(x-1)^2$.
   2. Set $x=1$: LHS $=-1+6+1=6$; RHS $=0+B(3)+0=3B$. So $B=2$.
   3. Expand the RHS fully with $B=2$: $A(x^3-x^2+2x-2)+2x^2+4+(Cx-3)(x^2-2x+1)$. $(Cx-3)(x^2-2x+1)=Cx^3-2Cx^2+Cx-3x^2+6x-3$.
   4. Collect by power of $x$: $x^3$: $A+C$ $x^2$: $-A+2-2C-3=-A-2C-1$ $x^1$: $2A+C+6$ $x^0$: $-2A+4-3=-2A+1$
   5. Match to LHS coefficients ($0,-1,6,1$ for $x^3,x^2,x,1$): $x^3$: $A+C=0$ $x^0$: $-2A+1=1\Rightarrow A=0\Rightarrow C=0$ (from the $x^3$ equation). Check $x^2$: $-0-0-1=-1$ ✓. Check $x^1$: $0+0+6=6$ ✓ — all consistent.
   6. $A+B+C=0+2+0=2$.

   Common Mistakes

   • Forgetting the shortcut of plugging $x=1$ to find $B$ instantly, and instead trying to solve a full $3\times3$ system.
   • Sign errors expanding $(Cx-3)(x-1)^2$.

   ✓Final answer

   The correct option is (D) — $2$.

   ANSWER: D

9. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.If $\dfrac{x^2}{(x^2+2)(x^4-1)}=\dfrac{A}{x^2-1}+\dfrac{B}{x^2+1}+\dfrac{C}{x^2+2}$, then $A+B-C=$ (A) $0$ (B) $\dfrac{4}{3}$ (C) $\dfrac{3}{4}$ (D) $2$

   ›Reveal solutionSolution

   A partial-fractions problem in disguise (substitute $y=x^2$); solving gives $A+B-C=\frac{4}{3}$.

   Concept and Intuition

   Since $x^4-1=(x^2-1)(x^2+1)$, the whole expression is a rational function purely in $y=x^2$. Substituting $y=x^2$ converts it into an ordinary partial-fractions decomposition with three distinct linear factors $(y-1),(y+1),(y+2)$, solvable by the cover-up (Heaviside) method.

   Step-by-Step Solution

   1. Let $y=x^2$. The equation becomes $\dfrac{y}{(y+2)(y-1)(y+1)}=\dfrac{A}{y-1}+\dfrac{B}{y+1}+\dfrac{C}{y+2}$.
   2. Clear denominators: $y=A(y+1)(y+2)+B(y-1)(y+2)+C(y-1)(y+1)$.
   3. At $y=1$: $1=A(2)(3)=6A\Rightarrow A=\dfrac16$.
   4. At $y=-1$: $-1=B(-2)(1)=-2B\Rightarrow B=\dfrac12$.
   5. At $y=-2$: $-2=C(-3)(-1)=3C\Rightarrow C=-\dfrac23$.
   6. $A+B-C=\dfrac16+\dfrac12-\left(-\dfrac23\right)=\dfrac16+\dfrac36+\dfrac46=\dfrac86=\dfrac43$.

   Common Mistakes

   • Forgetting the substitution $y=x^2$ and trying to partial-fraction a quartic denominator directly in $x$.
   • Sign error when subtracting $C$ (note the question asks for $A+B-C$, not $A+B+C$).

   ✓Final answer

   The correct option is (B) — $\dfrac{4}{3}$.

   ANSWER: B

10. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
    Q.If $\dfrac{9}{(x-1)(x+2)^2} = \dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$ then $A-B-C$ is equal to (A) 3 (B) 5 (C) $-1$ (D) 0

    ›Reveal solutionSolution

    This is a standard partial-fractions problem; plugging in convenient roots quickly isolates each constant, giving $A-B-C=5$.

    Concept and Intuition

    For a repeated linear factor $(x+2)^2$, the partial fraction decomposition needs both a $\dfrac{B}{x+2}$ and a $\dfrac{C}{(x+2)^2}$ term. The fastest way to find the constants is to clear denominators and substitute the roots of the linear factors directly (this instantly kills all but one term).

    Step-by-Step Solution

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

    $9 = A(x+2)^2 + B(x-1)(x+2) + C(x-1)$

    2. Put $x=1$: $9 = A(3)^2 = 9A \Rightarrow A=1$.
    3. Put $x=-2$: $9 = C(-2-1) = -3C \Rightarrow C=-3$.
    4. Compare coefficients of $x^2$ on both sides: RHS gives $A+B$ (from $A(x+2)^2$ contributing $Ax^2$ and $B(x-1)(x+2)$ contributing $Bx^2$), LHS has 0. So $A+B=0 \Rightarrow B=-1$.
    5. Check constant term: $4A-2B-C = 4(1)-2(-1)-(-3)=4+2+3=9$ ✓, confirming the values.
    6. $A-B-C = 1-(-1)-(-3) = 1+1+3 = 5$.

    Common Mistakes

    • Forgetting the extra $\dfrac{C}{(x+2)^2}$ term for the repeated factor.
    • Sign errors when substituting $x=-2$ or $x=1$.

    ✓Final answer

    The correct option is (B) — 5.

    ANSWER: B

11. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.If $\dfrac{x^4}{(x-1)(x-2)} = f(x) + \dfrac{A}{x-1} + \dfrac{B}{x-2}$, then (A) $f(x) = x^2 - 3x + 7$ (B) $f(x) = x^2 + 3x + 7$ (C) $A + B = 17$ (D) $A - B = -18$

    ›Reveal solutionSolution

    Dividing $x^4$ by $(x-1)(x-2)$ gives quotient $f(x)=x^2+3x+7$ with remainder $15x-14$, and partial-fractioning the remainder gives $A=-1,\,B=16$ — so only the $f(x)$ option is correct.

    Concept and Intuition

    When a rational function's numerator degree exceeds the denominator's, you must first do polynomial long division to extract the polynomial part $f(x)$ before decomposing the proper-fraction remainder into partial fractions. Skipping the division step and partial-fractioning directly would be invalid since $\frac{x^4}{(x-1)(x-2)}$ isn't a proper fraction.

    Step-by-Step Solution

    1. Divide $x^4$ by $x^2-3x+2$ (the denominator's product):
       • $x^4 \div x^2 = x^2$; $x^2(x^2-3x+2)=x^4-3x^3+2x^2$; subtract: $3x^3-2x^2$.
       • $3x^3 \div x^2 = 3x$; $3x(x^2-3x+2)=3x^3-9x^2+6x$; subtract: $7x^2-6x$.
       • $7x^2 \div x^2 = 7$; $7(x^2-3x+2)=7x^2-21x+14$; subtract: $15x-14$.
    2. So $x^4 = (x^2-3x+2)(x^2+3x+7) + (15x-14)$, giving $f(x)=x^2+3x+7$ and remainder $15x-14$.
    3. Decompose $\dfrac{15x-14}{(x-1)(x-2)} = \dfrac{A}{x-1}+\dfrac{B}{x-2}$: so $15x-14=A(x-2)+B(x-1)$.
    4. At $x=1$: $15-14=1=A(-1) \Rightarrow A=-1$.
    5. At $x=2$: $30-14=16=B(1) \Rightarrow B=16$.
    6. Check: $A+B=-1+16=15$ (not 17, so (C) is wrong); $A-B=-1-16=-17$ (not $-18$, so (D) is wrong). Only $f(x)=x^2+3x+7$ (option B) holds.

    Common Mistakes

    • Sign errors during long division subtraction steps.
    • Assuming any answer option involving $A,B$ must also be checked as "possibly also correct" — but the question asks which statement is true, and only one is.

    ✓Final answer

    The correct option is (B) — $f(x) = x^2 + 3x + 7$.

    ANSWER: B

12. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.Given $\dfrac{3x-2}{(x+1)^2(x+3)} = \dfrac{A}{x+1} + \dfrac{B}{(x+1)^2} + \dfrac{C}{x+3}$ then $4A + 2B + 4C$ (A) $5$ (B) $-5$ (C) $-3$ (D) $3$

    ›Reveal solutionSolution

    Standard partial-fraction cover-up plus a coefficient match pins down $A,B,C$, giving $4A+2B+4C=-5$.

    Concept and Intuition

    For a repeated linear factor $(x+1)^2$ together with a simple factor $(x+3)$, the cover-up (Heaviside) method quickly gives $B$ and $C$ by substituting the roots that make each factor vanish. The remaining constant $A$ is then found by matching a coefficient (here, the $x^2$ coefficient, since the numerator on the left has no $x^2$ term).

    Step-by-Step Solution

    1. Clear denominators: $3x-2 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$.
    2. Set $x=-1$ (kills the $A$ and $C$ terms): $3(-1)-2 = B(-1+3) \Rightarrow -5 = 2B \Rightarrow B=-\dfrac{5}{2}$.
    3. Set $x=-3$ (kills the $A$ and $B$ terms): $3(-3)-2 = C(-3+1)^2 \Rightarrow -11 = 4C \Rightarrow C=-\dfrac{11}{4}$.
    4. Match the coefficient of $x^2$ on both sides (LHS has none): $0 = A + C \Rightarrow A = \dfrac{11}{4}$.
    5. Check with the constant term: $3A+3B+C = \dfrac{33}{4}-\dfrac{15}{2}-\dfrac{11}{4} = -2$ ✓, matching the LHS constant $-2$.
    6. Compute $4A+2B+4C = 4\left(\dfrac{11}{4}\right) + 2\left(-\dfrac52\right) + 4\left(-\dfrac{11}{4}\right) = 11 - 5 - 11 = -5$.

    Common Mistakes

    • Forgetting the repeated factor contributes two separate terms $\dfrac{B}{x+1}$ and $\dfrac{B'}{(x+1)^2}$ style constants — here correctly split as $A/(x+1)+B/(x+1)^2$.
    • Sign slips substituting negative roots.

    ✓Final answer

    The correct option is (B) — $-5$.

    ANSWER: B