Question

Find ‘a’ for which roots of the equation \[{{x}^{3}}-3x+a\] are real and distinct.

Answer 1

Hint: Assume the given equation as f (x). Differentiate f (x) to find f’(x) and substitute it equal to 0 to find the values of x. Now, differentiate the function again to find f’’(x). Find the point of maxima and minima by substituting the obtained value of x in f’’(x). If f’’(x) < 0 at any value of x then it will be a point of maxima and if f’’(x) > 0 then it will be a point of minima. At the point of maxima apply the condition f (x) > 0 and at the point of minima apply the condition f (x) < 0 and find the value of ‘a’ as some intervals and take the intersection of the two intervals and take the intersection of the two intervals obtained to get the answer.

Complete step-by-step solution
We have been provided with the equation: -
\[\Rightarrow f\left( x \right)={{x}^{3}}-3x+a\]
On differentiation, we get,
\[\Rightarrow f'\left( x \right)=3{{x}^{2}}-3\]
Substituting f’ (x) = 0, we get,
\[\begin{align}
  & \Rightarrow 3{{x}^{2}}-3=0 \\
 & \Rightarrow {{x}^{2}}=1 \\
 & \Rightarrow x=\pm 1 \\
\end{align}\]
Now, let us check the point of minima and maxima. On differentiating f’ (x), we get,
\[\Rightarrow \] f’’ (x) = 6x
(i) At x = 1, we have,
\[f''\left( x \right)=f''\left( 1 \right)=6\times 1=6>0\]
Hence, x = 1 is a point of minima.
(ii) At x = -1, we have,
\[f''\left( x \right)=f''\left( -1 \right)=6\times \left( -1 \right)=-6\]
Hence, x = -1 is a point of maxima.
Now, to find the minimum and maximum value of f (x) we have to substitute x = 1 and x = -1 in f (x) respectively. So, we have,
\[\begin{align}
  & \Rightarrow {{\left[ f\left( x \right) \right]}_{\max }}=f\left( -1 \right)={{\left( -1 \right)}^{3}}-3\times \left( -1 \right)+a=a+2 \\
 & \Rightarrow {{\left[ f\left( x \right) \right]}_{\min }}=f\left( 1 \right)={{\left( 1 \right)}^{3}}-3\times \left( 1 \right)+a=a-2 \\
\end{align}\]
We have been given that f (x) should have real and distinct roots. Since f (x) is a cubic equation, so it must have 3 roots. For these 3 roots to be distinct they must cut the x-axis at 3 different places and one of the roots must lie between -1 and 1. Hence, the function f (x) must change its sign from positive to negative from -1 to 1. Therefore, we have,
(a) f (-1) > 0, since it is a point of maxima.
\[\begin{align}
  & \Rightarrow a+2>0 \\
 & \Rightarrow a>-2 \\
\end{align}\]
(b) f (1) < 0, since it is a point of minima.
\[\begin{align}
  & \Rightarrow a-2<0 \\
 & \Rightarrow a>2 \\
\end{align}\]
From the above two cases we can say that: -
\[\Rightarrow a\in \left( -2,2 \right)\]

Note: One may note that we have used the double differentiation method to find the point of maxima and minima. One can use the single derivative test also. Since we have to satisfy both the conditions (a) and (b) and therefore, we have taken the intersection of the sets of values obtained in the two cases. One must remember that a cubic equation will have 3 real and distinct roots only when it will cut the x-axis at three different places.