CGPET 2015 Solved Paper

Three lines of triangle are given by $(x^2-y^2)(2x+3y-6)=0$ ⇒ $(x-y)(x+y)(2x+3y-6)=0$ ∴ Three lines of triangle are $x-y=0, x+y=0$ and $2x+3y-6=0$ From given lines of triangle, the required ∆OAB is formed. ∵ $(-2,\lambda)$ lies inside the triangle. ∴ $2(-2)+3(\lambda)-6<0 \text{ and } -2+\lambda>0$ ⇒$-4+3\lambda-6<0 \text{ and } \lambda>2$ ⇒$3\lambda<10 \text{ and } \lambda>2$ ⇒ $\lambda < \frac{10}{3} \text{ and } \lambda >2$ ∴$\lambda \in \left(2,\frac{10}{3}\right)$....(i) Now, (μ,1) lies outside the triangle. To find value of μ, we find the interval [M, N] for values of x. $x+1\ge 0 \text{ and } x-1\le 0$ $x \ge -1 \text{ and } x \le 1$ ∴ $x \in [-1,1]$ ∵ (μ,1) lies outside then triangle. ∴ $\mu \in (-\infty,-1) \cup (1,\infty)$ or $\mu \in \mathbb{R} - [-1,1]$