Let f(x) be a continuous function such that the area bounded by the curve y = f(x), x-axis and the lines x =0 and x = a is $\dfrac{{{a}^{2}}}{2}+\dfrac{a}{2}\sin a+\dfrac{\pi }{2}\cos a$ then $f\left( \dfrac{\pi }{2} \right)$
A . 1
B. $\dfrac{1}{2}$
C. $\dfrac{1}{3}$
D. None of these

Hint: In this question, we are given that f(x) be a continuous function and the area bounded by y = f(x) , x-axis and the lines x = 0 and x = a is $\dfrac{{{a}^{2}}}{2}+\dfrac{a}{2}\sin a+\dfrac{\pi }{2}\cos a$. We have to find the value of $f\left( \dfrac{\pi }{2} \right)$. For this we differentiate the given equation from a to 0. Then we substitute $a=\dfrac{\pi }{2}$ in the equation and get the desirable value of $f\left( \dfrac{\pi }{2} \right)$.

Formula Used: Formula for area under curve is = $\int_{a}^{b}{f(x)dx}$ where (b > a)
We use the formula to differentiate the given equation:-
$\int_{a}^{b}{f(x)dx}$$\dfrac{d}{dx}(fg)=fg'+gf'$

Complete step by step Solution:
Given that f(x) is a continuous function.
A function that does not have any discontinuity and arbitrarily small changes by restricting enough small changes is called the continuous function.
Area bounded by the lines x = 0 and x = a and y = f(x) will be
= $\int_{a}^{b}{f(x)dx}$ where (b > a)
And $\int_{0}^{a}{f(x)dx=\dfrac{{{a}^{2}}}{2}}+\dfrac{a}{2}\sin a+\dfrac{\pi }{2}\cos a$
First, we differentiate the above equation w.r.t a, and we get
$f(a)-f(0)=\dfrac{2a}{2}+\dfrac{1}{2}((1)\sin a+a\cos a)+\dfrac{\pi }{2}(-\sin a)$
$f(a)=a+\dfrac{1}{2}(\sin a+a\cos a)-\dfrac{\pi }{2}\sin a$
Now we put $a=\dfrac{\pi }{2}$, then the equation becomes
$f\left( \dfrac{\pi }{2} \right)=\dfrac{\pi }{2}+\dfrac{1}{2}-\dfrac{\pi }{2}$
Then $f\left( \dfrac{\pi }{2} \right)=\dfrac{1}{2}$

Therefore, the correct option is (B).

Note: Area under the curve between two points is finding out by doing integral between two points. To find the area under the curve y = f(x) between x = a and x = b, we integrate y = f(x) between the points a and b.