Standard +0.3 This is a standard C3 question testing routine techniques: solving exponential equations, differentiation to find turning points, integration (requiring substitution u = e^x - 2), and finding an inverse function with restricted domain. All parts follow predictable methods with no novel problem-solving required, making it slightly easier than the average A-level question.
Fig. 9 shows the curve \(y = f(x)\), where
$$f(x) = (e^x - 2)^2 - 1, x \in \mathbb{R}.$$
The curve crosses the x-axis at O and P, and has a turning point at Q.
\includegraphics{figure_9}
Find the exact x-coordinate of P. [2]
Show that the x-coordinate of Q is \(\ln 2\) and find its y-coordinate. [4]
Find the exact area of the region enclosed by the curve and the x-axis. [5]
The domain of f(x) is now restricted to \(x \geqslant \ln 2\).
Find the inverse function \(f^{-1}(x)\). Write down its domain and range, and sketch its graph on the copy of Fig. 9. [7]
one step is enough, i.e. y 1 = 2arctan x or x 1 = 2arctan y
need not have interchanged x and y at this stage
allow y = …
curves must cross on y = x line if present (or close enough to imply intention)
curves shouldn’t touch or cross in the third quadrant
4 gf(x) = e2ln x
= elnx2
Answer
Marks
= x2
M1
M1
A1
Answer
Marks
[3]
Forming gf(x)
(soi)
Doing fg: 2ln(ex) = 2x SC1
Allow x2 (but not 2x) unsupported
5 f(–x) = –f(x), g(–x) = g(x)
g f(–x) = g [–f (x)]
= g f (x)
Answer
Marks
g f is even
B1B1
M1
E1
Answer
Marks
[4]
condone f and g interchanged
forming gf(−x) or gf(x) and using
f(−x) = −f(x)
www
6 f(x) = 1 + 2 sin 3x = y x ↔ y
x = 1 + 2 sin 3y
sin 3y = (x – 1)/2
3y = arcsin [(x – 1)/2]
1 x−1 1 x−1
y= arcsin so f−1(x)= arcsin
3 2 3 2
Range of f is −1 to 3
Answer
Marks
–1 ≤ x ≤ 3
M1
A1
A1
A1
M1
A1
Answer
Marks
[6]
attempt to invert
must be y = … or f−1(x) = …
or –1 ≤ (x – 1)/2 ≤ 1
Answer
Marks
must be ‘x’, not y or f(x)
at least one step attempted, or reasonable attempt at flow chart
inversion
(or any other variable provided same used on each side)
condone <’s for M1
allow unsupported correct answers; −1 to 3 is M1 A0
1
7 Either y= ln(x−1) x ↔ y
2
1
⇒ x= ln(y−1)
2
⇒ 2x = ln (y – 1)
⇒ e2x = y – 1
⇒ 1 + e2x = y
⇒ g(x) = 1 + e2x
or gf(x) = g( ½ ln (x – 1))
= 1 + eln(x – 1)
= 1 + x – 1
Answer
Marks
= x
M1
M1
E1
M1
M1
E1
Answer
Marks
[3]
or y = e(x – 1)/2
attempt to invert and interchanging x with y o.e.
(at any stage)
eln y–1 = y – 1 or ln (ey) = y used
www
or fg(x) = … (correct way round)
eln(x – 1) = x – 1 or ln(e2x) = 2x
www
8
2π
–
1
Answer
Marks
1
M1
B1
A1
Answer
Marks
[3]
Can use degrees or radians
reasonable shape (condone extra range)
passes through (–1, 2π), (0, π) and (1, 0)
good sketches – look for curve reasonably
vertical at (–1, 2π) and (1, 0), negative gradient
at (0, π). Domain and range must be clearly
marked and correct.
Question 2:
2 | (i) | At P, (ex – 2)2 – 1 = 0
ex – 2 = [±]1,
ex = [1 or] 3 | M1 | square rooting – condone no ±
or (ex)2 – 4 ex + 3 = 0
(ex – 1)(ex – 3) = 0, ex = 1 or 3 | M1 | expanding to correct quadratic and solve by
factorising or using quadratic formula | condone e^x^2
x = [0 or] ln 3 | A1
[2] | x-coordinate of P is ln 3; must be exact | condone P = ln 3, but not y = ln 3
2 | (ii) | f (x) = 2(ex – 2)ex
= 0 when ex = 2, x = ln 2 * | M1
A1
A1 | chain rule
correct derivative
not from wrong working NB AG | e.g. 2 u × their deriv of ex
2(ex – 2)x is M0
or verified by substitution
or f(x) = e2x – 4ex + 3
f’(x) = 2e2x – 4ex | M1
A1 | expanding to 3 term quadratic with (ex)2 or e2x
correct derivative, not from wrong working | condone e^x^2
= 0 when 2e2x = 4ex, ex = 2, x = ln 2 * | A1 | or 2ex(ex – 2) = 0 ex = 2, x = ln 2
not from wrong working NB AG | or verified by substitution
y = f(ln(2)) = −1 | B1
[4]
2 | (iii) | ln3 ln3
[(ex 2)2 1]dx [(ex)2 4ex 41]dx
0 0
ln3
[e2x 4ex 3]dx
0
ln3
1
e2x 4ex 3x
2
0
= (4.5 – 12 + 3ln3) – (0.5 – 4)
= 3ln3 – 4 [so area = 4 – 3ln3] | M1
A1
B1
A1ft
A1
[5] | expanding brackets
must have 3 terms: (ex)2 – 4 is M0,
condone e^x^2
e2x – 4ex + 3 [dx] (condone no dx)
e2x = ½ e2x
[½ e2x – 4ex + 3x]
condone 3ln3 – 4 as final ans; mark final ans | or if u = ex, 3 [u2 4u41]/udu
1
= u – 4 + 3/u du
= [ ½ u2 – 4u + 3ln u]
2 | (iv) | y = (ex – 2)2 – 1 x y
x = (ey – 2)2 − 1
x + 1 = (ey – 2)2
±√(x + 1) = ey – 2 (+ for y ≥ ln 2)
2 + √(x + 1) = ey
y = ln(2 + √(x + 1)) = f−1(x)
Domain is x ≥ −1
Range is y ≥ ln 2
f(x)
f−1(x)
(−1,ln2)
(ln2,−1) | M1
A1
A1
B1
B1
M1
A1
[7] | attempt to solve for y (might be indicated by
expanding and then taking lns)
condone no ±
must have interchanged x and y in final ans
must be ≥ and x (not y)
or f−1(x) ≥ ln 2, must be ≥ (not x or f(x))
if x > −1 and y > ln 2 SCB1
recognisable attempt to reflect curve, or any
part of curve, in y = x
good shape, cross on y = x (if shown), correct
domain and range indicated.
[see extra sheet for examples] | or x if x and y not interchanged yet
or adding (or subtracting) 1
if not specified, assume first ans is
domain and second range
y = x shown indicative but not
essential
e.g. −1 and ln 2 marked on axes
3 (i) bounds + 1, + 1
– + 1 < f(x) < + 1
(ii) y = 2arctan x + 1 x y
x = 2arctan y + 1
x1
arctany
2
x1 x1
ytan( ) f1(x)tan( )
2 2
1
1 | B1B1
B1cao
[3]
M1
A1
A1
B1
B1
[5] | or … < y < … or ( + 1, + 1)
attempt to invert formula
y1
or arctanx
2
reasonable reflection in y = x
(1, 0) intercept indicated. | not … < x < …, not ‘between …’
one step is enough, i.e. y 1 = 2arctan x or x 1 = 2arctan y
need not have interchanged x and y at this stage
allow y = …
curves must cross on y = x line if present (or close enough to imply intention)
curves shouldn’t touch or cross in the third quadrant
4 gf(x) = e2ln x
= elnx2
= x2 | M1
M1
A1
[3] | Forming gf(x)
(soi) | Doing fg: 2ln(ex) = 2x SC1
Allow x2 (but not 2x) unsupported
5 f(–x) = –f(x), g(–x) = g(x)
g f(–x) = g [–f (x)]
= g f (x)
g f is even | B1B1
M1
E1
[4] | condone f and g interchanged
forming gf(−x) or gf(x) and using
f(−x) = −f(x)
www
6 f(x) = 1 + 2 sin 3x = y x ↔ y
x = 1 + 2 sin 3y
sin 3y = (x – 1)/2
3y = arcsin [(x – 1)/2]
1 x−1 1 x−1
y= arcsin so f−1(x)= arcsin
3 2 3 2
Range of f is −1 to 3
–1 ≤ x ≤ 3 | M1
A1
A1
A1
M1
A1
[6] | attempt to invert
must be y = … or f−1(x) = …
or –1 ≤ (x – 1)/2 ≤ 1
must be ‘x’, not y or f(x) | at least one step attempted, or reasonable attempt at flow chart
inversion
(or any other variable provided same used on each side)
condone <’s for M1
allow unsupported correct answers; −1 to 3 is M1 A0
1
7 Either y= ln(x−1) x ↔ y
2
1
⇒ x= ln(y−1)
2
⇒ 2x = ln (y – 1)
⇒ e2x = y – 1
⇒ 1 + e2x = y
⇒ g(x) = 1 + e2x
or gf(x) = g( ½ ln (x – 1))
= 1 + eln(x – 1)
= 1 + x – 1
= x | M1
M1
E1
M1
M1
E1
[3] | or y = e(x – 1)/2
attempt to invert and interchanging x with y o.e.
(at any stage)
eln y–1 = y – 1 or ln (ey) = y used
www
or fg(x) = … (correct way round)
eln(x – 1) = x – 1 or ln(e2x) = 2x
www
8
2π
–
1
1 | M1
B1
A1
[3] | Can use degrees or radians
reasonable shape (condone extra range)
passes through (–1, 2π), (0, π) and (1, 0)
good sketches – look for curve reasonably
vertical at (–1, 2π) and (1, 0), negative gradient
at (0, π). Domain and range must be clearly
marked and correct.
Fig. 9 shows the curve $y = f(x)$, where
$$f(x) = (e^x - 2)^2 - 1, x \in \mathbb{R}.$$
The curve crosses the x-axis at O and P, and has a turning point at Q.
\includegraphics{figure_9}
\begin{enumerate}[label=(\roman*)]
\item Find the exact x-coordinate of P. [2]
\item Show that the x-coordinate of Q is $\ln 2$ and find its y-coordinate. [4]
\item Find the exact area of the region enclosed by the curve and the x-axis. [5]
\end{enumerate}
The domain of f(x) is now restricted to $x \geqslant \ln 2$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item Find the inverse function $f^{-1}(x)$. Write down its domain and range, and sketch its graph on the copy of Fig. 9. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 Q2 [18]}}