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 problem involves analyzing the tangent to the curve \(y = x^3 - ax^2 + x + 1\) and determining the set of values for \(a\) such that the tangent is inclined at an acute angle to the positive direction of the \(X\)-axis.

1. Find the derivative: The derivative of the curve with respect to \(x\) gives the slope of the tangent.

\[ \frac{dy}{dx} = 3x^2 - 2ax + 1 \]

2. Acute angle condition: A line is inclined at an acute angle to the positive \(X\)-axis if its slope is positive.

\( \frac{dy}{dx} > 0 \Rightarrow 3x^2 - 2ax + 1 > 0 \)

3. Discriminant condition for positivity: For the quadratic equation \(3x^2 - 2ax + 1 = 0\) to have no real roots, its discriminant must be non-positive.

Discriminant, \(D = (2a)^2 - 4 \cdot 3 \cdot 1 = 4a^2 - 12\)

For positivity of the entire quadratic: \(D \leq 0\)

\[ 4a^2 - 12 \leq 0 \]

\[ 4a^2 \leq 12 \]

\[ a^2 \leq 3 \]

\[ -\sqrt{3} \leq a \leq \sqrt{3} \]

4. Complement the interval: Since the slopes must be positive everywhere, choose values of \(a\) outside this interval.

5. Resultant set: Hence, the set of all possible values of \(a\) is \(\mathbb{R} - \left( -\sqrt{3}, \sqrt{3} \right)\).

This represents excluding the interval where \(D\) would cause real roots of the derivative equation, ensuring positive slope throughout.

Answer 2

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).