JEE Main2013MathematicsDeterminantsActual
The number of values of k , for which the system of equations : k + 1 x + 8 y = 4 k k x + k + 3 y = 3 k - 1 has no solution, is :
Options
- A2
- B3
- CInfinite
- D1
Correct answer
D. 1
Step-by-step solution
For a 1 x + b 1 y = c 1 and a 2 x + b 2 y = c 2 to have no solution, a 1 a 2 = b 1 b 2 ≠ c 1 c 2 ∴ k + 1 k = 8 k + 3 ≠ 4 k 3 k - 1 k + 1 k + 3 = 8 k ⇒ k 2 - 4 k + 3 = 0 ⇒ k = 1 , 3 but, for k = 1 , 8 k + 3 = 4 k 3 k - 1 ∴ k = 3 Hence, only one value