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A-Level CAIE Mathematics A23.4 DifferentiationQuestion Bank

[Maximum number: 5]

Find the exact coordinates of the stationary point on the curve with equation y=5xe12xy=5 x \mathrm{e}^{\frac{1}{2} x}.

The equation of a curve is y=e2x1+e2xy=\frac{\mathrm{e}^{2 x}}{1+\mathrm{e}^{2 x}}. Show that the gradient of the curve at the point for which x=ln3x=\ln 3 is 950\frac{9}{50}.

[Maximum number: 4]

The parametric equations of a curve are

x=3(1+sin2t),y=2cos3tx=3\left(1+\sin ^{2} t\right), \quad y=2 \cos ^{3} t

Find dy dx\frac{\mathrm{d} y}{\mathrm{~d} x} in terms of t, simplifying your answer as far as possible.

The curve y=lnxx3y=\frac{\ln x}{x^{3}} has one stationary point. Find the x-coordinate of this point.

[Maximum number: 4]

Find dy dx\frac{\mathrm{d} y}{\mathrm{~d} x} in each of the following cases:

(a)

y=ln(1+sin2x)y=\ln (1+\sin 2 x),

[ 2 ]
(b)

y=tanxxy=\frac{\tan x}{x}.

[ 2 ]

The parametric equations of a curve are

x=t2t+3,y=e2tx=\frac{t}{2 t+3}, \quad y=\mathrm{e}^{-2 t}

Find the gradient of the curve at the point for which t=0.

[Maximum number: 5]

The equation of a curve is cos3x+5siny=3\cos 3 x+5 \sin y=3.
Find the gradient of the curve at the point (19π,16π)\left(\frac{1}{9} \pi, \frac{1}{6} \pi\right).

[Maximum number: 5]
Question image

The diagram shows the curve with equation

y=5sin2x3tan2xy=5 \sin 2 x-3 \tan 2 x

for values of x such that 0x<14π0 \leqslant x<\frac{1}{4} \pi. Find the x-coordinate of the stationary point M, giving your answer correct to 3 significant figures.

[Maximum number: 7]

The parametric equations of a curve are

x=t+ln(t+2),y=(t1)e2tx=t+\ln (t+2), \quad y=(t-1) \mathrm{e}^{-2 t}

where t>-2.

(a)

Express dy dx\frac{\mathrm{d} y}{\mathrm{~d} x} in terms of t, simplifying your answer.

[ 5 ]
(b)

Find the exact y-coordinate of the stationary point of the curve.

[ 2 ]
[Maximum number: 5]

The parametric equations of a curve are

x=3cos2θ,y=2θ+sin2θ,x=3-\cos 2 \theta, \quad y=2 \theta+\sin 2 \theta,

for 0<θ<12π0<\theta<\frac{1}{2} \pi.
Show that dy dx=cotθ\frac{\mathrm{d} y}{\mathrm{~d} x}=\cot \theta.

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