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Step-by-Step Solution
Step 1: Identify the General Term in the First Expression
Consider the expression
$$(a x^2 + \tfrac{1}{b x})^{11}.$$
Its general term (the $(r+1)$-th term) is given by the binomial expansion:
$$
T_{r+1} = \binom{11}{r} (a x^2)^{11-r}\left(\tfrac{1}{b x}\right)^r.
$$
Simplify the powers of $a$, $b$, and $x$:
$$
T_{r+1} = \binom{11}{r} a^{\,11-r} \, b^{-r} \, x^{2(11-r)} \, x^{-r}
= \binom{11}{r} a^{\,11-r} \, b^{-r} \, x^{22 - 3r}.
$$
Step 2: Find the Condition for the Power of $x$ to be $x^7$
We want the coefficient of $x^7$. Therefore, we set
$$22 - 3r = 7.$$
Solving for $r$:
$$
22 - 3r = 7 \quad \Longrightarrow \quad 3r = 15 \quad \Longrightarrow \quad r = 5.
$$
So, the coefficient of $x^7$ in $(a x^2 + \tfrac{1}{b x})^{11}$ is
$$
\binom{11}{5} a^{\,11-5} \, b^{-5}
= \binom{11}{5} a^6 \, b^{-5}.
$$
Step 3: Identify the General Term in the Second Expression
Now consider the expression
$$(a x - \tfrac{1}{b x^2})^{11}.$$
Its general term (the $(r+1)$-th term) is given by:
$$
T_{r+1}
= \binom{11}{r} (a x)^{11-r} \Bigl(-\tfrac{1}{b x^2}\Bigr)^{r}.
$$
Note that
$(-1)^r$ will appear from the negative sign. Simplifying:
$$
T_{r+1}
= \binom{11}{r} a^{\,11-r} x^{11-r} \, (-1)^r \, b^{-r} \, x^{-2r}
= \binom{11}{r} a^{\,11-r} (-1)^r b^{-r} \, x^{\,11 - 3r}.
$$
Step 4: Find the Condition for the Power of $x$ to be $x^{-7}$
We want the coefficient of $x^{-7}$. Set
$$11 - 3r = -7.$$
Solving for $r$:
$$
11 - 3r = -7 \quad \Longrightarrow \quad 3r = 18 \quad \Longrightarrow \quad r = 6.
$$
So, the coefficient of $x^{-7}$ in $(a x - \tfrac{1}{b x^2})^{11}$ is
$$
\binom{11}{6} a^{\,11-6} (-1)^6 b^{-6}
= \binom{11}{6} a^5 b^{-6},
$$
since $(-1)^6 = 1.$
Step 5: Equate the Two Coefficients and Solve for the Relation Between $a$ and $b$
According to the problem statement, these two coefficients are equal:
$$
\binom{11}{5} a^6 b^{-5} = \binom{11}{6} a^5 b^{-6}.
$$
Recall that $\binom{11}{5} = \binom{11}{6}$; specifically, $\binom{11}{5} = \binom{11}{11-5} = \binom{11}{6}.$ Hence they cancel out on both sides, leaving:
$$
a^6 b^{-5} = a^5 b^{-6}.
$$
Rearrange to get:
$$
\frac{a^6}{b^5} = \frac{a^5}{b^6}
\quad \Longrightarrow \quad
a \, b = 1.
$$
Thus, the required relation is
$$ab = 1.$$
Final Answer
The values of $a$ and $b$ satisfy the relation
$$\boxed{ab = 1}.$$
