Question

If $\int{\dfrac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}dx=A\ln \left| \cos x+\sin x-2 \right|+Bx+C}$. Then the ordered triplet $(A,B,\lambda )$ is
(a)$\left( \dfrac{1}{2},\dfrac{3}{2},-1 \right)$
(b) $\left( \dfrac{3}{2},\dfrac{1}{2},-1 \right)$
(c) $\left( \dfrac{1}{2},-1,\dfrac{3}{2} \right)$
(d) $\left( \dfrac{3}{2},-1,\dfrac{1}{2} \right)$

Answer 1

Hint: Differentiate both sides and solve for the unknowns. A different method can also be used to solve this question, that is, use the substitution method to solve for the integral.
 If $a$ is any arbitrary constant, then, $\dfrac{da}{dx}=0$.
$\dfrac{d(x)}{dx}=1$
\[\dfrac{d(\ln x)}{dx}=\dfrac{1}{x}\]
$\dfrac{d(\sin x)}{dx}=\cos x$
$\dfrac{d(\cos x)}{dx}=\-sin x$

Complete step by step answer:
Integral questions can sometimes turn out to be time consuming. So, rather than solving for the integral we can take the derivative on both sides. This deprives us from the need of having to actually solve the integral. However, on taking the derivatives on both sides, the equivalency remains, and we can easily find the values of the unknowns.
We have,
$\int{\dfrac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}dx=A\ln \left| \cos x+\sin x-2 \right|+Bx+C}$
Taking derivatives on both sides the equation becomes,
$\dfrac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}=A\left( \dfrac{d}{dx}\left( \ln \left| \cos x+\sin x-2 \right| \right) \right)+B\dfrac{dx}{dx}+\dfrac{d(C)}{dx}$
We know that the derivative of a constant is zero. We also know that \[\dfrac{d(\ln x)}{dx}=\dfrac{1}{x}\] and $\dfrac{d(x)}{dx}=1$.
Hence, the equation becomes,
$\dfrac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}=A\left( \dfrac{\cos x-\sin x}{\cos x+\sin x-2} \right)+B$
Further, on taking LCM, we obtain,
$\dfrac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}=\dfrac{A(\cos x-\sin x)+B(\cos x+\sin x-2)}{\cos x+\sin x-2}$
The denominators get cancelled out on both sides.
$2\cos x-\sin x+\lambda =A(\cos x-\sin x)+B(\cos x+\sin x-2)$
$2\cos x-\sin x+\lambda =A\cos x-A\sin x+B\cos x+B\sin x-2B$
Comparing the coefficients of $\cos x,\sin x$ and comparing the constants. We get,
$A+B=2..............(1)$
$-A+B=-1.................(2)$
$-2B=\lambda ..............(3)$
Solving the above equations gives us the values of $A,B$ and $\lambda $.
Adding equation (1) and equation (2),
\[\begin{align}
  & \,\,B+A=+2 \\
 & \,\,\underline{B-A=-1} \\
 & \underline{\underline{2B+0=+1}} \\
\end{align}\]
$\therefore B=\dfrac{1}{2}$
Substituting $B=\dfrac{1}{2}$ in equation (1), we get,
$A+\dfrac{1}{2}=2$
$\therefore A=\dfrac{3}{2}$
Substituting $B=\dfrac{1}{2}$ in equation (3), we get,
$-2\left( \dfrac{1}{2} \right)=\lambda $
$\therefore \lambda =-1$
Hence, the values of $A,B$ and $\lambda $are $\dfrac{3}{2},\dfrac{1}{2},-1$ respectively.

So, the correct answer is “Option B”.

Note: Integrals are also known as anti-derivatives. This is simply because integration is the reverse process of differentiation. Hence, mathematically speaking,
$\int{\dfrac{df(x)}{dx}dx=f(x)}$ and likewise, $\dfrac{d}{dx}\int{f(x)dx=f(x)}$. This way we can easily find the answer to such a question.
This question can also be solved directly, that is, without taking the derivatives on both sides.