Question

If \( f(x) = \sqrt{x} + \sin x \), then all the points of the set \( \left( x, f(x) \right)/f'(x) = 0 \) lie on:

• A. a circle
• B. a straight line
• C. an ellipse
• D. a parabola

Hint:

For functions involving square roots and trigonometric components, the equation \( f'(x) = 0 \) can often yield a parabolic relationship.

Answer 1

The Correct Option is D

We are given \( f(x) = \sqrt{x} + \sin x \). We need to find the points where \( f'(x) = 0 \). The first step is to calculate \( f'(x) \). \[ f'(x) = \frac{d}{dx} \left( \sqrt{x} + \sin x \right) = \frac{1}{2\sqrt{x}} + \cos x \] We now need to find the points where \( f'(x) = 0 \), i.e., \[ \frac{1}{2\sqrt{x}} + \cos x = 0 \] This equation implies that the points satisfying this equation lie on a parabola. The key part is the relationship between \( x \) and \( \sqrt{x} \), which is characteristic of a parabola. Hence, the points lie on a parabola.

Answer 2

To find all the points where \((x, f(x))\) lie such that \( f'(x) = 0 \), we first need to compute the derivative of \( f(x) = \sqrt{x} + \sin x \).

Step 1: Differentiate \( f(x) \)

\( f'(x) = \frac{d}{dx}(\sqrt{x}) + \frac{d}{dx}(\sin x) = \frac{1}{2\sqrt{x}} + \cos x \)

Step 2: Set the derivative to zero

\(\frac{1}{2\sqrt{x}} + \cos x = 0\)

Step 3: Solve for x

\(\frac{1}{2\sqrt{x}} = -\cos x\)

Given the domain of \(\sqrt{x}\) is \(x \geq 0\) and \(-1 \leq \cos x \leq 1\), analyze below feasible solutions:

When \( \cos x = -1 \), \( x = \frac{1}{4} \), solving gives \( 2\sqrt{\frac{1}{4}} = 2 \times \frac{1}{2} = 1\). Thus \( f'(x) = 0 \) holds true.

Step 4: Evaluate \( f(x) \) at found x-value

\( f\left(\frac{1}{4}\right) = \sqrt{\frac{1}{4}} + \sin \left(\frac{1}{4}\right) = \frac{1}{2} + \sin \left(\frac{1}{4}\right) \)

Conclusion: The problem implies that all such points \((x, f(x))\) where \( f'(x) = 0 \) lie as solutions forming a parabola. Therefore, selecting & examining multiple x-values based on any derived constraint reveals the property as parabola-inclined points.