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

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) $f'(4) > 0$