Q.Integrate the function $\frac{1}{x-x^3}$

Showing the 12 most recent of 33 on this concept.

1. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $\dfrac{d}{dx} \left( \dfrac{2x+1}{(x+1)^2 (x-2)} \right) = \dfrac{A}{(x-2)^2} + \dfrac{B}{(x+2)^3} + \dfrac{C}{(x+1)^2}$ then $A+B+C=$ (A) $\dfrac{-2}{3}$ (B) $\dfrac{2}{3}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{-1}{3}$

   ›Reveal solutionSolution

   The problem asks for the sum of constants in a partial-fraction-like decomposition of a derivative. By differentiating the given rational function and matching the form, we find $A+B+C = -\frac{2}{3}$, which corresponds to option (A).

   The key insight here is that the derivative of a rational function can be expressed as a sum of simpler fractions, but the given form is not a standard partial fraction decomposition — it’s a template for the derivative itself. The constants $A$, $B$, $C$ are not from splitting the original function, but from rewriting its derivative in that specific pattern. So we must differentiate first, then compare.

   Let’s work through it.

   1. Differentiate the given function.

      Let $f(x) = \dfrac{2x+1}{(x+1)^2 (x-2)}$.

      Use the quotient rule: $f'(x) = \frac{u'v - uv'}{v^2}$, where $u = 2x+1$ and $v = (x+1)^2 (x-2)$.

      First, $u' = 2$.

      For $v'$, use the product rule: $v = (x+1)^2 \cdot (x-2)$.

      Derivative: $v' = 2(x+1)(1) \cdot (x-2) + (x+1)^2 \cdot 1 = 2(x+1)(x-2) + (x+1)^2$.

      Simplify: $v' = (x+1)[2(x-2) + (x+1)] = (x+1)(2x - 4 + x + 1) = (x+1)(3x - 3) = 3(x+1)(x-1)$.

      So $f'(x) = \frac{2 \cdot (x+1)^2 (x-2) - (2x+1) \cdot 3(x+1)(x-1)}{(x+1)^4 (x-2)^2}$.

   2. Simplify the numerator.

      Factor out $(x+1)$ from the numerator (common factor):

      Numerator $= (x+1) \left[ 2(x+1)(x-2) - 3(2x+1)(x-1) \right]$.

      Expand inside:

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

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

      Subtract: $(2x^2 - 2x - 4) - (6x^2 - 3x - 3) = -4x^2 + x - 1$.

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

      Thus $f'(x) = \frac{(x+1)(-4x^2 + x - 1)}{(x+1)^4 (x-2)^2} = \frac{-4x^2 + x - 1}{(x+1)^3 (x-2)^2}$.

   3. Match to the given form.

      We are told $f'(x) = \frac{A}{(x-2)^2} + \frac{B}{(x+2)^3} + \frac{C}{(x+1)^2}$.

      But our denominator has factors $(x+1)^3$ and $(x-2)^2$ — note the $(x+2)^3$ term in the given form is suspicious. That’s likely a typo in the problem statement (common in such questions); it should be $(x+1)^3$ instead of $(x+2)^3$. We’ll proceed assuming the intended form is $\frac{A}{(x-2)^2} + \frac{B}{(x+1)^3} + \frac{C}{(x+1)^2}$.

      So we need to decompose $\frac{-4x^2 + x - 1}{(x+1)^3 (x-2)^2}$ into partial fractions with denominators as above.

   4. Set up the decomposition.

      Write:

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

   Multiply both sides by $(x+1)^3 (x-2)^2$:

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

   5. Find constants by substitution.

      • Put $x = 2$:

        $-4(4) + 2 - 1 = -16 + 1 = -15$.

        Right side: $A (3)^3 + B(0) + C(0) = 27A$.

        So $27A = -15 \Rightarrow A = -\frac{15}{27} = -\frac{5}{9}$.

      • Put $x = -1$:

        $-4(1) -1 -1 = -4 -2 = -6$.

        Right side: $A(0) + B(-3)^2 + C(0) = 9B$.

        So $9B = -6 \Rightarrow B = -\frac{2}{3}$.

      • To find $C$, put $x = 0$ (a convenient value):

        Left: $-4(0) + 0 - 1 = -1$.

        Right: $A(1)^3 + B(-2)^2 + C(1)(-2)^2 = A + 4B + 4C$.

        Substitute $A = -\frac{5}{9}$, $B = -\frac{2}{3}$:

        $-\frac{5}{9} + 4\left(-\frac{2}{3}\right) + 4C = -\frac{5}{9} - \frac{8}{3} + 4C$.

        $\frac{8}{3} = \frac{24}{9}$, so $-\frac{5}{9} - \frac{24}{9} = -\frac{29}{9}$.

        Equation: $-\frac{29}{9} + 4C = -1 \Rightarrow 4C = -1 + \frac{29}{9} = \frac{-9 + 29}{9} = \frac{20}{9}$.

        So $C = \frac{5}{9}$.

   6. Sum the constants.

      $A + B + C = -\frac{5}{9} + \left(-\frac{2}{3}\right) + \frac{5}{9} = -\frac{2}{3}$.

   Watch out

   The given form has $(x+2)^3$, but the derivative’s denominator has $(x+1)^3$. This is almost certainly a misprint; the intended term is $(x+1)^3$. Without this correction, the problem has no solution.

   ✓Final answer

   The value is $\boxed{-\frac{2}{3}}$, which corresponds to option (A).

2. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.$\int \frac{1}{(x-2)(x^2+1)} \, dx =$ (A) $\log \frac{\sqrt{x^2+1}}{|x-2|} + 2 \tan^{-1} x + c$ (B) $\log \frac{|x-2|}{x^2+1} + 2 \tan^{-1} x + c$ (C) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{7+x^2}} + 2 \tan^{-1} x \right] + c$ (D) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{1+x^2}} - 2 \tan^{-1} x \right] + c$

   ›Reveal solutionSolution

   The integral is solved by partial fraction decomposition, leading to a combination of a logarithmic term and an arctangent term. The correct result is $\frac{1}{5}\left[\log\frac{|x-2|}{\sqrt{x^2+1}} + 2\tan^{-1}x\right] + c$, which matches option (C).

   We have a rational function whose denominator factors into a linear term $(x-2)$ and an irreducible quadratic $(x^2+1)$. The natural strategy is partial fractions, because the integrand is a proper rational function (degree of numerator < degree of denominator). This will break the integral into simpler pieces we can integrate directly.

   ---

   1. Set up the partial fraction decomposition

   We write:

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

   The numerator for the quadratic term is linear because $x^2+1$ cannot be factored further over the reals.

   2. Clear denominators

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

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

   3. Expand and collect like terms

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

   $= A x^2 + A + Bx^2 - 2Bx + Cx - 2C$

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

   4. Equate coefficients

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

   $\begin{cases} A + B = 0 \\ -2B + C = 0 \\ A - 2C = 1 \end{cases}$

   5. Solve the system

   From the first equation: $B = -A$.

   From the second: $C = 2B = -2A$.

   Substitute into the third: $A - 2(-2A) = A + 4A = 5A = 1$, so $A = \frac{1}{5}$.

   Then $B = -\frac{1}{5}$, $C = -\frac{2}{5}$.

   6. Write the decomposed integrand

   $\frac{1}{(x-2)(x^2+1)} = \frac{1/5}{x-2} + \frac{-\frac{1}{5}x - \frac{2}{5}}{x^2+1}$

   7. Integrate term by term

   $\int \frac{1}{(x-2)(x^2+1)} \, dx = \frac{1}{5} \int \frac{dx}{x-2} - \frac{1}{5} \int \frac{x}{x^2+1} \, dx - \frac{2}{5} \int \frac{dx}{x^2+1}$

   • First term: $\int \frac{dx}{x-2} = \log|x-2| + c_1$
   • Second term: Let $u = x^2+1$, $du = 2x\,dx$, so $\int \frac{x}{x^2+1} dx = \frac{1}{2} \log(x^2+1) + c_2$
   • Third term: $\int \frac{dx}{x^2+1} = \tan^{-1} x + c_3$

   8. Combine results

   $= \frac{1}{5} \log|x-2| - \frac{1}{5} \cdot \frac{1}{2} \log(x^2+1) - \frac{2}{5} \tan^{-1} x + C$

   $= \frac{1}{5} \left[ \log|x-2| - \frac{1}{2} \log(x^2+1) - 2 \tan^{-1} x \right] + C$

   9. Simplify the logarithmic part

   Recall $-\frac{1}{2}\log(x^2+1) = \log\left( (x^2+1)^{-1/2} \right) = \log\frac{1}{\sqrt{x^2+1}}$.

   So:

   $\log|x-2| + \log\frac{1}{\sqrt{x^2+1}} = \log\frac{|x-2|}{\sqrt{x^2+1}}$

   Thus:

   $\int \frac{1}{(x-2)(x^2+1)} \, dx = \frac{1}{5} \left[ \log\frac{|x-2|}{\sqrt{x^2+1}} - 2 \tan^{-1} x \right] + C$

   Watch out

   Option (D) has a minus sign before $2\tan^{-1}x$, but our result has a minus sign as well — wait, check carefully: Our result is $\frac{1}{5}[\log\frac{|x-2|}{\sqrt{x^2+1}} - 2\tan^{-1}x]$, which is exactly option (D)? No, look again: Option (D) says $-2\tan^{-1}x$, but option (C) says $+2\tan^{-1}x$. Did we get a sign error? Let's verify the sign of the arctangent term.

   We had $-\frac{2}{5}\int \frac{dx}{x^2+1} = -\frac{2}{5}\tan^{-1}x$. So indeed the coefficient is negative. That matches option (D), not (C). But wait — let's re-check the original options:

   (A) $\log \frac{\sqrt{x^2+1}}{|x-2|} + 2 \tan^{-1} x + c$

   (B) $\log \frac{|x-2|}{x^2+1} + 2 \tan^{-1} x + c$

   (C) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{7+x^2}} + 2 \tan^{-1} x \right] + c$

   (D) $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{1+x^2}} - 2 \tan^{-1} x \right] + c$

   Our result: $\frac{1}{5} \left[ \log \frac{|x-2|}{\sqrt{x^2+1}} - 2 \tan^{-1} x \right] + c$ matches (D) exactly. But option (C) has $\sqrt{7+x^2}$ (a typo? likely meant $\sqrt{1+x^2}$) and a plus sign. So the correct match is (D).

   Tip

   Always double-check the sign of the arctangent term: the partial fraction gave $-\frac{2}{5}\tan^{-1}x$, so the minus sign is correct.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

3. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.If $\dfrac{3x^2 + ax + 3}{(2x+3)(x^2+2)} = \dfrac{3}{2x+3} + \dfrac{Bx+C}{x^2+2}$ then $a\ (B+C) =$ (A) $-2$ (B) $3$ (C) $-3$ (D) $2$

   ›Reveal solutionSolution

   Clear the denominators and match coefficients: $B=0$, $C=-1$, $a=-2$, so $a(B+C)=(-2)(-1)=2$.

   Multiply both sides by $(2x+3)(x^2+2)$:

   $3x^2 + ax + 3 = 3(x^2+2) + (Bx+C)(2x+3).$

   Expand the right side:

   $3(x^2+2) + (Bx+C)(2x+3) = 3x^2 + 6 + 2Bx^2 + 3Bx + 2Cx + 3C = (3+2B)x^2 + (3B+2C)x + (6+3C).$

   Compare coefficients with $3x^2 + ax + 3$:

   • $x^2$: $\;3 = 3 + 2B \Rightarrow B = 0.$
   • constant: $\;3 = 6 + 3C \Rightarrow C = -1.$
   • $x$: $\;a = 3B + 2C = 3(0) + 2(-1) = -2.$

   Therefore $B + C = 0 + (-1) = -1$ and

   $a(B+C) = (-2)(-1) = 2.$

   ✓Final answer

   $a(B+C) = 2$ — option (D).

4. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If $\dfrac{x^2+3}{(x^2+1)(x^2+2)} = \dfrac{A x + B}{x^2+1} + \dfrac{C x + D}{x^2+2}$ then $A + B + C + D =$ (A) 3 (B) 2 (C) 0 (D) 1

   ›Reveal solutionSolution

   Clearing denominators and matching coefficients gives $A=0,\ B=2,\ C=0,\ D=-1$, so $A+B+C+D=1$ — option (D).

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

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

   Expand and collect powers of $x$:

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

   Match coefficients:

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

   Solve. Subtracting $A+C=0$ from $2A+C=0$ gives $A=0$, hence $C=0$. Subtracting $B+D=1$ from $2B+D=3$ gives $B=2$, hence $D=-1$.

   Sum: $A+B+C+D=0+2+0-1=1.$

   ✓Final answer

   $A+B+C+D=1$ — option (D).

5. 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:

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

   Our result has no $\frac{2}{x-1}$ term, but note that we found $A=0$, so that term is zero. However, option (C) includes an extra $\frac{2}{x-1}$ which is not present in our decomposition. Wait—check carefully: Our result is $-\frac{1}{x+1} + \frac{5}{4(x-2)} + \frac{7}{4(x+2)}$. Option (C) has an additional $\frac{2}{x-1}$. That seems inconsistent. Let’s re-evaluate: Did we miss a term?

   Actually, the numerator $2x^3 + x - 3$ at $x=1$ gave 0, so $A=0$ is correct. But option (C) includes $\frac{2}{x-1}$. That suggests option (C) is not exactly our result. Let’s check the other options:

   • (A) and (B) have denominators like $x^2-3x+2$ which factor as $(x-1)(x-2)$, so they are different forms.
   • (D) is missing the $-\frac{1}{x+1}$ term? Actually (D) is $\frac{5}{4(x-2)} - \frac{1}{x+1} + \frac{7}{4(x+2)}$, which matches exactly our result!

   So the correct option is (D), not (C). Let’s verify: (D) has no $\frac{2}{x-1}$ term, and the signs match.

   Watch out

   A common mistake is to assume all factors yield nonzero coefficients. Here, the numerator vanishes at $x=1$, so the coefficient for $1/(x-1)$ is zero. Always check each root.

   Tip

   When a factor’s coefficient turns out zero, the decomposition simplifies. Option (D) is the clean result.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.If $\dfrac{1}{(x-1)(x-2)(x-3)} = \dfrac{A}{x-1} + \dfrac{B}{x-2} + \dfrac{C}{x-3}$, and $\dfrac{x}{(x-1)(x-2)(x-3)} = \dfrac{P}{x-1} + \dfrac{Q}{x-2} + \dfrac{R}{x-3}$ then $A + 2B + 3C =$ (A) $P + Q + R$ (B) $P + 2Q + 3R$ (C) $3P + 2Q + R$ (D) $AP + BQ + CR$

   ›Reveal solutionSolution

   The key idea is to find the partial fraction constants by comparing numerators after clearing denominators, then compute $A+2B+3C$ and match it to a combination of $P, Q, R$. The result is $A+2B+3C = P+2Q+3R$, so option (B) is correct.

   The problem gives two partial fraction expansions for very similar rational functions. The first has numerator $1$, the second has numerator $x$. The constants $A, B, C$ and $P, Q, R$ are determined uniquely by the denominators. The question asks for a linear combination of $A, B, C$ and wants it expressed in terms of $P, Q, R$. The natural approach is to compute each constant explicitly, then evaluate the required sum.

   1. Find $A, B, C$. Start with

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

   Multiply both sides by $(x-1)(x-2)(x-3)$:

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

   This identity holds for all $x$. To isolate each constant, substitute the roots of the denominators.

   • For $x=1$: the terms with $B$ and $C$ vanish because $(x-1)$ factors become zero.

   $1 = A(1-2)(1-3) = A(-1)(-2) = 2A \quad\Rightarrow\quad A = \frac12.$

   • For $x=2$:

   $1 = B(2-1)(2-3) = B(1)(-1) = -B \quad\Rightarrow\quad B = -1.$

   • For $x=3$:

   $1 = C(3-1)(3-2) = C(2)(1) = 2C \quad\Rightarrow\quad C = \frac12.$

   So $A = \frac12$, $B = -1$, $C = \frac12$.

   2. Find $P, Q, R$. Now

   $\frac{x}{(x-1)(x-2)(x-3)} = \frac{P}{x-1} + \frac{Q}{x-2} + \frac{R}{x-3}.$

   Multiply through by the denominator:

   $x = P(x-2)(x-3) + Q(x-1)(x-3) + R(x-1)(x-2).$

   Again substitute the roots.

   • For $x=1$:

   $1 = P(1-2)(1-3) = P(-1)(-2) = 2P \quad\Rightarrow\quad P = \frac12.$

   • For $x=2$:

   $2 = Q(2-1)(2-3) = Q(1)(-1) = -Q \quad\Rightarrow\quad Q = -2.$

   • For $x=3$:

   $3 = R(3-1)(3-2) = R(2)(1) = 2R \quad\Rightarrow\quad R = \frac32.$

   So $P = \frac12$, $Q = -2$, $R = \frac32$.

   3. Compute $A + 2B + 3C$.

   $A + 2B + 3C = \frac12 + 2(-1) + 3\left(\frac12\right) = \frac12 - 2 + \frac32 = \frac{1+3}{2} - 2 = 2 - 2 = 0.$

   4. Check each option.

      • (A) $P+Q+R = \frac12 - 2 + \frac32 = 0$. This matches.
      • (B) $P+2Q+3R = \frac12 + 2(-2) + 3\left(\frac32\right) = \frac12 - 4 + \frac92 = \frac{1+9}{2} - 4 = 5 - 4 = 1$. Not 0.
      • (C) $3P+2Q+R = 3\left(\frac12\right) + 2(-2) + \frac32 = \frac32 - 4 + \frac32 = 3 - 4 = -1$. Not 0.
      • (D) $AP+BQ+CR = \frac12\cdot\frac12 + (-1)(-2) + \frac12\cdot\frac32 = \frac14 + 2 + \frac34 = 3$. Not 0.

      Only option (A) gives 0.

   Watch out

   A common mistake is to forget that $A+2B+3C$ is a number (here 0), and then to try to match it to an expression in $P, Q, R$ without computing the constants. Always compute the constants explicitly — the substitution method is fast and reliable.

   Thus $A+2B+3C = P+Q+R$.

   ✓Final answer

   The correct option is (A).

7. 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 $C = 0$.

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

   Plug into $3B + 2D = 1$: $3B + 2(1 - B) = 3B + 2 - 2B = B + 2 = 1$ → $B = -1$.

   Then $D = 1 - (-1) = 2$.

   So: $A = 0,\; B = -1,\; C = 0,\; D = 2$.

   ---

   5. Compute the sum.

   $A + B + C + D = 0 + (-1) + 0 + 2 = 1$

   Watch out

   A common mistake is to forget that the numerators are linear ($Ax+B$), not just constants. Here it turned out $A=C=0$, so the numerators are actually constants, but that’s not always the case.

   Tip

   Notice that the $x^3$ and $x$ coefficients gave $A=C=0$ immediately — this shortcut saves time.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

8. 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 $D = 1-B$.

      Substitute into $3B+2D=1$: $3B + 2(1-B) = 3B + 2 - 2B = B+2 = 1$ → $B = -1$, then $D = 1-(-1)=2$.

   6. Sum the constants

   $A+B+C+D = 0 + (-1) + 0 + 2 = 1$

   Watch out

   A common mistake is forgetting that the numerators for irreducible quadratics must be linear ($Ax+B$), not just constants. Using constants would give a wrong system.

   Tip

   Notice that the $x^3$ and $x$ coefficients forced $A=C=0$ immediately, so the sum reduces to $B+D$, which from the $x^2$ coefficient is already 1 — no need to solve fully!

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.The partial fraction decomposition of $\dfrac{9x-7}{(x+3)(x^2+1)}$ is (A) $\dfrac{17}{5(x+3)} - \dfrac{(17x-6)}{5(x^2+1)}$ (B) $\dfrac{-17}{5(x+3)} - \dfrac{(17x-6)}{5(x^2+1)}$ (C) $\dfrac{17}{5(x+3)} + \dfrac{(17x-6)}{5(x^2+1)}$ (D) $\dfrac{-17}{5(x+3)} + \dfrac{(17x-6)}{5(x^2+1)}$

   ›Reveal solutionSolution

   The key idea is to decompose a rational function with an irreducible quadratic factor into a sum of a linear-over-quadratic term and a constant-over-linear term, then solve for the unknown coefficients. The correct decomposition is option (D).

   When you see a denominator with a linear factor $(x+3)$ and an irreducible quadratic factor $(x^2+1)$, the standard partial fraction form is:

   $\frac{9x-7}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+1}$

   The numerator over the quadratic must be linear (degree 1) because the denominator is degree 2 — a common point where students mistakenly put just a constant. The goal is to find $A$, $B$, and $C$ by clearing denominators and equating coefficients.

   1. Set up the equation Multiply both sides by $(x+3)(x^2+1)$:

   $9x - 7 = A(x^2+1) + (Bx+C)(x+3)$

   2. Expand the right-hand side

   $A(x^2+1) = Ax^2 + A$

   $(Bx+C)(x+3) = Bx^2 + 3Bx + Cx + 3C = Bx^2 + (3B+C)x + 3C$

   Adding them:

   $9x - 7 = (A+B)x^2 + (3B+C)x + (A + 3C)$

   3. Equate coefficients Comparing coefficients of $x^2$, $x$, and the constant term gives three equations:

   $\begin{cases} A + B = 0 & \text{(coefficient of }x^2\text{)} \\ 3B + C = 9 & \text{(coefficient of }x\text{)} \\ A + 3C = -7 & \text{(constant term)} \end{cases}$

   4. Solve the system From the first equation: $B = -A$. Substitute into the second: $3(-A) + C = 9 \implies -3A + C = 9$. The third equation is $A + 3C = -7$. Solve these two:

   $\begin{aligned} -3A + C &= 9 \\ A + 3C &= -7 \end{aligned}$

   Multiply the first equation by 3: $-9A + 3C = 27$. Subtract the second equation from this:

   $(-9A + 3C) - (A + 3C) = 27 - (-7) \implies -10A = 34 \implies A = -\frac{34}{10} = -\frac{17}{5}$

   Then $B = -A = \frac{17}{5}$.

   Substitute $A$ into $A + 3C = -7$:

   $-\frac{17}{5} + 3C = -7 \implies 3C = -7 + \frac{17}{5} = -\frac{35}{5} + \frac{17}{5} = -\frac{18}{5} \implies C = -\frac{6}{5}$

   5. Write the decomposition With $A = -\frac{17}{5}$, $B = \frac{17}{5}$, $C = -\frac{6}{5}$, we have:

   $\frac{9x-7}{(x+3)(x^2+1)} = \frac{-\frac{17}{5}}{x+3} + \frac{\frac{17}{5}x - \frac{6}{5}}{x^2+1}$

   Factor out $\frac{1}{5}$:

   $= -\frac{17}{5(x+3)} + \frac{17x - 6}{5(x^2+1)}$

   Watch out

   A common mistake is to forget the minus sign on the first term or to put a minus sign in the numerator of the second term. Check by combining the right-hand side back — it must give the original numerator $9x-7$.

   ✓Final answer

   The correct option is (D): $\displaystyle -\frac{17}{5(x+3)} + \frac{17x-6}{5(x^2+1)}$.

10. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.If $\dfrac{3x+2}{(x+1)(2x^2+3)} = \dfrac{A}{x+1} + \dfrac{Bx+C}{2x^2+3}$, then $A-B+C=$ (A) 2 (B) 1 (C) 3 (D) 6

    ›Reveal solutionSolution

    We find the constants in the partial fraction decomposition by clearing denominators and equating coefficients, then compute $A - B + C = 1$.

    The problem gives a rational function and its partial fraction decomposition. The key idea: multiply both sides by the common denominator to get a polynomial identity, then match coefficients to solve for $A$, $B$, and $C$. Once we have them, the expression $A - B + C$ is straightforward.

    ---

    1. Set up the equation We have

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

    Multiply both sides by $(x+1)(2x^2+3)$ to clear denominators:

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

    2. Expand the right-hand side First term: $A(2x^2+3) = 2A x^2 + 3A$. Second term: $(Bx+C)(x+1) = Bx^2 + Bx + Cx + C = Bx^2 + (B+C)x + C$. Adding them:

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

    3. Equate coefficients Since the left side has no $x^2$ term, its coefficient is 0. So:

    $\begin{cases} \text{Coefficient of } x^2: & 2A + B = 0 \quad (1) \\ \text{Coefficient of } x: & B + C = 3 \quad (2) \\ \text{Constant term:} & 3A + C = 2 \quad (3) \end{cases}$

    4. Solve the system

       From (1): $B = -2A$.

       Substitute into (2): $-2A + C = 3 \Rightarrow C = 3 + 2A$.

       Substitute into (3): $3A + (3 + 2A) = 2 \Rightarrow 5A + 3 = 2 \Rightarrow 5A = -1 \Rightarrow A = -\frac{1}{5}$.

       Then $B = -2\left(-\frac{1}{5}\right) = \frac{2}{5}$, and $C = 3 + 2\left(-\frac{1}{5}\right) = 3 - \frac{2}{5} = \frac{13}{5}$.

    5. Compute $A - B + C$

    $A - B + C = -\frac{1}{5} - \frac{2}{5} + \frac{13}{5} = \frac{-1 - 2 + 13}{5} = \frac{10}{5} = 2.$

    Watch out

    A common mistake is forgetting that the numerator of the second fraction is $Bx + C$, not just a constant. Also, when equating coefficients, note that the left side has no $x^2$ term — that gives $2A + B = 0$, not something else.

    Tip

    You can also solve by plugging in convenient $x$ values (like $x = -1$ to get $A$ directly), but the coefficient method is systematic and avoids fractions until the end.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

11. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.If $\dfrac{6x^4 + 13x^3 + 2x^2 - x + 3}{2x^2 + 3x - 2} = f(x) + \dfrac{A}{ax-1} + \dfrac{B}{x+b}$, then $f(1) + a \cdot B + b \cdot A =$ (A) 8 (B) 12 (C) 4 (D) 6

    ›Reveal solutionSolution

    Long division gives $f(x)=3x^2+2x+1$ with $a=2,b=2,A=2,B=-1$, so $f(1)+aB+bA = 6-2+4 = 8$ (option A).

    Factor the denominator: $2x^2+3x-2 = (2x-1)(x+2)$, so $ax-1 = 2x-1 \Rightarrow a=2$ and $x+b = x+2 \Rightarrow b=2$.

    Polynomial division of $6x^4+13x^3+2x^2-x+3$ by $2x^2+3x-2$:

    $f(x) = 3x^2 + 2x + 1, \qquad \text{remainder } = 5.$

    Partial fractions of the remainder:

    $\frac{5}{(2x-1)(x+2)} = \frac{A}{2x-1} + \frac{B}{x+2} \ \Rightarrow\ 5 = A(x+2) + B(2x-1).$

    • $x = \tfrac12:\ 5 = \tfrac52 A \Rightarrow A = 2.$
    • $x = -2:\ 5 = -5B \Rightarrow B = -1.$

    Evaluate: $f(1) = 3+2+1 = 6$, so

    $f(1) + a\cdot B + b\cdot A = 6 + 2(-1) + 2(2) = 6 - 2 + 4 = 8.$

    ✓Final answer

    $f(1) + a\cdot B + b\cdot A = 8$ — option (A).

12. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.If $\dfrac{x-2}{x^2(2x-3)} = \dfrac{A}{x} + \dfrac{B}{x^2} + \dfrac{C}{2x-3}$ then $2(A-C)=$ (A) 3B (B) 2B (C) 0 (D) B

    ›Reveal solutionSolution

    To find the coefficients A, B, and C in the partial fraction decomposition, we equate the numerators after combining the terms on the right-hand side. By substituting specific values of $x$ or comparing coefficients, we find $A = 1/9$, $B = 2/3$, and $C = -2/9$. The expression $2(A-C)$ then evaluates to $2/3$, which is equal to $B$.

    Partial fraction decomposition is a technique used to break down a complex rational function into a sum of simpler fractions. This is particularly useful in calculus for integration, but also in algebra for manipulating expressions. The core idea is that any proper rational function (where the degree of the numerator is less than the degree of the denominator) can be expressed as a sum of fractions whose denominators are the factors of the original denominator.

    When the denominator has repeated linear factors, like $x^2$ in this problem, the decomposition must include a term for each power of the factor up to its multiplicity. For a factor $(ax+b)^n$, we include terms $\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \dots + \frac{A_n}{(ax+b)^n}$. For distinct linear factors, like $(2x-3)$, we simply have a term $\frac{C}{2x-3}$.

    The strategy is to combine the partial fractions on the right-hand side, equate the resulting numerator to the original numerator, and then solve for the unknown coefficients (A, B, C) by either substituting convenient values of $x$ or by comparing coefficients of like powers of $x$.

    1. Set up the equation and clear denominators: We are given the partial fraction decomposition:

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

    To find the coefficients $A$, $B$, and $C$, we first combine the terms on the right-hand side by finding a common denominator, which is $x^2(2x-3)$.
    

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

    Since the denominators are now identical, the numerators must be equal:
    

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

    Expand the right-hand side:
    

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

    Rearrange the terms by powers of $x$:
    

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

    2. Determine the coefficients using strategic substitution and comparison:

       We can find the coefficients by substituting values of $x$ that make certain terms zero, or by comparing the coefficients of $x^2$, $x$, and the constant term on both sides of the equation.

       • Find B: Substitute $x=0$ into the equation $x-2 = A(x)(2x-3) + B(2x-3) + C(x^2)$. This eliminates the terms with $A$ and $C$:

    $(0)-2 = A(0)(2(0)-3) + B(2(0)-3) + C(0)^2$

    $-2 = 0 + B(-3) + 0$

    $-2 = -3B \implies B = \frac{2}{3}$

    *   **Find C:** Substitute $x=\frac{3}{2}$ (which makes $2x-3=0$) into the equation $x-2 = A(x)(2x-3) + B(2x-3) + C(x^2)$. This eliminates the terms with $A$ and $B$:
    

    $\frac{3}{2}-2 = A\left(\frac{3}{2}\right)(0) + B(0) + C\left(\frac{3}{2}\right)^2$

    $\frac{3-4}{2} = 0 + 0 + C\left(\frac{9}{4}\right)$

    $-\frac{1}{2} = \frac{9}{4}C \implies C = -\frac{1}{2} \times \frac{4}{9} = -\frac{2}{9}$

    *   **Find A:** Now that we have $B$ and $C$, we can find $A$ by comparing the coefficients of $x^2$ from the expanded equation:
    

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

        On the left side, the coefficient of $x^2$ is $0$. On the right side, it is $(2A+C)$.
    

    $0 = 2A+C$

        Substitute the value of $C = -\frac{2}{9}$:
    

    $0 = 2A - \frac{2}{9}$

    $2A = \frac{2}{9} \implies A = \frac{1}{9}$

    > [!TIP]
    > For repeated linear factors like $x^n$, the coefficient of the highest power term (e.g., $B$ for $x^2$) can often be found by substituting $x=0$. For distinct linear factors like $(ax+b)$, the coefficient (e.g., $C$ for $2x-3$) can be found by substituting $x = -b/a$. For the remaining coefficients (like $A$), you can either compare coefficients of other powers of $x$ or substitute another convenient value for $x$ (e.g., $x=1$).
    

    3. Calculate $2(A-C)$:

    We have the values:

    $A = \frac{1}{9}$

    $B = \frac{2}{3}$

    $C = -\frac{2}{9}$

    First, calculate $A-C$:
    

    $A-C = \frac{1}{9} - \left(-\frac{2}{9}\right) = \frac{1}{9} + \frac{2}{9} = \frac{3}{9} = \frac{1}{3}$

    Now, calculate $2(A-C)$:
    

    $2(A-C) = 2 \times \frac{1}{3} = \frac{2}{3}$

    4. Compare with options:

       The calculated value is $\frac{2}{3}$. Let's check the given options:

       (A) $3B = 3 \times \frac{2}{3} = 2$

       (B) $2B = 2 \times \frac{2}{3} = \frac{4}{3}$

       (C) $0$

       (D) $B = \frac{2}{3}$

       Our result $2(A-C) = \frac{2}{3}$ matches option (D).

    ✓Final answer

    The value of $2(A-C)$ is $\boxed{\text{B}}$.