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Question : 74 of 160
Marks:
+1,
-0
Solution:
Given,
f(x)=x3+ax2+bx Also,
f(1)−f(3)=0 ⇒‌‌f(1)=f(3)⇒1+a+b=27+9a+3b ⇒‌‌8a+2b=−26 ⇒‌‌4a+b=−13. . . (i)
Now,
‌‌f′(x)=3x2+2ax+b ∴‌‌f′(‌)=3(‌)2+2a(‌)+b=0 ⇒‌‌12+1+4√3+‌+b=0 ⇒‌‌13√3+12+4a√3+2a+b√3=0 ⇒‌‌13√3+12+4a√3+2a+(−13−4a)√3=0 ⇒‌‌a=‌=−6 ∴‌b=−13−4a=−13+24=11 ∴‌a−b=−6−11=−17
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