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प्रश्न
The function f(x) = 4 sin3x – 6 sin2x + 12 sinx + 100 is strictly ______.
विकल्प
Increasing in `(pi, (3pi)/2)`
Decreasing in `(pi/2, pi)`
Decreasing in `[(-pi)/2, pi/2]`
Decreasing in `[0, pi/2]`
उत्तर
The function f(x) = 4 sin3x – 6 sin2x + 12 sinx + 100 is strictly decreasing in `(pi/2, pi)`.
Explanation:
Here, f(x) = 4 sin3x – 6 sin2x + 12 sin x + 100
f'(x) = 12 sin2x · cos x – 12 sin x cos x + 12 cos
= 12 cos x [sin2x – sin x + 1]
= 12 cos x [sin2x + (1 – sin x)]
∵ 1 – sin x ≥ 0 and sin2x ≥ 0
∴ sin2x + 1 – sin x ≥ 0 .....(when cos x > 0)
Hence, f'(x) > 0, when cos x > 0 i.e., `x ∈ ((-pi)/2, pi/2)`
So, f(x) is increasing where `x ∈ ((-pi)/2, pi/2)` and f'(x) < 0
When cos x < 0 i.e. `x ∈ (pi/2, (3pi)/2)`
Hence, (x) is decreasing when `x ∈ (pi/2, (3pi)/2)`
As `(pi/2, pi) ∈ (pi/2, (3pi)/2)`
So f(x) is decreasing in `(pi/2, pi)`.
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