RRB NTPC · Question 56
Given w = -2, x = 3, y = 0 & \(z = - \frac{1}{2},\) find the value of \({w^2}\sqrt {\left( {{z^2} + {y^2}} \right)} \) A. -4 B. 8 C. 2 D. 4
Show answer & explanation
Answer: (C) C
Explanation
w² = 4 and √(z² + y²) = √(1/4 + 0) = 1/2. So the value is 4 × 1/2 = 2.