Question

Jika [tex]\displaystyle f(x)=\sqrt{x^2+ax+b}-\sqrt{x^2+bx+a},f(0)=-1[/tex] dan [tex]\displaystyle \lim_{x\to\infty}f(x)=\frac{5}{2}[/tex], maka [tex]\displaystyle a+b=...[/tex]

Answer 1

Penjelasan dengan langkah-langkah:

limit fungsi

• [tex]\displaystyle\sf \lim_{x\to\infty}(\sqrt{ax^2+bx+c}-\sqrt{px^2+q^2+r}=\frac{b-q}{2\sqrt{a}}[/tex] , dengan syarat a = p

diket :

• [tex]\displaystyle f(x)=\sqrt{x^2+ax+b}-\sqrt{x^2+bx+a}[/tex]
• f(0) = -1
• [tex]\displaystyle \lim_{x\to\infty}f(x)=\frac{5}{2}[/tex]

a + b = ....?

f(0) = -1

√b - √a = -1

.

[tex]\begin{aligned}\sf\displaystyle \lim_{x\to\infty}f(x)&\sf=\frac{5}{2} \\\sf \frac{a - b}{\cancel{2 \sqrt{1}} } &\sf = \frac{5}{\cancel2} \\ \sf {a - b} &\sf = 5\end{aligned}[/tex]

Cari nilai b

b = -5 + a

, sehingga

[tex]\begin{aligned}\sf \sqrt{ - 5 + a} - \sqrt{a} &\sf = - 1 \\ \sf (\sqrt{ - 5 + a} {)}^{2} &\sf = ( \sqrt{a} - 1) ^{2} \\ \sf - 5 + \cancel a&\sf = \cancel a - 2 \sqrt{a} + 1 \\ \sf2 \sqrt{a} &\sf = 5 + 1 \\ \sf2 \sqrt{a} &\sf = 6 \\ \sf\sqrt{a} &\sf = 3 \\ \sf \: a&\sf = 9\cdots \: diperoleh \: nilai \: a\end{aligned}[/tex]

untuk nilai b

b = -5 + a

b = -5 + 9

b = 4

Maka nilai a + b adalah 9 + 4 = 13

Answer 2

Verified answer

LIMIT TAK HINGGA

f(x) = √(x² + ax + b) - √(x² + bx + a)

lim x⇢∞ √(x² + ax + b) - √(x² + bx + a) = 5/2

(a - b)/2 = 5/2

a - b = 5 ... (1)

f(0) = -1

√b - √a = -1

(√a - √b)(√a + √b) = 5

√a + √b = 5

√a + √b = 5

√a - √b = 1

2√a = 6

a = 9

√b = 5 - √a

√b = 2

b = 4

a + b = 9 + 4 = 13