Multiple choice

Let the function $f(x)$, defined as $f(x)= \begin{cases} 3ax+b\quad & x<1 \ 11\quad \quad \quad \quad & x=1\quad \ 5ax-2b & x>1 \end{cases} $ be continuous at $x=1$. $a$ and $b$ are the roots of a quadratic equation, then the equation is

  1. ${ x }^{ 2 }-5x+6=0$
  2. ${ x }^{ 2 }+5x+6=0$
  3. ${ x }^{ 2 }-5x-6=0$
  4. ${ 3x }^{ 2 }-5x+2=0$
  5. ${ 2x }^{ 2 }-5x+3=0$
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A Correct answer
Explanation

For continuity at x=1, 3a(1)+b = 11 and 5a(1)-2b = 11. Solving: 6a+2b=22 and 5a-2b=11. Adding gives 11a=33, a=3. Then 9+b=11, b=2. The quadratic equation with roots 3 and 2 is (x-3)(x-2) = x^2-5x+6=0.

AI explanation

For the function to be continuous at x = 1, the left-hand limit must equal the right-hand limit, so we set 3a(1) + b = 5a(1) - 2b. Solving this gives 3a + b = 5a - 2b, which simplifies to 2a = 3b, or a = 3b/2. Additionally, this continuous limit value must equal the function's defined value at x = 1, meaning f(1) = 11, so we set 3a + b = 11. Substituting a = 3b/2 into this equation yields 3(3b/2) + b = 11, which simplifies to 11b/2 = 11, giving b = 2, and subsequently a = 3. The quadratic equation with roots a = 3 and b = 2 is x^2 - (3+2)x + (3)(2) = 0, which is x^2 - 5x + 6 = 0.

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