JEE Advanced 2023 — Paper 1 — MATH-III solution

JEE Advanced2023Paper 1MATH-III

Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^{5}$ in the expansion of $(a x^{2} + \frac{70}{27 b x})^{4}$ is equal to the coefficient of $x^{- 5}$ in the expansion of $(a x - \frac{1}{b x ^{2}})^{7}$ , then the value of $2 b$ is

Your answer

Solution

For the first expansion, the general term is $(r 4) (a x^{2})^{4 - r} (\frac{70}{27 b x})^{r} = (r 4) a^{4 - r} (\frac{70}{27 b})^{r} x^{8 - 3 r}$ .

Setting $8 - 3 r = 5$ gives $r = 1$ . Coefficient of $x^{5}$ :

$(1 4) a^{3} \cdot \frac{70}{27 b} = 4 a^{3} \cdot \frac{70}{27 b} = \frac{280 a ^{3}}{27 b}$ .

For the second expansion, the general term is $(r 7) (a x)^{7 - r} (- \frac{1}{b x ^{2}})^{r} = (r 7) a^{7 - r} (\frac{- 1}{b})^{r} x^{7 - 3 r}$ .

Setting $7 - 3 r = - 5$ gives $r = 4$ . Coefficient of $x^{- 5}$ :

$(4 7) a^{3} \cdot \frac{1}{b ^{4}} = \frac{35 a ^{3}}{b ^{4}}$ .

Equate the two coefficients:

$\frac{280 a ^{3}}{27 b} = \frac{35 a ^{3}}{b ^{4}} \Rightarrow b^{3} = \frac{27 \cdot 35}{280} = \frac{27}{8} \Rightarrow b = \frac{3}{2}$ .

Therefore $2 b = 3$ .

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