Home / IB Mathematics SL 2.2 Concept of a function, domain, range and graph AI SL Paper 2 – Exam Style Questions

IB Mathematics SL 2.2 Concept of a function, domain, range and graph AI SL Paper 2 - Exam Style Questions - New Syllabus

IB DP Maths AI SL Paper 2 - All Topics

Question

Examine the function \(f(x)=3x-1+4x^{-2}\). Part of the graph of \(y=f(x)\) is shown below.

The function is defined for every real \(x\) except \(x=a\).
(a) Write down the value of \(a\). [1]
(b) Use your graphing display calculator to find the coordinates of the local minimum. [2]
The equation \(f(x)=w\), where \(w\in\mathbb{R}\), has three solutions.
(c) Give one possible value for \(w\). [1]
The line \(y=mx+\dfrac{1}{4}\) is tangent to \(f(x)\) when \(x=-4\).
(d) State whether the value of \(m\) is positive or negative. Justify your answer. [2]
A second function is given by \(g(x)=kp^{x}-9\), where \(p>0\). The graph of \(y=g(x)\) cuts the \(y\)-axis at point \(A(0,-5)\) and passes through point \(B(3,4.5)\).

(e) Find the value of

(i) \(k\);
(ii) \(p\). [4]
(f) State the equation of the horizontal asymptote of \(y=g(x)\). [2]
(g) Determine the solution of \(f(x)=g(x)\) when \(x>0\). [2]
Consider a third function, \(h\), where \(h(x)=f(x)+g(x)\). The point \(C(-1,q)\) lies on the graph of \(g(x)\).
(h) Decide whether \(C\) also lies on the graph of \(h(x)\). Justify your answer. [2]
▶️ Answer/Explanation
Markscheme (with detailed working)

(a)

\(x=0\) is excluded by the term \(4x^{-2}\), so \(\boxed{a=0}\). A1

(b)

On a graphing calculator, use a minimum command on the curve \(y=3x-1+4x^{-2}\) (for \(x>0\)). The local minimum occurs at \[ \boxed{(1.39,\; 5.24)} \quad \text{(exact GDC values }(1.38672\ldots,\; 5.24025\ldots)\text{)}. \] A1 A1

(Award a mark for each correct coordinate; avoid 2 s.f.)

(c)

The line \(y=w\) cuts the curve in three points when \(w>5.24025\ldots\) (above the local minimum). Any specific such value earns the mark, e.g. \(\boxed{w=6}\). A1

(d)

Differentiate \(f\): \(f'(x)=3-8x^{-3}\). Then \[ f'(-4)=3-\frac{8}{(-4)^3}=3-\frac{8}{-64}=3+\frac18=\frac{25}{8}=3.125>0. \] Therefore the tangent gradient \(m=f'(-4)\) is positive. \(\;\boxed{m>0}\) R1 A1

(e)

Using \(A(0,-5)\): \(g(0)=k\cdot p^{0}-9=k-9=-5\Rightarrow \boxed{k=4}.\) (M1) A1
Using \(B(3,4.5)\): \(g(3)=4p^{3}-9=4.5\Rightarrow 4p^3=13.5\Rightarrow p^3=3.375\Rightarrow \boxed{p=1.5}.\) (M1) A1

(f)

For \(g(x)=4p^{x}-9\) with \(p>1\), the horizontal asymptote is \(\boxed{y=-9}\). A2

(Award A1 for seeing “\(-9\)”, A2 for the full equation \(y=-9\).)

(g)

Solve \(f(x)=g(x)\) for \(x>0\): \[ 3x-1+\frac{4}{x^2}=4(1.5)^x-9. \] Checking \(x=4\): \(f(4)=12-1+\frac{4}{16}=11.25\), and \(g(4)=4(1.5)^4-9=4(5.0625)-9=11.25\). Hence \(\boxed{x=4}\). (M1) A1

(h)

Since \(h(x)=f(x)+g(x)\) and \(C(-1,q)\) lies on \(g\), we have \(q=g(-1)\). Now \[ f(-1)=3(-1)-1+4(-1)^{-2}=-3-1+4=0. \] Therefore \(h(-1)=f(-1)+g(-1)=0+q=q\), so \(C\) also lies on the graph of \(h(x)\). \(\;\boxed{\text{Yes.}}\) R1 A1

(Equivalent reasoning: \(x=-1\) is the \(x\)-intercept of \(f\), so adding \(f\) does not change the \(y\)-value on \(g\) at \(x=-1\).)
[Total: 16 marks]
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