Question

If the tangent drawn to the curve \( y = x^3 - ax^2 + x + 1 \) at each point \( x \in \mathbb{R} \), is inclined at an acute angle with the positive direction of \( X \)-axis, then the set of all possible values of \( a \) is:

• A. \( \mathbb{R} - \left( -\sqrt{3}, \sqrt{3} \right) \)
• B. \( [-3, 3] \)
• C. \( \mathbb{R} \)
• D. \( \mathbb{R} - \left( -\sqrt{3}, \sqrt{3} \right) \)

Hint:

When solving for the possible values of a parameter, always check the discriminant of quadratic equations to determine the conditions for real roots.

Answer 1

The Correct Option is D

The equation of the curve is given by \( y = x^3 - ax^2 + x + 1 \). The slope of the tangent at any point \( x \) on the curve is given by the first derivative of the equation of the curve with respect to \( x \). The derivative of \( y \) is: \[ \frac{dy}{dx} = 3x^2 - 2ax + 1 \] For the tangent to make an acute angle with the positive direction of the X-axis, the slope of the tangent (i.e., \( \frac{dy}{dx} \)) should be positive. Hence, we need to solve the inequality: \[ 3x^2 - 2ax + 1>0 \] for all \( x \). The discriminant of the quadratic equation \( 3x^2 - 2ax + 1 = 0 \) is: \[ \Delta = (-2a)^2 - 4 \cdot 3 \cdot 1 = 4a^2 - 12 \] For the quadratic to have real roots, the discriminant must be non-negative, i.e., \( 4a^2 - 12 \geq 0 \), which simplifies to: \[ a^2 \geq 3 \] Thus, \( a \) can take values outside the interval \( (-\sqrt{3}, \sqrt{3}) \), i.e., \( a \in \mathbb{R} - \left( -\sqrt{3}, \sqrt{3} \right) \). Hence, the correct answer is option (4).