Verified Solution • 2025–26 Session moderate

QJEEM24Maafc1: JEE Main 2024 Mathematics - Class 11 JEEM24Maafc1 Solution: If the term independent of in the expansion of is 105 , then is …

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 :

Multiple Choice Options

  • A 6
  • B 4
  • C 2
  • D 9

Step-by-Step Verified Solution

amp;(ax2+12x3)10amp;Tr+1=10Cr(ax2)10−r(12x3)r\begin{aligned}& \left(\sqrt{a} x^2+\frac{1}{2 x^3}\right)^{10} \\& T_{r+1}={ }^{10} C_r\left(\sqrt{a} x^2\right)^{10-r}\left(\frac{1}{2 x^3}\right)^r\end{aligned}

Independent of x⇒20−2r−3r=0x \Rightarrow 20-2 r-3 r=0

r=4r=4

Independent of xx is 10C4(a)6(12)4=105{ }^{10} C_4(\sqrt{a})^6\left(\frac{1}{2}\right)^4=105

2102×8a3=105⇒a=2a2=4\begin{gathered}\frac{210}{2 \times 8} a^3=105 \\\Rightarrow \quad a=2 \\a^2=4\end{gathered}

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