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NCERT Exemplar · Q67

Q.Let f(x)=x−[x];∈Rf(x) = x - [x]; \in \mathbf{R}, then f′(12)f'\left(\dfrac{1}{2}\right) is
(A) 32\dfrac{3}{2}
(B) 11
(C) 00
(D) −1-1

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The fractional-part function f(x)=x−[x]f(x) = x - [x] is constant on any interval not containing an integer, so its derivative at x=12x = \frac{1}{2} (which lies strictly between 00 and 11) is 11.

The function f(x)=x−[x]f(x) = x - [x] is the fractional part of xx, often written {x}\{x\}. Here [x][x] denotes the greatest integer less than or equal to xx (the floor function). Understanding how this function behaves is the key to finding its derivative.

On any interval (n,n+1)(n, n+1) where nn is an integer, the floor function [x][x] is constant: [x]=n[x] = n for all x∈(n,n+1)x \in (n, n+1). Therefore on that interval,

f(x)=x−n,f(x) = x - n,

which is simply a linear function with slope 11.

Since 12\frac{1}{2} lies in the interval (0,1)(0, 1), we have [x]=0[x] = 0 for all xx in a neighborhood of 12\frac{1}{2}. Let's verify this using the definition of the derivative.

Finding f′(12)f'\left(\frac{1}{2}\right) from first principles

The derivative at x=12x = \frac{1}{2} is defined as

f′(12)=lim⁡h→0f(12+h)−f(12)h.f'\left(\frac{1}{2}\right) = \lim_{h \to 0} \frac{f\left(\frac{1}{2} + h\right) - f\left(\frac{1}{2}\right)}{h}.

  1. Evaluate f(12)f\left(\frac{1}{2}\right). Since 12∈(0,1)\frac{1}{2} \in (0,1), we have [12]=0\left[\frac{1}{2}\right] = 0, so

f(12)=12−0=12.f\left(\frac{1}{2}\right) = \frac{1}{2} - 0 = \frac{1}{2}.

  1. Evaluate f(12+h)f\left(\frac{1}{2} + h\right) for small hh. For ∣h∣|h| sufficiently small (specifically ∣h∣<12|h| < \frac{1}{2}), the point 12+h\frac{1}{2} + h still lies in (0,1)(0, 1), so [12+h]=0\left[\frac{1}{2} + h\right] = 0. Thus

f(12+h)=12+h−0=12+h.f\left(\frac{1}{2} + h\right) = \frac{1}{2} + h - 0 = \frac{1}{2} + h.

  1. Compute the difference quotient. …

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