Bab 2 Persamaan Kuadratik

Posted on May 15, 2016 by user

2.5 Persamaan Kuadratik, SPM Praktis (Soalan Pendek)

Soalan 1:

Diberi persamaan kuadratik

mx ² + (3 – 2m)x+ m – 5 = 0.

Cari nilai m atau julat nilai m bagi setiap kes yang berikut.

(i) Jika persamaan kuadratik mempunyai dua punca yang nyata dan sama.

(ii) Jika persamaan kuadratik tidak mempunyai punca yang nyata.

Penyelesaian:

(i)

mx ² + (3 – 2m)x+ m – 5 = 0

a = m, b = 3 – 2m, dan c = m – 5

Bagi dua punca yang nyata dan sama,

b²– 4ac = 0

(3 – 2m)² – 4m (m – 5) = 0

9 – 12m + 4m²– 4m² + 20m = 0

8m= –9

m = - \frac{9}{8}

(ii)

Bagi dua punca nyata yang tidak wujud

b²– 4ac < 0

(3 – 2m)² – 4m (m – 5) < 0

9 – 12m + 4m²– 4m² + 20m < 0

8m + 9 < 0

m < - \frac{9}{8}

Soalan 2:

Cari nilai m jika garis lurus y = 5x – m ialah satu tangen kepada lengkung y = x²+ 2x + 1.

Penyelesaian:

Diberi

y = 5x – m -------- (1)

y = x² + 2x + 1 --- (2)

Gantikan (1) ke dalam (2)

5x – m = x² + 2x + 1

x ² – 3x + 1 + m = 0 ----- (3)

a = 1, b = –3, dan c = 1 + m

Tangen kepada lengkung mempunyai satu punca, iaitu

b²– 4ac = 0

(–3)²– 4(1) (1 + m) = 0

9 – 4 – 4m = 0

5 – 4m = 0

4m = 5

m = \frac{5}{4}

Bab 1 Fungsi

Posted on May 15, 2016 by user

1.5 Fungsi, SPM Praktis (Soalan Pendek)

Soalan 10:

Jika

g : x \to \frac{m - x}{x - 3}, x \neq 3

dan g ⁻ ¹ (5) = 14. Cari nilai m.

Penyelesaian:

Soalan 11 (Kaedah Bandingan):

Jika f : x \to \frac{m x - n}{x - 2}, x \neq 2 dan f^{- 1} : x \to \frac{5 - 2 x}{2 - x}, x \neq 2.

Cari nilai m dan n.

Penyelesaian:

\begin{array}{l} f (x) = \frac{m x - n}{x - 2} \\ Katakan y = \frac{m x - n}{x - 2} \to cari f^{- 1} (x) \\ y (x - 2) = m x - n \\ x y - 2 y = m x - n \\ x y - m x = 2 y - n \to \begin{array}{l} pindah x ke \\ sebelah kiri \end{array} \end{array}

\begin{array}{l} x (y - m) = 2 y - n \\ x = \frac{2 y - n}{y - m} \\ f^{- 1} (x) = \frac{2 x - n}{x - m} \\ \frac{2 x - n}{x - m} = \frac{5 - 2 x}{2 - x} \to \begin{array}{l} bandingkan dengan \\ f^{- 1} (x) yang diberi \end{array} \\ \times \frac{- 1}{- 1} \frac{n - 2 x}{m - x} = \frac{5 - 2 x}{2 - x} \\ maka, n = 5, m = 2 \end{array}

Soalan 12 (Kaedah Bandingan):

Diberi bahawa

f : x \to \frac{2 h}{x - 3 k}, x \neq 3 k,

dengan keadaan h dan k ialah pemalar dan

f^{- 1} : x \to \frac{14 + 24 x}{x}, x \neq 0.

Cari nilai h dan k .

Penyelesaian:

\begin{array}{l} f (x) = \frac{2 h}{x - 3 k} \\ Katakan y = \frac{2 h}{x - 3 k} \to cari f^{- 1} (x) \\ y (x - 3 k) = 2 h \\ x y - 3 k y = 2 h \\ x y = 2 h + 3 k y \\ x = \frac{2 h + 3 k y}{y} \end{array}

\begin{array}{l} f^{- 1} (x) = \frac{2 h + 3 k x}{x} \\ \frac{2 h + 3 k x}{x} = \frac{14 + 24 x}{x} \to \begin{array}{l} bandingkan dengan \\ f^{- 1} (x) yang diberi \end{array} \\ 2 h = 14 3 k = 24 \\ h = 7 k = 8 \end{array}

Bab 1 Fungsi

Posted on May 14, 2016 by user

1.5 Fungsi, SPM Praktis (Soalan Pendek)

Soalan 8:

Cari fungsi songsangan

f (x) = \frac{3 x + 2}{5 x + 3}

Penyelesaian:

\begin{array}{l} f (x) = \frac{3 x + 2}{5 x + 3} \\ katakan y = \frac{3 x + 2}{5 x + 3} \\ y (5 x + 3) = 3 x + 2 \\ 5 x y + 3 y = 3 x + 2 \\ 5 x y - 3 x = 2 - 3 y \\ x (5 y - 3) = 2 - 3 y \\ x = \frac{2 - 3 y}{5 y - 3} \\ f^{- 1} (x) = \frac{2 - 3 x}{5 x - 3} \to \begin{array}{l} tukar y kepada x \\ untuk cari f^{- 1} (x) \end{array} \end{array}

Soalan 9:

(a) Jika f : x → x – 2, cari f ^-1 (5),

(b) Jika f : x \to \frac{x + 9}{x - 5}, x \neq 5, cari f^{- 1} (3) .

Penyelesaian:

(a)

f (x) = x– 2

Katakan y = f ^-1(5)

f (y) = 5

y – 2 = 5

y = 7

oleh itu, f ^-1(5) = 7

(b)

\begin{array}{l} f (x) = \frac{x + 9}{x - 5} \\ katakan y = f^{- 1} (3) \\ f (y) = 3 \\ \frac{y + 9}{y - 5} = 3 \\ y + 9 = 3 y - 15 \\ 2 y = 24 \\ y = 12 \\ ∴ f^{- 1} (3) = 7 \end{array}

Bab 1 Fungsi

Posted on May 14, 2016 by user

1.5 Fungsi, SPM Praktis (Soalan Pendek)

Soalan 5:

Jika f : x → 2x + 1 dan

g : x \to \frac{5}{x}, x \neq 0

. Cari fungsi gubahan gf, fg dan nilai gf (4).

Penyelesaian:

Soalan 6:

Suatu fungsi f ditakrifkan oleh f : x → 2x + 1.
Cari fungsi gjika

f g : x \to \frac{5 x - 2}{x + 5}, x \neq - 5.

Penyelesaian:

[ Perhatian: Fungsi pertama diberi]

Soalan 7:

Suatu fungsi f ditakrifkan oleh

f : x \to \frac{7}{x} .

Cari fungsi g jika

g f : x \to \frac{10}{2 x + 3}, x \neq - \frac{3}{2} .

Penyelesaian:

[Perhatian: Fungsi kedua diberi, guna kaedah penggantian]

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}