JEE Advanced 2023 Paper 1 · Q13 · Binomial Theorem

Question

Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^5$ in the expansion of $\left(ax^2 + \dfrac{70}{27 bx}\right)^4$ 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

SolveKar AI · Accepted Answer

From $(ax^2 + 70/(27bx))^4$: general term $T_{r+1} = \binom{4}{r}(ax^2)^{4-r}(70/(27bx))^r$ has $x$-power $2(4-r) - r = 8 - 3r$. For $x^5$: $r = 1$, coefficient $= 4 \cdot a^3 \cdot \dfrac{70}{27b} = \dfrac{280 a^3}{27b}$. From $(ax - 1/(bx^2))^7$: general term $\binom{7}{r}(ax)^{7-r}(-1/(bx^2))^r$ has $x$-power $7 - 3r$. For $x^{-5}$: $r = 4$, coefficient $= \binom{7}{4} a^3 (-1)^4/b^4 = 35 a^3/b^4$. Equating: $\dfrac{280}{27b} = \dfrac{35}{b^4} \Rightarrow b^3 = \dfrac{35 \cdot 27}{280} = \dfrac{27}{8} \Rightarrow b = 3/2$. So $2b = 3$.

JEE Advanced 2023 Paper 1 · Q13 · Binomial Theorem

Solution