Given: BC ≅ CD and AC bisects ∠BCD
1 answer:
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Answer: c) .
Step-by-step explanation:
Mean value theorem : If f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that
Given function :
Interval : [0,3]
Then, by the mean value theorem, there is at least one number c in the interval (0,3) such that
Since
then, at x=c,
From (i) and (ii), we have
Hence, the correct option is c) .