Question

If $\int \frac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}dx=A\ln |\cos x+\sin x-2|+Bx+C,$ then ordered triplet $(A,B,\lambda )$ is
(Here, $C$ is a constant of integration)

• A. $(\frac{1}{2},\frac{3}{2},-1)$
• B. $(\frac{3}{2},\frac{1}{2},-1)$
• C. $(\frac{1}{2},-1,-\frac{3}{2})$
• D. $(\frac{3}{2},-1,\frac{1}{2})$

Answer 1

The correct option is B $(\frac{3}{2},\frac{1}{2},-1)$

$\int \frac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}dx=A\ln |\cos x+\sin x-2|+Bx+C$
Differentiating both sides w.r.t. $x,$
$\frac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}=\frac{A(\cos x-\sin x)}{\cos x+\sin -2}+B$

$\Rightarrow \frac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}=\frac{A\cos x-A\sin x+B\cos x+B\sin x-2B}{\cos x+\sin x-2}$

$\Rightarrow \frac{2\cos x-\sin x+\lambda }{\cos x+\sin x-2}=\frac{(A+B)\cos x+(B-A)\sin x-2B}{\cos x+\sin x-2}$

Comparing the numerators,
$A+B=2,B-A=-1,\lambda =-2B$
$\Rightarrow A=\frac{3}{2},B=\frac{1}{2},\lambda =-1$