JEE Advanced 2023 · Paper 1 · Mathematics Q13

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MathematicsInteger+4Question 13 of 51

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

Your answer

Answer and worked solution

Official answer: 3

First expansion. In $\left(ax^2+\frac{70}{27bx}\right)^4$ the general term has power $x^{8-3r}$ . For $x^5$ , $r=1$ :
$\binom41a^3\cdot\frac{70}{27b}=\frac{280a^3}{27b}.$

Second expansion. In $\left(ax-\frac1{bx^2}\right)^7$ the power is $x^{7-3r}$ . For $x^{-5}$ , $r=4$ :
$\binom74a^3\cdot\frac{1}{b^4}=\frac{35a^3}{b^4}.$

Equate.
$\frac{280}{27b}=\frac{35}{b^4}\implies b^3=\frac{27}{8}\implies b=\frac32,\quad 2b=3.$

Topics:General and middle terms

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JEE Advanced 2023 · Paper 1, Mathematics question 13