8 Fig. 8 shows the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\). P is the point on this curve with \(x\)-coordinate 1 , and R is the point \(\left( 0 , - \frac { 7 } { 8 } \right)\).
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\includegraphics[alt={},max width=\textwidth]{56672660-b7dc-4e10-8039-1c041e75b598-3_1022_995_479_575}
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\caption{Fig. 8}
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- Find the gradient of PR.
- Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that PR is a tangent to the curve.
- Find the exact coordinates of the turning point Q .
- Differentiate \(x \ln x - x\).
Hence, or otherwise, show that the area of the region enclosed by the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).