JEE MAIN - Mathematics (2025 - 29th January Morning Shift - No. 8)
The number of solutions of the equation
$ \left( \frac{9}{x} - \frac{9}{\sqrt{x}} + 2 \right) \left( \frac{2}{x} - \frac{7}{\sqrt{x}} + 3 \right) = 0 $ is :
$ \left( \frac{9}{x} - \frac{9}{\sqrt{x}} + 2 \right) \left( \frac{2}{x} - \frac{7}{\sqrt{x}} + 3 \right) = 0 $ is :
3
2
1
4
Explanation
Consider $\frac{1}{\sqrt{\mathrm{x}}}=\alpha \quad \mathrm{x}>0$
$$\begin{aligned} & \left(9 \alpha^2-9 \alpha+2\right)\left(2 \alpha^2-7 \alpha+3\right)=0 \\ & (3 \alpha-2)(3 \alpha-1)(\alpha-3)(2 \alpha-1)=0 \\ & \alpha=\frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 3 \\ & x=9,4, \frac{9}{4}, \frac{1}{9} \end{aligned}$$
So, no. of solutions $=4$
Comments (0)
