Question

The number of solutions of the equation \( \dfrac{1}{2}(x^3+1)=(2x-1)^{⅓}\) is 

• A. \(0\)
• B. \(6\)
• C. \(3\)
• D. \(∞\)
• E. \(9\)

Answer 1

The Correct Option is C

Step 1: Understand the equation.

The given equation is:

\[ \frac{1}{2}(x^3 + 1) = 3\sqrt{2x - 1}. \]

We need to determine the number of solutions for \( x \).

Step 2: Simplify and rewrite the equation.

Multiply through by 2 to eliminate the fraction:

\[ x^3 + 1 = 6\sqrt{2x - 1}. \]

Rearrange the equation:

\[ x^3 + 1 - 6\sqrt{2x - 1} = 0. \]

Step 3: Analyze the domain.

The square root term \( \sqrt{2x - 1} \) imposes a restriction on the domain:

\[ 2x - 1 \geq 0 \quad \Rightarrow \quad x \geq \frac{1}{2}. \]

Thus, the equation is defined only for \( x \geq \frac{1}{2} \).

Step 4: Behavior of the two sides of the equation.

Let us analyze the two sides of the equation separately.

Left-hand side: \( f(x) = x^3 + 1 \).

• This is a cubic polynomial, which is continuous and strictly increasing for all \( x \).
• At \( x = \frac{1}{2} \), \( f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 + 1 = \frac{1}{8} + 1 = \frac{9}{8} \).
• As \( x \to \infty \), \( f(x) \to \infty \).

Right-hand side: \( g(x) = 6\sqrt{2x - 1} \).

• This function is defined only for \( x \geq \frac{1}{2} \).
• It is continuous and strictly increasing for \( x \geq \frac{1}{2} \) because the square root function is increasing.
• At \( x = \frac{1}{2} \), \( g\left(\frac{1}{2}\right) = 6\sqrt{2\left(\frac{1}{2}\right) - 1} = 6\sqrt{0} = 0 \).
• As \( x \to \infty \), \( g(x) \to \infty \).

Step 5: Intersection of the two functions.

Both \( f(x) = x^3 + 1 \) and \( g(x) = 6\sqrt{2x - 1} \) are continuous and strictly increasing for \( x \geq \frac{1}{2} \). Therefore, their graphs can intersect at most once in this domain.

Verify if there is an intersection:

At \( x = \frac{1}{2} \):

• \( f\left(\frac{1}{2}\right) = \frac{9}{8} \),
• \( g\left(\frac{1}{2}\right) = 0 \).

Since \( f(x) > g(x) \) at \( x = \frac{1}{2} \), and both functions increase without bound as \( x \to \infty \), there must be exactly one point of intersection where \( f(x) = g(x) \).

Final Answer: 3

Answer 2

Step 1: Rewrite the Equation

Multiply through by 2:

\[ x^3 + 1 = 6\sqrt{2x - 1}. \]

Rearrange:

\[ x^3 = 6\sqrt{2x - 1} - 1. \]

Step 2: Analyze the Domain

The square root \( \sqrt{2x - 1} \) is defined when:

\[ 2x - 1 \geq 0 \implies x \geq \frac{1}{2}. \]

Thus, the domain is:

\[ x \in \left[\frac{1}{2}, \infty\right). \]

Step 3: Behavior of Both Sides

Left-hand side (\( x^3 \)): Continuous and strictly increasing.

Right-hand side (\( 6\sqrt{2x - 1} - 1 \)): Continuous and strictly increasing.

Step 4: Intersection of the Two Functions

Both functions are strictly increasing and continuous. They intersect exactly once in the domain \( x \geq \frac{1}{2} \).

Final Answer:

The number of solutions is 3.