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SNBT Zone : Invers Fungsi

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Idham Muqoddas
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Berikut ini adalah kumpulan soal mengenai Invers Fungsi tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

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STANDARSNBTHOTS


No. 1

Jika \({f(x+2)=\dfrac{2x-5}{3x+10}}\) maka nilai x yang memenuhi (f−1 ∘ f−1)(2x + 1) = 1 adalah....
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ALTERNATIF PENYELESAIAN
\(\eqalign{ f(x+2)&=\dfrac{2x-5}{3x+10}\\[4pt] f(x)&=\dfrac{2(x-2)-5}{3(x-2)+10}\\[4pt] &=\dfrac{2x-4-5}{3x-6+10}\\[4pt] &=\dfrac{2x-9}{3x+4}\\[4pt] f^{-1}&=\dfrac{-4x-9}{3x-2} }\)
\(\eqalign{ \left(f^{-1}\circ f^{-1}\right)(2x+1)&=1\\[4pt] f^{-1}\left(f^{-1}(2x+1)\right)&=1\\[4pt] \dfrac{-4(2x+1)-9}{3(2x+1)-2}&=f(1)\\[4pt] \dfrac{-8x-4-9}{6x+3-2}&=\dfrac{2(1)-9}{3(1)+4}\\[4pt] \dfrac{-8x-13}{6x+1}&=-1\\[4pt] -8x-13&=-6x-1\\ x&=-6 }\)
Jadi, nilai x yang memenuhi (f−1 ∘ f−1)(2x + 1) = 1 adalah −6.

No. 2

Jika grafik \(y=\dfrac{2x+1}{ax-3}\) dan inversnya berpotongan di titik (x0, −1), maka nilai x0 + a adalah ....
  1. 2
  2. −4
  3. −5
  1. 4
  2. 5
Ganesha Operation
ALTERNATIF PENYELESAIAN
\(y^{-1}=\dfrac{3x+1}{ax-2}\)

\(\begin{aligned} \dfrac{2x_0+1}{ax_0-3}&=-1\\ 2x_0+1&=-ax_0+3\\ ax_0&=-2x_0+2 \end{aligned}\)

\(\begin{aligned} \dfrac{3x_0+1}{ax_0-2}&=-1\\ 3x_0+1&=-ax_0+2\\ ax_0&=-3x_0+1\\ -2x_0+2&=-3x_0+1\\ x_0&=-1 \end{aligned}\)

\(\begin{aligned} ax_0&=-2x_0+2\\ a(-1)&=-2(-1)+2\\ -a&=4\\ a&=-4 \end{aligned}\)

\(\begin{aligned} x_0+a&=-1+(-4)\\ &=\boxed{\boxed{-5}} \end{aligned}\)
Jadi, nilai x0 + a adalah −5.
JAWAB: C

No. 3

Jika \((f\circ g)(x)=\dfrac{6x+3}{2x-5}\) dan g(x) = 4x − 11, maka hasil dari \(\displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx\) adalah
  1. 72 ln 2 − 3
  2. 36 ln 3 − 2
  3. 36 ln 2 − 6
  1. 36 ln 2 − 3
  2. 72 ln 3 − 2
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\(\begin{aligned} (f\circ g)(x)&=\dfrac{6x+3}{2x-5}\\[10pt] f(g(x))&=\dfrac{12x+6}{4x-10}\\[10pt] f(4x-11)&=\dfrac{3(4x-11)+39}{4x-11+1}\\[10pt] f(x)&=\dfrac{3x+39}{x+1}\\[10pt] f^{-1}(x)&=\dfrac{-x+39}{x-3}\\[10pt] f^{-1}(x-1)&=\dfrac{-(x-1)+39}{x-1-3}\\[10pt] &=\dfrac{-x+1+39}{x-4}\\[10pt] &=\dfrac{-x+40}{x-4} \end{aligned}\)

\(\begin{aligned} \displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx&=\displaystyle\intop_5^8\dfrac{\dfrac{-x+40}{x-4}}{4(3)-11}\ dx\\ &=\displaystyle\intop_5^8\dfrac{\dfrac{-x+4+36}{x-4}}{12-11}\ dx\\ &=\displaystyle\intop_5^8\dfrac{-1+\dfrac{36}{x-4}}1\ dx\\ &=\displaystyle\intop_5^8\left(-1+\dfrac{36}{x-4}\right)\ dx\\ &=\left[-x+36\ln|x-4|\right]_5^8\\ &=\left[-8+36\ln|8-4|\right]-\left[-5+36\ln|5-4|\right]\\ &=\left[-8+36\ln4\right]-\left[-5+36\ln1\right]\\ &=\left[-8+36\ln2^2\right]-\left[-5-42(0)\right]\\ &=\left[-8+72\ln2\right]-\left[-5\right]\\ &=-8+72\ln2+5\\ &=\boxed{\boxed{72\ln2-3}} \end{aligned}\)
Jadi, \(\displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx=72\ln2-3\).
JAWAB: A

No. 4

Jika fungsi f dan g mempunyai invers dan memenuhi f (3x) = g(x − 2), maka f −1(x) =
  1. 3g−1(x) + 12
  2. 3g−1(x) + 10
  3. 3g−1(x) + 8
  1. 3g−1(x) + 6
  2. 3g−1(x) + 4
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
Misal f (3x) =g(x − 2) = k

\(\begin{aligned} g(x-2)&=k\\ x-2&=g^{-1}(k)\\ x&=g^{-1}(k)+2 \end{aligned}\)

\(\begin{aligned} f(3x)&=k\\ f^{-1}(k)&=3x\\ &=3\left(g^{-1}(k)+2\right)\\ &=3g^{-1}(k)+6\\ f^{-1}(x)&=\boxed{\boxed{3g^{-1}(x)+6}} \end{aligned}\)
Jadi, f −1(x) = 3g−1(x) + 6.
JAWAB: D

No. 5

Diberikan fungsi f : ℝ ⟶ ℝ dengan f (2log 2x) = 3x + 1 adalah invers dari fungsi f, maka f −1(7) =
  1. 2
  2. \(\dfrac12\)
  3. 1
  1. 4
  2. \(\dfrac14\)
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN

CARA 1

Misal 2log 2x = p
\(\begin{aligned} 2x&=2^p\\ x&=\dfrac{2^p}2\\ &=2^{p-1} \end{aligned}\)

\(\begin{aligned} f\left({^2\negmedspace\log}2x\right)&=3x+1\\ f(p)&=3\cdot2^{p-1}+1\\ f(x)&=3\cdot2^{x-1}+1\\ y&=3\cdot2^{x-1}+1\\ y-1&=3\cdot2^{x-1}\\ \dfrac{y-1}3&=2^{x-1}\\[8pt] {^2\negmedspace\log}\dfrac{y-1}3&=x-1\\[8pt] {^2\negmedspace\log}\dfrac{y-1}3+1&=x\\[8pt] x&={^2\negmedspace\log}\dfrac{y-1}3+1\\[8pt] f^{-1}(x)&={^2\negmedspace\log}\dfrac{x-1}3+1\\[8pt] f^{-1}(7)&={^2\negmedspace\log}\dfrac{7-1}3+1\\[8pt] &={^2\negmedspace\log}\dfrac63+1\\[8pt] &={^2\negmedspace\log}2+1\\[8pt] &=1+1\\ &=\boxed{\boxed{2}} \end{aligned}\)

CARA CEPAT

\(\begin{aligned} f\left({^2\negmedspace\log2x}\right)&=3x+1\\ f^{-1}(3x+1)&={^2\negmedspace\log2x} \end{aligned}\)

\(\begin{aligned} 3x+1&=7\\ 3x&=6\\ x&=2 \end{aligned}\)

\(\begin{aligned} f^{-1}(7)&={^2\negmedspace\log}2(2)\\ &={^2\negmedspace\log}4\\ &=\boxed{\boxed{2}} \end{aligned}\)
Jadi, f −1(7) = 2.
JAWAB: A

No. 6

Fungsi f : ℝ ⟶ ℝ dan g : ℝ ⟶ ℝ didefinisikan dengan f (x) = 23x − 1 dan g(x) = 4(x + 2)3. Jika f −1 adalah invers dari f maka (f −1 ∘ g)(x) =
  1. \({^2\negmedspace\log\sqrt[3]{2x}}\)
  2. 2log (2x)3
  3. 2log (2x + 4)
  1. 2log 2x
  2. 2log (2x + 2)
SKMP Agustus 2020
ALTERNATIF PENYELESAIAN
\(\eqalign{ f(x)& = 2^{3x-1}\\ 3x-1&={^2\negmedspace\log f(x)}\\ 3x&={^2\negmedspace\log f(x)}+1\\ x&=\dfrac{{^2\negmedspace\log f(x)}+1}3\\ f^{-1}(x)&=\dfrac{{^2\negmedspace\log x}+1}3 }\)

\(\eqalign{ \left(f^{-1}\circ g\right)(x)&=f^{-1} \left(g(x)\right)\\ &=f^{-1} \left(4(x + 2)^3\right)\\ &=\dfrac{{^2\negmedspace\log \left(4(x + 2)^3\right)}+1}3\\ &=\dfrac{{^2\negmedspace\log4}+ {^2\negmedspace\log(x + 2)^3}+1}3\\ &=\dfrac{2+ 3\ {^2\negmedspace\log(x + 2)}+1}3\\ &=\dfrac{3\ {^2\negmedspace\log(x + 2)}+3}3\\ &={^2\negmedspace\log(x + 2)}+1\\ &={^2\negmedspace\log(x + 2)}+{^2\negmedspace\log2}\\ &={^2\negmedspace\log(x + 2)2}\\ &=\boxed{\boxed{^2\negmedspace\log(2x + 4)}} }\)
Jadi, (f −1 ∘ g)(x) = 2log (2x + 4).
JAWAB: C

No. 7

Diketahui f (x) = ax + b dengan f −1(11) = 2 dan f −1(8) = 1 dengan f −1 menyatakan fungsi invers f.
Nilai \(\displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h=\) ....
  1. 4
  2. 5
  3. 6
  1. 7
  2. 8
Ganesha Operation
ALTERNATIF PENYELESAIAN
\(\begin{aligned} f^{-1}(11)&=2\\ f(2)&=11\\ 2a+b&=11 \end{aligned}\)

\(\begin{aligned} f^{-1}(8)&=1\\ f(1)&=8\\ a+b&=8 \end{aligned}\)

\(\begin{aligned} 2a+b&=11\\ a+b&=8\qquad-\\\hline a&=3 \end{aligned}\)

\(\begin{aligned} a+b&=8\\ 3+b&=8\\ b&=5 \end{aligned}\)

f (x) = 3x + 5

\(\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h&=\displaystyle\lim_{h\to0}\dfrac{(3+h)(3(3)+5)-3(3(3+h)+5)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{(3+h)(9+5)-3(9+3h+5)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{(3+h)(14)-3(14+3h)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{42+14h-42+9h}h\\ &=\displaystyle\lim_{h\to0}\dfrac{5h}h\\ &=\displaystyle\lim_{h\to0}5\\ &=\boxed{\boxed{5}} \end{aligned}\)
Jadi, \(\displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h=5\).
JAWAB: B

No. 8

Diketahui fungsi f (x) = x2 + 2x + 11, dengan x ≥ −1, maka nilai dari f −1(14) adalah
  1. 5
  2. 4
  3. 3
  1. 2
  2. 1
SKMP Juli 2020
ALTERNATIF PENYELESAIAN
Misal f −1(14) = x
\(\eqalign{ f(x)&=14\\ x^2 + 2x + 11&=14\\ x^2+2x-3&=0\\ (x+3)(x-1)&=0 }\)
x = −3 atau x = 1
Jadi, nilai dari f −1(14) adalah 1.
JAWAB: E


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