Question

Let \( f, g : \mathbb{R} \to \mathbb{R} \) be defined by \[ f(x) = x^2 - \frac{\cos(x)}{2}, g(x) = \frac{x \sin(x)}{2}. \] Then

• A. \( f(x) = g(x) \) for more than two values of \( x \)
• B. \( f(x) \neq g(x) \), for all \( x \text{ in } \mathbb{R} \)
• C. \( f(x) = g(x) \) for exactly one value of \( x \)
• D. \( f(x) = g(x) \) for exactly two values of \( x \)

Hint:

Transcendental equations involving both algebraic and trigonometric functions can often be solved by testing specific values of \( x \).

Answer 1

The Correct Option is D

Given: $f(x) = x^2 - \frac{\cos x}{2}$ and $g(x) = \frac{x \sin x}{2}$

Set up equation: $f(x) = g(x)$ $$x^2 - \frac{\cos x}{2} = \frac{x \sin x}{2}$$ $$2x^2 - \cos x = x \sin x$$ $$2x^2 = x \sin x + \cos x$$

Define: $h(x) = 2x^2 - x \sin x - \cos x$

We need to find zeros of $h(x)$.

Check $x = 0$: $$h(0) = 0 - 0 - 1 = -1 \neq 0$$

Analyze behavior:

• As $x \to \infty$: $h(x) \approx 2x^2 \to +\infty$
• As $x \to -\infty$: $h(x) \approx 2x^2 \to +\infty$
• $h(0) = -1 < 0$

Check derivative: $h'(x) = 4x - \sin x - x \cos x + \sin x = 4x - x \cos x$

At $x = 0$: $h'(0) = 0$

For small positive $x$: $h'(x) \approx 4x - x = 3x > 0$ For small negative $x$: $h'(x) \approx 4x - x = 3x < 0$

So $x = 0$ is a local minimum with $h(0) = -1 < 0$.

By Intermediate Value Theorem:

• Since $h(0) = -1 < 0$ and $h(x) \to +\infty$ as $x \to +\infty$, there exists at least one root for $x > 0$.
• Since $h(0) = -1 < 0$ and $h(x) \to +\infty$ as $x \to -\infty$, there exists at least one root for $x < 0$.

Check for multiple roots: For large $|x|$, the $2x^2$ term dominates, and since $h$ is continuous with $h(0) < 0$ and $h(x) \to +\infty$ for $|x| \to \infty$, by symmetry considerations and the behavior of the function, there are exactly two intersection points.

Answer: (D) $f(x) = g(x)$ for exactly two values of $x$