If the tangent drawn to the curve y=x^3-ax^2+ | vedclass

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(D) Given the curve $y = x^3 - ax^2 + x + 1$. The slope of the tangent is given by $\frac{dy}{dx} = 3x^2 - 2ax + 1$. Since the tangent makes an acute

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$(-\sqrt{3}, \sqrt{3})$

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(D) Given the curve $y = x^3 - ax^2 + x + 1$. The slope of the tangent is given by $\frac{dy}{dx} = 3x^2 - 2ax + 1$. Since the tangent makes an acute angle with the positive direction of the $X$-axis,the slope must be positive for all $x \in R$. Thus,$3x^2 - 2ax + 1 > 0$ for all $x \in R$. For a quadratic expression $Ax^2 + Bx + C > 0$ to be true for all $x \in R$,we must have $A > 0$ and the discriminant $D Here,$A = 3 > 0$,which is satisfied. Now,we require $D = B^2 - 4AC Substituting the values,$(-2a)^2 - 4(3)(1) $4a^2 - 12 $a^2 - 3 $(a - \sqrt{3})(a + \sqrt{3}) Therefore,$a \in (-\sqrt{3}, \sqrt{3})$.

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

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