Question

Given that for each $a \in \left( {0,1} \right)$ ,$\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^a}{{\left( {1 - t} \right)}^{a - 1}}dt} $ exists.
 Let this limit be \[g\left( a \right)\] . In addition, it is given that the function $g\left( a \right)$ is differentiable on $\left( {0,1} \right)$, then the value of $g'\left( {\dfrac{1}{2}} \right)$ is?
A. $\dfrac{\pi }{2}$
B. $\pi $
C. $ - \dfrac{\pi }{2}$
D. 0

Answer 1

Hint: In the integral calculus, we are to find a function whose differential is given. Thus, integration is a process that is the inverse of differentiation.
Let $\dfrac{{dF\left( x \right)}}{{dx}} = f\left( x \right)$, then $\smallint f\left( x \right)dx = F\left( x \right) + C$ , where C is the integration constant.
The given limit \[g\left( a \right)\] is consists of definite integral, use the property of definite integral to find the relation between \[g\left( a \right)\] and \[g\left( {1 - a} \right)\] , so that the differential calculation on \[g\left( a \right)\] will be easy.

Complete step-by-step answer:
Simplifying given integral:
\[g\left( a \right) = \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}} dt\]
For $a = 1 - h + h - a$,
\[g\left( {1 - h + h - a} \right) = \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - \left( {1 - h + h - a} \right)}}{{\left( {1 - t} \right)}^{\left( {1 - h + h - a - 1} \right)}}} dt\]
\[g\left( {1 - a} \right) = \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{a - 1}}{{\left( {1 - t} \right)}^{ - a}}} dt\]
Using the property of definite integral \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} \]
i.e. replacing $t$ by $\left( {h + 1 - h - t} \right)$
\[g\left( {1 - a} \right) = \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{{\left( {h + 1 - h - t} \right)}^{a - 1}}{{\left( {1 - \left( {h + 1 - h - t} \right)} \right)}^{ - a}}} dt\]
\[
   \Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{{\left( {1 - t} \right)}^{a - 1}}{{\left( {1 - 1 + t} \right)}^{ - a}}} dt \\
   \Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{{\left( {1 - t} \right)}^{a - 1}}{{\left( t \right)}^{ - a}}} dt = g\left( a \right) \\
 \]
\[ \Rightarrow g\left( a \right) = g\left( {1 - a} \right)\]
On differentiating both sides
\[g'\left( a \right)da = g'\left( {1 - a} \right)\left( { - 1} \right)da\]
\[g'\left( a \right) = - g'\left( {1 - a} \right)\]
\[g'\left( a \right) + g'\left( {1 - a} \right) = 0\]
calculating required \[g'\left( {\dfrac{1}{2}} \right)\]:
For \[a = \dfrac{1}{2}\]
\[g'\left( {\dfrac{1}{2}} \right) + g'\left( {1 - \dfrac{1}{2}} \right) = 0\]
\[ \Rightarrow g'\left( {\dfrac{1}{2}} \right) + g'\left( {\dfrac{1}{2}} \right) = 0\]
\[ \Rightarrow 2g'\left( {\dfrac{1}{2}} \right) = 0\]
\[ \Rightarrow g'\left( {\dfrac{1}{2}} \right) = 0\]

So, the correct answer is “Option D”.

Additional Information: If the function $f\left( x \right)$ is differentiable in \[x \in \left( {a,b} \right)\], then the function $f\left( x \right)$ must be continuous in \[x \in \left[ {a,b} \right]\].
Square brackets on interval $\left[ {a,b} \right]$ imply that the extreme points a, b are included along with the points between a and b.
Curve brackets on interval \[\left( {a,b} \right)\] implies that only the points between a and b are included.

Note: The $\mathop {\lim }\limits_{h \to {0^{{\text{ }} + }}} $i.e., limit h tends to ${0^ + }$because, for $h = 0$, ${t^{ - a}}$ or $\dfrac{1}{{{t^a}}}$ will be $\dfrac{1}{{{0^a}}}$ which is an indeterminate form, to avoid this situation limit h tends to ${0^ + }$ instead of 0.
$\mathop {\lim }\limits_{x \to {a^{{\text{ }} + }}} f\left( x \right)$ is the right-hand limit of the function $f\left( x \right)$ which is the value of $f\left( x \right)$ when $x$ tends to $a$ from the right.
$\mathop {\lim }\limits_{x \to {a^{{\text{ }} - }}} f\left( x \right)$ is the left-hand limit of the function $f\left( x \right)$ which is the value of $f\left( x \right)$ when $x$ tends to $a$ from the left.
 The following are some important properties of definite integral, those will be useful in calculation definite integral more easily.
*$\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx$
*$\int\limits_a^b {f\left( x \right)} dx = - \int\limits_b^a {f\left( x \right)} dx$
*$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx$