OCR Pure 1 2018 September — Question 12

Exam BoardOCR
ModulePure 1 (Pure Mathematics 1)
Year2018
SessionSeptember
TopicDifferential equations

12 The gradient function of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ^ { 2 } \sin 2 x } { 2 \cos ^ { 2 } 4 y - 1 }\).
  1. Find an equation for the curve in the form \(\mathrm { f } ( y ) = g ( x )\). The curve passes through the point \(\left( \frac { 1 } { 4 } \pi , \frac { 1 } { 12 } \pi \right)\).
  2. Find the smallest positive value of \(y\) for which \(x = 0\). \section*{END OF QUESTION PAPER}