Question

\(f:\mathbb{R}\to\mathbb{R}\) is a function such that \[ |f(x)-f(y)|\leq \frac{1}{2}|x-y| \quad \forall x,y\in \mathbb{R} \] and \[ f'(x)\geq \frac{1}{2}\quad \forall x\in \mathbb{R},\quad f(1)=\frac{1}{2} \] Then the number of points of intersection of the curve \(y=f(x)\) and the curve \[ y=x^2-2x-5 \] is:

• A. \(1\)
• B. \(0\)
• C. \(2\)
• D. infinite

Hint:

If a function satisfies \[ |f(x)-f(y)|\leq k|x-y|, \] then its derivative, wherever it exists, satisfies \[ |f'(x)|\leq k. \] Combine this with any given lower or upper bound on \(f'(x)\).

Answer 1

The Correct Option is C

Step 1: Determine the slope of $f(x)$.
From $|f(x)-f(y)| \leq \frac{1}{2}|x-y|$, dividing by $|x-y|$ and taking the limit gives $|f'(x)| \leq \frac{1}{2}$. Combined with $f'(x) \geq \frac{1}{2}$, we get $f'(x) = \frac{1}{2}$ for all $x$.

Step 2: Determine $f(x)$ explicitly.
Since $f'(x) = 1/2$ everywhere and $f(1) = 1/2$: \[ f(x) = \frac{1}{2}x + c. \] At $x=1$: $\frac{1}{2} + c = \frac{1}{2} \Rightarrow c = 0$. So $f(x) = \frac{x}{2}$.

Step 3: Set up the intersection equation.
Find intersections of $y = \frac{x}{2}$ and $y = x^2 - 2x - 5$: \[ \frac{x}{2} = x^2 - 2x - 5 \Rightarrow x^2 - \frac{5x}{2} - 5 = 0 \Rightarrow 2x^2 - 5x - 10 = 0. \]

Step 4: Check the discriminant.
$D = 25 + 80 = 105 > 0$. So there are two distinct real roots and hence two intersection points.

Step 5: Verify both roots are real.
$x = \dfrac{5 \pm \sqrt{105}}{4}$. Both values are real, giving 2 intersection points.

Step 6: State the answer.
The number of points of intersection is 2.
\[ \boxed{2} \]