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<p>A curve is such that&nbsp;\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=kx+2\), where&nbsp;\(k\) is a constant. The gradient is \(-1\) when&nbsp;\(x=0\) and \(7\) when&nbsp;\(x=2\). Given that the curve passes through \((1,3)\), find the value of \(k\) and the equation of the curve.<span class="inline-marks">[8]</span></p>

A curve is such that \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=kx+2\), where \(k\) is a constant. The gradient is \(-1\) when \(x=0\) and \(7\) when \(x=2\). Given that the curve passes through \((1,3)\), find the value of \(k\) and the equation of the curve.

Finding Equation of Curve from Second Derivatives

Parameterised second derivative and gradient data

Integration

<p>A curve is such that&nbsp;\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=6x-4\). The gradient of the curve at the point where&nbsp;\(x=1\) is \(2\), and the curve passes through \((2,7)\). Find the equation of the curve.<span class="inline-marks">[7]</span></p>

A curve is such that \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=6x-4\). The gradient of the curve at the point where \(x=1\) is \(2\), and the curve passes through \((2,7)\). Find the equation of the curve.

Cubic curve from a second derivative

<p>A curve is such that&nbsp;\(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=4\cos2x\). The gradient of the curve when&nbsp;\(x=0\) is \(1\), and the curve passes through \(\left(\dfrac{\pi}{4},2\right)\). Find the equation of the curve.<span class="inline-marks">[7]</span></p>

A curve is such that \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}=4\cos2x\). The gradient of the curve when \(x=0\) is \(1\), and the curve passes through \(\left(\dfrac{\pi}{4},2\right)\). Find the equation of the curve.

Trigonometric curve from a second derivative

<p>A curve is such that&nbsp;\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=3e^{2x}-e^{-x}\). The curve cuts the y-axis at \(P (0,1)\) and has a gradient of \(4\) at \(P\).</p>
<ol class="abc-list"><li data-list-item-id="item-1773464662797-ow3lqjvav"><div class="list-item-paragraph">Show that the curve has no stationary points.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1773464663273-5sy4ovusl"><div class="list-item-paragraph">Find the equation of the curve.<span class="inline-marks">[4]</span></div></li></ol>

A curve is such that \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=3e^{2x}-e^{-x}\). The curve cuts the \(y\)-axis at \(P(0,1)\) and has a gradient of 4 at \(P\).

Show that the curve has no stationary points.

Stationary Points and Their Nature

2025 MSHS PRELIMS P1 Q9

Differentiation

<p>The curve \(y = \mathrm{f}(x)\) is such that&nbsp;\(\mathrm{f}''(x) = 2e^x + e^{-2x}\).</p>
<p>The \(y\)-axis is a normal to the curve at the point \(Q\), where&nbsp;\(y = 1\).</p>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-content"><span class="list-item-text">Show that&nbsp;\(Q\) is a stationary point.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b"><div class="list-item-content"><span class="list-item-text">Hence, find an expression for&nbsp;\(\mathrm{f}(x)\).</span><span class="inline-marks">[6]</span></div></li></ol>

The curve \(y=\mathrm{f}(x)\) is such that \(\mathrm{f}''(x)=2e^x+e^{-2x}\).

The \(y\)-axis is a normal to the curve at the point \(Q\), where \(y=1\).

Hence, find an expression for \(\mathrm{f}(x)\).

Tangents and Normals Parallel to the Axes

2025 AHS PRELIMS P2 Q6

<p>Assume the same two integration constants are used in the resulting formula on both intervals \(x&lt;2\) and \(x&gt;2\).</p>
<p>At any point \((x, y)\) on a curve, \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{18}{(x-2)^3}\). The gradient of the curve at \((5, 20)\) is \(2\).</p>
<p>Find the equation of the tangent to the curve when it cuts the \(y\)-axis.</p>

At any point \((x, y)\) on a curve, \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{18}{(x-2)^3}\). The gradient of the curve at \((5, 20)\) is \(2\).

Find the equation of the tangent to the curve when it cuts the \(y\)-axis.

Assume the same two integration constants are used in the resulting formula on both intervals \(x<2\) and \(x>2\).

Tangents and Normals at a Point

<p>A curve is such that \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{1}{(2x-1)^2}\). The gradient of the curve at the point \(\left(1, \frac{1}{4}\right)\) is \(-\frac{1}{2}\).</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Express \(y\) in terms of \(x\).</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence, find the coordinates of the point where the curve cuts the \(x\)-axis, giving your answer in exact form.</span></div></li></ol>

A curve is such that \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{1}{(2x-1)^2}\). The gradient of the curve at the point \(\left(1, \frac{1}{4}\right)\) is \(-\frac{1}{2}\).

Hence, find the coordinates of the point where the curve cuts the \(x\)-axis, giving your answer in exact form.

<p>A curve is such that \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 36x^2\). The curve has a turning point \((-1, -9)\).</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find the equation of the curve.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Determine, with working, whether there are any other turning points on the curve.</span></div></li></ol>

A curve is such that \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 36x^2\). The curve has a turning point \((-1, -9)\).

Determine, with working, whether there are any other turning points on the curve.

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