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Question
At x = `(5pi)/6`, f(x) = 2 sin3x + 3 cos3x is ______.
Options
Maximum
Minimum
Zero
Neither maximum nor minimum
Solution
At x = `(5pi)/6`, f(x) = 2 sin3x + 3 cos3x is maximum.
Explanation:
We have f(x) = 2 sin 3x + 3 cos 3x
f'(x) = 2 cos 3x · 3 – 3 sin 3x·3 = 6 cos 3x – 9 sin 3x
f'(x) = – 6 sin 3x · 3 – 9 cos 3x · 3
= – 18 sin 3x – 27 cos 3x
`"f''"((5pi)/6) = - 18 sin 3((5pi)/6) - 27 cos 3((5pi)/6)`
= `- 18 sin ((5pi)/2) - 27 cos((5pi)/2)`
= `-18 sin(2pi + pi/2) - 27 cos(2pi + pi/2)`
= `-18sin pi/2 - 27 cos pi/2`
= – 18 · 1 – 27 · 0
= – 18 < 0 maxima
Maximum value of f(x) at x = `(5pi)/6`
`"f"((5pi)/6) = 2 sin 3((5pi)/6) + 3 cos 3((5pi)/6)`
= `2 sin (5pi)/2 + 3 cos (5pi)/2`
= `2 sin (2x + pi/2) + 3cos(2pi + pi/2)`
= `2 sin pi/2 + 3 cos pi/2`
= 2
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