Berikut ini adalah kumpulan soal mengenai Fungsi tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 1
Diberikan fungsi
f memenuhi persamaan
2f(x+2)+f(-x)=x-3 untuk setiap bilangan bulat
x. Nilai
f(2) adalah ....
Untuk x=0,
\(\begin{aligned}
2f(0+2)+f(-0)&=0-3\\
2f(2)+f(0)&=-3\\
4f(2)+2f(0)&=-6
\end{aligned}\)
Untuk x=-2,
\(\begin{aligned}
2f(-2+2)+f(-(-2))&=-2-3\\
2f(0)+f(2)&=-5\\
f(2)+2f(0)&=-5
\end{aligned}\)
\(\begin{aligned}
4f(2)+2f(0)&=-6\\
f(2)+2f(0)&=-5\qquad-\\\hline
3f(2)&=-1\\
f(2)&=\boxed{\boxed{-\dfrac13}}
\end{aligned}\)
No. 2
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= ....
\(\begin{aligned}
f^{-1}(11)&=2\\
f(2)&=11\\
2a+b&=11
\end{aligned}\)
\(\begin{aligned}
f^{-1}(8)&=1\\
f(1)&=8\\
a+b&=8
\end{aligned}\)
\(\begin{aligned}
2a+b&=11\\
a+b&=8\qquad-\\\hline
a&=3
\end{aligned}\)
\(\begin{aligned}
a+b&=8\\
3+b&=8\\
b&=5
\end{aligned}\)
f(x)=3x+5
\(\begin{aligned}
\displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h&=\displaystyle\lim_{h\to0}\dfrac{(3+h)(3(3)+5)-3(3(3+h)+5)}h\\
&=\displaystyle\lim_{h\to0}\dfrac{(3+h)(9+5)-3(9+3h+5)}h\\
&=\displaystyle\lim_{h\to0}\dfrac{(3+h)(14)-3(14+3h)}h\\
&=\displaystyle\lim_{h\to0}\dfrac{42+14h-42+9h}h\\
&=\displaystyle\lim_{h\to0}\dfrac{5h}h\\
&=\displaystyle\lim_{h\to0}5\\
&=\boxed{\boxed{5}}
\end{aligned}\)
No. 3
{f\left(x^2+3ax+1\right)=2x-1} dan
{f(5)=3}. Hitung nilai
a
\(\begin{aligned}
2x-1&=3\\
2x&=4\\
x&=2
\end{aligned}\)
\(\begin{aligned}
x^2+3ax+1&=5\\
2^2+3a(2)+1&=5\\
4+6a+1&=5\\
6a+5&=5\\
6a&=0\\
a&=\boxed{\boxed{0}}
\end{aligned}\)
No. 4
Jika
{f(x)=3h(x)+4};
{g(x)=\dfrac{\sqrt{f(x)}}9}; dan
{h(x)=3x^2 + 2x-1}, maka nilai dari
{2f(3)\cdot g(2)-f^2(3)-g^2(2)} adalah ....
- -\dfrac{797449}{81}
- \dfrac{797449}{81}
- -\dfrac{893}9
- \dfrac{893749}9
- \dfrac{893749}{81}
\(\begin{aligned}
2f(3)\cdot g(2)-f^2(3)-g^2(2)&=-\left(f(3)-g(2)\right)^2\\
&=-\left(3h(3)+4-\dfrac{\sqrt{f(2)}}9\right)^2\\
&=-\left(3\left(3(3)^2+2(3)-1\right)+4-\dfrac{\sqrt{3h(2)+4}}9\right)^2\\
&=-\left(3\left(27+6-1\right)+4-\dfrac{\sqrt{3\left(3(2)^2+2(2)-1\right)+4}}9\right)^2\\
&=-\left(3\left(32\right)+4-\dfrac{\sqrt{3\left(12+4-1\right)+4}}9\right)^2\\
&=-\left(96+4-\dfrac{\sqrt{3\left(15\right)+4}}9\right)^2\\
&=-\left(100-\dfrac{\sqrt{45+4}}9\right)^2\\
&=-\left(100-\dfrac{\sqrt{49}}9\right)^2\\
&=-\left(100-\dfrac79\right)^2\\
&=-\left(\dfrac{893}9\right)^2\\
&=-\dfrac{797449}{81}
\end{aligned}\)
No. 5
Jika
f(x)=a^x, maka untuk setiap
x dan
y berlaku
- {f(x)f(y)=f(xy)}
- {f(x)f(y)=f(x+y)}
- {f(x)f(y)=f(x)+f(y)}
- {f(x)+f(y)=f(xy)}
- {f(x)+f(y)=f(x+y)}
\begin{aligned}
f(x)f(y)&=a^xa^y\\
&=a^{x+y}\\
&=f(x+y)
\end{aligned}
No. 6
Bila
f(x) memenuhi
{2f(x)+f(1-x)=x^2} untuk semua nilai real
x, maka
f(x) sama dengan
- {\dfrac12x^2-\dfrac32x+\dfrac12}
- {\dfrac19x^2+\dfrac89x-\dfrac13}
- {\dfrac23x^2+\dfrac12x-\dfrac13}
- {\dfrac13x^2+\dfrac23x-\dfrac13}
- {\dfrac19x^2+x-\dfrac49}
\(\begin{aligned}
2f(1-x)+f(1-(1-x))&=(1-x)^2\\
2f(1-x)+f(1-1+x)&=1-2x+x^2\\
2f(1-x)+f(x)&=x^2-2x+1\\
f(x)+2f(1-x)&=x^2-2x+1
\end{aligned}\)
\(\begin{aligned}
2f(x)+f(1-x)&=x^2&\color{red}{\times 2}\\
f(x)+2f(1-x)&=x^2-2x+1&
\end{aligned}\)
\(\begin{aligned}
4f(x)+2f(1-x)&=2x^2\\
f(x)+2f(1-x)&=x^2-2x+1&\color{red}{-}\\\hline
3f(x)&=x^2+2x-1\\
f(x)&=\dfrac13x^2+\dfrac23x-\dfrac13
\end{aligned}\)
No. 7
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
- {72\ln2-3}
- {36\ln 3-2}
- {36\ln 2-6}
- {36\ln 2 - 3}
- {72\ln 3-2}
\(\begin{aligned}
(f\circ g)(x)&=\dfrac{6x+3}{2x-5}\\[10pt]
f(g(x))&=\dfrac{12x+6}{4x-10}\\[10pt]
f(4x-11)&=\dfrac{3(4x-11)+39}{4x-11+1}\\[10pt]
f(x)&=\dfrac{3x+39}{x+1}\\[10pt]
f^{-1}(x)&=\dfrac{-x+39}{x-3}\\[10pt]
f^{-1}(x-1)&=\dfrac{-(x-1)+39}{x-1-3}\\[10pt]
&=\dfrac{-x+1+39}{x-4}\\[10pt]
&=\dfrac{-x+40}{x-4}
\end{aligned}\)
\(\begin{aligned}
\displaystyle\intop_5^8\dfrac{f^{-1}(x-1)}{g(3)}\ dx&=\displaystyle\intop_5^8\dfrac{\dfrac{-x+40}{x-4}}{4(3)-11}\ dx\\
&=\displaystyle\intop_5^8\dfrac{\dfrac{-x+4+36}{x-4}}{12-11}\ dx\\
&=\displaystyle\intop_5^8\dfrac{-1+\dfrac{36}{x-4}}1\ dx\\
&=\displaystyle\intop_5^8\left(-1+\dfrac{36}{x-4}\right)\ dx\\
&=\left[-x+36\ln|x-4|\right]_5^8\\
&=\left[-8+36\ln|8-4|\right]-\left[-5+36\ln|5-4|\right]\\
&=\left[-8+36\ln4\right]-\left[-5+36\ln1\right]\\
&=\left[-8+36\ln2^2\right]-\left[-5-42(0)\right]\\
&=\left[-8+72\ln2\right]-\left[-5\right]\\
&=-8+72\ln2+5\\
&=\boxed{\boxed{72\ln2-3}}
\end{aligned}\)
No. 8
Jika
{f\left(\dfrac2{\sqrt{2x-1}}\right)=x} dengan
x\geq\dfrac12, maka
f(1)=
\(\begin{aligned}
\dfrac2{\sqrt{2x-1}}&=1\\
\sqrt{2x-1}&=2\\
2x-1&=4\\
2x&=5\\
x&=\dfrac52
\end{aligned}\)
f(1)=\boxed{\boxed{\dfrac52}}
No. 9
Fungsi
f terdefinisi pada bilangan real kecuali
2 sehingga
{f\left(\dfrac{2x}{x-5}\right)=2x-1},
x\neq5. Nilai dari
{f(3) + f(1)} adalah ....
CARA 1
Misal \left(\dfrac{2x}{x-5}\right)=t
\(\begin{aligned}
2x&=tx-5t\\
2x-tx&=-5t\\
tx-2x&=5t\\
(t-2)x&=5t\\
x&=\dfrac{5t}{t-2}
\end{aligned}\)
\(\begin{aligned}
f\left(\dfrac{2x}{x-5}\right)&=2x-1\\
f(t)&=2\left(\dfrac{5t}{t-2}\right)-1\\
&=\dfrac{10t}{t-2}-1\\
&=\dfrac{10t-(t-2)}{t-2}\\
&=\dfrac{10t-t+2}{t-2}\\
&=\dfrac{9t+2}{t-2}
\end{aligned}\)
\(\begin{aligned}
f(3)+f(1)&=\dfrac{9(3)+2}{3-2}+\dfrac{9(1)+2}{1-2}\\
&=\dfrac{29}1+\dfrac{11}{-1}\\
&=29-11\\
&=\boxed{\boxed{18}}
\end{aligned}\)
CARA 2
\(\begin{aligned}
\dfrac{2x}{x-5}&=3\\
2x&=3x-15\\
-x&=-15\\
x&=15
\end{aligned}\)
\(\begin{aligned}
f(3)&=2(15)-1\\
&=29
\end{aligned}\)
\(\begin{aligned}
\dfrac{2x}{x-5}&=1\\
2x&=x-5\\
x&=-5
\end{aligned}\)
\(\begin{aligned}
f(1)&=2(-5)-1\\
&=-11
\end{aligned}\)
\(\begin{aligned}
f(3)+f(1)&=29+(-11)\\
&=\boxed{\boxed{18}}
\end{aligned}\)
No. 10
Diberikan fungsi
f memenuhi persamaan
{2f(-x)+f(x+3)=x+5} untuk setiap bilangan real
x. Nilai
3f(1) adalah
Untuk x=-1,
\(\begin{aligned}
2f(-(-1))+f(-1+3)&=-1+5\\
2f(1)+f(2)&=4\\
4f(1)+2f(2)&=8
\end{aligned}\)
Untuk x=-2,
\(\begin{aligned}
2f(-(-2))+f(-2+3)&=-2+5\\
2f(2)+f(1)&=3\\
f(1)+2f(2)&=3
\end{aligned}\)
\(\begin{aligned}
4f(1)+2f(2)&=8\\
f(1)+2f(2)&=3\qquad-\\\hline
3f(1)&=5
\end{aligned}\)
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