Question

Which of the following is differentiable in the interval\[(-1,2)\]\[?\]
\[(A)\int\limits_{x}^{2x}{{{(logx)}^{2}}dx}\] $(B)\int\limits_{x}^{2x}{\dfrac{\sin x}{x}dx}$
$(C)\int\limits_{x}^{2x}{\dfrac{1-t+{{t}^{2}}}{1+t+{{t}^{2}}}dt}$ D. None of these

Answer 1

Hint: In this type of question you can check first if the function is continuous in the given interval or not , and then check if the domain of its derivative is having the range of the given interval , and if both the conditions are satisfied for a given function then it is differentiable in the given interval.

Complete answer:
We will check differentiability of first option i.e. option \[(A)\]:
\[f(x)=\int\limits_{x}^{2x}{{{(\log x)}^{2}}dx}\] \[x\in (-1,2)\]
Since the function \[f(x)\] containing \[\log x\] and we know that \[\log x\] is defined in \[x>0\] that is, it is not defined for negative numbers, in other words \[x\notin (-1,0)\] and this interval is in the given interval.
Therefore, \[f(x)=\int\limits_{x}^{2x}{{{(\log x)}^{2}}dx}\] is not differentiable in the interval \[(-1,2)\] ,
So, option \[(A)\]is wrong.
Now, we will check differentiability of second option i.e. option\[(B)\]:
\[g(x)=\int\limits_{x}^{2x}{\dfrac{\sin x}{x}dx}\] \[x\in (-1,2)\]
Since it is clearly visible that function \[g(x)\] is not defined at \[x=0\] i.e.\[x\notin 0\]
Although we can show it by evaluating further \[\int\limits_{x}^{2x}{\dfrac{\sin x}{x}dx}\]
So, by using Leibniz Integral Rule we got,
\[g(x)'=2\left( \dfrac{\sin 2x}{2x} \right)-1\left( \dfrac{\sin x}{x} \right)\]
\[=\dfrac{\sin 2x-\sin x}{x}\] \[........(i)\]
Hence, it is clearly visible above equation \[(i)\] that is derivative of function \[g(x)\] cannot be defined for \[x=0\].And we have \[x=0\] in the given interval.
Therefore, the derivative of the function \[g(x)\] cannot be defined in the interval \[(-1,2)\].
So, option \[(B)\] is wrong.
Now, we will check differentiability of second option i.e. option \[(C)\]:
Let \[h(x)=\int\limits_{x}^{2x}{\dfrac{1-t+{{t}^{2}}}{1+t+{{t}^{2}}}dt}\] \[x\in (-1,2)\]
Now, we will check if the denominator of \[h(x)\] can be zero or not for \[x\in (-1,2)\].
So, we will find it by evaluating the denominator of \[h(x)\] since it is a quadratic function.
Here we have equation \[1+t+{{t}^{2}}=0\]
On comparing it with \[a{{x}^{2}}+bx+c=0\]
We have \[a=1,b=1,c=1\]
Discriminant:
\[(D)={{b}^{2}}-4ac\]
 \[\begin{align}
  & =(1)(1)-4(1)(1) \\
 & =1-4 \\
 & =-3 \\
\end{align}\]
That is we got \[D<0\].
Hence we can conclude that the denominator i.e. \[1+t+{{t}^{2}}\] cannot be zero for any real value of \[x\].
Hence we can conclude for now that the function \[h(x)\] is defined in the interval \[x\in (-1,2)\].
Now, we will check that the derivative of the function \[h(x)\] is defined in the interval \[x\in (-1,2)\].
So to check this we again use Leibniz Integral Rule:
\[h(x)'=2\left( \dfrac{1-2x+4{{x}^{2}}}{1+2x+4{{x}^{2}}} \right)-1\left( \dfrac{1-x+{{x}^{2}}}{1+x+{{x}^{2}}} \right)\]
Now, we need to check whether \[h(x)'\] is defined in the interval \[x\in (-1,2)\].
So as it is a polynomial function, therefore it is defined for all real values of \[x\],we only need to check for any value of \[x\] in the given interval whether the denominators of \[h(x)'\] are becoming zero or not. If for any value of \[x\] in the given interval if these denominators below became zero then the derivative of \[h(x)\] will not be defined in the interval \[x\in (-1,2)\] and for all values of \[x\] in the given interval these denominators cannot be zero then the function \[h(x)\]
Is differentiable in the interval \[x\in (-1,2)\].
Now, we have denominator functions as:
\[1+2x+4{{x}^{2}}\] and \[1+x+{{x}^{2}}\]
On comparing with the equations \[a_1{{x}^{2}}+b_1x+c1=0\] and \[a_2{{x}^{2}}+b_2x+c2=0\] respectively,
We have, \[a_1=4,b_1=2,c1=1\] and \[a_2=1,b_2=1,c2=1\]
Now calculating discriminants of above functions, we got:
\[\begin{align}
  & D_1=(2)(2)-4(4)(1) \\
 & =4-16 \\
 & =-12 \\
\end{align}\]
\[\begin{align}
  & D_2=(1)(1)-4(1)(1) \\
 & =1-4 \\
 & =-3 \\
\end{align}\]
Since, \[D_1<0,D_2<0\] both are negative .
Hence we can conclude that for any real value of \[x\] denominators the derivative of \[h(x)\] will not become zero.
Therefore, \[h(x)\] is differentiable in the interval\[(-1,2)\].
So, option \[(C)\] is correct.

Note:
A differentiable function is necessarily continuous (at every point where it is differentiable). It is continuously differentiable if its derivative is also a continuous function. But it must be remembered that converse may not always be true, that is a function that is continuous in a specific interval may not be differentiable in that interval.