2012 functions tutorial solutions barely passed

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.

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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

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 =

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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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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 ,).

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1 0 1

DfRg

2012 functions tutorial solutions barely passed

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