MonotonicityHard
Question
Let φ (x) = (f (x))3 -3(f (x))2 + 4f (x) + 5x + 3sin x + 4cos x∀ x ∈ R, then -
Options
A.φ is increasing whenever f is increasing
B.φ is increasing whenever f is decreasing
C.φ is decreasing whenever f is decreasing
D.φ is decreasing if f′(x) = - 11
Solution
φ(x) = f3 (x) - 3f2 (x) + 4f (x) + 5x + 3sin x + 4cos x
φ′(x) = (3f2 (x) - 6f (x) + 4f′(x) + 5 + 3cos x - 4sin x ...(i)
3cos x - 4sin x ≥ - 5
5 + (3cos x - 4sin x) ≥ 0
also 3f2 (x) - 6f (x) + 4 > 0 ∵ D < 0
φ′ (x) > 0 ∀ f′(x) > 0
Now let f′(x) = -11
φ′(x) = - 11(3f2 (x) - 6f (x) + 4) + 5 + 3cos x - 4sin x
Now 3f2 (x) - 6f (x) + 4 ≥ 1
⇒ - 11(3f2 (x) - 6f (x) + 4) ≤ - 11 - (ii)
3cos x - 4sin x ≤ 5
⇒ 5 + (3cos x - 4sin x) ≤ 10 - (iii)
(ii) + (iii)
⇒ -11(3f2 (x) - 6f (x) + 4) + 5 + (3cos x - 4sin x) ≤ - 1
⇒ φ′(x) ≤ - 1
φ′(x) = (3f2 (x) - 6f (x) + 4f′(x) + 5 + 3cos x - 4sin x ...(i)
3cos x - 4sin x ≥ - 5
5 + (3cos x - 4sin x) ≥ 0
also 3f2 (x) - 6f (x) + 4 > 0 ∵ D < 0
φ′ (x) > 0 ∀ f′(x) > 0
Now let f′(x) = -11
φ′(x) = - 11(3f2 (x) - 6f (x) + 4) + 5 + 3cos x - 4sin x
Now 3f2 (x) - 6f (x) + 4 ≥ 1
⇒ - 11(3f2 (x) - 6f (x) + 4) ≤ - 11 - (ii)
3cos x - 4sin x ≤ 5
⇒ 5 + (3cos x - 4sin x) ≤ 10 - (iii)
(ii) + (iii)
⇒ -11(3f2 (x) - 6f (x) + 4) + 5 + (3cos x - 4sin x) ≤ - 1
⇒ φ′(x) ≤ - 1
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