Question

For what value of a and b will
f(x)={█((x^2-4)/(x-2); x≤2@〖ax〗^2-bx+3; 2≤x≤3@2x-a+b; x≥3)┤
Be continuous at R

Answer 1

Given function is continuous on R.

So, function f(x) is continuous at x=2 and x=3 also.

So, f(2^-) = f(2⁺) and f(3^-)=f(3⁺).

Now by continuity of f(x) at x=2, we get

$\lim_{x \to 2^-} f(x) = \lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} (x+2) = 4. \\ \lim_{x \to 2^+} f(x) = (2a)^2-2b+3.\\ \implies 4a^2-2b+3 = 4 \\ \implies 4a^2-2b -1= 0.$

Similarly by continuity of f(x) at x=3, we have

$f(3^-)=f(3^+)\\ \implies 9a^2-3b+3=6-a+b \\ \implies 9a^2+a-4b-3=0.$

So we get $a^2+a-1=0. \implies a=\frac{-1\pm\sqrt{5}}{2}.$

And

$b = (4a^2-1)/2 = (4-4a-1)/2 = (3-4a)/2.$