What will be the value of x for which the value of cosx is minimum?
(d) Cannot be determined
I have been asked this question in an international level competition.
This question is from Calculus Application topic in chapter Application of Calculus of Mathematics – Class 12
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Correct option is (b) -1
Easiest explanation: Let, f(x) = cosx
Then, f’(x) = -sinx and f”(x) = -cosx
At an extreme point of f(x) we must have,
f’(x) = 0
Or -sinx = 0
Or x = nπ where, n – any integer.
If n is an odd integer i.e., n = 2m + 1 where m is any integer, then at,
x = (2m + 1)π we have, f”(x) = [(2m + 1)π] = -cos(2mπ + π) = -cosπ = -1(-1) = 1
So, f”(x) is positive at x = (2m + 1)π
Hence, f(x) = cosx is minimum at x = (2m + 1)π.
So, the minimum value of cosx is cos(2mπ + π) = cosπ = -1.