Question

The function y = f(x) is represented parametrically by

x = t⁵ – 5t³ – 20t + 7 and y = 4t³ – 3t² – 18t + 3, (–2 < t < 2). The minimum of y = f(x) occurs at

• A. t = – 1
• B. t = 0
• C. t = ½
• D. ) t = 3/2

Answer 1

)

Sol. x = f(t) = t⁵ – 5t³ – 20t + 7

= f '(t) = 5t⁴ – 15 t² – 20 = 5(t² – 4) (t² + 1) ¹ 0

If – 2 < t < 2

y = y(t) = 4t³ – 3t² – 18t + 3

= y'(t) = 12 t² – 6t – 18

= 0 Ž t = – 1 or 3/2

$\frac { \mathrm { d } ^ { 2 } \mathrm { y } } { \mathrm { dt } ^ { 2 } }$ = y"(t) = 24t – 6 = y"(–1) = – 30

and y"(3/2) = 30

y = f(x) is minimum at t = 3/2