JEE MAIN - Mathematics (2023 - 24th January Evening Shift - No. 2)
The number of real solutions of the equation $$3\left( {{x^2} + {1 \over {{x^2}}}} \right) - 2\left( {x + {1 \over x}} \right) + 5 = 0$$, is
3
4
0
2
Explanation
$3\left(x^{2}+\frac{1}{x^{2}}\right)-2\left(x+\frac{1}{x}\right)+5=0$
$3\left[\left(x+\frac{1}{x}\right)^{2}-2\right]-2\left(x+\frac{1}{x}\right)+5=0$
Put $x+\frac{1}{x}=t \Rightarrow t \in(-\infty,-2] \cup[2, \infty)$
$$ \begin{aligned} & 3 t^{2}-2 t-1=0 \\\\ & 3 t^{2}-3 t+t-1=0 \\\\ & \Rightarrow 3 t(t-1)+1(t-1)=0 \Rightarrow t=1,=-\frac{1}{3} \\\\ & \Rightarrow \quad t=1,-\frac{1}{3} \\\\ & \because t \in(-\infty,-2] \cup[2, \infty) \end{aligned} $$
No real value of $t \Rightarrow$ no real value of $x$.
$3\left[\left(x+\frac{1}{x}\right)^{2}-2\right]-2\left(x+\frac{1}{x}\right)+5=0$
Put $x+\frac{1}{x}=t \Rightarrow t \in(-\infty,-2] \cup[2, \infty)$
$$ \begin{aligned} & 3 t^{2}-2 t-1=0 \\\\ & 3 t^{2}-3 t+t-1=0 \\\\ & \Rightarrow 3 t(t-1)+1(t-1)=0 \Rightarrow t=1,=-\frac{1}{3} \\\\ & \Rightarrow \quad t=1,-\frac{1}{3} \\\\ & \because t \in(-\infty,-2] \cup[2, \infty) \end{aligned} $$
No real value of $t \Rightarrow$ no real value of $x$.
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