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Question: Given that for each \(a \in \left( {0,1} \right)\) ,\(\mathop {\lim }\limits_{h \to {0^ + }} \int\li...

Given that for each a∈(0,1)a \in \left( {0,1} \right) ,lim⁡h→0+∫h1−hta(1−t)a−1dt\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(a)g\left( a \right) . In addition, it is given that the function g(a)g\left( a \right) is differentiable on (0,1)\left( {0,1} \right), then the value of g′(12)g'\left( {\dfrac{1}{2}} \right) is?
A. π2\dfrac{\pi }{2}
B. π\pi
C. −π2 - \dfrac{\pi }{2}
D. 0

Explanation

Solution

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 dF(x)dx=f(x)\dfrac{{dF\left( x \right)}}{{dx}} = f\left( x \right), then ∫f(x)dx=F(x)+C\smallint f\left( x \right)dx = F\left( x \right) + C , where C is the integration constant.
The given limit g(a)g\left( a \right) is consists of definite integral, use the property of definite integral to find the relation between g(a)g\left( a \right) and g(1−a)g\left( {1 - a} \right) , so that the differential calculation on g(a)g\left( a \right) will be easy.

Complete step-by-step answer:
Simplifying given integral:
g(a)=lim⁡h→0+∫h1−ht−a(1−t)a−1dtg\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−aa = 1 - h + h - a,
g(1−h+h−a)=lim⁡h→0+∫h1−ht−(1−h+h−a)(1−t)(1−h+h−a−1)dtg\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(1−a)=lim⁡h→0+∫h1−hta−1(1−t)−adtg\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 ∫abf(x)dx=∫abf(a+b−x)dx\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx}
i.e. replacing tt by (h+1−h−t)\left( {h + 1 - h - t} \right)
g(1−a)=lim⁡h→0+∫h1−h(h+1−h−t)a−1(1−(h+1−h−t))−adtg\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

⇒lim⁡h→0+∫h1−h(1−t)a−1(1−1+t)−adt ⇒lim⁡h→0+∫h1−h(1−t)a−1(t)−adt=g(a)  \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) \\\

⇒g(a)=g(1−a) \Rightarrow g\left( a \right) = g\left( {1 - a} \right)
On differentiating both sides
g′(a)da=g′(1−a)(−1)dag'\left( a \right)da = g'\left( {1 - a} \right)\left( { - 1} \right)da
g′(a)=−g′(1−a)g'\left( a \right) = - g'\left( {1 - a} \right)
g′(a)+g′(1−a)=0g'\left( a \right) + g'\left( {1 - a} \right) = 0
calculating required g′(12)g'\left( {\dfrac{1}{2}} \right):
For a=12a = \dfrac{1}{2}
g′(12)+g′(1−12)=0g'\left( {\dfrac{1}{2}} \right) + g'\left( {1 - \dfrac{1}{2}} \right) = 0
⇒g′(12)+g′(12)=0\Rightarrow g'\left( {\dfrac{1}{2}} \right) + g'\left( {\dfrac{1}{2}} \right) = 0
⇒2g′(12)=0\Rightarrow 2g'\left( {\dfrac{1}{2}} \right) = 0
⇒g′(12)=0\Rightarrow g'\left( {\dfrac{1}{2}} \right) = 0

So, the correct answer is “Option D”.

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

Note: The lim⁡h→0 +\mathop {\lim }\limits_{h \to {0^{{\text{ }} + }}} i.e., limit h tends to 0+{0^ + }because, for h=0h = 0, t−a{t^{ - a}} or 1ta\dfrac{1}{{{t^a}}} will be 10a\dfrac{1}{{{0^a}}} which is an indeterminate form, to avoid this situation limit h tends to 0+{0^ + } instead of 0.
lim⁡x→a +f(x)\mathop {\lim }\limits_{x \to {a^{{\text{ }} + }}} f\left( x \right) is the right-hand limit of the function f(x)f\left( x \right) which is the value of f(x)f\left( x \right) when xx tends to aa from the right.
lim⁡x→a −f(x)\mathop {\lim }\limits_{x \to {a^{{\text{ }} - }}} f\left( x \right) is the left-hand limit of the function f(x)f\left( x \right) which is the value of f(x)f\left( x \right) when xx tends to aa from the left.
The following are some important properties of definite integral, those will be useful in calculation definite integral more easily.
*∫0af(x)dx=∫0af(a−x)dx\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx
*∫abf(x)dx=−∫baf(x)dx\int\limits_a^b {f\left( x \right)} dx = - \int\limits_b^a {f\left( x \right)} dx
*∫abf(x)dx=∫acf(x)dx+∫cbf(x)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