Question

Let \(k\) be a constant. The equations \(kx + y = 3\) and \(4x + ky = 4\) have a unique solution if and only if

• A. \(|k|≠2\)
• B. \(|k|=2\)
• C. \(k≠2\)
• D. \(k=2\)

Answer 1

The Correct Option is A

Simultaneous equation have a unique solution only if \(\frac{a_1}{a_2}≠\frac{b_1}{b_2}\)
From the given equations, a unique solution would exist only if \(\frac{k}{2}≠\frac{2}{k}\)
\(⇒ k^2≠4⇒|k|≠2\)
So, the correct option is (A): \(|k|≠2\)

Answer 2

Now , we have two linears equations having unique solution if \(\frac{a}{d}\ne\frac{b}{e}\) which are as follows :
ax+by= c and dx+ ey = f
∴ \(\frac{k}{4}\ne\frac{1}{k}\)
k² ≠ 4
⇒ |k| ≠ 2
Therefore, the correct option is (A) : \(|k|≠2\)