Pembahasan Soal Matematika IPA SIMAK UI 2015

Salah satu soal Matematika IPA SIMAK UI 2015 adalah sebagai berikut.

Soal

Diketahui [latex]f(x) = \sin{ \left( \sin^3{ \left( \cos{x} \right) } \right)}[/latex]. Jika [latex]\frac{f”(0)}{\sin{(2)}} = A \cos{ \left( \sin^3{(1)} \right) } \sin{(1)}[/latex], maka [latex]A= \ldots[/latex]

A. [latex]\frac{3}{2}[/latex]

B. [latex]\frac{1}{2}[/latex]

C. 0

D. [latex]-\frac{1}{2}[/latex]

E. [latex]-\frac{3}{2}[/latex]

Jawab

Ingat kembali bahwa untuk menentukan turunan dari [latex]y = f(u)[/latex] terhadap variabel [latex]x[/latex], maka digunakan aturan rantai, yaitu [latex]y’ = f'(u) \cdot u'[/latex]. Biar mudah, anggap [latex]f(x) = \sin{u}[/latex], dengan [latex]u = \sin^3{t}[/latex], dan [latex]t = \cos{x}[/latex] Dengan demikian diperoleh, \begin{align} f(x) &= \sin{ \left( \sin^3{ \left( \cos{x} \right) } \right)}\\ f'(x) = \frac{df}{dx} &= \frac{df}{du} \cdot \frac{du}{dt} \cdot \frac{dt}{dx}\\ &= \cos{u} \cdot 3 \sin^2{t} \cos{t} \cdot (-\sin{x})\\ &= \cos{ \left( \sin^3{ \left( \cos{x} \right) } \right)} \cdot \left( 3 \sin^2{ \left( \cos{x} \right) } \cdot \cos{ (\cos{x})} \right) \cdot ( -\sin{x} )\\ &= -3 \cos{ \left( \sin^3{ \left( \cos{x} \right) } \right)} \cdot \sin^2{ \left( \cos{x} \right) } \cdot \cos{ (\cos{x})} \cdot \sin{x} \\ \end{align}

Sekarang, anggap [latex]f'(x) = a \cdot b \cdot c \cdot d[/latex], dengan [latex]a(x) = -3 \cos{ \left( \sin^3{ \left( \cos{x} \right) } \right)}[/latex], [latex]b(x) = \sin^2{ \left( \cos{x} \right) }[/latex], [latex]c(x) = \cos{ (\cos{x})}[/latex] dan [latex]d(x) = \sin{x}[/latex] Kemudian, \begin{align} f”(x) &= (abcd)’\\ &= (ab)’cd + ab(cd)’ \Rightarrow (u\cdot v)’ = u’ \cdot v + u \cdot v’\\ &= \left( a’b + ab’ \right) cd + ab \left( c’d + cd’ \right)\\ &= a’bcd + ab’cd + abc’d + abcd’ \\\\ f”(0) &= a'(0) \cdot b(0) \cdot c(0) \cdot d(0) + a(0) \cdot b'(0) \cdot c(0) \cdot d(0) +\\ &\qquad a(0) \cdot b(0) \cdot c'(0) \cdot d(0) + a(0) \cdot b(0) \cdot c (0) \cdot d'(0) \end{align}

Perhatikan bahwa jika [latex]x = 0[/latex], [latex]d(0) = \sin{0} = 0[/latex], sehingga [latex]f”(0) = a(0) \cdot b(0) \cdot c (0) \cdot d'(0)[/latex]

Kemudian \begin{align} a(0) &= -3 \cos{ \left( \sin^3{ \left( \cos{0} \right) } \right)}\\ &= -3 \cos{ \left( \sin^3{ (1) } \right)}\\ b(0) &= \sin^2{ \left( \cos{0} \right) }\\ &= \sin^2{ \left( 1 \right) }\\ c(0) &= \cos{ (\cos{0})}\\ &= \cos{(1)}\\ d(x) &= \sin{x}\\ d'(x) &= \cos{x}\\ &= \cos{0} = 1 \end{align}

Sehingga diperoleh \begin{align} f”(0) &= a(0) \cdot b(0) \cdot c (0) \cdot d'(0)\\ &= -3 \cos{ \left( \sin^3{ (1) } \right)} \cdot \sin^2{ \left( 1 \right) } \cdot \cos{(1)} \cdot 1\\ &= -3 \cos{ \left( \sin^3{ (1) } \right)} \sin^2{ \left( 1 \right) } \cos{(1)} \end{align}

Oleh karena itu, \begin{align} \frac{f”(0)}{\sin{(2)}} &= \frac{ -3 \cos{ \left( \sin^3{ (1) } \right)} \sin^2{ \left( 1 \right) } \cos{(1)} }{2 \sin{(1)} \cos{(1)} } & (\sin{2\alpha} = 2 \sin{\alpha} \cos{\alpha} )\\ &= \frac{ -3 \cos{ \left( \sin^3{ (1) } \right)} \sin{ \left( 1 \right) } \sin{ (1) } \cos{(1)}}{2 \sin{(1)} \cos{(1)}} \\ &= -\frac{3}{2} \cos{ \left( \sin^3{ (1) } \right)} \sin{ \left( 1 \right) } \end{align}

Jadi [latex]A = -\frac{3}{2}[/latex]

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