Question

If $f\left( x \right) = \cos \left\{ {\left( {\dfrac{\pi }{2}} \right)\left[ x \right] - {x^3}} \right\},1 < x < 2$ and $\left[ x \right] = $ greatest integer $ \leqslant $, then $f'\left( {\sqrt[3]{{\dfrac{\pi }{2}}}} \right)$
a)$0$
b)${\left( {\dfrac{\pi }{2}} \right)^{\dfrac{1}{3}}}$
c) $3{\left( {\dfrac{\pi }{2}} \right)^{\dfrac{1}{3}}}$
d) $ - 3{\left( {\dfrac{\pi }{2}} \right)^{\dfrac{3}{2}}}$

Answer 1

Hint: First we need to simplify the given function in a form in which it is differentiable within the given domain, as due to the greatest integer function it is not differentiable, as the greatest integer function is not a differentiable function. Then we have to differentiate the function, then find the required value as mentioned in the question.

Complete step by step answer:
The given function is, $f\left( x \right) = \cos \left\{ {\left( {\dfrac{\pi }{2}} \right)\left[ x \right] - {x^3}} \right\}$ in the given domain $1 < x < 2$.
Given, $\left[ x \right] = $greatest integer function.
Since, the domain is $1 < x < 2$, so for the greatest integer function every value of the domain will be more than $1$ and less than $2$.
So, in the greatest integer function it will be $1$ in the given domain.
So, the given function can be re-written as,
$f\left( x \right) = \cos \left\{ {\left( {\dfrac{\pi }{2}} \right).1 - {x^3}} \right\}$
$ \Rightarrow f\left( x \right) = \cos \left\{ {\left( {\dfrac{\pi }{2}} \right) - {x^3}} \right\}$
Now, we know, $\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta $, so, using this property, we get,
$ \Rightarrow f\left( x \right) = \sin {x^3}$
It is now in the form of a differentiable function.
Now, differentiating the function with respect to $x$, we get,
$f'\left( x \right) = \cos {x^3}.\dfrac{d}{{dx}}\left( {{x^3}} \right)$
[Using chain rule]
$ \Rightarrow f'\left( x \right) = 3{x^2}\cos {x^3}$
Therefore, as per the question, we are to find $f'\left( {\sqrt[3]{{\dfrac{\pi }{2}}}} \right)$.
So, $f'\left( {\sqrt[3]{{\dfrac{\pi }{2}}}} \right) = 3{\left( {\sqrt[3]{{\dfrac{\pi }{2}}}} \right)^2}\cos {\left( {\sqrt[3]{{\dfrac{\pi }{2}}}} \right)^3}$
$ = 3{\left( {\dfrac{\pi }{2}} \right)^{\dfrac{2}{3}}}\cos {\left( {\dfrac{\pi }{2}} \right)^{\dfrac{3}{3}}}$
$ = 3{\left( {\dfrac{\pi }{2}} \right)^{\dfrac{2}{3}}}\cos \left( {\dfrac{\pi }{2}} \right)$
Now, we know, $\cos \left( {\dfrac{\pi }{2}} \right) = 0$, substituting this in above equation, we get,
$f'\left( {\sqrt[3]{{\dfrac{\pi }{2}}}} \right) = 3{\left( {\dfrac{\pi }{2}} \right)^{\dfrac{2}{3}}}.0$
$ \Rightarrow f'\left( {\sqrt[3]{{\dfrac{\pi }{2}}}} \right) = 0$
Therefore, the required solution is $0$, the correct option is (a).

Note:
Some of the functions we see and use daily are not differentiable, may be continuous but they are not differentiable, like in this problem the greatest integer function, the most commonly used modulus function, these are not differentiable functions. So, for making a function differentiable involving these, we have to operate these functions first.