If c = 16xy + 49yx for some non-zero real numbers x and y, then c cannot take the value:

Question

UPSC Quizroom · Accepted Answer

Step 1: Let k = x/y

Then y/x = 1/k

So c = 16(x/y) + 49(y/x) = 16k + 49/k

Multiply through by k: 16k^2 - c k + 49 = 0

Step 2: Condition for real x and y

For x and y real and non-zero, k = x/y must be real and non-zero.

So the quadratic 16k^2 - c k + 49 = 0 must have real solutions.

Real solutions exist if discriminant ≥ 0:

Δ = c^2 - 4 *16 *49 = c^2 - 3136 ≥ 0

c^2 ≥ 3136 → |c| ≥ 56

Step 3: Check options

-70 → |c| = 70 ≥ 56 ✅ possible

60 → |c| = 60 ≥ 56 ✅ possible

-50 → |c| = 50 < 56 ❌ impossible

-60 → |c| = 60 ≥ 56 ✅ possible

✅ Answer:

c cannot take the value -50

🎯 UPSC Quiz Room