JEE Main 2024 · 5 April · evening · Single correct · +4 / −1

The values of m, n, for which the system of equations
x+y+z=4, ; 2 x+5 y+5 z=17, ; x+2 y+m z=n
has infinitely many solutions, satisfy the equation :

  1. (a) m²+n²-m-n=46
  2. (b) m²+n²+mn=68
  3. (c) m²+n²-mn=39
  4. (d) m²+n²+m+n=64
Answer and solution

Answer (c) m²+n²-mn=39

Solution

Maths sir · 4 steps

Eliminate x, and the whole question shrinks to one line: (m-2)z = n-7. Make it read 0 = 0.

  1. Remove x. Subtract twice the first equation from the second: 3y + 3z = 9, so y + z = 3. Subtract the first equation from the third: y + (m-1)z = n - 4.
  2. Remove y. Subtract y + z = 3 from y + (m-1)z = n - 4. Only one equation in z is left.

    (m-2) z = n - 7

  3. Infinitely many solutions. If m ≠ 2, z has exactly one value, so the solution is unique. If m = 2 and n ≠ 7, you get 0 = n - 7 ≠ 0, so there is no solution. Only m = 2 and n = 7 give 0 = 0, which leaves z free. This matches Δ = 3m - 6 = 0 and Δ_z = 3n - 21 = 0.
  4. So. With m = 2 and n = 7: m² + n² - mn = 4 + 49 - 14 = 39. The other three options give 44, 67 and 62. The answer is (c).

The idea to keep. For infinitely many solutions, eliminate down to one equation like az = b. You need both a = 0 and b = 0. Only a = 0 with b ≠ 0 means no solution at all.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 190 times in JEE Main since 2020. This question is from 2024.