A Tibetan monk leaves the monastery at 7:00 A. M. and takes his usual path to the top of the mountain, arriving at 7:00 P. M. The following morning, he starts at 7:00 A. M. at the top and takes the same path back, arriving at the monastery at 7:00 P. M. Is there a point on the path that the monk will cross at exactly the same time of day on both days.

There is a point the monk passes same time on both days.and this is explained by the Intermediate value theorem.

Explanation: The intermediate value theorem states that If f is a real-valued continuous function on the interval [a, b], and u is a number between f(a) and f(b), then there is a c ∈ [a, b] such that f(c) = u

Let us assume that f1(t) represents the distance that the monk has traveled away from the monastery on the first day where t represents any time between 7 a.m and 7 p.m.

Then assume that f2(t) represents the distance that the monk has traveled from the monastery on day 2 and assume that the distance between the top of the mountain and the monastery is d.

The monk begins to walk at 7 a.m. So, f1(7a.m)=0 and at 7 p.m he reaches the monastery f1(7p.m)=d

Similarly, f2(7a.m)=d and f2(7p.m)=0

The path taken by the monk is a continuous function. Then consider the function g(t)=f1(t)−f2(t).

Given that both the functions are continuous, so g(7p.m)=d.

Thus, by the Intermediate Value Theorem, there must exist a time

c between 7 a.m and 7 p.m such that g(c)=0.

So g(c)=0

f1(c)−f2(c)=0

f1(c)=f2(c)

So c is the point in time in which the monk is at the same distance from the monastery towards the top of the mountain as he is from the top of the mountain towards the monastery.

me no say

answer; beware of exaggerated claims without supporting evidence;