SKEMA
JAWAPAN BAHAGIAN 3
a)
katakan y
= x
+ 3
x
= y – 3
f -1 (x) = x – 3
b)
katakana y
= x2
x
=
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g -1 (x) =
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c)
fg ( x) = x + 3
= (x2
) + 3
= x2 + 3
gf ( x ) = x2
= (x + 3 ) 2
= x2 + 6x + 9
f 2 ( x ) = ff ( x
) = x + 3
= (x + 3 ) + 3
= x + 6
g 2 ( x ) = gg ( x
) = x2
= (x2) 2
= x 4
d)
![]() ![]()
=
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( fg ) -1 ( x ) → let y = x2 + 3
x2 = y – 3
x =
![]()
( fg ) -1 ( x
) =
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kesimpulan :
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e)
ff -1 ( x ) = x + 3
= ( x – 3 ) + 3
= x
gg -1 ( x + 1) =
![]()
= x + 1
So, ff -1 ( x ) ≠
gg -1 ( x + 1)
ff -1 ( k 2 + 2 ) = k 2
+ 2
Soalan Lanjutan
i)
![]()
g 2 ( x) = gg(x) =
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=
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g 3 ( x) =
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=
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g 4 ( x) =
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=
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g 5 ( x) =
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g 6 ( x) =
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g 30 ( x)=
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ii) g 2n ( x) =
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g 2n ( x) = x bila n adalah integer genap
g 2n+1( x) =
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g 2n+1 ( x) =
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b.
![]()
i)
![]() ![]() ![]()
ii)
![]() ![]() ![]()
iii)
![]() ![]() ![]()
Semak menggunakan kaedah lain
i)
![]() ![]() ![]() ![]()
ii)
![]() ![]() ![]() ![]()
iii)
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