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Question
Find the maximum and minimum values of f(x) = secx + log cos2x, 0 < x < 2π
Solution
f(x) = secx + 2 log cosx
Therefore, f'(x) = secx tanx – 2 tanx = tanx (secx –2)
f'(x) = 0
⇒ tanx = 0 or secx = 2 or cosx = `1/2`
Therefore, possible values of x are x = 0
or x = π and x = `pi/3` or x = `(5pi)/3`
Again, f′(x) = sec2x (secx –2) + tanx (secx tanx)
= sec3x + secx tan2x – 2sec2x
= secx (sec2x + tan2x – 2secx).
We note that
f′(0) = 1(1 + 0 – 2) = –1 < 0. Therefore, x = 0 is a point of maxima.
f′(π) = –1(1 + 0 + 2) = –3 < 0. Therefore, x = π is a point of maxima.
`"f'"(pi/3)` = 2(4 + 3 – 4) = 6 > 0. Therefore, x = `pi/3` is a point of minima.
`"f'"((5pi)/3)` = 2(4 + 3 – 4) = 6 > 0. Therefore, x = `(5pi)/3` is a point of minima.
Maximum Value of y at x = 0 is 1 + 0 = 1
Maximum Value of y at x = π is –1 + 0 = –1
Minimum Value of y at x = `pi/3` is `2 + 2 log 1/2` = 2(1 – log2)
Minimum Value of y at x = `(5pi)/3` is `2 + 2 log 1/2` = 2(1 – log2)
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