For the equations ax + (a 2 + 1)y = 4 and 4x + ay = a 2 , which of the following statements is TRUE?

Correct — B. Each statement hands you a value of a, so substitute it and see whether the offered pair satisfies both equations.

With a = −12 the pair becomes
−12x + 145y = 4 (since a² + 1 = 145)
4x − 12y = 144 (since a² = 144)

From the second, x = 36 + 3y.
Substituting: −432 − 36y + 145y = 4
109y = 436 → y = 4
x = 36 + 3(4) = 48 → option (b)

Why the others are wrong

• (a)
  At a = 6 the first equation is 6x + 37y = 4, and x = 48, y = 4 give 288 + 148 = 436. The pair produces 436 where the equation needs 4.
• (c)
  At a = 6 the second equation is 4x + 6y = 36, but x = 325⁄28 and y = −25⁄28 give (1300 − 150)⁄28 ≈ 41.1. It fails the easier of the two equations.
• (d)
  At a = −12 the system already has the unique solution x = 48, y = 4, because its determinant −3a² − 4 = −436 is non-zero. A second, different pair cannot solve the same system.

For a₁x + b₁y = c₁ and a₂x + b₂y = c₂, the solution is unique when the determinant a₁b₂ − a₂b₁ is non-zero.

Here that determinant is a·a − 4(a² + 1) = −3a² − 4, which is negative for every real a. So no value of a can make these lines parallel or coincident: whatever a is, there is exactly one (x, y).

That is why the question is settled by substitution rather than by theory. Take the a an option offers, solve, and compare — and start with 4x + ay = a², the equation with the smaller numbers.

Key facts

• The coefficient determinant of this pair is a² − 4(a² + 1) = −3a² − 4, non-zero for every real a.
• At a = −12 the equations read −12x + 145y = 4 and 4x − 12y = 144.
• Their unique solution is x = 48, y = 4.
• A proposed (x, y) must satisfy both equations, so fitting just one of them proves nothing.

Common traps

• Verifying one equation only and accepting a pair that happens to fit it.
• Reading a² + 1 as (a + 1)², which turns 145 into 121 at a = −12.
• Writing a² as −144 for a = −12 and wrecking the right-hand side.

No directly related past PYQ was found.