2012 functions tutorial solutions barely passed
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7/29/2019 2012 Functions Tutorial Solutions Barely Passed
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CJC MATHEMATICS DEPARTMENT
2012 JC1 H2 MATHEMATICS
TOPIC: FUNCTIONS
QUESTIONS
1(a) 1,,12:f2
+ xxxxx
12)(f 2 += xxx 2)1( = x
From the graph, any horizontal line cuts the
graph ofy = f(x) at most once. Thus, the
function is a one one function.
(b) [ ]2,2,4:f 2 xxx
From the graph, the horizontal liney = 1 cuts the graph ofy = f(x) at 2 points.
Thus, the function is not a one to one function.
2(a) 0,,2)(f2
+= xxxxx
)2( += xx
),0[R f =
Since every horizontal line cuts through f(x) at
most once, therefore f is one-one and f1 exists.
11
1)1(
1)1(
2
2
2
2
+=+
+=+
+=
+=
yx
yx
xy
xxy
11 ++= yx
or11 += yx (N.A. 0xQ )
),0[RD ff 1- ==
0,,11)(f1
++=
xxxx
(b)121,),1cos()(h+=
xxxx
]1,0[R h =
Since every horizontal line cuts through h(x) at
most once, therefore h is one-one and h1
exists.
1cos
1cos
)1cos(
1
1
=
+=
+=
yx
xy
xy
]1,0[RD hh 1- ==
10,,1cos)(h11
=
xxxx
(c) 0,,1)(s += xxxx
x
),2[]2,(R s =
Since the horizontal liney = 3 cuts through the
graph more than once, therefore s is not one-oneand s
1does not exist.
____________________________________________
3. [1999/NJC//I/8]
(i) g :x ln(x + 1), x > 1
2 20
2
y= 1
y= f(x)
0
1
2
2
y= 3
1
11
0
x = 1
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1
),(R),1(D gg ==
h : x 22
1
++x , x < 2
)2,(R)2,(D hh ==
(ii) For gh to exist, gh DR
),1()2,(
Since gh DR , gh does not exist.
)2,1(RNew h =
122
1,1When =++
=x
y
32
1 =+x
3
7 =x
)3
7,(DNew h =
____________________________________________
4. (i) Write 3)12ln( ++= xy
Then, ( ).1e2
1e12 33 ==+ yy xx
Thus, ( ).1e2
1)(f 31 = xx
),(D 1-f =
= ,
2
1R, 1-f
(ii)
The coordinates where the curves intersect the axes
are (0,3),( )
0,1e
2
1 3, ( )
1e
2
1,0
3and
(3, 0).
(iii) When the two curves intersect, they also
intersect at the line y = x.
That is )(f)(f 1 xx = is equivalent to f(x) = x.
Thus, we have xx =++ 3)12ln(
From GC, the values ofx are 0.4847 and
5.482 correct to 4 SF.__________________________________________
5. [2009/CJC/II/4]
(i)
Since any horizontal line y = k, k R will cutthe graph of f at most once, hence f is one-one.
Thus, f-1
exists.
(ii) Letx
xy 1=
12 = xxy
4
4
2
0142
01
22
22
2
+=
=
=
yyx
yyx
xyx
42
1
2
2+= y
yx
42
1
2or4
2
1
2
22+=++= y
yxy
yx
(N.A. sincex > 0)
),(RD ff 1- ==
Hence, f-1
: x 42
1
2
2++ x
x, x (- , )
(iii)
(iv) For fg to exist, Rg DfDf = (0,) and Rg = [1 , 1].
Since Rg Df , therefore fg does not exist.
2.5
x = 2
y = 2
1 2
DgR
h
y = f(x)
y = f(x)
y = f1
(x)
1
y = x
y = x
y = f(x)
y = f-1
(x)
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0 2
1
1
(v)
New Rg = (0, 1]
Largest Dg = (0 ,)
fg(x) = f(sin x)
=x
xsin
1sin
Hence, fg : xx
xsin
1sin , x (0 ,).
____________________________________________
1 0 1
DfRg
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