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Question : 141 of 160
Marks:
+1,
-0
Solution:
We have,
1=a(1−2x)(1−3x)+b(1−x)(1‌−3x)‌+c(1−x)(1−2x)Put
x=‌,
‌1=0+b(1−‌)(1−‌)+0⇒‌‌1‌=b(‌)(−‌)⇒b=−4Put
x=1,
1‌=a(1−2)(1−3)+0+0⇒‌‌1‌=a(−1)(−2)⇒a=‌Put
x=‌,
‌1=0+0+c(1−‌)(1−‌)‌⇒‌‌1=c(‌)(‌)⇒c=‌Now,
‌+‌+‌‌=‌+‌+‌‌=‌−‌+‌‌=‌=‌=‌
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