If the tangent at any point on the curve `y=x^3-lamdax^2+x+1` makes an acute angle with the +ive direction of x-axis, then

Options

A. `lamdagt0`
B. `lamdalesqrt3`
C. `-sqrt3lelamdalesqrt3`
D. None

Answer

To solve the problem, we need to find the conditions under which the tangent to the curve \( y = x^3 - \lambda x^2 + x + 1 \) makes an acute angle with the positive direction of the x-axis.

### Step-by-Step Solution:

1. **Understanding the Condition for Acute Angle:**
The tangent to the curve makes an acute angle with the positive x-axis if the slope \( m \) of the tangent line is greater than 0. This means we need to ensure that:
\[ m = \frac{dy}{dx} > 0 \]

2. **Finding the Derivative:**
We will differentiate the function \( y = x^3 - \lambda x^2 + x + 1 \) with respect to \( x \):
\[ \frac{dy}{dx} = 3x^2 - 2\lambda x + 1 \]

3. **Setting Up the Inequality:**
We want to find when:
\[ 3x^2 - 2\lambda x + 1 > 0 \]
This is a quadratic inequality in \( x \).

4. **Analyzing the Quadratic Inequality:**
A quadratic expression \( ax^2 + bx + c \) is positive for all \( x \) if its discriminant is less than or equal to zero. Here, \( a = 3 \), \( b = -2\lambda \), and \( c = 1 \).

5. **Calculating the Discriminant:**
The discriminant \( D \) is given by:
\[ D = b^2 - 4ac = (-2\lambda)^2 - 4 \cdot 3 \cdot 1 = 4\lambda^2 - 12 \]

6. **Setting the Discriminant Condition:**
For the quadratic to be non-negative for all \( x \), we require:
\[ 4\lambda^2 - 12 \leq 0 \]

7. **Solving the Inequality:**
Rearranging gives:
\[ 4\lambda^2 \leq 12 \implies \lambda^2 \leq 3 \]
Taking the square root of both sides, we find:
\[ -\sqrt{3} \leq \lambda \leq \sqrt{3} \]

### Conclusion:
The values of \( \lambda \) for which the tangent at any point on the curve makes an acute angle with the positive direction of the x-axis are:
\[ \lambda \in [-\sqrt{3}, \sqrt{3}] \]