A function f(x) is continuous in the interval [0, 2]. It is known that f(0) = f(2) =-1 and f(1) = 1. Which one of the following statements must be true?

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  1. There exists a y in the interval (0, 1) such that f(y) = f(y + 1)
  2. For every y in the interval (0, 1), f(y) = f(2 – y)
  3. The maximum value of the function in the interval (0, 2) is 1
  4. There exists a y in the interval (0, 1) such that f(y) = -f(2 – y)

Answer (Detailed Solution Below)

Option 1 : There exists a y in the interval (0, 1) such that f(y) = f(y + 1)
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Detailed Solution

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Let’s define a new function t : t(y) = f(y) –  f(y+1)

We know that f is continuous in [ 0 , 2 ], t will also be continuous in [ 0 , 1 ].

In question it is given that f(0)= - 1 , f(2) = -1 and f(1) = 1, putting into above function  we get  t(0) and t(1)

t(0) = f(0) - f(1) = -1 - 1 = - 2 and t(1) = f(1)- f(2) = 1+1 = 2

so we get  t(0)=-2  and  t(1) = 2 since function is changing its value from -2 negative to +2 positive, it means function t is a increasing function and there exists a point where t will be 0 in (0,1)

t=0 then f(y)= f(y+1)

Hence option 1 is the correct answer.

Same logic can be applied to option d:

h(y) = f(y) + f(y+2)

We know that function f is continuous in [0,2], function h will also be continuous in [0,1].

In question it is given that f(0)=-1, f(2)=-1 and f(1)=1, putting into above function.

h(0) = f(0) - f(1) = -1 - 1 = - 2 and h(1) = f(1)- f(2) = 1+1 = 2

So we get h(0)=-2  and  h(1) = 2, since function is changing its value from -2 to +2. It means function h is a increasing function and there exists a point where h would be 0 in (0,1)

h=0 then f(y) = -f(y+2)

Hence option 1 and 4 both are correct answer.
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