QJEEM23Maa43a: JEE Main 2023 Mathematics - Class 11 JEEM23Maa43a Solution: Let be the constant term in the binomial expansion of . If the…

Given expression $(\sqrt{x}-\frac{6}{x^{3/2}})^n$ . Here, $a = \sqrt{x}$ and $b = -\frac{6}{x^{3/2}}$ .

The $r$ -th term of the binomial expansion of $(a+b)^n$ is given by

$T_{r} = {}^n{C_r}a^{n-r}b^{r}$ .

Substitute $a$ and $b$ in this formula, we get:

$T_{r} = {}^n{C_r}(\sqrt{x})^{n-r}\left(-\frac{6}{x^{3/2}}\right)^r = {}^n{C_r}(-6)^r x^{\frac{n-4r}{2}}$ .

The constant term in the binomial expansion is obtained when the power of $x$ in the terms equals zero.

This happens when $\frac{n-4r}{2} = 0$ , which gives $n = 4r$ .

$\begin{aligned}& { }^n C_{\frac{n}{4}}(-6)^{\frac{n}{4}}=\alpha \\\\& (-5)^n-{ }^n C_{\frac{n}{4}}(-6)^{n / 4}=649 \\\\& \text { By observation (625 + 24 = 649) , we get n = 4 } \\\\& \therefore \alpha=-24\end{aligned}$

Now, for coefficient of $x^{-4}$

$\begin{aligned}& \frac{n-4 r}{2}=-4 \\\\& n=4 r-8 \Rightarrow r=3 \\\\& \lambda(-24)=(-6)^3 \cdot{ }^4 C_3 \\\\& \Rightarrow \lambda=36\end{aligned}$