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If at each point of the curve y=x^3-a x^...

If at each point of the curve `y=x^3-a x^2+x+1,` the tangent is inclined at an acute angle with the positive direction of the x-axis, then (a)`a >0` (b) `a<-sqrt(3)` (c)`-sqrt(3)< a< ` `sqrt(3)` (d) `non eoft h e s e`

A

`agt0`

B

`alesqrt3`

C

`-sqrt3lealesqrt3`

D

none of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the curve given by the equation: \[ y = x^3 - ax^2 + x + 1 \] We want to find the conditions under which the tangent to this curve at any point makes an acute angle with the positive direction of the x-axis. ### Step 1: Find the derivative of the curve The slope of the tangent to the curve at any point is given by the derivative \( \frac{dy}{dx} \). We will differentiate the equation with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx}(x^3 - ax^2 + x + 1) \] Using the power rule of differentiation: \[ \frac{dy}{dx} = 3x^2 - 2ax + 1 \] ### Step 2: Set the condition for the slope For the tangent to make an acute angle with the positive x-axis, the slope must be positive: \[ 3x^2 - 2ax + 1 > 0 \] ### Step 3: Analyze the quadratic inequality The expression \( 3x^2 - 2ax + 1 \) is a quadratic in \( x \). For this quadratic to be positive for all \( x \) (i.e., for all points on the curve), its discriminant must be less than zero. The discriminant \( D \) of a quadratic \( Ax^2 + Bx + C \) is given by: \[ D = B^2 - 4AC \] Here, \( A = 3 \), \( B = -2a \), and \( C = 1 \). Thus, the discriminant is: \[ D = (-2a)^2 - 4 \cdot 3 \cdot 1 \] \[ D = 4a^2 - 12 \] ### Step 4: Set the discriminant less than zero For the quadratic to be positive for all \( x \): \[ 4a^2 - 12 < 0 \] ### Step 5: Solve the inequality Rearranging gives: \[ 4a^2 < 12 \] \[ a^2 < 3 \] Taking the square root of both sides, we find: \[ -\sqrt{3} < a < \sqrt{3} \] ### Conclusion The range of \( a \) for which the tangent to the curve makes an acute angle with the positive direction of the x-axis is: \[ -\sqrt{3} < a < \sqrt{3} \] Thus, the correct option is: (c) \(-\sqrt{3} < a < \sqrt{3}\)
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