Jika \(x = (p^{-\frac{1}{2}} \:-\: q^{-\frac{1}{2}})(p^{-1} + q^{-1} + 2(pq)^{-\frac{1}{2}})^{\frac{1}{2}}\) dan \(y = (p + q)^{-2}(p^{-1}\:-\:q^{-1})\) dengan \(p, q > 0\), \(p \neq q\), maka \(\frac{x}{y} = \dotso\)
Jawaban: C
Langkah 1: Menyederhanakan nilai \(x\)
Perhatikan bagian ini:
\(p^{-1} + q^{-1} + 2(pq)^{-\frac{1}{2}}\)
\(p^{-1} + 2p^{-\frac{1}{2}}q^{-\frac{1}{2}} + q^{-1}\)
Bentuk ini sama dengan \((p^{-\frac{1}{2}} + q^{-\frac{1}{2}})^2\)
\(x = (p^{-\frac{1}{2}} \:-\: q^{-\frac{1}{2}})(p^{-1} + q^{-1} + 2(pq)^{-\frac{1}{2}})^{\frac{1}{2}}\)
\(x = (p^{-\frac{1}{2}} \:-\: q^{-\frac{1}{2}})[(p^{-\frac{1}{2}} + q^{-\frac{1}{2}})^2]^{\frac{1}{2}}\)
\(x = (p^{-\frac{1}{2}} \:-\: q^{-\frac{1}{2}})(p^{-\frac{1}{2}} + q^{-\frac{1}{2}})\)
\(x = (p^{-\frac{1}{2}})^2 \:-\: (q^{-\frac{1}{2}})^2\)
\(x = p^{-1} \:-\: q^{-1}\)
Langkah 2: Penyelesaian
\(\dfrac{x}{y} = \dfrac{p^{-1} \:-\: q^{-1}}{(p + q)^{-2}(p^{-1}\:-\:q^{-1})}\)
\(\dfrac{x}{y} = \dfrac{\cancel{p^{-1} \:-\: q^{-1}}}{(p + q)^{-2}\cancel{(p^{-1}\:-\:q^{-1})}}\)
\(\dfrac{x}{y} = \dfrac{1}{(p + q)^{-2}}\)
\(\dfrac{x}{y} =(p + q)^2\)
Jawaban: C
Syarat fungsi turun: \(\color{blue} f'(x) < 0\)
\(f(x) = x^3 + 3x^2 + 5 < 0\)
\(f'(x) = 3x^2 + 6x < 0\)
\(3x(x + 2) < 0\)
Pembuat nol:
\(x_1 = 0\)
\(x + 2 = 0 \rightarrow x_2 = -2\)
Himpunan penyelesaian: \(\lbrace x| -2 < x < 0, \: x \in R \rbrace\)
Jika persamaan kuadrat 3x² + x − 3 = 0 mempunyai akar-akar persamaan α dan β, maka persamaan kuadrat yang akar-akarnya \(2 + \dfrac{1}{\alpha + 1}\) dan \(2 + \dfrac{1}{\beta + 1}\) adalah …
Jawaban: A
Langkah 1: Menentukan jumlah dan hasil kali akar persamaan kuadrat 3x² + x − 3 = 0
\(\color{blue}\alpha + \beta = -\dfrac{b}{a}\)
\(\alpha + \beta = -\dfrac{1}{3}\)
\(\color{blue}\alpha \times \beta = \dfrac{c}{a}\)
\(\alpha \times \beta = \dfrac{-3}{3} = -1\)
Langkah 2: Menentukan jumlah dan hasil kali akar persamaan yang baru
Jumlah Akar (JA):
\(2 + \dfrac{1}{\alpha + 1} + 2 + \dfrac{1}{\beta + 1}\)
\(\dfrac{2\alpha + 2 + 1}{\alpha + 1} + \dfrac{2\beta + 2 + 1}{\beta + 1}\)
\(\dfrac{2\alpha + 3}{\alpha + 1} + \dfrac{2\beta +3}{\beta + 1}\)
\(\dfrac{(2\alpha + 3)(\beta + 1) + (2\beta +3)(\alpha + 1)}{(\alpha + 1)(\beta + 1)}\)
\(\dfrac{2\alpha\cdot \beta + 2\alpha + 3 \beta + 3 + 2\alpha\cdot \beta + 2\beta + 3 \alpha + 3}{\alpha\cdot \beta + \alpha + \beta + 1}\)
\(\dfrac{4\alpha\cdot \beta + 5\alpha + 5\beta + 6}{\alpha\cdot \beta + \alpha + \beta + 1}\)
\(\dfrac{4\alpha\cdot \beta + 5(\alpha + \beta) + 6}{\alpha\cdot \beta + \alpha + \beta + 1}\)
Substitusikan nilai jumlah akar dan hasil kali akar pada langkah 1
\(\dfrac{4(-1) + 5(-\frac{1}{3}) + 6}{-1 -\frac{1}{3} + 1}\)
\(\dfrac{\frac{1}{3}}{-\frac{1}{3}}\)
\(\textbf{JA} = -1\)
Hasil Kali Akar (HA):
\(\left(2 + \dfrac{1}{\alpha + 1}\right)\left(2 + \dfrac{1}{\beta + 1}\right)\)
\(4 + \dfrac{2}{\beta + 1} + \dfrac{2}{\alpha + 1} + \dfrac{1}{\alpha + 1}\cdot \dfrac{1}{\beta + 1}\)
\(4 + \dfrac{2(\alpha + 1) + 2(\beta + 1)}{(\beta + 1)(\alpha + 1)} + \dfrac{1}{(\alpha + 1)(\beta + 1)}\)
\(4 + \dfrac{2(\alpha + \beta) + 4}{\alpha\cdot \beta + (\alpha + \beta) +1} + \dfrac{1}{\alpha\cdot \beta + (\alpha + \beta) +1 }\)
\(4 + \dfrac{2(-\frac{1}{3}) + 4}{-1+ (-\frac{1}{3}) +1} + \dfrac{1}{-1 + (-\frac{1}{3}) +1 }\)
\(4 + \dfrac{\frac{10}{3}}{-\frac{1}{3}} + \dfrac{1}{-\frac{1}{3}}\)
\(4 + (-10) + (-3)\)
\(\textbf{HA} = -9\)
Langkah 3: Menentukan persamaan kuadrat baru
\(\color{blue} x^2 \:-\:\textbf{JA}x + \textbf{HA} = 0\)
\(x^2 \:-\:(-1)x + (-9) = 0\)
\(x^2 + x \:-\:9 = 0\)
