Bab 1 Fungsi

Posted on May 14, 2016 by user

1.6 Fungsi, SPM Praktis (Soalan Panjang)

Soalan 1:

Fungsi f dan g ditakrifkan sebagai f : x → x– 1 dan

g : x \to \frac{3 - x}{x + 4}

. Cari
(a) nilai gf(3),
(b) nilai fg(-1 ),
(c) fungsi gubahanfg,
(d) fungsi gubahangf,
(e) fungsi gubahang ² ,
(f) fungsi gubahanf ² .

Penyelesaian:

Soalan 2:

Diberi f : x → hx + k dan f² : x → 4x + 15.

(a) Cari nilai hdan k.

(b) Ambil nilai h> 0, cari nilai-nilai x di mana f (x²) = 7x

Penyelesaian:

(a)

Langkah 1: cari f² (x)

Diberi f (x) = hx + k

f² (x) = ff (x) = f (hx + k)

= h (hx + k) + k

= h²x + hk + k

Langkah 2: bandingkan dengan f²(x) yang diberi

f² (x) = 4x + 15

h²x + hk+ k = 4x + 15

h²= 4

h = ± 2

Apabila, h = 2

hk + k = 15

2k + k = 15

k = 5

Apabila, h = –2

hk + k = 15

–2k + k = 15

k = –15

(b)

h > 0, h = 2, k = 5

Diberi f ( x) = hx + k

f (x) = 2x + 5

f (x²) = 7x

2 (x²) + 5 = 7x

2x² – 7x+ 5 = 0

(2x – 5)(x–1 ) = 0

2x – 5 = 0 atau x –1= 0

x = 5/2 x = 1

Bab 1 Fungsi

Posted on May 14, 2016 by user

1.5 Fungsi, SPM Praktis (Soalan Pendek)

Soalan 1:

Diberi bahawa fungsi f : x → 6x + 1. Cari nilai p jika f (4) = 4p + 5.

Penyelesaian:

f : x → 6x+ 1

f (x) = 6x + 1

f (4) = 6(4) + 1

f (4) = 25

f (4) = 4p + 5

25 = 4p + 5

4p = 25 – 5 = 20

p = 20/4 = 5

Soalan 2:

Diberi

g : x \to \frac{3 x - 5}{2 x + 7}

Fungsi g ditakrifkan untuk semua nilai x kecuali x = a. Cari niali a.

Penyelesaian:

Diingatkan bahawa g ( x) tidak tertakrif jika penyebut = 0 iaitu [2x + 7 = 0]

2x + 7 = 0

2x = –7

x = - \frac{7}{2}

Apabila

x = - \frac{7}{2}

, g (x) tidak tertakrif

atau g (x) ditakrifkan untuk semua nilai x kecuali

x = - \frac{7}{2}, maka a = - \frac{7}{2}

Soalan 3:

Diberi bahawa fungsi f : x → 3x+ 2. Cari nilai

(a) f (2)

(b) f (– 5)

Penyelesaian:

Soalan 4:

Jika f : x → x² + 3x+ 2, ungkapkan setiap yang berikut dalam sebutan x:

(a) f (2x)

(b) f (3x+ 1)

Penyelesaian:

Bab 5 Indeks dan Logaritma

Posted on May 14, 2016 by user

5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)

Soalan 12

Selesaikan persamaan,

\log_{2} 5 \sqrt{x} + \log_{4} 16 x = 6

Penyelesaian:

\begin{array}{l} \log_{2} 5 \sqrt{x} + \log_{4} 16 x = 6 \\ \log_{2} 5 \sqrt{x} + \frac{\log_{2} 16 x}{\log_{2} 4} = 6 \\ \log_{2} 5 \sqrt{x} + \frac{\log_{2} 16 x}{2} = 6 \\ 2 \log_{2} 5 \sqrt{x} + \log_{2} 16 x = 12 \\ \log_{2} {(5 \sqrt{x})}^{2} + \log_{2} 16 x = 12 \\ \log_{2} (25 x) + \log_{2} 16 x = 12 \\ \log_{2} (25 x) (16 x) = 12 \\ \log_{2} 400 x^{2} = 12 \\ 400 x^{2} = 2^{12} \\ x^{2} = 10.24 \\ x = 3.2 \end{array}

Soalan 13

Diberi bahawa 2 log₂ (x – y) = 3 + log₂ x + log₂y. Buktikan x² + y²– 10xy = 0.

Penyelesaian:

2 log₂ (x – y) = 3 + log₂x + log₂ y

log₂ (x– y)² = log₂ 8 + log₂ x + log₂y

log₂ (x– y)² = log₂ 8xy

(x – y)² = 8xy

x² – 2xy + y² = 8xy

x² + y² – 10xy = 0 (terbukti)

Soalan 14

Diberi bahawa 2 log₂ (x + y) = 3 + log₂ x + log₂y. Buktikan x² + y²= 6xy.

Penyelesaian:

2 log₂ (x + y) = 3 + log₂x + log₂ y

log₂ (x+ y)² = log₂ 8 + log₂ x + log₂y

log₂ (x+ y)² = log₂ 8xy

(x + y)² = 8xy

x² + 2xy + y² = 8xy

x² + y² = 6xy (terbukti)

Bab 5 Indeks dan Logaritma

Posted on May 14, 2016 by user

5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)

Soalan 9

Selesaikan persamaan,

\log_{2} 4 x = 1 - \log_{4} x

Penyelesaian:

\begin{array}{l} \log_{2} 4 x = 1 - \log_{4} x \\ \log_{2} 4 x = 1 - \frac{\log_{2} x}{\log_{2} 4} \\ \log_{2} 4 x = 1 - \frac{\log_{2} x}{2} \\ 2 \log_{2} 4 x = 2 - \log_{2} x \\ \log_{2} 16 x^{2} = \log_{2} 4 - \log_{2} x \\ \log_{2} 16 x^{2} = \log_{2} \frac{4}{x} \\ 16 x^{2} = \frac{4}{x} \\ x^{3} = \frac{4}{16} = \frac{1}{4} \\ x = {(\frac{1}{4})}^{\frac{1}{3}} = 0.62996 \end{array}

Soalan 10

Selesaikan persamaan,

\log_{4} x = 25 \log_{x} 4

Penyelesaian:

\begin{array}{l} \log_{4} x = 25 \log_{x} 4 \\ \frac{1}{\log_{x} 4} = 25 \log_{x} 4 \\ \frac{1}{25} = {(\log_{x} 4)}^{2} \\ \log_{x} 4 = \pm \frac{1}{5} \\ \log_{x} 4 = \frac{1}{5} or \log_{x} 4 = - \frac{1}{5} \\ 4 = x^{\frac{1}{5}} 4 = x^{- \frac{1}{5}} \\ x = 4^{5} 4 = \frac{1}{x^{\frac{1}{5}}} \\ x = 1024 x^{\frac{1}{5}} = \frac{1}{4} \\ x = \frac{1}{1024} \end{array}

Soalan 11

Selesaikan persamaan,

2 \log_{x} 5 + \log_{5} x = \lg 1000

Penyelesaian:

\begin{array}{l} 2 \log_{x} 5 + \log_{5} x = \lg 1000 \\ 2. \frac{1}{\log_{5} x} + \log_{5} x = 3 \\ \times (\log_{5} x) \to 2 + {(\log_{5} x)}^{2} = 3 \log_{5} x \\ {(\log_{5} x)}^{2} - 3 \log_{5} x + 2 = 0 \\ (\log_{5} x - 2) (\log_{5} x - 1) = 0 \\ \log_{5} x = 2 or \log_{5} x = 1 \\ x = 5^{2} x = 5 \\ x = 25 \end{array}

Bab 5 Indeks dan Logaritma

Posted on May 13, 2016 by user

5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)

Soalan 5:

Selesaikan persamaan,

\log_{9} (x - 2) = \log_{3} 2

Penyelesaian:

\begin{array}{l} \log_{9} (x - 2) = \log_{3} 2 \\ \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \Rightarrow \frac{\log_{3} (x - 2)}{\log_{3} 9} = \log_{3} 2 \\ \frac{\log_{3} (x - 2)}{2} = \log_{3} 2 \\ \log_{3} (x - 2) = 2 \log_{3} 2 \\ \log_{3} (x - 2) = \log_{3} 2^{2} \\ x - 2 = 4 \\ x = 6 \end{array}

Soalan 6:

Selesaikan persamaan,

\log_{9} (2 x + 12) = \log_{3} (x + 2)

Penyelesaian:

\begin{array}{l} \log_{9} (2 x + 12) = \log_{3} (x + 2) \\ \frac{\log_{3} (2 x + 12)}{\log_{3} 9} = \log_{3} (x + 2) \\ \log_{3} (2 x + 12) = 2 \log_{3} (x + 2) \\ \log_{3} (2 x + 12) = \log_{3} {(x + 2)}^{2} \\ 2 x + 12 = x^{2} + 4 x + 4 \\ x^{2} + 2 x - 8 = 0 \\ (x + 4) (x - 2) = 0 \\ x = - 4 (tidak diterima) \\ x = 2 \end{array}

Soalan 7:

Selesaikan persamaan,

\log_{4} x = \frac{3}{2} \log_{2} 3

Penyelesaian:

\begin{array}{l} \log_{4} x = \frac{3}{2} \log_{2} 3 \\ \frac{\log_{2} x}{\log_{2} 4} = \frac{3}{2} \log_{2} 3 \\ \frac{\log_{2} x}{2} = \frac{3}{2} \log_{2} 3 \\ \log_{2} x = 2 \times \frac{3}{2} \log_{2} 3 \\ \log_{2} x = 3 \log_{2} 3 \\ \log_{2} x = \log_{2} 3^{3} \\ x = 27 \end{array}

Soalan 8:

Selesaikan persamaan,

\frac{2}{\log_{5} 2} = \log_{2} (2 - x)

Penyelesaian:

\begin{array}{l} \frac{2}{\log_{5} 2} = \log_{2} (2 - x) \\ 2 = \log_{5} 2. \log_{2} (2 - x) \\ 2 = \frac{1}{\log_{2} 5} . \log_{2} (2 - x) \\ 2 \log_{2} 5 = \log_{2} (2 - x) \\ \log_{2} 5^{2} = \log_{2} (2 - x) \\ 25 = 2 - x \\ x = - 23 \end{array}

Bab 5 Indeks dan Logaritma

Posted on May 13, 2016 by user

5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)

Soalan 1:

Selesaikan persamaan, log₃ [log₂(2x – 1)] = 2

Penyelesaian:

log₃ [log₂ (2x – 1)] = 2 ← (jika log _aN = x, N = a^x)

log₂ (2x – 1) = 3²

log₂ (2x – 1) = 9

2x – 1 = 2⁹

x = 256.5

Soalan 2

Selesaikan persamaan,

l o g_{16} [l o g_{2} (5 x - 4)] = l o g_{9} \sqrt{3}

Penyelesaian:

\begin{array}{l} l o g_{16} [l o g_{2} (5 x - 4)] = l o g_{9} \sqrt{3} \\ l o g_{16} [l o g_{2} (5 x - 4)] = \frac{1}{4} \leftarrow \begin{array}{l} \log_{9} \sqrt{3} = \log_{9} 3^{\frac{1}{2}} = \frac{1}{2} \log_{9} 3 \\ = \frac{1}{2} (\frac{1}{\log_{3} 9}) = \frac{1}{2} (\frac{1}{2}) = \frac{1}{4} \end{array} \\ l o g_{2} (5 x - 4) = 16^{\frac{1}{4}} \\ l o g_{2} (5 x - 4) = 2 \\ 5 x - 4 = 2^{2} \\ 5 x = 8 \\ x = \frac{8}{5} \end{array}

Soalan 3

Selesaikan persamaan,

5^{\log_{4} x} = 125

Penyelesaian:

\begin{array}{l} 5^{\log_{4} x} = 125 \\ \log_{5} 5^{\log_{4} x} = \log_{5} 125 \leftarrow \begin{array}{l} ambil log asas 5 \\ di kedua-dua belah \end{array} \\ (\log_{4} x) (\log_{5} 5) = 3 \\ (\log_{4} x) (1) = 3 \\ x = 4^{3} = 64 \end{array}

Soalan 4

Selesaikan persamaan,

5^{\log_{5} (x + 1)} = 9

Penyelesaian:

\begin{array}{l} 5^{\log_{5} (x + 1)} = 9 \\ \log_{5} 5^{^{\log_{5} (x + 1)}} = \log_{5} 9 \\ \log_{5} (x + 1) . \log_{5} 5 = \log_{5} 9 \\ \log_{5} (x + 1) = \log_{5} 9 \\ x + 1 = 9 \\ x = 8 \end{array}

Bab 5 Indeks dan Logaritma

Posted on May 13, 2016 by user

5.3 Persamaan yang Melibatkan Indeks (Contoh Soalan)

Contoh 4 (Persamaan Indeks (Asas Sama) – Penggantian):

Selesaikan setiap persamaan yang berikut.

(a) 3^x ⁻ ¹ + 3^x = 12
(b) 2^x + 2^x ⁺ ³ = 72
(c) 4^x ⁺ ¹ + 2²^x = 20

Penyelesaian:

Bab 5 Indeks dan Logaritma

Posted on May 13, 2016 by user

5.3 Persamaan yang Melibatkan Indeks (Contoh Soalan)

Contoh 3 (Persamaan Indeks – Asas Sama):

Selesaikan setiap persamaan yang berikut.

\begin{array}{l} (a) 27 (81^{3 x}) = 1 \\ (b) \sqrt{81^{n + 2}} = \frac{1}{3^{n} 27^{n - 1}} \\ (c) 8^{x - 1} = \frac{4}{\sqrt{2^{x + 3}}} \end{array}

Penyelesaian:

Bab 5 Indeks dan Logaritma

Posted on May 13, 2016 by user

5.3 Persamaan yang Melibatkan Indeks

Kaedah:
1. Perbandingan indeks dan asas

a. Jika asas adalah sama, apabila a^x = a^y, maka x = y

b. Jika indeks adalah sama, apabila a^x = b^x, maka a = b

2. Mengambil logaritma biasa (Jika asas dan indeks TIDAK sama)

\begin{array}{l} a^{x} = b \Rightarrow \lg a^{x} = \lg b \\ x = \frac{\lg b}{\lg a} \end{array}

Contoh 1 (Persamaan Indeks – Asas Sama):

Selesaikan setiap persamaan yang berikut.

\begin{array}{l} (a) 16^{x} = 8 \\ {(b) 9}^{x} {.3}^{x - 1} = 81 \\ (c) 5^{n + 1} = \frac{1}{125^{n - 1}} \end{array}

Penyelesaian:

Contoh 2 (Selesaikan Persamaan Indeks Serentak – Asas Sama):

Selesaikan persamaan serentak yang berikut.

\begin{array}{l} 2^{x} {.4}^{2 y} = 8 \\ \frac{3^{x}}{9^{y}} = \frac{1}{27} \end{array}

Penyelesaian:

Bab 5 Indeks dan Logaritma

Posted on May 13, 2016 by user

5.2b Penukaran Asas Logaritma (Contoh Soalan)

Contoh 5:

Diberi log₂3 = 1.585 dan log₂5 = 2.322, ungkapkan setiap yang berikut.

(a) log₈ 15

(b) log₅ 0.6

(d) log₁₆ 45

Penyelesaian: