Find the coefficient of $x^{-3}$ in the expansion of $(ax^2 - \frac{1}{bx})^{12}$.

Subject: Mathematics | Topic: Binomial Theorem | Difficulty: MEDIUM

Options

1. (A) $220 \frac{a^3}{b^9}$
2. (B) $-165 \frac{a^4}{b^8}$
3. (C) $-220 \frac{a^3}{b^9}$
4. (D) $165 \frac{a^4}{b^8}$

Verified Answer

Option (C): $-220 \frac{a^3}{b^9}$

10-Second Shortcut Trick

⚡ 10-Second Exam Shortcut: Eliminate options ending in incorrect units or calculate unit digits first before solving full arithmetic.

Step-by-Step Derivation

1. Write the general term $T_{r+1} = \binom{11}{r} (ax^2)^{11-r} (-b^{-1}x^{-1})^r = \binom{11}{r} a^{11-r} (-b)^{-r} x^{22 - 3r}.$
2. Set $22 - 3r = -7 \implies 3r = 29$ (Let's adjust index so it works cleanly: say $22 - 3r = -2 \implies 3r = 24 \implies r = 8$). Let's use $x^{-2}$ or adjust power to $12$ so $24 - 3r = -3 \implies 3r = 27 \implies r = 9$.
3. Substitute $r = 9$ to get $\binom{12}{9} a^3 (-b)^{-9} = -220 \frac{a^3}{b^9}$.

Common Trap to Avoid

⚠️ Common Trap: Misreading inverse proportions or rushing calculation in the last algebraic step.

Exam Tip

💡 Exam Tip: Always verify if question asks for perimeter, area, or remaining fraction.

Related Practice Questions in Binomial Theorem

• What is the sum of the series $\sum_{r=0}^{n} r \binom{n}{r}$?
• Find the coefficient of $x^5$ in the expansion of $(1 + x^2)^5 (1 + x)^4$.
• If $C_r$ denotes $\binom{n}{r}$, what is the value of $C_0 + 2C_1 + 3C_2 + \dots + (n+1)C_n$?
• Find the coefficient of $x^{18}$ in the expansion of $(x^2 - \frac{1}{x})^{15}$.

Practice more Binomial Theorem questions in SSC CGL Complete Test Series 2026.

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