Question

Consider the curves: \[ y = 2x^3 + 3x^2 + 4, y = 3x^2 - 2x + 8 \] How many times do they intersect for \( -3 \leq x \leq 2 \)?

• A. The two curves intersect thrice.
• B. The two curves intersect twice.
• C. The two curves intersect once.
• D. The two curves do not intersect.

Hint:

Always subtract the functions and factor to find points of intersection. Cubic equations can have max 3 roots, but not all real.

Answer 1

The Correct Option is A

Let’s equate the two expressions: \[ 2x^3 + 3x^2 + 4 = 3x^2 - 2x + 8 \Rightarrow 2x^3 + 3x^2 - 3x^2 + 2x + 4 - 8 = 0 \Rightarrow 2x^3 + 2x - 4 = 0 \Rightarrow x^3 + x - 2 = 0 \] Now solve: \[ x^3 + x - 2 = 0 \] Try rational root theorem: - \( x = 1 \): \( 1 + 1 - 2 = 0 \Rightarrow x = 1 \) is a root. Now divide: \[ x^3 + x - 2 = (x - 1)(x^2 + x + 2) \] Quadratic part: \[ x^2 + x + 2 \Rightarrow \text{Discriminant} = 1^2 - 4(1)(2) = -7<0 \Rightarrow \text{Two complex roots} \] So total intersections: \(\boxed{1 \text{ real root}} \Rightarrow\) contradicts (A)? Wait! Mistake: \[ 2x^3 + 3x^2 + 4 - (3x^2 - 2x + 8) = 2x^3 + 3x^2 - 3x^2 + 2x + 4 - 8 = 2x^3 + 2x - 4 \Rightarrow 2x^3 + 2x - 4 = 0 \Rightarrow x^3 + x - 2 = 0 \Rightarrow \text{Only 1 real root} \] So Final Answer: \( \boxed{1} \Rightarrow \text{
(c) once} \) But option (A) was earlier considered correct? No. Actually,
(c) is correct.