Question

If $\int\limits_a^b {\dfrac{{f(x)}}{{f(x) + f(a + b - x)}}dx = 10} $ , then
A) $b = 22,a = 2$
B) $b = 15,a = - 5$
C) $b = 10,a = - 10$
D) $b = 10,a = - 2$

Answer 1

Hint: We can see that this is an integration question and we have to find the value of $b,a$ . We will solve this question with the help of this formula i.e.
$\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)dx} $ . We will put this value in the place of $f(x)$ and then solve it.

Complete step by step answer:
Let us assume that
\[I = \int\limits_a^b {\dfrac{{f(x)}}{{f(x) + f(a + b - x)}}dx} ...(i)\]
This is our first equation.
Now from the formula, by substituting the value of
$\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)dx} $ .
We can write the expression as:
\[I = \int\limits_a^b {\dfrac{{f(a + b - x)}}{{f(a + b - x) + f(a + b - \left( {a + b - x} \right))}}dx} \]
On simplifying we have :
\[I = \int\limits_a^b {\dfrac{{f(a + b - x)}}{{f(a + b - x) + f(a + b - a - b + x)}}dx} \]
\[I = \int\limits_a^b {\dfrac{{f(a + b - x)}}{{f(a + b - x) + f(x)}}dx} ...(ii)\] This is our second equation
We will add both the equation i.e.
\[I + I = \int\limits_a^b {\dfrac{{f(a + b - x)}}{{f(a + b - x) + f(x)}}dx} + \int\limits_a^b {\dfrac{{f(x)}}{{f(x) + f(a + b - x)}}dx} \]
We can see that the L.C.M in both the terms are same, so by adding we have :
\[2I = \int\limits_a^b {\dfrac{{f(a + b - x) + f(x)}}{{f(a + b - x) + f(x)}}dx} \]
Both the numerators and denominators are same, so they will get cancelled out i.e.
\[2I = \int\limits_a^b {1dx} \]
Now we can write the above expression also as
$2I = \left[ x \right]_a^b$
$2I = b - a$
By isolating the term $I$ , we have
 $I = \dfrac{{b - a}}{2}$
From the question it is given that $I = 10$ , since we have assume $I$ as $\int\limits_a^b {\dfrac{{f(x)}}{{f(x) + f(a + b - x)}}dx = 10} $ .
So we can write the expression as :
 $\dfrac{{b - a}}{2} = 10$
And,
 $b - a = 10 \times 2 \Rightarrow b - a = 20$
Now we will use the options to see which satisfies the above expression.
In first case we have
$b = 22,a = 2$ ,
By putting the values in the expression
$22 - 2 = 20$ .
This option is correct.
In the second option we have
$b = 15,a = - 5$
It can be written as
$b - a = 15 - ( - 5)$
It gives the value
 $b - a = 15 + 5 = 20$
Again in the third option, we have
 $b = 10,a = - 10$
By putting the values in the expression it gives,
 $10 - ( - 10) = 10 + 10$
So we have
 $b - a = 20$
In the fourth option, we have
$b = 10,a = - 2$
By substituting the values
  $10 - ( - 2) = 10 + 2$
It gives
$b - a = 12$ .
We can see that this does not satisfy the above expression.
Hence we can say that (A), (B) and (C) are the correct options.

Note:
We should note that \[\int\limits_a^b {1dx} \] can also be written as
\[2I = \int\limits_a^b {1dx} \] because any base with the exponential power as $0$ equals to one.
So we can say that
${x^0} = 1$ .
Now we will apply the formula:
$\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$
Here we have $n = 0$ , so by applying the formula we can write:
$\dfrac{{{x^{0 + 1}}}}{{0 + 1}}$
So it gives us $x$ .
Now we know the formula of upper limit and lower limit i.e.
$\int\limits_a^b {f(x)dx} = f(b) - f(a)$
So by applying this formula and putting the terms together, we have
$\left[ x \right]_a^b = b - a$