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Grade: 12th pass
        
plese answer the question with explanation...................................
2 years ago

Answers : (1)

Partha Math Expert - askIITians
askIITians Faculty
25 Points
							The extremum of the codomain are 1 and 13. Thus the minimum value of f(x) is 1 and maximum is 13.
Also, we know that:
m(b-a)\leq \int_{a}^{b}f(x)dx \leq M (b-a) where m and M are the mimimum and maximum values of the function.
Since this equality holds in either of the case, f(x) = 1 between 3p and (3p+3)
and f(x) = 13 between 3q and 3q+3.
p2+q2is an odd integer implies that one of p & q is odd and the other is even.
So f(x) can be defined in 2 ways i.e. :f(x) = \left\{\begin{matrix} 1; 6k < x < lt 6k+3 \\ 13 ; 6k+3 < x < 6k + 6 \end{matrix}\right.

OR
f(x) = \left\{\begin{matrix} 13; 6k < x < 6k+3 \\ 1 ; 6k+3 < x < 6k + 6 \end{matrix}\right..

Thus there are only 2 such functions and clearly, both have a period of 6since f(x+6) = f(x).
We know that for a periodic function f(x) with period p,\int_{0}^{kp}f(x)dx = k\int_{0}^{p}f(x)dx , k\in \mathbb{N}
Thus,
\int_{0}^{900}f(x)dx = 150\int_{0}^{6}f(x)dx = 150 (3 + 39) = 150*42 = \mathbf{6300}
At x = 3, the Left Hand Limit = k1and Right Hand Limit = k2
From the definition of f(x), either k1 = 13k2 or k2 = 13k1
2 years ago
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