Question

If at each point of the curve \[y = {x^3} - a{x^2} + x + 1\], tangent is inclined at an acute angle with the positive direction of the x-axis then
1) \[a \leqslant \sqrt 3 \]
 2) \[a > 0\]
3) \[ - \sqrt 3 < a \leqslant \sqrt 3 \]
4) None of these

Answer 1

Hint: We are given an equation of the curve. Since the tangent is inclined at an acute angle, this means that \[\theta < {90^\circ }\]. Next, we will apply a derivative on both sides with respect to x and so, the slope of the tangent \[\dfrac{{dy}}{{dx}} = \tan \theta > 0\] . Next, we will find the value of \[\Delta \] which is less than zero. We know that, \[\Delta = {b^2} - 4ac\] and using this, we will find the final output.

Complete step-by-step answer:
Given that, the tangent is inclined at an acute angle, \[\theta < {90^\circ }\]
Also, given that, the curve is \[y = {x^3} - a{x^2} + x + 1\].
Applying derivate on both the sides with respect to x, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 3{x^2} - 2ax + 1\]
As we know that, slope of the tangent is \[\dfrac{{dy}}{{dx}} = \tan \theta \]
Since, \[\theta < {90^\circ }\] then, \[\tan \theta > 0\].
\[ \Rightarrow \dfrac{{dy}}{{dx}} > 0\]
\[ \Rightarrow 3{x^2} - 2ax + 1 > 0\] for all x which belong to the real number.
This will be true if and only if \[a > 0\] and \[\Delta < 0\]
Now, we will find this value,
\[\Delta < 0\]
\[ \Rightarrow {( - 2a)^2} - 4(3)(1) < 0\] \[(\because \Delta = {b^2} - 4ac)\]
\[ \Rightarrow 4{a^2} - 12 < 0\]
\[ \Rightarrow 4{a^2} < 12\]
\[ \Rightarrow {a^2} < 3\]
\[\therefore - \sqrt 3 < a < \sqrt 3 \]
Hence, for each point of the curve, the tangent is inclined at the acute angle with positive direction of x-axis then the value is \[ - \sqrt 3 < a < \sqrt 3 \] .
So, the correct answer is “Option B”.

Note: A tangent is a line which represents the slope of a curve at that point. A slope of a line is calculated by dividing the change in height by the change in horizontal distance. Tangent is a straight line (or smooth curve) that touches a given curve at one point and at that point the slope of the curve is equal to that of the tangent.