Question

boleh dijawab ya kak...​

Answer 1

Jawab:

C

Penjelasan dengan langkah-langkah:

Ubah f(x + 5) menjadi f(x)

[tex]\begin{aligned}f(x+5)&=x^2+4x-3\\f(a)&=(a-5)^2+4(a-5)-3\\f(a)&=a^2-10a+25+4a-20-3\\f(x)&=x^2-6x+2\end{aligned}[/tex]

Inverskan

[tex]\begin{aligned}y&=x^2-6x+2\\y+7&=x^2-6x+9\\y+7&=(x-3)^2\\\pm\sqrt{y+7}&=x-3\\x&=3\pm\sqrt{y+7}\\f^{-1}(x)&=3\pm\sqrt{x+7}\end{aligned}[/tex]

Maka

[tex]\begin{aligned}f^{-1}(2a-3)&=3\pm\sqrt{2a-3+7}\\6&=3\pm\sqrt{2a+4}\\9&=2a+4\\a&=\frac{5}{2}\end{aligned}[/tex]

sehingga

[tex]\displaystyle 4a=4\left ( \frac{5}{2} \right )=10[/tex]