NIMCET 2011 — Mathematics PYQ

1. Analyze the Equations:

• Equation 1: $x^2=y^2$. This represents two straight lines passing through the origin: $y=x$ and $y=-x$.

• Equation 2: $(x-a{)}^2+y^2=1$. This represents a circle with radius 1 and center at $(a,0)$ on the x-axis.

2. Substitution:

Substitute $y^2=x^2$ from the first equation into the second:

This is a quadratic equation in $x$. Let the roots be $x_1$ and $x_2$. For every valid $x$:

• If $x\ne 0$, then $y^2=x^2$ gives two values of $y$ ($y=x$ and $y=-x$).

• If $x=0$, then $y^2=0$, so $y=0$ (only one value).

3. Case Analysis based on the Discriminant ($D$):

The discriminant of $2x^2-2ax+(a^2-1)=0$ is:

• Case 1: D < 0 (8 - 4a^2 < 0 \implies |a| > \sqrt{2})

  There are no real values for $x$.

  Number of solutions = 0.

• Case 2: $D=0$ ($a^2=2⟹a=\pm \sqrt{2}$)

  There is one value for $x$. Since $x=\frac{2a}{2(2)}=\frac{a}{2}$, and $a=\pm \sqrt{2}$, $x$ is not zero.

  Each $x$ gives two $y$ values.

  Number of solutions = 2.

• Case 3: D > 0 (a^2 < 2)

  There are two distinct real values for $x$, say $x_1$ and $x_2$.

  • If neither $x_1$ nor $x_2$ is zero: Each $x$ gives two $y$ values. Total = 4 solutions.

  • If one root is zero ($x=0$): For $x=0$ to be a root, substitute $x=0$ into the quadratic: $2(0{)}^2-2a(0)+a^2-1=0⟹a^2=1⟹a=\pm 1$.

    • If $a=1$, the quadratic is $2x^2-2x=0⟹2x(x-1)=0$. Roots are $x=0$ and $x=1$.

    • $x=0⟹y=0$ (1 point: $(0,0)$).

    • $x=1⟹y=\pm 1$ (2 points: $(1,1),(1,-1)$).

      Total solutions = 3.

Conclusion:

The possible number of distinct solutions are 0, 2, 3, or 4.