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If the tangent drawn to the curve $y=x^3-a x^2+x+1$ at each point $x \in \mathbb{R}$, is inclined at an acute angle with the positive direction of $\mathrm{X}$ - axis, then the set of all possible values of ' $a$ ' is
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Verified Answer
The correct answer is:
$(-\sqrt{3}, \sqrt{3})$
Given $y=x^3-a x^2+x+1$
$$
\Rightarrow \frac{d y}{d x}=3 x^2-2 a x+1
$$
Since $\theta$ is an acute angle. Hence $\tan \theta>0$
$$
\Rightarrow 3 x^2-2 a x+1>0
$$
Which is true off $A>0$ and $D < 0$
$$
\begin{aligned}
& \text { Here } A=3>0 \\
& \Rightarrow D < 0 \\
& \Rightarrow B^2-4 A C < 0 \\
& \Rightarrow(2 a)^2-4(3)(1) < 0 \\
& \Rightarrow(a-\sqrt{3})(a+\sqrt{3}) < 0 \\
& \Rightarrow a \in(-\sqrt{3}, \sqrt{3})
\end{aligned}
$$
$$
\Rightarrow \frac{d y}{d x}=3 x^2-2 a x+1
$$
Since $\theta$ is an acute angle. Hence $\tan \theta>0$
$$
\Rightarrow 3 x^2-2 a x+1>0
$$
Which is true off $A>0$ and $D < 0$
$$
\begin{aligned}
& \text { Here } A=3>0 \\
& \Rightarrow D < 0 \\
& \Rightarrow B^2-4 A C < 0 \\
& \Rightarrow(2 a)^2-4(3)(1) < 0 \\
& \Rightarrow(a-\sqrt{3})(a+\sqrt{3}) < 0 \\
& \Rightarrow a \in(-\sqrt{3}, \sqrt{3})
\end{aligned}
$$
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