Application of DerivativesmediumFree

Values of a Making (a+2)x^3 - 3ax^2 + 9ax - 1 Decreasing for All x | JEE

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Question
The set of values of aa for which the function
f(x)=(a+2)x3−3ax2+9ax−1f(x) = (a+2)x^{3} - 3ax^{2} + 9ax - 1
 is decreasing for all real xx, is
Aa≤−3a \le -3correct
Ba≤−2a \le -2
Ca≥−3a \ge -3
Da≥3a \ge 3
Solution
Step 1: Differentiate.
f′(x)=3(a+2)x2−6ax+9a.f'(x) = 3(a+2)x^{2} - 6ax + 9a .
 Step 2: Write the condition and divide by the common factor 33.
f′(x)≤0 ∀x  ⟹  (a+2)x2−2ax+3a≤0  ∀x∈R.f'(x) \le 0 \ \forall x \implies (a+2)x^{2} - 2ax + 3a \le 0 \ \ \forall x \in \mathbb{R} .
 Step 3: Recall when a quadratic Ax2+Bx+CAx^{2}+Bx+C is ≤0\le 0 for every xx: its parabola must open downwards and not cross the axis, i.e.
A<0andD=B2−4AC≤0.A < 0 \quad \text{and} \quad D = B^{2}-4AC \le 0 .
 Step 4: Apply the first condition.
a+2<0  ⟹  a<−2.a + 2 < 0 \implies a < -2 .
 Step 5: Apply the second, with A=a+2A = a+2, B=−2aB = -2a, C=3aC = 3a.
D=4a2−4(a+2)(3a)=4a2−12a2−24a=−8a2−24a=−8a(a+3)≤0.D = 4a^{2} - 4(a+2)(3a) = 4a^{2} - 12a^{2} - 24a = -8a^{2} - 24a = -8a(a+3) \le 0 .
 Step 6: Divide by −8-8, which flips the inequality.
a(a+3)≥0  ⟹  a≤−3  or  a≥0.a(a+3) \ge 0 \implies a \le -3 \ \text{ or } \ a \ge 0 .
 Step 7: Intersect Steps 4 and 6. The branch a≥0a \ge 0 is incompatible with a<−2a < -2, leaving
a≤−3.a \le -3 .
 Step 8: (Check the boundary.) At a=−3a = -3: f′(x)=3(−1)x2+18x−27=−3(x2−6x+9)=−3(x−3)2≤0f'(x) = 3(-1)x^{2}+18x-27 = -3\left(x^{2}-6x+9\right) = -3(x-3)^{2} \le 0, so a=−3a = -3 is included. Answer: (1).
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