PUNE MCA Exam 2017 Question Paper with answers for online practice

Concept:Unique solution requires that the system yields exactly one valid triple (x,y,z). For each integer $k\neq0$, the quadratic in $x$ gives two potential roots, but a root making a denominator zero is invalid.Explanation:1. Rewrite the equations: $2x+3y-z=6$ ... (1) $ky+\frac{z}{x}=1\Rightarrow kxy+z=x$ ... (2) (since $x\neq0$) $k^2y+k^2z=1\Rightarrow y+z=\frac{1}{k^2}$ ... (3)2. From (2), $z=x(1-ky)$. Substitute into (1) and (3) to eliminate $z$, obtaining a quadratic in $x$: $2k^3x^2+(2k^2-6k^3-k)x+(6k^2-3)=0$.3. For a unique solution, exactly one of the two roots must be valid, i.e., the other root must be extraneous because it makes a denominator zero. The denominator in the expression for $y$ is $kx-1$. So one root must satisfy $kx=1$ (i.e., $x=1/k$).4. Substitute $x=1/k$ into the quadratic: $2k^3\left(\frac{1}{k^2}\right)+(2k^2-6k^3-k)\left(\frac{1}{k}\right)+(6k^2-3)=2k+(2k-6k^2-1)+(6k^2-3)=4k-4=0\Rightarrow k=1$.5. For $k=1$, the roots are $x=1$ (invalid) and $x=\frac{3}{2}$ (valid). The valid root gives a single solution: $x=\frac{3}{2}, y=1, z=0$. No other integer $k$ yields an extraneous root or a repeated root.Answer:The system has a unique solution for exactly one value of $k$.