001 - ISI Questions & Solutions

Step 1: Define a helper function

Define

$$g(x)=f(x)-x.$$

Then $g$ is differentiable on $\mathbb{R}$, and

$$g'(x)=f'(x)-1.$$

---

Step 2: Analyze the derivative of $g$

Since $|f'(x)|<\tfrac12$, we have

$$f'(x)-1 \le \tfrac12 -1 = -\tfrac12,$$

and also

$$f'(x)-1 \ge -\tfrac12 -1 = -\tfrac32.$$

Hence,

$$g'(x)<0 \quad \text{for all } x\in\mathbb{R}.$$

So $g$ is strictly decreasing on $\mathbb{R}$.

---

Step 3: Look at the behavior at infinity

Fix any $x\in\mathbb{R}$. By the Mean Value Theorem applied between $0$ and $x$, there exists $\xi$ between $0$ and $x$ such that

$$f(x)-f(0)=f'(\xi)x.$$

Taking absolute values,

$$|f(x)-f(0)| \le \tfrac12 |x|.$$

Thus,

$$f(x) \le f(0)+\tfrac12 |x| \quad \text{and} \quad f(x) \ge f(0)-\tfrac12 |x|.$$

Now consider $g(x)=f(x)-x$:

• As $x\to +\infty$,

  $$g(x) \le f(0)+\tfrac12 x - x = f(0) - \tfrac12 x \to -\infty.$$

• As $x\to -\infty$,

  $$g(x) \ge f(0) - \tfrac12 |x| - x = f(0) + \tfrac12 x \to +\infty.$$

So,

$$\lim_{x\to -\infty} g(x)=+\infty, \qquad \lim_{x\to +\infty} g(x)=-\infty.$$

---

Step 4: Apply the Intermediate Value Theorem

Since $g$ is continuous, strictly decreasing, and changes sign from positive to negative, there exists a unique $x_0\in\mathbb{R}$ such that

$$g(x_0)=0.$$

That is,

$$f(x_0)-x_0=0 \quad \Rightarrow \quad f(x_0)=x_0.$$

---

Conclusion

There exists (in fact, exactly one) $x_0\in\mathbb{R}$ such that

$$\boxed{f(x_0)=x_0.}$$

This completes the proof.

for information related to ISI Entrance Exam Read Further