Question

If \[f\left( x \right) = \left[ {x - 2} \right]\], where \[\left[ x \right]\] denotes the greatest integer less than or equal to \[x\], then \[f'\left( {2.5} \right)\] is equal to
A. \[\dfrac{1}{2}\]
B. \[0\]
C. \[1\]
D. Does not exist

Answer 1

Hint: The function \[\left[ x \right]\] denotes the greatest integer less than or equal to \[x\]. The value of this type of function varies on each interval of two successive integers. Since the given point \[2.5\] lies between two successive integers \[2\] and \[3\]. Find the value of the given function in the interval \[\left[ {2,3} \right)\] and then differentiate it.

Formula Used:
\[x \in \left[ {a,b} \right) \Rightarrow a \le x < b\]
Derivative of a constant is zero.

Complete step-by-step solution:
The given function is \[f\left( x \right) = \left[ {x - 2} \right]\], where \[\left[ x \right]\] denotes the greatest integer less than or equal to \[x\]
Putting \[x = 2.5\], we get
\[f\left( {2.5} \right) = \left[ {2.5 - 2} \right] = \left[ {0.5} \right]\]
According to the definition of \[\left[ x \right]\],
\[\left[ x \right] = 0\], when \[x \in \left[ {0,1} \right)\] i.e. when \[0 \le x < 1\]
Similarly, \[\left[ {x - 2} \right] = 0\], when \[0 \le x - 2 < 1\]
Adding \[2\] on all the sides of the inequation, we get
\[2 \le x < 3\] i.e. \[x \in \left[ {2,3} \right)\]
So, \[f\left( x \right) = \left[ {x - 2} \right] = 0\], when \[x \in \left[ {2,3} \right)\]
, so \[f\left( {2.5} \right) = \left[ {0.5} \right] = 0\]
Differentiating \[f\left( x \right)\] with respect to \[x\], we get \[f'\left( x \right) = 0\], the derivative of a constant being zero.

Note: To solve the question, first find the sample value of \[f(x-2)\] at \[x=2.5\]. We know that,\[\left[ x \right]\] gives the greatest integer less than or equal to x . The value of \[2.5-2=0.5\]. The value of \[[0.5] = 0\].