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QJEEAdv23Mac1aa: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Mac1aa Solution: Let and be two nonzero real numbers. If the coefficient of in…

Let aa and bb be two nonzero real numbers. If the coefficient of x5x^5 in the expansion of (ax2+7027bx)4\left(a x^2+\frac{70}{27 b x}\right)^4 is equal to the coefficient of x−5x^{-5} in the expansion of (ax−1bx2)7\left(a x-\frac{1}{b x^2}\right)^7 , then the value of 2b2 b is :

Step-by-Step Verified Solution

amp;Tr+1=4Cr(a⋅x2)4−r⋅(7027bx)ramp;=4Cr⋅a4−r⋅70r(27b)r⋅x8−3r\begin{aligned}& \mathrm{T}_{\mathrm{r}+1}={ }^4 \mathrm{C}_{\mathrm{r}}\left(\mathrm{a} \cdot \mathrm{x}^2\right)^{4-\mathrm{r}} \cdot\left(\frac{70}{27 \mathrm{bx}}\right)^{\mathrm{r}} \\\\& ={ }^4 \mathrm{C}_{\mathrm{r}} \cdot \mathrm{a}^{4-\mathrm{r}} \cdot \frac{70^{\mathrm{r}}}{(27 b)^{\mathrm{r}}} \cdot x^{8-3 \mathrm{r}}\end{aligned}

amp; here 8−3r=5amp;8−5=3r⇒r=1amp;∴ coeff. =4⋅a3⋅7027b\begin{aligned}& \text { here } 8-3 r=5 \\\\& 8-5=3 r \Rightarrow r=1 \\\\& \therefore \text { coeff. }=4 \cdot a^3 \cdot \frac{70}{27 b}\end{aligned}

amp;Tr+1=7Cr(ax)7−r(−1bx2)ramp;=7Cr⋅a7−r(−1 b)r⋅x7−3ramp;7−3r=−5⇒12=3r⇒r=4\begin{aligned}& \mathrm{T}_{\mathrm{r}+1}={ }^7 \mathrm{C}_{\mathrm{r}}(\mathrm{ax})^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{bx}^2}\right)^{\mathrm{r}} \\\\& ={ }^7 \mathrm{C}_{\mathrm{r}} \cdot a^{7-\mathrm{r}}\left(\frac{-1}{\mathrm{~b}}\right)^{\mathrm{r}} \cdot \mathrm{x}^{7-3 \mathrm{r}} \\\\& 7-3 \mathrm{r}=-5 \Rightarrow 12=3 \mathrm{r} \Rightarrow \mathrm{r}=4\end{aligned}

amp; now 35a3 b4=280a327 bamp; b3=35×27280=b=32⇒2 b=3\begin{aligned}& \text { now } \frac{35 \mathrm{a}^3}{\mathrm{~b}^4}=\frac{280 \mathrm{a}^3}{27 \mathrm{~b}} \\\\& \mathrm{~b}^3=\frac{35 \times 27}{280}=\mathrm{b}=\frac{3}{2} \Rightarrow 2 \mathrm{~b}=3\end{aligned}

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