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Question: If the points of local extremum of \[f\left( x \right)={{x}^{3}}-3a{{x}^{2}}+3\left( {{a}^{2}}-1 \ri...

If the points of local extremum of f(x)=x3−3ax2+3(a2−1)x+1f\left( x \right)={{x}^{3}}-3a{{x}^{2}}+3\left( {{a}^{2}}-1 \right)x+1 lies between −2&4-2\And 4 , then a belongs to-
A.(−2,2)\left( -2,2 \right)
B.(−∞,−1)∪(3,∞)\left( -\infty ,-1 \right)\cup \left( 3,\infty \right)
C.(−1,3)\left( -1,3 \right)
D.(3,∞)\left( 3,\infty \right)

Explanation

Solution

Simplify the given equation and find the first derivative. Then find the discriminate D=b2−4ac{{b}^{2}}-4ac from the equation of the first derivative. Find whether D>0>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(x)=x3−3ax2+3(a2−1)x+1f\left( x \right)={{x}^{3}}-3a{{x}^{2}}+3\left( {{a}^{2}}-1 \right)x+1
On simplifying we get,
⇒f(x)=x3−3ax2+3a2x−3x+1\Rightarrow f\left( x \right)={{x}^{3}}-3a{{x}^{2}}+3{{a}^{2}}x-3x+1
On differentiating the given eq. we get,
⇒f′(x)=ddx[x3−3ax2+3(a2−1)x+1]\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 d(xn)dx=nxn−1\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}
On applying this we get,
⇒f′(x)=3x3−1−3×2ax2−1+3(a2−1)x1−1+0\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,
f′(x)=3x2−6ax+3(a2−1){{\text{f}}^{'}}\left( \text{x} \right)=3{{x}^{2}}-6ax+3\left( {{a}^{2}}-1 \right)
On taking 33 common from the eq. we get,
⇒f′(x)=3[x2−2ax+(a2−1)]\Rightarrow {{\text{f}}^{'}}\left( \text{x} \right)=3\left[ {{x}^{2}}-2ax+\left( {{a}^{2}}-1 \right) \right]
On using the formula D=b2−4ac{{b}^{2}}-4ac we get,
⇒D= (−2a)2−4(a2−1)\Rightarrow D=~{{\left( -2a \right)}^{2}}-4\left( {{a}^{2}}-1 \right)
On simplifying we get,
⇒D= 4a2−4a2+4\Rightarrow D=~4{{a}^{2}}-4{{a}^{2}}+4
⇒D= 4>0\Rightarrow D=~4>0
This means a∈Ra\in R
Now here D=b2−4ac{{b}^{2}}-4ac>00 which means the equation is a parabola.
It is given that the local points of extremum are −2&4-2\And 4 on putting these on the equation, the equation must be greater than zero.
So for f′(−2)>0{{\text{f}}^{'}}\left( -2 \right)>0 , we get
⇒4+4a+a2−1>0\Rightarrow 4+4a+{{a}^{2}}-1>0
⇒a2+4a+3>0\Rightarrow {{a}^{2}}+4a+3>0
On factorizing we get,
⇒a2+3a+a+3>0\Rightarrow {{a}^{2}}+3a+a+3>0
⇒a(a+3)+1(a+3)>0\Rightarrow a\left( a+3 \right)+1\left( a+3 \right)>0
⇒(a+3)(a+1)>0\Rightarrow \left( a+3 \right)\left( a+1 \right)>0
This means that a<−3&a>−1a<-3\And a>-1 --- (i)
And for f′(4)>0{{\text{f}}^{'}}\left( 4 \right)>0 , we get,
⇒16−8a+a2−1>0\Rightarrow 16-8a+{{a}^{2}}-1>0
⇒a2−8a+15>0\Rightarrow {{a}^{2}}-8a+15>0
On factorizing we get,
(a−5)(a−3)>0\left( a-5 \right)\left( a-3 \right)>0
This means that a<3 Or a>5a<3\text{ Or }a>5 --- (ii)
So since it lies in range −2&4-2\And 4 then
⇒−2<−b2a<4\Rightarrow -2<\dfrac{-b}{2a}<4
On putting values we get,
⇒−2<−(−2a)2<4\Rightarrow -2<\dfrac{-\left( -2a \right)}{2}<4
On simplifying we get,
⇒−2<2a2<4\Rightarrow -2<\dfrac{2a}{2}<4
⇒−2Fromeq.(i)(ii)and(iii)itisclearthat\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&4-2\And 4but is also given that a>−1&a<3a>-1\And a<3 so we conclude that the range belongs to (−1,3)\left( -1,3 \right) .