Q70 / 125 · BITSAT 2009 · Mathematics

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 -

  • A
    \(\left(\frac{1}{2}, \frac{3}{2},-1\right)\)
  • Correct answer
    B
    \(\left(\frac{3}{2}, \frac{1}{2},-1\right)\)
  • C
    \(\left(\frac{1}{2},-1, \frac{3}{2}\right)\)
  • D
    \(\left(\frac{3}{2},-1, \frac{1}{2}\right)\)

Solution

\(\begin{aligned} & \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{A} \ln |\cos \mathrm{x}+\sin \mathrm{x}-2|+\mathrm{Bx}+\mathrm{C}) \\ &= \mathrm{A} \frac{\cos \mathrm{x}-\sin \mathrm{x}}{\cos \mathrm{x}+\sin \mathrm{x}-2}+\mathrm{B} \\ &= \frac{\mathrm{A} \cos \mathrm{x}-\mathrm{A} \sin \mathrm{x}+\mathrm{B} \cos \mathrm{x}+\mathrm{B} \sin \mathrm{x}-2 \mathrm{~B}}{\cos \mathrm{x}+\sin \mathrm{x}-2} \\ & \therefore 2=\mathrm{A}+\mathrm{B} \text { or }-1=-\mathrm{A}+\mathrm{B} ; \lambda=-2 \mathrm{~B} \\ & \therefore \mathrm{A}=3 / 2, \mathrm{~B}=1 / 2, \lambda=-1 \end{aligned}\)

Hence, the correct answer is option: 'B'.

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