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No.
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\\[-12pt]
3f(2)&=-1\\
f(2)&=\boxed{\boxed{-\dfrac13}}
\end{aligned}
No.
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&=0
\end{aligned}
No.
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 ....
\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.
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.
Bila
f(x) memenuhi
2f(x)+f(1-x)=x^2 untuk semua nilai real
x, maka
f(x) sama dengan
\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\qquad&\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\\[-12pt]
3f(x)&=x^2+2x-1\\
f(x)&=\dfrac13x^2+\dfrac23x-\dfrac13
\end{aligned}
No.
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
\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.
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.
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.
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\\[-12pt]
3f(1)&=5
\end{aligned}
No.
Jika fungsi
{f\left(x^2 + 3x + 5\right) = {^3\log}\left(20x^2 + 3x + 4\right)}, dengan
{x\geq0}, maka nilai
f(9) =
\begin{aligned}
x^2 + 3x + 5&=9\\
x^2+3x-4&=0\\
(x+4)(x-1)&=0
\end{aligned}
x=-4 (TM) atau x=\boxed{1}
\begin{aligned}
f(9)&={^3\log}\left(20(1)^2 + 3(1) + 4\right)\\
&={^3\log}\left(27\right)\\
&=\boxed{\boxed{3}}
\end{aligned}
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