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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Binomial Theorem

If the term independent of x in the expansion of ( a x^2+(1)/(2...

If the term independent of xx in the expansion of (ax2+12x3)10\left(\sqrt{\mathrm{a}} x^2+\frac{1}{2 x^3}\right)^{10} is 105 , then a2\mathrm{a}^2 is equal to :

Options

  1. A2
  2. B4
  3. C6
  4. D9

Correct answer

B. 4

Step-by-step solution

(ax2+12x3)10\left(\sqrt{\mathrm{a}} \mathrm{x}^2+\frac{1}{2 \mathrm{x}^3}\right)^{10} General term =10Cr(ax2)10−r(12x3)r={ }^{10} \mathrm{C}_{\mathrm{r}}\left(\sqrt{\mathrm{a}} \mathrm{x}^2\right)^{10-\mathrm{r}}\left(\frac{1}{2 \mathrm{x}^3}\right)^{\mathrm{r}}20−2r−3r=0r=410C4a3⋅116=105a3=8a2=4\begin{aligned} & 20-2 r-3 r=0 \\ & \mathrm{r}=4 \\ & { }^{10} \mathrm{C}_4 \mathrm{a}^3 \cdot \frac{1}{16}=105 \\ & \mathrm{a}^3=8 \\ & \mathrm{a}^2=4\end{aligned}$$
SECTION - B
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