Resolve (2+cosx)(1−cosx)(3+cosx)(1+cosx)2 into partial fractions.
A
38(1−cosx)−18(3+cosx)+14(1+cosx)+44(1+cosx)2
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B
316(1−cosx)−116(3+cosx)+18(1+cosx)+44(1+cosx)2
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C
316(1−cosx)−116(3+cosx)+14(1+cosx)+44(1+cosx)2
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D
316(1−cosx)−116(3+cosx)+14(1+cosx)+48(1+cosx)2
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Solution
The correct option is D316(1−cosx)−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)(1−cosx)(3+cosx)(1+cosx)2=(2+t)(1−t)(3+t)(1+t)2 Let(2+t)(1−t)(3+t)(1+t)2=A1−t+B3+t+C1+t=D(1+t)2 ...(i) ⇒2+t=A(3+t)(1+t)2+B(1−t)(1+t)2
+C(1−t)(1+t)(3+t)+D(1−t)(3+t) ...(ii) Puttitng 1−t=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 2−1=0+0+0+0+D(1+1)(3−1) ∴D=14 Putting 3+t=0 or t=−3 in (ii), we obtain 2−3=0+B(1+3)(1−3)2+0+0 ∴B=−116 Comparing the coefficient of t3 in (ii), we obtain 0=A−B−C ⇒0=316+116−C⇒C=14 Substituting the values of A,B,C,D in (i), then (2+t)(1−t)(3+t)(1+t)2=316(1−t)−116(3+t)+14(1+t)+44(1+t)2 Putting t=cosx in last then,