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SBMPTN Zone : Fungsi

Written By Anonymous Tuesday, July 14, 2020 Add Comment Edit
Berikut ini adalah kumpulan soal mengenai Fungsi tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
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  • 2

No. 1

Diberikan fungsi ff memenuhi persamaan 2f(x+2)+f(−x)=x−32f(x+2)+f(-x)=x-3 untuk setiap bilangan bulat xx. Nilai f(2)f(2) adalah ....
  1. 13\dfrac13
  2. −13-\dfrac13
  3. −3-3
  1. 33
  2. −2-2
Ganesha Operation
Untuk x=0x=0,
2f(0+2)+f(−0)=0−32f(2)+f(0)=−34f(2)+2f(0)=−6\begin{aligned}2f(0+2)+f(-0)&=0-3\\2f(2)+f(0)&=-3\\4f(2)+2f(0)&=-6\end{aligned}

Untuk x=−2x=-2,
2f(−2+2)+f(−(−2))=−2−32f(0)+f(2)=−5f(2)+2f(0)=−5\begin{aligned}2f(-2+2)+f(-(-2))&=-2-3\\2f(0)+f(2)&=-5\\f(2)+2f(0)&=-5\end{aligned}

4f(2)+2f(0)=−6f(2)+2f(0)=−5−3f(2)=−1f(2)=−13\begin{aligned}4f(2)+2f(0)&=-6\\f(2)+2f(0)&=-5\qquad-\\\hline3f(2)&=-1\\f(2)&=\boxed{\boxed{-\dfrac13}}\end{aligned}
Jadi, f(2)=−13f(2)=-\dfrac13.
JAWAB: B

No. 2

Diketahui f(x)=ax+b{f(x)=ax+b} dengan f−1(11)=2{f^{-1}(11)=2} dan f−1(8)=1{f^{-1}(8)=1} dengan f−1f^{-1} menyatakan fungsi invers ff. Nilai lim⁡h→0(3+h)f(3)−3f(3+h)h=\displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h= ....
  1. 44
  2. 55
  3. 66
  1. 77
  2. 88
Ganesha Operation
f−1(11)=2f(2)=112a+b=11\begin{aligned}f^{-1}(11)&=2\\f(2)&=11\\2a+b&=11\end{aligned}

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

2a+b=11a+b=8−a=3\begin{aligned}2a+b&=11\\a+b&=8\qquad-\\\hlinea&=3\end{aligned}

a+b=83+b=8b=5\begin{aligned}a+b&=8\\3+b&=8\\b&=5\end{aligned}

f(x)=3x+5f(x)=3x+5

lim ⁡h→0(3+h)f(3)−3f(3+h)h= lim ⁡h→0(3+h)(3(3)+5)−3(3(3+h)+5)h= lim ⁡h→0(3+h)(9+5)−3(9+3h+5)h= lim ⁡h→0(3+h)(14)−3(14+3h)h= lim ⁡h→042+14h−42+9hh= lim ⁡h→05hh= lim ⁡h→05=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, lim⁡h→0(3+h)f(3)−3f(3+h)h=5\displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h=5.

No. 3

f(x2+3ax+1)=2x−1{f\left(x^2+3ax+1\right)=2x-1} dan f(5)=3{f(5)=3}. Hitung nilai aa
SUMBER
2x−1=32x=4x=2\begin{aligned}2x-1&=3\\2x&=4\\x&=2\end{aligned}
x2+3ax+1=522+3a(2)+1=54+6a+1=56a+5=56a=0a=0\begin{aligned}x^2+3ax+1&=5\\2^2+3a(2)+1&=5\\4+6a+1&=5\\6a+5&=5\\6a&=0\\a&=\boxed{\boxed{0}}\end{aligned}
Jadi, a=0a=0.

No. 4

Jika f(x)=3h(x)+4{f(x)=3h(x)+4}; g(x)=f(x)9{g(x)=\dfrac{\sqrt{f(x)}}9}; dan h(x)=3x2+2x−1{h(x)=3x^2 + 2x-1}, maka nilai dari 2f(3)⋅g(2)−f2(3)−g2(2){2f(3)\cdot g(2)-f^2(3)-g^2(2)} adalah ....
  1. −79744981-\dfrac{797449}{81}
  2. 79744981\dfrac{797449}{81}
  3. −8939-\dfrac{893}9
  1. 8937499\dfrac{893749}9
  2. 89374981\dfrac{893749}{81}
SUMBER
2f(3)⋅g(2)−f2(3)−g2(2)=−(f(3)−g(2))2=−(3h(3)+4−f(2)9)2=−(3(3(3)2+2(3)−1)+4−3h(2)+49)2=−(3(27+6−1)+4−3(3(2)2+2(2)−1)+49)2=−(3(32)+4−3(12+4−1)+49)2=−(96+4−3(15)+49)2=−(100−45+49)2=−(100−499)2=−(100−79)2=−(8939)2=−79744981\begin{aligned}2f(3)\cdot g(2)-f^2(3)-g^2(2)&=-\left(f(3)-g(2)\right)^2\\&=-\left(3h(3)+4-\dfrac{\sqrt{f(2)}}9\right)^2\\&=-\left(3\left(3(3)^2+2(3)-1\right)+4-\dfrac{\sqrt{3h(2)+4}}9\right)^2\\&=-\left(3\left(27+6-1\right)+4-\dfrac{\sqrt{3\left(3(2)^2+2(2)-1\right)+4}}9\right)^2\\&=-\left(3\left(32\right)+4-\dfrac{\sqrt{3\left(12+4-1\right)+4}}9\right)^2\\&=-\left(96+4-\dfrac{\sqrt{3\left(15\right)+4}}9\right)^2\\&=-\left(100-\dfrac{\sqrt{45+4}}9\right)^2\\&=-\left(100-\dfrac{\sqrt{49}}9\right)^2\\&=-\left(100-\dfrac79\right)^2\\&=-\left(\dfrac{893}9\right)^2\\&=-\dfrac{797449}{81}\end{aligned}
Jadi, 2f(3)⋅g(2)−f2(3)−g2(2)=−797449812f(3)\cdot g(2)-f^2(3)-g^2(2)=-\dfrac{797449}{81}.
JAWAB: A

No. 5

Jika f(x)=axf(x)=a^x, maka untuk setiap xx dan yy berlaku
  1. f(x)f(y)=f(xy){f(x)f(y)=f(xy)}
  2. f(x)f(y)=f(x+y){f(x)f(y)=f(x+y)}
  3. f(x)f(y)=f(x)+f(y){f(x)f(y)=f(x)+f(y)}
  1. f(x)+f(y)=f(xy){f(x)+f(y)=f(xy)}
  2. f(x)+f(y)=f(x+y){f(x)+f(y)=f(x+y)}
SPMB '02 (Regional 1)
f(x)f(y)=axay=ax+y=f(x+y)
Jadi, berlaku f(x)f(y)=f(x+y)f(x)f(y)=f(x+y).
JAWAB: B

No. 6

Bila f(x)f(x) memenuhi 2f(x)+f(1−x)=x2{2f(x)+f(1-x)=x^2} untuk semua nilai real xx, maka f(x)f(x) sama dengan
  1. 12x2−32x+12{\dfrac12x^2-\dfrac32x+\dfrac12}
  2. 19x2+89x−13{\dfrac19x^2+\dfrac89x-\dfrac13}
  3. 23x2+12x−13{\dfrac23x^2+\dfrac12x-\dfrac13}
  1. 13x2+23x−13{\dfrac13x^2+\dfrac23x-\dfrac13}
  2. 19x2+x−49{\dfrac19x^2+x-\dfrac49}
UMB '08 Kode 270
2f(1−x)+f(1−(1−x))=(1−x)22f(1−x)+f(1−1+x)=1−2x+x22f(1−x)+f(x)=x2−2x+1f(x)+2f(1−x)=x2−2x+1\begin{aligned}2f(1-x)+f(1-(1-x))&=(1-x)^2\\2f(1-x)+f(1-1+x)&=1-2x+x^2\\2f(1-x)+f(x)&=x^2-2x+1\\f(x)+2f(1-x)&=x^2-2x+1\end{aligned}

2f(x)+f(1−x)=x2×2f(x)+2f(1−x)=x2−2x+1\begin{aligned}2f(x)+f(1-x)&=x^2&\color{red}{\times 2}\\f(x)+2f(1-x)&=x^2-2x+1&\end{aligned}

4f(x)+2f(1−x)=2x2f(x)+2f(1−x)=x2−2x+1−3f(x)=x2+2x−1f(x)=13x2+23x−13\begin{aligned}4f(x)+2f(1-x)&=2x^2\\f(x)+2f(1-x)&=x^2-2x+1&\color{red}{-}\\\hline3f(x)&=x^2+2x-1\\f(x)&=\dfrac13x^2+\dfrac23x-\dfrac13\end{aligned}
Jadi, f(x)=13x2+23x−13f(x)=\dfrac13x^2+\dfrac23x-\dfrac13.
JAWAB: D

No. 7

Jika (f∘g)(x)=6x+32x−5{(f\circ g)(x)=\dfrac{6x+3}{2x-5}} dan g(x)=4x−11{g(x)=4x-11}, maka hasil dari ∫58f−1(x−1)g(3) dx{\displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx} adalah
  1. 72ln⁡2−3{72\ln2-3}
  2. 36ln⁡3−2{36\ln 3-2}
  3. 36ln⁡2−6{36\ln 2-6}
  1. 36ln⁡2−3{36\ln 2 - 3}
  2. 72ln⁡3−2{72\ln 3-2}
(f∘g)(x)=6x+32x−5f(g(x))=12x+64x−10f(4x−11)=3(4x−11)+394x−11+1f(x)=3x+39x+1f−1(x)=−x+39x−3f−1(x−1)=−(x−1)+39x−1−3=−x+1+39x−4=−x+40x−4\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}

∫58f−1(x−1)g(3) dx=∫58−x+40x−44(3)−11 dx=∫58−x+4+36x−412−11 dx=∫58−1+36x−41 dx=∫58(−1+36x−4) dx=[−x+36ln⁡∣x−4∣]58=[−8+36ln⁡∣8−4∣]−[−5+36ln⁡∣5−4∣]=[−8+36ln⁡4]−[−5+36ln⁡1]=[−8+36ln⁡22]−[−5−42(0)]=[−8+72ln⁡2]−[−5]=−8+72ln⁡2+5=72ln⁡2−3\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, ∫58f−1(x−1)g(3) dx=72ln⁡2−3\displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx=72\ln2-3.
JAWAB: A

No. 8

Jika f(22x−1)=x{f\left(\dfrac2{\sqrt{2x-1}}\right)=x} dengan x≥12x\geq\dfrac12, maka f(1)=f(1)=
  1. 44
  2. 22
  3. 12\dfrac12
  1. 52\dfrac52
  2. 32\dfrac32
22x−1=12x−1=22x−1=42x=5x=52\begin{aligned}\dfrac2{\sqrt{2x-1}}&=1\\\sqrt{2x-1}&=2\\2x-1&=4\\2x&=5\\x&=\dfrac52\end{aligned}

f(1)=52f(1)=\boxed{\boxed{\dfrac52}}
Jadi, f(1)=52f(1)=\dfrac52.
JAWAB : D

No. 9

Fungsi ff terdefinisi pada bilangan real kecuali 22 sehingga f(2xx−5)=2x−1{f\left(\dfrac{2x}{x-5}\right)=2x-1}, x≠5x\neq5. Nilai dari f(3)+f(1){f(3) + f(1)} adalah ....
  1. −2-2
  2. 88
  1. 1313
  2. 1818

CARA 1

Misal (2xx−5)=t\left(\dfrac{2x}{x-5}\right)=t
2x=tx−5t2x−tx=−5ttx−2x=5t(t−2)x=5tx=5tt−2\begin{aligned}2x&=tx-5t\\2x-tx&=-5t\\tx-2x&=5t\\(t-2)x&=5t\\x&=\dfrac{5t}{t-2}\end{aligned}

f(2xx−5)=2x−1f(t)=2(5tt−2)−1=10tt−2−1=10t−(t−2)t−2=10t−t+2t−2=9t+2t−2\begin{aligned}f\left(\dfrac{2x}{x-5}\right)&=2x-1\\f(t)&=2\left(\dfrac{5t}{t-2}\right)-1\\&=\dfrac{10t}{t-2}-1\\&=\dfrac{10t-(t-2)}{t-2}\\&=\dfrac{10t-t+2}{t-2}\\&=\dfrac{9t+2}{t-2}\end{aligned}

f(3)+f(1)=9(3)+23−2+9(1)+21−2=291+11−1=29−11=18\begin{aligned}f(3)+f(1)&=\dfrac{9(3)+2}{3-2}+\dfrac{9(1)+2}{1-2}\\&=\dfrac{29}1+\dfrac{11}{-1}\\&=29-11\\&=\boxed{\boxed{18}}\end{aligned}

CARA 2

2xx−5=32x=3x−15−x=−15x=15\begin{aligned}\dfrac{2x}{x-5}&=3\\2x&=3x-15\\-x&=-15\\x&=15\end{aligned}

Related:


f(3)=2(15)−1=29\begin{aligned}f(3)&=2(15)-1\\&=29\end{aligned}

2xx−5=12x=x−5x=−5\begin{aligned}\dfrac{2x}{x-5}&=1\\2x&=x-5\\x&=-5\end{aligned}

f(1)=2(−5)−1=−11\begin{aligned}f(1)&=2(-5)-1\\&=-11\end{aligned}

f(3)+f(1)=29+(−11)=18\begin{aligned}f(3)+f(1)&=29+(-11)\\&=\boxed{\boxed{18}}\end{aligned}
Jadi, f(3)+f(1)=18f(3) + f(1)=18.
JAWAB: D

No. 10

Diberikan fungsi ff memenuhi persamaan 2f(−x)+f(x+3)=x+5{2f(-x)+f(x+3)=x+5} untuk setiap bilangan real xx. Nilai 3f(1)3f(1) adalah
  1. 55
  2. 44
  3. 33
  1. 22
  2. 00
Untuk x=−1x=-1,
2f(−(−1))+f(−1+3)=−1+52f(1)+f(2)=44f(1)+2f(2)=8\begin{aligned}2f(-(-1))+f(-1+3)&=-1+5\\2f(1)+f(2)&=4\\4f(1)+2f(2)&=8\end{aligned}

Untuk x=−2x=-2,
2f(−(−2))+f(−2+3)=−2+52f(2)+f(1)=3f(1)+2f(2)=3\begin{aligned}2f(-(-2))+f(-2+3)&=-2+5\\2f(2)+f(1)&=3\\f(1)+2f(2)&=3\end{aligned}
4f(1)+2f(2)=8f(1)+2f(2)=3−3f(1)=5\begin{aligned}4f(1)+2f(2)&=8\\f(1)+2f(2)&=3\qquad-\\\hline3f(1)&=5\end{aligned}
Jadi, 3f(1)=53f(1)=5.
JAWAB: A

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