Application of DerivativesmediumFree

Values of a When Local Extrema Lie Between -2 and 4 | JEE Advanced

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Question
If the points of local extremum of f(x)=x3−3ax2+3(a2−1)x+1f(x) = x^{3}-3ax^{2}+3\left(a^{2}-1\right)x+1 lie between −2-2 and 44, then aa belongs to
A(−2,2)(-2, 2)
B(−∞,−1)∪(3,∞)(-\infty, -1) \cup (3, \infty)
C(−1,3)(-1, 3)correct
D(3,∞)(3, \infty)
Solution
Step 1: Differentiate.
f′(x)=3x2−6ax+3(a2−1)=3[x2−2ax+a2−1].f'(x) = 3x^{2} - 6ax + 3\left(a^{2}-1\right) = 3\left[x^{2}-2ax+a^{2}-1\right].
 Step 2: Complete the square inside the bracket - this is much quicker than the quadratic formula.
x2−2ax+a2−1=(x−a)2−1.x^{2}-2ax+a^{2}-1 = (x-a)^{2}-1 .
 Step 3: Factorise as a difference of squares.
f′(x)=3[(x−a)−1][(x−a)+1]=3(x−a−1)(x−a+1).f'(x) = 3\left[(x-a)-1\right]\left[(x-a)+1\right] = 3\left(x-a-1\right)\left(x-a+1\right).
 Step 4: Read off the points of local extremum - the two roots of f′(x)=0f'(x) = 0.
x=a+1andx=a−1.x = a+1 \qquad \text{and} \qquad x = a-1 .
 They are always distinct, so both really are extremum points (one maximum, one minimum). Step 5: Impose that both lie strictly between −2-2 and 44. The smaller is a−1a-1 and the larger is a+1a+1, so it is enough that
a−1>−2anda+1<4.a-1 > -2 \qquad \text{and} \qquad a+1 < 4 .
 Step 6: Solve each.
a>−1anda<3.a > -1 \qquad \text{and} \qquad a < 3 .
 Step 7: Combine.
a∈(−1, 3).a \in (-1,\ 3) .
 Answer: (3).
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