Let I(m)=∫0πln(1−2mcosx+m2)dxCOLUMN - ICOLUMN – II1I(m)I(−m)=P12I(9)I(3)=Q23I(81)I(3)=R34I(25)I(5)=S4

I(m)=∫0πln(1−2mcosx+m2)dx I(m)=∫0πln(1+2mcosx+m2)dx (By king’s rule) I(m)=∫0πln(1+2mcosx+m2)dx =I(−m)2I(m)=∫0πln((1−2mcosx+m2)(1+2mcosx+m2))dx 2I(m)=∫0πln(1+m4+2m2(1−2cos2x))dx Put t = 2x ⇒2I(m)=12∫0πln(1+m4+2m2(1−2cos2t2))dt =12∫0πln(1+m4+2m2(−cost))dt Now I(m2)I(m)=∫0πln(1+m4−2m2cost)dt12∫0πln(1+m4+2m2(−cost))dt=2

Let I(m)=∫0πln(1−2mcosx+m2)dxCOLUMN - ICOLUMN – II1I(m)I(−m)=P12I(9)I(3)=Q23I(81)I(3)=R34I(25)I(5)=S4

I(m)=∫0πln(1−2mcosx+m2)dx I(m)=∫0πln(1+2mcosx+m2)dx (By king’s rule) I(m)=∫0πln(1+2mcosx+m2)dx =I(−m)2I(m)=∫0πln((1−2mcosx+m2)(1+2mcosx+m2))dx 2I(m)=∫0πln(1+m4+2m2(1−2cos2x))dx Put t = 2x ⇒2I(m)=12∫0πln(1+m4+2m2(1−2cos2t2))dt =12∫0πln(1+m4+2m2(−cost))dt Now I(m2)I(m)=∫0πln(1+m4−2m2cost)dt12∫0πln(1+m4+2m2(−cost))dt=2

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Let $I (m) = \int_{0}^{π} \ln (1 - 2 m \cos x + m^{2}) d x$
COLUMN - I
COLUMN – II
1 $\frac{I (m)}{I (- m)} =$ P 1
2 $\frac{I (9)}{I (3)} =$ Q 2
3 $\frac{I (81)}{I (3)} =$ R 3
4 $\frac{I (25)}{I (5)} =$ S 4

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1 - Q, 2 - P, 3 - S, 4 - Q

1- P, 2 - Q , 3 - S, 4 - Q

1 - P, 2 - P, 3 - S, 4 - Q

1 - P, 2 - S, 3 - R, 4 - P

answer is A.

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Detailed Solution

$I (m) = \int_{0}^{π} \ln (1 - 2 m \cos x + m^{2}) d x I (m) = \int_{0}^{π} \ln (1 + 2 m \cos x + m^{2}) d x$

(By king’s rule) $I (m) = \int_{0}^{π} \ln (1 + 2 m \cos x + m^{2}) d x = I (- m)$

$2 I (m) = \int_{0}^{π} \ln ((1 - 2 m \cos x + m^{2}) (1 + 2 m \cos x + m^{2})) d x 2 I (m) = \int_{0}^{π} \ln (1 + m^{4} + 2 m^{2} (1 - 2 \cos^{2} x)) d x P u t t = 2 x \Rightarrow 2 I (m) = \frac{1}{2} \int_{0}^{π} \ln (1 + m^{4} + 2 m^{2} (1 - 2 \cos^{2} \frac{t}{2})) d t = \frac{1}{2} \int_{0}^{π} \ln (1 + m^{4} + 2 m^{2} (- \cos t)) d t N o w \frac{I (m^{2})}{I (m)} = \frac{\int_{0}^{π} \ln (1 + m^{4} - 2 m^{2} \cos t) d t}{\frac{1}{2} \int_{0}^{π} \ln (1 + m^{4} + 2 m^{2} (- \cos t)) d t} = 2$