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Question

Resolve (2+cosx)(1−cosx)(3+cosx)(1+cosx)2 into partial fractions.

A
38(1cosx)18(3+cosx)+14(1+cosx)+44(1+cosx)2
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B
316(1cosx)116(3+cosx)+18(1+cosx)+44(1+cosx)2
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C
316(1cosx)116(3+cosx)+14(1+cosx)+44(1+cosx)2
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D
316(1cosx)116(3+cosx)+14(1+cosx)+48(1+cosx)2
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Solution

The correct option is D 316(1cosx)116(3+cosx)+14(1+cosx)+44(1+cosx)2
In the given fraction everwhere put cosx=t for the sake of partial fractions.
(2+cosx)(1cosx)(3+cosx)(1+cosx)2=(2+t)(1t)(3+t)(1+t)2
Let(2+t)(1t)(3+t)(1+t)2=A1t+B3+t+C1+t=D(1+t)2 ...(i)
2+t=A(3+t)(1+t)2+B(1t)(1+t)2
+C(1t)(1+t)(3+t)+D(1t)(3+t) ...(ii)
Puttitng 1t=0 or t=1
In (ii), we obtan 2+1=A(3+1)(1+1)2+0+0+0
A=316
Putting 1+t=0 or t=1 in (ii), we obtain
21=0+0+0+0+D(1+1)(31)
D=14
Putting 3+t=0 or t=3 in (ii), we obtain
23=0+B(1+3)(13)2+0+0
B=116
Comparing the coefficient of t3 in (ii), we obtain
0=ABC
0=316+116CC=14
Substituting the values of A,B,C,D in (i), then
(2+t)(1t)(3+t)(1+t)2=316(1t)116(3+t)+14(1+t)+44(1+t)2
Putting t=cosx in last then,
(2+cosx)(1cosx)(3+cosx)(1+cosx)2=316(1cosx)116(3+cosx)
+14(1+cosx)+44(1+cosx)2
which are the required partial fractions.
Hence, option 'C' is correct.

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