Question

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

Answer 1

Hint: Simplify the given equation and find the first derivative. Then find the discriminate D=${{b}^{2}}-4ac$ from the equation of the first derivative. Find whether D$>0$ or equal to or less than zero. Then, put the values of its points of local extremum in the equation to find the range of a.

Complete step-by-step answer:

Given, \[f\left( x \right)={{x}^{3}}-3a{{x}^{2}}+3\left( {{a}^{2}}-1 \right)x+1\]
On simplifying we get,
\[\Rightarrow f\left( x \right)={{x}^{3}}-3a{{x}^{2}}+3{{a}^{2}}x-3x+1\]
On differentiating the given eq. we get,
\[\Rightarrow {{\text{f}}^{'}}\left( \text{x} \right)=\dfrac{d}{dx}\left[ {{x}^{3}}-3a{{x}^{2}}+3\left( {{a}^{2}}-1 \right)x+1 \right]\]
We know that differentiation of constant is zero and $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$
On applying this we get,
$\Rightarrow {{\text{f}}^{'}}\left( \text{x} \right)=3{{x}^{3-1}}-3\times 2a{{x}^{2-1}}+3\left( {{a}^{2}}-1 \right){{x}^{1-1}}+0$
On simplifying we get,
${{\text{f}}^{'}}\left( \text{x} \right)=3{{x}^{2}}-6ax+3\left( {{a}^{2}}-1 \right)$
On taking $3$ common from the eq. we get,
$\Rightarrow {{\text{f}}^{'}}\left( \text{x} \right)=3\left[ {{x}^{2}}-2ax+\left( {{a}^{2}}-1 \right) \right]$
On using the formula D=${{b}^{2}}-4ac$ we get,
\[\Rightarrow D=~{{\left( -2a \right)}^{2}}-4\left( {{a}^{2}}-1 \right)\]
On simplifying we get,
\[\Rightarrow D=~4{{a}^{2}}-4{{a}^{2}}+4\]
\[\Rightarrow D=~4>0\]
This means $a\in R$
Now here D=${{b}^{2}}-4ac$>$0$ which means the equation is a parabola.
It is given that the local points of extremum are $-2\And 4$ on putting these on the equation, the equation must be greater than zero.
So for ${{\text{f}}^{'}}\left( -2 \right)>0$ , we get
$\Rightarrow 4+4a+{{a}^{2}}-1>0$
$\Rightarrow {{a}^{2}}+4a+3>0$
On factorizing we get,
$\Rightarrow {{a}^{2}}+3a+a+3>0$
$\Rightarrow a\left( a+3 \right)+1\left( a+3 \right)>0$
$\Rightarrow \left( a+3 \right)\left( a+1 \right)>0$
This means that $a<-3\And a>-1$ --- (i)
And for ${{\text{f}}^{'}}\left( 4 \right)>0$ , we get,
$\Rightarrow 16-8a+{{a}^{2}}-1>0$
$\Rightarrow {{a}^{2}}-8a+15>0$
On factorizing we get,
$\left( a-5 \right)\left( a-3 \right)>0$
This means that $a<3\text{ Or }a>5$ --- (ii)
So since it lies in range $-2\And 4$ then
$\Rightarrow -2<\dfrac{-b}{2a}<4$
On putting values we get,
$\Rightarrow -2<\dfrac{-\left( -2a \right)}{2}<4$
On simplifying we get,
$\Rightarrow -2<\dfrac{2a}{2}<4$
$\Rightarrow -2From eq. (i) (ii) and (iii) it is clear that $a\in \left( -1,3 \right)$
Hence the correct answer is C.

Note: Here the student may get confused between the range. It is clear that the range of a lies between$-2\And 4$but is also given that $a>-1\And a<3$ so we conclude that the range belongs to $\left( -1,3 \right)$ .