JEE Main · 2013 · 07 Apr 2013 · Offline · Matrices

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

Answer: (A)

Solution

The two lines fail to meet when their coefficients are proportional but the constants are not:

\frac{k + 1}{k} = \frac{8}{k + 3} \neq = \frac{4 k}{3 k - 1}

. The first equality gives

(k + 1) (k + 3) = 8 k

, i.e.

k^{2} - 4 k + 3 = 0

, so

k = 1

k = 3

. Checking the constants,

k = 1

makes the two sides equal and the lines coincide with infinitely many solutions, while

k = 3

keeps them distinct and parallel. Only

k = 3

gives no solution, so there is exactly one such value.

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option