Question

If $f\left( {\dfrac{{3x - 4}}{{3x + 4}}} \right) = x + 2$, $x \ne - \dfrac{4}{3},$ and $\int {f(x)dx = A\log \left| {1 - x} \right|} + Bx + C$, then the ordered pair $(A,B)$ is equal to (where C is the constant of integration)
(A). $\left( {\dfrac{8}{3},\dfrac{2}{3}} \right)$
(B). $\left( { - \dfrac{8}{3},\dfrac{2}{3}} \right)$
(C). $\left( { - \dfrac{8}{3}, - \dfrac{2}{3}} \right)$
(D). $\left( {\dfrac{8}{3}, - \dfrac{2}{3}} \right)$

Answer 1

To solve this question at first we have to differentiate the integral with respect to x to evaluate $f(x)$. Then we must substitute $\dfrac{{3x - 4}}{{3x + 4}}$ in place of x in the obtained expression and determine the value of A and B by the method of equating the coefficients.

Complete step-by-step answer :
The integral is given by
$\int {f(x)dx = A\log \left| {1 - x} \right|} + Bx + C$ ……………………………… (1)
Now we have to get the value of $f(x)$ in terms of A and B. differentiating both the sides with respect to x, we will get,

\Rightarrow \dfrac{d}{{dx}}\int {f(x)dx = \dfrac{d}{{dx}}\left[ {A\log \left| {1 - x} \right| + Bx + C} \right]} \\\ \Rightarrow f(x) = \dfrac{A}{{\left( {1 - x} \right)}}\dfrac{d}{{dx}}\left( {1 - x} \right) + B\dfrac{{dx}}{{dx}} + 0 \\\ \Rightarrow f(x) = \dfrac{A}{{x - 1}} + B \\\ $$ ………………………………… (2) Let’s substitute $$\left( {\dfrac{{3x - 4}}{{3x + 4}}} \right)$$in place of x in eq. (2), we will get,

\Rightarrow f\left( {\dfrac{{3x - 4}}{{3x + 4}}} \right) = \dfrac{A}{{\dfrac{{3x - 4}}{{3x + 4}} - 1}} + B \\
\Rightarrow f\left( {\dfrac{{3x - 4}}{{3x + 4}}} \right) = \dfrac{{A\left( {3x + 4} \right)}}{{3x - 4 - \left( {3x + 4} \right)}} + B \\
\Rightarrow f\left( {\dfrac{{3x - 4}}{{3x + 4}}} \right) = - \dfrac{{A\left( {3x + 4} \right)}}{8} + B \\

But in the question it is given that $$f\left( {\dfrac{{3x - 4}}{{3x + 4}}} \right) = x + 2$$ ………………………... (4) Substituting the value of eq. (4) in eq. (3) we will get,

\Rightarrow x + 2 = - \dfrac{{A\left( {3x + 4} \right)}}{8} + B \\
\Rightarrow 8\left( {x + 2} \right) = - A\left( {3x + 4} \right) + 8B \\
\Rightarrow 8x + 16 = ( - 3A)x + (8B - 4A) \\

$$Here applying the method of equating the coefficient, we have to equate the coefficient of x on the both sides we will get,$$

\Rightarrow 8 = - 3A \\
\Rightarrow A = - \dfrac{8}{3} \\

$$………………………………….. (6) Similarly equating the constant terms on both the sides of eq. (5) we will get,$$

\Rightarrow 8B - 4A = 16 \\
\Rightarrow B = \dfrac{{16 + 4A}}{8} \\

$$…………………………….. (7) Substituting the value of eq. (6) in eq. (7) we will get,$$

\Rightarrow B = \dfrac{{16 + 4\left( { - \dfrac{8}{3}} \right)}}{8} \\
\Rightarrow B = \dfrac{{16}}{{24}} = \dfrac{2}{3} \\

…………………………………….. (8) $$(A,B) = \left( { - \dfrac{8}{3},\dfrac{2}{3}} \right)$$ **Therefore option (B) is correct.** **Note** : The method of equating the coefficients needs the fact that two expressions are identical precisely when corresponding coefficients are equal for each different type of them. While equating one should focus on the degree of the polynomials of the term of either side must be the same.