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Question

Let $ f(x) = x - [x] $, where $ x \ge 0 $ and $ [x] $ is the greatest integer not larger than x. Then $ f(x) $ is a

The correct answer is
linearly increasing function between two integers

Understanding the Fractional Part Function

The function given is \( f(x) = x - [x] \), where \( x \ge 0 \). The term \( [x] \) denotes the greatest integer function, which returns the largest integer less than or equal to \( x \). The function \( f(x) \) specifically calculates the fractional part of \( x \).

Analyzing Behavior Between Integers

Consider the function's behavior over an interval between two consecutive integers, say \( [n, n+1) \), where \( n \) is any non-negative integer.

  • Within the interval \( [n, n+1) \), the value of the greatest integer function is constant: \( [x] = n \).
  • Substituting this into the function definition, we get \( f(x) = x - n \) for \( x \in [n, n+1) \).
  • This expression \( f(x) = x - n \) is a linear equation with a slope of \( 1 \).

Interpreting the Function's Nature

Because \( f(x) = x - n \) has a positive slope (\( 1 \)) within the interval \( [n, n+1) \), the function increases linearly as \( x \) increases in this range.

  • At \( x = n \), \( f(x) = n - n = 0 \).
  • As \( x \) increases towards \( n+1 \), \( f(x) \) increases. For example, if \( x \) is just below \( n+1 \), \( f(x) \) will be just below \( (n+1) - n = 1 \).

While the function drops to 0 at every integer value of \( x \) (since \( f(n+1) = (n+1) - [n+1] = (n+1) - (n+1) = 0 \)), its behavior *between* any two consecutive integers is consistently linear and increasing.

Therefore, the function \( f(x) = x - [x] \) is a linearly increasing function between two integers.

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Important Questions from Functions Of Single Variable

  1. The gradient of $y = 3x^2 \sin(2x)$ at (0.2, 1) is __________ (rounded off to three decimal places).
  2. Consider the hyperbolic functions in Group – 1 and their definitions in Group - 2.

        Group - 1     Group - 2
    P$\tanh x$I$\frac{e^x + e^{-x}}{e^x - e^{-x}}$
    Q$\coth x$II$\frac{2}{e^x + e^{-x}}$
    R$\text{sech } x$III$\frac{2}{e^x - e^{-x}}$
    S$\text{cosech } x$IV$\frac{e^x - e^{-x}}{e^x + e^{-x}}$

    The correct combination is

  3. The equation of the straight line representing the tangent to the curve $y = x^2$ at the point $(1,1)$ is
  4. The figure which represents $y = \frac{\sin x}{x}$ for $x > 0$ (x in radians) is
  5. Consider the function $y = e^x$. The slope of this function at $x = 10$ is
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