Consider the function f(x) = [x+1] - [x] + 1/x^2 - 1 , where [×] represents the greatest integer function. The function f(x) is decreasing in the inte - Mathematics MCQ Mathematics Mock Test - 10 2026 | ExamTest.live

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Consider the function f (x) = [x + 1] - [x] + \frac{1}{x ^{2} - 1}, where [\cdot] represents the greatest integer function. The function f (x) is decreasing in the interval:

Solution & Step-by-step Explanation

Using property [x + I] = [x] + I, we have [x + 1] - [x] = 1 . f (x) = 1 + \frac{1}{x ^{2} - 1} = \frac{x ^{2} - 1 + 1}{x ^{2} - 1} = \frac{x ^{2}}{x ^{2} - 1} . f^{'} (x) = \frac{( x ^{2} - 1 ) ( 2 x ) - ( x ^{2} ) ( 2 x )}{( x ^{2} - 1 ) ^{2}} = \frac{2 x ^{3} - 2 x - 2 x ^{3}}{( x ^{2} - 1 ) ^{2}} = \frac{- 2 x}{( x ^{2} - 1 ) ^{2}} .For the function to be decreasing, f^{'} (x) < 0 . \frac{- 2 x}{( x ^{2} - 1 ) ^{2}} < 0 . Since the denominator is always positive (where defined), we need - 2 x < 0 ⟹ x > 0 .The function is not defined at x = 1 .Thus, x \in (0, 1) \cup (1, \infty) .

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Consider the function f (x) = [x + 1] - [x] + \frac{1}{x ^{2} - 1}, where [\cdot] represents the greatest integer function. The function f (x) is decreasing in the interval:

(- 1, 1)

(- \infty, - 1) \cup (- 1, 0)

(- \infty, \infty) - {- 1, 1}

(0, 1) \cup (1, \infty)

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