IB Mathematics AHL 2.8 Transformations of graphs AI HL Paper 2- Exam Style Questions- New Syllabus

Question

Consider the curve \( y = \sqrt{x} \).

(a) (i) Find \(\frac{dy}{dx}\).
(ii) Hence show that the equation of the tangent to the curve at the point (0.16, 0.4) is \( y = 1.25x + 0.2 \).

The shape of a piece of metal can be modelled by the region bounded by the functions \( f \), \( g \), the x-axis and the line segment [AB], as shown in the following diagram. The units on the x and y axes are measured in metres.

The function \( f \) is defined as:

\[ f(x) = \begin{cases} \sqrt{x}, & 0 \leq x \leq 0.16 \\ 1.25x + 0.2, & 0.16 < x \leq 0.5 \end{cases} \]

The graph of \( g \) is obtained from the graph of \( f \) by:

a sketch scale factor of \(\frac{1}{2}\) in the x direction,
followed by a stretch scale factor \(\frac{1}{2}\) in the y direction,
followed by a translation of 0.2 units to the right.

Point A lies on the graph of \( f \) and has coordinates (0.5, 0.825). Point B is the image of A under the given transformations and has coordinates \((p, q)\).

(b) Find the value of \( p \) and the value of \( q \).

The piecewise function \( g \) is given by \( g(x)=\left\{\begin{matrix} h(x) & 0.2 \leq x \leq a \\ 1.25x + b & a < x \leq p \end{matrix}\right.\)

(c) Find
(i) an expression for \( h(x) \).
(ii) the value of \( a \).
(iii) the value of \( b \).

(d) (i) Find the area enclosed by \( y = f(x) \), the x-axis and the line \( x = 0.5 \).
The area enclosed by \( y = g(x) \), the x-axis and the line \( x = p \) is 0.0627292 \(\text{m}^2\) correct to six significant figures.
(ii) Find the area of the shaded region on the diagram.

▶️ Answer/Explanation

Markscheme

(a)(i)
For \( y = \sqrt{x} = x^{1/2} \):
\(\frac{dy}{dx} = \frac{1}{2} x^{1/2 – 1} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}\) (A1)
Result:
\(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\)

(a)(ii)
    At \( x = 0.16 \), \( y = \sqrt{0.16} = 0.4 \).
    Slope:
    \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\)
    \( m = \frac{1}{2\sqrt{0.16}} = \frac{1}{2 \times 0.4} = 1.25 \) (M1A1)
    Point-slope form at (0.16, 0.4):
    \( y – 0.4 = 1.25 (x – 0.16) \) (M1)
    \( y – 0.4 = 1.25x – 0.2 \)
    \( y = 1.25x – 0.2 + 0.4 \)
    \( y = 1.25x + 0.2 \) (AG)
Result:
\( y = 1.25x + 0.2 \)

(b)
    Point A: (0.5, 0.825)
    Transformations on (0.5, 0.825):
    1. Horizontal scale by \(\frac{1}{2}\):
    \( x’ = 0.5 \times \frac{1}{2} = 0.25 \), \( y’ = 0.825 \)
    2. Vertical scale by \(\frac{1}{2}\):
    \( x” = 0.25 \), \( y” = 0.825 \times \frac{1}{2} = 0.4125 \)
    3. Translate right by 0.2:
    \( x”’ = 0.25 + 0.2 = 0.45 \), \( y”’ = 0.4125 \)
    Thus:
    \( p = 0.45 \), \( q = 0.4125 \) (A1A1)
Result:
\( p = 0.45 \), \( q = 0.4125 \)

(c)(i)
    Start with \( f(x) = \sqrt{x} \).
    1. Horizontal scale by \(\frac{1}{2}\): Replace \( x \) with \( 2x \):
    \( f(2x) = \sqrt{2x} \)
    2. Vertical scale by \(\frac{1}{2}\):
    \(\frac{1}{2} f(2x) = \frac{1}{2} \sqrt{2x} \)
    3. Translate right by 0.2: Replace \( x \) with \( x – 0.2 \):
    \( h(x) = \frac{1}{2} \sqrt{2 (x – 0.2)} \) (A2)
Result:
\( h(x) = \frac{1}{2} \sqrt{2 (x – 0.2)} \)

(c)(ii)
    Point of intersection between \( h(x) = \frac{1}{2} \sqrt{2 (x – 0.2)} \) and \( y = 1.25x + b \).
    From (c)(iii), \( b = -0.15 \).
    Solve:
    \(\frac{1}{2} \sqrt{2 (x – 0.2)} = 1.25x – 0.15\)
    \(\sqrt{2 (x – 0.2)} = 2.5x – 0.3\)
    Square both sides:
    \(2 (x – 0.2) = (2.5x – 0.3)^2\)
    \(2x – 0.4 = 6.25x^2 – 1.5x + 0.09\)
    \(0 = 6.25x^2 – 3.5x + 0.49\)
    Quadratic formula:
    \(x = \frac{3.5 \pm \sqrt{(-3.5)^2 – 4 \times 6.25 \times 0.49}}{2 \times 6.25}\)
    \( = \frac{3.5 \pm \sqrt{12.25 – 12.25}}{12.5}\)
    Since discriminant is near zero, approximate numerically:
    \(x \approx 0.28\)
    Verify:
    \(h(0.28) = \frac{1}{2} \sqrt{2 (0.28 – 0.2)} \approx 0.2\)
    \(1.25 \times 0.28 – 0.15 = 0.35 – 0.15 = 0.2\)
    \(a = 0.28\) (A1)
Result:
\( a = 0.28 \)

(c)(iii)
    The line passes through point B \((0.45, 0.4125)\):
    \(y = 1.25x + b\)
    \(0.4125 = 1.25 \times 0.45 + b\)
    \(0.4125 = 0.5625 + b\)
    \(b = 0.4125 – 0.5625 = -0.15\) (M1A1)
Result:
\( b = -0.15 \)

(d)(i)
    Area from \( x = 0 \) to \( x = 0.5 \):
    \(\text{Area} = \int_0^{0.16} \sqrt{x} \, dx + \int_{0.16}^{0.5} (1.25x + 0.2) \, dx\) (M1A1)
    First integral:
    \(\int \sqrt{x} \, dx = \frac{2}{3} x^{3/2}\)
    \(\left[ \frac{2}{3} x^{3/2} \right]_0^{0.16} = \frac{2}{3} (0.16)^{3/2} – 0\)
    \((0.16)^{1/2} = 0.4\), \( (0.16)^{3/2} = 0.4 \times 0.16 = 0.064\)
    \(\frac{2}{3} \times 0.064 \approx 0.0426666666667\)
    Second integral:
    \(\int (1.25x + 0.2) \, dx = 0.625x^2 + 0.2x\)
    \(\left[ 0.625x^2 + 0.2x \right]_{0.16}^{0.5}\)
    At \( x = 0.5 \):
    \(0.625 \times 0.25 + 0.2 \times 0.5 = 0.15625 + 0.1 = 0.25625\)
    At \( x = 0.16 \):
    \(0.625 \times 0.0256 + 0.2 \times 0.16 = 0.016 + 0.032 = 0.048\)
    \(0.25625 – 0.048 = 0.20825\)
    Total area:
    \(0.0426666666667 + 0.20825 \approx 0.250916666667\)
    \(\approx 0.251 \text{ m}^2\) (A1)
Result:
\( 0.251 \text{ m}^2 \)

(d)(ii)
    Either:
    Area of trapezoid = \(\frac{1}{2} \times 0.05 \times (0.4125 + 0.825) = 0.0309375\) (M1A1)
    Or:
    \(\int_{0.45}^{0.5} (8.25x – 3.3) \, dx = 0.0309375\) (M1A1)
    Note: If the rounded answer of 0.413 from part (b) is used, the integral is
    \(\int_{0.45}^{0.5} (8.24x – 3.295) \, dx = 0.03095\) (M1A1)
    Then:
    Shaded area = \( 0.250916\ldots – 0.0627292\ldots – 0.0309375 \) (M1)
    \( = 0.157 \, \text{m}^2 \) (0.15725) (A1)
Result:
\( 0.157 \, \text{m}^2 \)

IB Mathematics AHL 2.8 Transformations of graphs AI HL Paper 2- Exam Style Questions- New Syllabus

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