IB Mathematics SL 5.4 Tangents and normal AI SL Paper 1- Exam Style Questions- New Syllabus
Question
▶️Answer/Explanation
(a)
Since \(L_2\) is perpendicular to \(L_1\) and has gradient \(-\tfrac{1}{2}\), we must have \[ m_{L_1}\cdot m_{L_2} = -1 \;\Rightarrow\; m_{L_1}\left(-\tfrac{1}{2}\right)=-1 \quad\Rightarrow\quad m_{L_1}=2. \] A1
[1 mark]
(b)
Using point–slope with \(m=2\) through \(P(3,-1)\): \[ y-(-1)=2\,(x-3)\;\Rightarrow\;y+1=2x-6\;\Rightarrow\;y=2x-7. \] Equation of \(L_1\): \(y=2x-7\). M1 A1
[2 marks]
(c)
A normal at \(P\) has gradient perpendicular to the tangent: \(m_{\text{normal}}=-\dfrac{1}{m_{L_1}}=-\dfrac{1}{2}\). Through \(P(3,-1)\) this normal is \[ y-(-1)=-\tfrac{1}{2}(x-3)\;\Rightarrow\;y=-\tfrac{1}{2}x+\tfrac{1}{2}. \] But \(L_2\) is \(y=-\tfrac{1}{2}x-\tfrac{5}{2}\). Although the gradients match, \(P(3,-1)\) is not on \(L_2\) since \[ y_{L_2}(3)=-\tfrac{1}{2}\cdot 3-\tfrac{5}{2}=-\tfrac{3}{2}-\tfrac{5}{2}=-4\neq -1. \] Therefore \(L_2\) is not the normal to \(f(x)\) at \(P\). R1 A1
[2 marks]
Question
▶️ Answer/Explanation
Differentiate \( f(x) = \frac{2}{x} + 3x^2 – 3 \):
\[ \begin{aligned} f(x) &= 2x^{-1} + 3x^2 – 3 \\ f'(x) &= 2 \cdot (-1)x^{-2} + 3 \cdot 2x – 0 \\ &= -2x^{-2} + 6x \\ &= -\frac{2}{x^2} + 6x \end{aligned} \] A1(M1)A1
[3 marks]
Find the gradient at \( x = 1 \):
\[ \begin{aligned} f'(x) &= -\frac{2}{x^2} + 6x \\ f'(1) &= -\frac{2}{1^2} + 6 \cdot 1 = -2 + 6 = 4 \end{aligned} \] A1
Find the perpendicular gradient:
\[ m_{\perp} = -\frac{1}{4} \] M1
Find the equation of the normal at \( (1, 2) \):
\[ \begin{aligned} y – 2 &= -\frac{1}{4}(x – 1) \\ y – 2 &= -\frac{1}{4}x + \frac{1}{4} \\ y &= -\frac{1}{4}x + \frac{1}{4} + 2 \\ y &= -\frac{1}{4}x + \frac{9}{4} \\ \frac{1}{4}x + y – \frac{9}{4} &= 0 \\ x + 4y – 9 &= 0 \end{aligned} \] M1, A1
Final equation: \( x + 4y – 9 = 0 \). [4 marks]