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If \(\int \frac{2 \cos x-\sin x+\lambda}{\cos x+\sin x-2} d x\) \(=A \ln |\cos x+\sin x-2|+B x+C\).
Then the ordered triplet \(A, B, \lambda\) is -
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Then the ordered triplet \(A, B, \lambda\) is -
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Verified Answer
The correct answer is:
\(\left(\frac{3}{2}, \frac{1}{2},-1\right)\)
\(\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{A} \ln |\cos \mathrm{x}+\sin \mathrm{x}-2|+\mathrm{Bx}+\mathrm{C}) \\
&= \mathrm{A} \frac{\cos \mathrm{x}-\sin \mathrm{x}}{\cos \mathrm{x}+\sin \mathrm{x}-2}+\mathrm{B} \\
&= \frac{\mathrm{A} \cos \mathrm{x}-\mathrm{A} \sin \mathrm{x}+\mathrm{B} \cos \mathrm{x}+\mathrm{B} \sin \mathrm{x}-2 \mathrm{~B}}{\cos \mathrm{x}+\sin \mathrm{x}-2} \\
& \therefore 2=\mathrm{A}+\mathrm{B} \text { or }-1=-\mathrm{A}+\mathrm{B} ; \lambda=-2 \mathrm{~B} \\
& \therefore \mathrm{A}=3 / 2, \mathrm{~B}=1 / 2, \lambda=-1
\end{aligned}\)
& \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{A} \ln |\cos \mathrm{x}+\sin \mathrm{x}-2|+\mathrm{Bx}+\mathrm{C}) \\
&= \mathrm{A} \frac{\cos \mathrm{x}-\sin \mathrm{x}}{\cos \mathrm{x}+\sin \mathrm{x}-2}+\mathrm{B} \\
&= \frac{\mathrm{A} \cos \mathrm{x}-\mathrm{A} \sin \mathrm{x}+\mathrm{B} \cos \mathrm{x}+\mathrm{B} \sin \mathrm{x}-2 \mathrm{~B}}{\cos \mathrm{x}+\sin \mathrm{x}-2} \\
& \therefore 2=\mathrm{A}+\mathrm{B} \text { or }-1=-\mathrm{A}+\mathrm{B} ; \lambda=-2 \mathrm{~B} \\
& \therefore \mathrm{A}=3 / 2, \mathrm{~B}=1 / 2, \lambda=-1
\end{aligned}\)
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