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Question : 68 of 160
Marks:
+1,
-0
Solution:
af(x)+bf(1∕x)=x+1Substitute
x by
1∕x af(‌)+bf(x)=‌+1 From Eq. (i)
×a− Eq. (ii)
×b‌e ‌ a2f(x)+abf(‌)=ax+a abf(‌)+b2f(x)=‌+b ‌(a2−b2)f(x)=ax+(a−b)−‌ ⇒x2f(x)=‌[ax3+(a−b)x2−bx] ‌‌(x2f(x))=‌[3ax2+2(a−b)x−b]According to the question,
‌‌‌=2,‌=1,−‌=‌ ⇒‌‌‌=a−b=−3b=a2−b2⇒‌‌3a=2a−2b⇒a=−2b ⇒‌‌‌ Also ‌‌=1‌ and ‌‌=1 ⇒‌‌a+b=1⇒−2a+b=1 ⇒b=−1⇒a=2 ∴‌‌a−b=2−(−1)=3
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