The function y = f(x) is represented parametrically by x = t^5 – 5t^3 – 20t + 7 and y = 4t^3 – 3t^2 – 18t + 3 (-2 < t < 2). At which point does y = f(x) has a minimum value?(a) t = -1(b) t = 0(c) t = 1/2(d) t = 3/2This question was addressed to me in homework.Origin of the question is Second Order Derivative in chapter Limits and Derivatives of Mathematics – Class 11

The function y = f(x) is represented parametrically by x = t^5 – 5t^

1.	The function y = f(x) is represented parametrically by x = t^5 – 5t^3 – 20t + 7 and y = 4t^3 – 3t^2 – 18t + 3 (-2 < t < 2). At which point does y = f(x) has a minimum value?(a) t = -1(b) t = 0(c) t = 1/2(d) t = 3/2This question was addressed to me in homework.Origin of the question is Second Order Derivative in chapter Limits and Derivatives of Mathematics – Class 11
Answer» RIGHT CHOICE is (d) t = 3/2 Best explanation: We have dx/dt = Φ’(t) Φ’(t) = 5(t^2 – 4)(t^2 + 1) ≠ 0 if -2 < t < 2 And dy/dt = 12t^2 – 6t – 18 Also, dy/dt = 0 => t = -1 or t = 3/2 Now, d^2y/dt^2 = 24t – 8 => d^2(-1)/dt^2 = -30 and, d^2(3/2)/dt^2 = 30 Consequently, y = f(x) has maximum at t = -1 minimum at t = 3/2

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