SBMPTN Zone : Invers Fungsi

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

Tipe :

Standar SBMPTN

No. 1

Jika {f(x+2)=\dfrac{2x-5}{3x+10}} maka nilai x yang memenuhi {\left(f^{-1}\circ f^{-1}\right)(2x+1)=1} adalah....

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\(\eqalign{ f(x+2)&=\dfrac{2x-5}{3x+10}\\[4pt] f(x)&=\dfrac{2(x-2)-5}{3(x-2)+10}\\[4pt] &=\dfrac{2x-4-5}{3x-6+10}\\[4pt] &=\dfrac{2x-9}{3x+4}\\[4pt] f^{-1}&=\dfrac{-4x-9}{3x-2} }\)

\(\eqalign{ \left(f^{-1}\circ f^{-1}\right)(2x+1)&=1\\[4pt] f^{-1}\left(f^{-1}(2x+1)\right)&=1\\[4pt] \dfrac{-4(2x+1)-9}{3(2x+1)-2}&=f(1)\\[4pt] \dfrac{-8x-4-9}{6x+3-2}&=\dfrac{2(1)-9}{3(1)+4}\\[4pt] \dfrac{-8x-13}{6x+1}&=-1\\[4pt] -8x-13&=-6x-1\\ x&=-6 }\)

Jadi, x=-6.

No. 2

Jika grafik y=\dfrac{2x+1}{ax-3} dan inversnya berpotongan di titik \left(x_0,-1\right), maka nilai x_0+a adalah ....

1. 2
2. -4
3. -5

4. 4
5. 5

Ganesha Operation

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 x_0+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\ln2-3}
2. {36\ln 3-2}
3. {36\ln 2-6}

4. {36\ln 2 - 3}
5. {72\ln 3-2}

\(\begin{aligned} (f\circ g)(x)&=\dfrac{6x+3}{2x-5}\\[10pt] f(g(x))&=\dfrac{12x+6}{4x-10}\\[10pt] f(4x-11)&=\dfrac{3(4x-11)+39}{4x-11+1}\\[10pt] f(x)&=\dfrac{3x+39}{x+1}\\[10pt] f^{-1}(x)&=\dfrac{-x+39}{x-3}\\[10pt] f^{-1}(x-1)&=\dfrac{-(x-1)+39}{x-1-3}\\[10pt] &=\dfrac{-x+1+39}{x-4}\\[10pt] &=\dfrac{-x+40}{x-4} \end{aligned}\)

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

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}

4. {3g^{-1}(x)+6}
5. {3g^{-1}(x)+4}

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,

No. 5

Diberikan fungsi {f:\mathbb{R}\to\mathbb{R}} dengan {f\left({^2\negmedspace\log2x}\right)=3x+1} adalah invers dari fungsi f, maka {f^{-1}(7)=}

1. 2
2. \dfrac12
3. 1

4. 4
5. \dfrac14

CARA 1

Misal {^2\negmedspace\log}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 \to R} dan {g : R \to R} didefinisikan dengan {f(x) = 2^{3x-1}} dan {g(x) = 4(x + 2)^3}. Jika f^{-1} adalah invers dari f maka \left(f^{-1}\circ g\right)(x) =

1. {^2\negmedspace\log\sqrt[3]{2x}}
2. {^2\negmedspace\log(2x)^3}
3. {^2\negmedspace\log(2x + 4)}

4. {^2\negmedspace\log 2x}
5. {^2\negmedspace\log(2x + 2)}

SKMP Agustus 2020

\(\eqalign{ f(x)& = 2^{3x-1}\\ 3x-1&={^2\negmedspace\log f(x)}\\ 3x&={^2\negmedspace\log f(x)}+1\\ x&=\dfrac{{^2\negmedspace\log f(x)}+1}3\\ f^{-1}(x)&=\dfrac{{^2\negmedspace\log x}+1}3 }\)

\(\eqalign{ \left(f^{-1}\circ g\right)(x)&=f^{-1} \left(g(x)\right)\\ &=f^{-1} \left(4(x + 2)^3\right)\\ &=\dfrac{{^2\negmedspace\log \left(4(x + 2)^3\right)}+1}3\\ &=\dfrac{{^2\negmedspace\log4}+ {^2\negmedspace\log(x + 2)^3}+1}3\\ &=\dfrac{2+ 3\ {^2\negmedspace\log(x + 2)}+1}3\\ &=\dfrac{3\ {^2\negmedspace\log(x + 2)}+3}3\\ &={^2\negmedspace\log(x + 2)}+1\\ &={^2\negmedspace\log(x + 2)}+{^2\negmedspace\log2}\\ &={^2\negmedspace\log(x + 2)2}\\ &=\boxed{\boxed{^2\negmedspace\log(2x + 4)}} }\)

Jadi, \left(f^{-1}\circ g\right)(x) ={^2\negmedspace\log(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

4. 7
5. 8

Ganesha Operation

\(\begin{aligned} f^{-1}(11)&=2\\ f(2)&=11\\ 2a+b&=11 \end{aligned}\)

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

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

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

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.

No. 8

Diketahui fungsi {f(x) = x^2 + 2x + 11}, dengan x\geq-1, maka nilai dari f^{-1}(14) adalah

1. 5
2. 4
3. 3

4. 2
5. 1

SKMP Juli 2020

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, f^{-1}(14)=1.
JAWAB: E