Let f(x) = 3sin?x + 10sin³x + 6sin²x - 3, x ∈ [-π/6, π/2]. Then, f is
Let f(x) = 3sin?x + 10sin³x + 6sin²x - 3, x ∈ [-π/6, π/2]. Then, f is
Option 1 -
decreasing in (-π/6, 0)
Option 2 -
increasing in (0, π/2)
Option 3 -
decreasing in (0, π/6)
Option 4 -
increasing in (-π/6, 0)
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1 Answer
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Correct Option - 2
Detailed Solution:f' (x) = 12sin³xcosx+30sin²xcosx+12sinxcosx = 3sin2x (2sin²x+5sinx+2) = 3sin2x (2sinx+1) (sinx+2).
In [-π/6, π/2], sinx+2>0. 2sinx+1>0 except at x=-π/6. sin2x>0 for x∈ (0, π/2), <0 for x∈ (-π/6,0).
So f' (x)<0 on (-π/6,0) (decreasing) and f' (x)>0 on (0, π/2) (increasing).
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->
->Area (D) = |xy| = |x (– 2x2 + 54x)|
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->at x = 0, minima
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