Multiple choice

Let $f\left ( x \right )=\begin{cases}x^{2}-1 ;0< x< 2 \ 2x+3; 2\leq x< 3\end{cases}$ The quadratic equation whose roots are $\displaystyle \lim_{x\to 2^-}f\left ( x \right )$ and $\displaystyle \lim_{x\to 2^+}f\left ( x \right )$ is

  1. $x^{2}-6x+9=0$
  2. $x^{2}-10x+21=0$
  3. $x^{2}-14x+49=0$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Limit x->2- of f(x) = 2^2 - 1 = 3. Limit x->2+ of f(x) = 2(2) + 3 = 7. Roots are 3 and 7. Equation: (x-3)(x-7) = x^2 - 10x + 21 = 0.

AI explanation

We evaluate the left-hand limit by substituting x = 2 into the first part of the piecewise function, x^2 - 1, which gives 2^2 - 1 = 3. We evaluate the right-hand limit by substituting x = 2 into the second part of the piecewise function, 2x + 3, which gives 2(2) + 3 = 7. The required quadratic equation has roots 3 and 7, so its sum is 10 and its product is 21. The equation is therefore x^2 - 10x + 21 = 0.

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