Question

If the points of local extremum of \[f(x) = {x^3} - 3a{x^2} + 3({a^2} - 1)x + 1\] lies between -2 and 4 , then ‘a’ belongs to
A. \[( - 2,2)\]
B. \[( - \infty , - 1) \cup (3,\infty )\]
C. \[( - 1,3)\]
D. \[(3,\infty )\]

Answer 1

Hint:
To find the points of the local extremum, first, we need to differentiate the given function \[f(x)\] . The differential equation we get by differentiating \[f(x)\] will lead us to 4 more new equations.
(1) \[D > 0\]
(2) \[f'( - 2) > 0\]
(3) \[f'(4) > 0\]
(4) \[ - 2 < \dfrac{{ - B}}{{2A}} < 4\]
We have to solve these equations and then compare all four results to observe the coinciding range of a to get to the final answer.

Complete step by step solution:
The given function \[f(x)\] is
 \[f(x) = {x^3} - 3a{x^2} + 3({a^2} - 1)x + 1\] … (1)

We know that,
 \[\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}\]

Hence, differentiating (1) with respect to x, we get
 \[ \Rightarrow f'(x) = 3{x^2} - 6ax + 3({a^2} - 1)\]
Taking 3 common we get,
 \[ \Rightarrow f'(x) = 3({x^2} - 2ax + {a^2} - 1)\] … (2)

Now, since roots of (2) are real and distinct and the points of local extremum lie between -2 and 4, we get
 \[D > 0\] … (3)
 \[f'( - 2) > 0\] … (4)
 \[f'(4) > 0\] … (5)
 \[ - 2 < \dfrac{{ - B}}{{2A}} < 4\] … (6)

We have \[{x^2} - 2ax + {a^2} - 1\] , where \[A = 1,\,B = ( - 2a)\,,\,C = {a^2} - 1\]

Using \[D > 0\] , we get
 \[ \Rightarrow {B^2} - 4AC > 0\]
On substituting the values of A, B, and C, we get,
 \[ \Rightarrow 4{a^2} - 4({a^2} - 1) > 0\]
On simplification we get,
 \[ \Rightarrow 4{a^2} - 4{a^2} + 4 > 0\]
Hence we get,
 \[ \Rightarrow 4 > 0\]

So for any real value of a, the equation is satisfied, hence
 \[ \Rightarrow a \in R\] …. (7)

Using \[f'( - 2) > 0\] , we get
 \[ \Rightarrow 3({( - 2)^2} - 2a( - 2) + {a^2} - 1) > 0\]
On simplification we get,
 \[ \Rightarrow 4 + 4a + {a^2} - 1 > 0\]
On adding like terms we get,
 \[ \Rightarrow {a^2} + 4a + 3 > 0\]

Solving by middle term split, we get
 \[ \Rightarrow {a^2} + a + 3a + 3 > 0\]
On taking factors common we get,
 \[ \Rightarrow a(a + 1) + 3(a + 1) > 0\]
On taking \[(a + 1)\] common we get,
 \[ \Rightarrow (a + 1)(a + 3) > 0\]

Hence, \[a \notin [ - 3, - 1]\] , so
 \[a < - 3\] and \[a > - 1\] … (8)

Using \[f'(4) > 0\] , we get
 \[ \Rightarrow 3({4^2} - 2a \times 4 + {a^2} - 1) > 0\]
On simplification we get,
 \[ \Rightarrow 16 - 8a + {a^2} - 1 > 0\]
On adding like terms we get,
 \[ \Rightarrow {a^2} - 8a + 15 > 0\]

Solving by middle term split, we get
 \[ \Rightarrow {a^2} - 3a - 5a + 15 > 0\]
On taking factors common we get,
 \[ \Rightarrow a(a - 3) - 5(a - 3) > 0\]
On taking \[(a - 3)\] common we get,
 \[ \Rightarrow (a - 3)(a - 5) > 0\] … (9)

Hence, \[a \notin [3,5]\] , so
 \[a < 3\] and \[a > 5\]

Using \[ - 2 < \dfrac{{ - B}}{{2A}} < 4\] , we get
On substituting the value of A and B we get,
 \[ \Rightarrow - 2 < \dfrac{{2a}}{2} < 4\]
On simplification we get,
 \[ \Rightarrow - 2 < a < 4\] … (10)

Now, using (7), (8), (9), (10), we get the coinciding range of a as
 \[ \Rightarrow - 1 < a < 3\]

Hence, the final answer is C.

Note:
In the above equation, we can get a rough idea of local extremums by trying to observe the equation of \[f'(x)\] . In the above question, if try to observe the equation \[f'(x) = 3({x^2} - 2ax + {a^2} - 1)\] , we can easily interpret that the given equation is an equation of parabola of the type \['{x^2} = 4ay'\] . Hence it opens upwards, so we can say that the curve at its extremums i.e -2 and 4 is positive. That’s how we got these two equations:
(1) \[f'( - 2) > 0\]
(2) \[f'(4) > 0\]