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2019
P1 Q7
Topical
Tangents
19 P1 Q7
2019 H2 Mathematics Paper 1 Question 7
Differentiation I: Tangents and Normals, Parametric Curves
Answers
(i)
Equation of tangent at
x
=
1
:
{x=1: \; }
x
=
1
:
y
=
e
−
1
{y=\mathrm{e}^{-1}}
y
=
e
−
1
Equation of tangent at
x
=
−
1
:
{x=-1: \; }
x
=
−
1
:
y
=
2
e
x
+
e
{y=2\mathrm{e}x+\mathrm{e}}
y
=
2
e
x
+
e
(ii)
79.6
∘
{79.6^\circ}
79.
6
∘
Full solutions
(i)
d
y
d
x
=
e
−
x
−
x
e
−
x
=
(
1
−
x
)
e
−
x
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \mathrm{e}^{- x}-x\mathrm{e}^{- x} \\ &= (1-x) \mathrm{e}^{- x} \end{align*}
d
x
d
y
=
e
−
x
−
x
e
−
x
=
(
1
−
x
)
e
−
x
When
x
=
1
,
{x=1,}
x
=
1
,
y
=
e
−
1
d
y
d
x
=
0
\begin{align*} y &= \mathrm{e}^{-1} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 0 \end{align*}
y
d
x
d
y
=
e
−
1
=
0
Equation of tangent at
x
=
1
{x=1}
x
=
1
is:
y
=
e
−
1
■
y= \mathrm{e}^{-1} \; \blacksquare
y
=
e
−
1
■
When
x
=
−
1
,
{x=-1,}
x
=
−
1
,
y
=
−
e
d
y
d
x
=
2
e
\begin{align*} y &= - \mathrm{e} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 2\mathrm{e} \end{align*}
y
d
x
d
y
=
−
e
=
2
e
Equation of tangent at
x
−
=
1
{x-=1}
x
−
=
1
is:
y
−
(
−
e
)
=
2
e
(
x
−
(
−
1
)
)
y
+
e
=
2
e
x
+
2
e
y
=
2
e
x
+
e
■
\begin{gather*} y - (-\mathrm{e}) = 2 \mathrm{e} \Big(x-(-1) \Big) \\ y + \mathrm{e} = 2\mathrm{e}x + 2\mathrm{e} \\ y = 2\mathrm{e}x + \mathrm{e} \; \blacksquare \end{gather*}
y
−
(
−
e
)
=
2
e
(
x
−
(
−
1
)
)
y
+
e
=
2
e
x
+
2
e
y
=
2
e
x
+
e
■
(ii)
Angle between tangents
=
tan
−
1
2
e
=
79.6
∘
(1 dp)
■
\begin{align*} & \textrm{Angle between tangents} \\ & = \tan^{-1} 2 \mathrm{e} \\ & = 79.6^{\circ} \textrm{ (1 dp)} \; \blacksquare \end{align*}
Angle between tangents
=
tan
−
1
2
e
=
79.
6
∘
(1 dp)
■
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