Double roots of an even quartic, its odd derivative, and the tangent at a point of inflexion

Let \(f(x)=2x^4+px^2+q\), where \(p\) and \(q\) are real constants and \(q\ne0\).

1. In this part, \(p=3\) and \(q=1\).
   1. Sketch the graph of \(y=2x^4+3x^2+1\), labelling any points where it meets the axes.[2]
   2. Hence, state a reason why \(2x^4+3x^2+1=0\) has no real roots.[1]
2. In this part, \(p=-8\) and \(q=8\).
   Sketch the graph of \(y=2x^4-8x^2+8\), labelling any points where it meets the axes.[2]
3. 1. Write down \(f'(x)\).[2]
   2. Show that \(f'\) is an odd function.[3]
4. If \(f(\beta)=0\) and \(f'(\beta)=0\), then \(\beta\) is called a double root of \(f(x)=0\).
   Describe how the graph of \(y=f(x)\) shows whether \(f(x)=0\) has a double root.[1]
5. For parts (e) to (g), assume that \(f(x)=0\) has a double root \(\beta\).
   Show, using part (c), that \(-\beta\) is a double root of \(f(x)=0\) as well.[4]
6. 1. Use \(f'(\beta)=0\), or another method, to show that \(\beta^2=-\frac p4\).[3]
   2. Hence, show that \(q=\frac{p^2}{8}\).[2]
7. Hence, deduce the values of \(p\) for which the double roots of \(f(x)=0\) are
   1. real;[1]
   2. purely imaginary.[1]
8. For parts (h) and (i), let \(f(x)=2x^4+px^2+q\) with \(p<0\) and \(q\ne0\).
   Show that \(f''(x)=0\) when \(x=\pm\sqrt{-\frac{p}{12}}\).[3]
9. The graph of \(y=f(x)\) has a point of inflexion where \(x=\sqrt{-\frac{p}{12}}\), and the tangent at this point has equation \(y=-\frac{2\sqrt3}{9}(-p)^{\frac32}x+c\).
   Show that \(c=\frac{p^2}{24}+q\).[5]