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The number of roots of e^x = 0.5x^2 - 2 in the range [-5, 0] is A. 0 B. 1 C. 2 D. 3

GATE 2017 · General Aptitude · Maxima Minima · medium

Answer: There are 2 roots of e^x = 0.5x^2 - 2 in [-5, 0].

  1. Evaluate h(x) at boundary and a midpoint: h(-5) = e^(-5) - 0.5(25) + 2 = 0.007 - 12.5 + 2 = -10.49 (negative). h(-3) = e^(-3) - 0.5(9) + 2 = 0.05 - 4.5 + 2 = -2.45 (negative). h(-1) = e^(-1) - 0.5(1) + 2 = 0.368 - 0.5 + 2 = 1.868 (positive). h(0) = 1 - 0 + 2 = 3 (positive).
  2. Check for a second sign change near x = -2: h(-2) = 0.135 - 2 + 2 = 0.135 (positive). Now check between -5 and -3: h(-4) = e^(-4) - 0.5(16)+2 = 0.018 - 8 + 2 = -5.98 (negative). So h goes from -10.49 at -5 to -2.45 at -3. Check h(-1.8) ≈ e^(-1.8) - 0.5(3.24)+2 ≈ 0.165 - 1.62 + 2 = 0.545 > 0. We need a finer check: h(-2.5) = e^(-2.5) - 0.5(6.25)+2 = 0.082 - 3.125+2 = -1.043 < 0. So between -2.5 and -2 the function goes from negative to positive — second root in (-2.5, -2).