If ƒ(x) = ax 3 + bx 2 + cx + d where a, b, c, d are real numbers and 3b 2 < c 2 , is an increasing cubic function and g(x) = aƒ ′ (x) + bƒ ′′ (x) + c 2 , then -
dt is a decreasing function
dt is an increasing function
dt is a neither increasing nor decreasing function
Text Solution
Verified by ExpertsB
ƒ ′ (x) = 3ax 2 + 2bx + c > 0 (since ƒ(x) is increasing)
⇒ a > 0 and b 2 – 3ac < 0
⇒ a > 0 and b 2 < 3ac
also, g(x) = aƒ ′ (x) + bƒ ′′ (x) + c 2
g(x) = 3a 2 x 2 + 2abx + ac + 6abx + 2b 2 + c 2
g(x) = 3a 2 x 2 + 8abx + (2b 2 + c 2 + ac)
where D = 64a 2 b 2 – 4 · 3a 2 · (2b 2 + c 2 + ac)
= 4a 2 (16b 2 – 6b 2 – 3c 2 – 3ac)
= 4a 2 (10b 2 – 3c 2 – 3ac) < 4a 2 (10b 2 – 3c 2 – b 2 )
{as 3ac > b 2 ⇒ – 3ac < – b 2 }
= 4a 2 (9b 2 – 3c 2 )
= 12a 2 (3b 2 – c 2 ) {given 3b 2 < c 2 }
∴ D > 0
⇒ g(x) < 0, ∀ x ∈ R
∴ 
dt is an increasing function.
Hence is the correct answer.
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