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: Use the given Lipschitz condition.
Given, \[ |f(x)-f(y)|\leq \frac{1}{2}|x-y| \] This implies that the rate of change of \(f(x)\) cannot exceed \(\frac{1}{2}\).
Therefore, \[ |f'(x)|\leq \frac{1}{2} \]

Step 2: Use the derivative condition.
Also given, \[ f'(x)\geq \frac{1}{2} \] Combining both conditions, \[ |f'(x)|\leq \frac{1}{2} \] and \[ f'(x)\geq \frac{1}{2} \] we get \[ f'(x)=\frac{1}{2} \]

Step 3: Find the function \(f(x)\).
Since \[ f'(x)=\frac{1}{2}, \] integrating, \[ f(x)=\frac{x}{2}+C \] Given, \[ f(1)=\frac{1}{2} \] So, \[ \frac{1}{2}+C=\frac{1}{2} \] \[ C=0 \] Hence, \[ f(x)=\frac{x}{2} \]

Step 4: Find the intersection points.
For intersection, \[ f(x)=x^2-2x-5 \] So, \[ \frac{x}{2}=x^2-2x-5 \] Multiplying by \(2\), \[ x=2x^2-4x-10 \] \[ 2x^2-5x-10=0 \]

Step 5: Check the number of real roots.
For \[ 2x^2-5x-10=0, \] the discriminant is \[ D=(-5)^2-4(2)(-10) \] \[ D=25+80 \] \[ D=105 \] Since \[ D\gt 0, \] the quadratic equation has two distinct real roots.

Step 6: Final conclusion.
Therefore, the number of points of intersection is \[ \boxed{2} \]