Differentiation – Year 12 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Write every term as a power of \(x\): \(-\dfrac5x=-5x^{-1}\) and \(\sqrt x=x^{1/2}\); in (b) divide each term of the numerator by \(x^2\) first.

   (a) \(12x^2+5x^{-2}+\tfrac12x^{-1/2}\) (2). (b) \(y=x+2x^{-1}\) so \(\dfrac{dy}{dx}=1-2x^{-2}\) (2).

   So (a) \(\dfrac{dy}{dx}=12x^2+\dfrac5{x^2}+\dfrac1{2\sqrt x}\) and (b) \(\dfrac{dy}{dx}=1-\dfrac2{x^2}\).

   A common mistake in (b) is to differentiate the numerator and the denominator separately; always divide by \(x^2\) first.

   In (a), the derivative of \(-5x^{-1}\) is \(+5x^{-2}\): the two minus signs combine, and this sign is where most marks are lost.

2. Answer to Question 2 (4 marks)

   Differentiate to find the gradient at \(x=2\), then use \(y-y_1=m(x-x_1)\); the normal has gradient \(-\dfrac1m\).

   \(\dfrac{dy}{dx}=3x^2-4x=4\) at \(x=2\) (1). Tangent: \(y-3=4(x-2)\), \(y=4x-5\) (1). Normal gradient \(-\tfrac14\) (1): \(y-3=-\tfrac14(x-2)\), \(x+4y=14\) (1).

   So the tangent is \(y=4x-5\) and the normal is \(x+4y=14\).

   Check: the point \((2,3)\) lies on both lines, since \(4\times2-5=3\) ✓ and \(2+4\times3=14\) ✓.

   The normal can also be written \(y=-\tfrac14x+\tfrac72\); any equivalent form scores full marks.

3. Answer to Question 3 (4 marks)

   Set \(\dfrac{dy}{dx}=0\), factorise, find the \(y\)-coordinates from the original equation and classify with the second derivative.

   \(y'=6x^2-18x+12=6(x-1)(x-2)\) (1), \(x=1\): \(y=5\); \(x=2\): \(y=4\) (1). \(y''=12x-18\): at \(x=1\), \(-6\) maximum \((1,5)\) (1); at \(x=2\), \(6\) minimum \((2,4)\) (1).

   So \((1,5)\) is a maximum and \((2,4)\) is a minimum.

   Check with nearby values: \(y(0.9)=4.968\) and \(y(1.1)=4.972\) are both less than 5, so \((1,5)\) is indeed a local maximum.

4. Answer to Question 4 (4 marks)

   A function is increasing when its derivative is positive for every \(x\).

   \(f'=3x^2+3\gt 0\) for all \(x\) (2). \(g'=3x^2+k\ge0\) for all \(x\) needs \(k\ge0\) (2).

   Since \(3x^2\ge0\) for all \(x\), we need \(k\ge0\); if \(k\lt 0\), \(g'(0)=k\lt 0\) and \(g\) decreases near 0. So \(k\ge0\).

   For example, \(k=-3\) gives \(g'(x)=3(x^2-1)\), which is negative for \(-1\lt x\lt 1\), so that \(g\) is not increasing everywhere.

   In the first part, a complete answer states that \(3x^2\ge0\), so \(3x^2+3\ge3\gt0\). For \(g\), the value \(k=0\) is allowed: \(g(x)=x^3\) has \(g'(0)=0\) but is still increasing everywhere.

5. Answer to Question 5 (4 marks)

   1. The wall forms one side, so the fencing covers two widths and one length. Length \(=80-2x\) so \(A=x(80-2x)\) (1) This is the required expression.
   2. Differentiate and set the derivative equal to zero, then check with the second derivative. \(A'=80-4x=0\), \(x=20\) (1), \(A''=-4\lt 0\) (1), \(A=800\) m\(^2\) (1) The enclosure should be 20 m wide and 40 m long, giving a maximum area of 800 m\(^2\). Check: \(20\times40=800\) and the fencing used is \(20+40+20=80\) m ✓.

   In (b), the second-derivative test earns the mark for justifying a maximum; showing that \(A(19)=798\) and \(A(21)=798\) are both below 800 is also accepted.

Total : 20 marks