Question

Rolle's theorem holds for monic quadratic polynomial \(f(x)\) on the interval \([α,α+3]\) where \(f(α) = 0\). Similarly, \(g(x) = f(x)+2\) also follows Rolle's theorem in the interval \([β,3]\) where \(g(3) = 0\), such that the value of \(c\) is the same for both \(f(x)\) and \(g(x)\). Then the value of \((f\circ g)(α)\) is...

• A. \(-4\)
• B. \(4\)
• C. \(-2\)
• D. \(2\)

Hint:

Rolle needs equal end values, so a monic quadratic has its vertex at the point c.

Answer 1

The Correct Option is C

Step 1: Parametrise
Let the common point be $c$. Then $f(x) = (x - c)^2 - \frac{9}{4}$ (roots $c \pm \frac{3}{2}$) and $g(x) = (x - c)^2 - \frac{1}{4}$ (roots $c \pm \frac{1}{2}$).

Step 2: Conditions
$g(3) = 0$ means $3$ is $c + \frac{1}{2}$ or $c - \frac{1}{2}$. The other root is $\beta$, which must be less than 3, so $3 = c + \frac{1}{2}$ and $c = \frac{5}{2}$, $\beta = 2$.

Step 3: Find alpha
$\alpha$ is the smaller root of $f$: $\alpha = c - \frac{3}{2} = 1$, and the interval is $[1, 4]$.

Step 4: Evaluate
$g(1) = \frac{9}{4} - \frac{1}{4} = 2$ and $f(2) = \left(\frac{-1}{2}\right)^2 - \frac{9}{4} = \frac{1}{4} - \frac{9}{4} = -2$.

Final Answer:
The value is -2. This is option (C). \[ \boxed{\text{(C) }-2} \]