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Question

If the function $$f(x) = x^4 - 2x^3 + ax^2 + bx$$ on $$[1,3]$$ satisfies the conditions of Rolle's Theorem with $$c = \dfrac{3}{2}$$, then find $$a$$ and $$b$$.


Solution

Given $$f(x)=x^4-2x^3+ax^2+bx$$ on $$[1,3]$$

$$\implies f'(x)=4x^3-6x^2+2ax+b$$

According to Rolle's theorem, if f(x) is continuous on $$[a,b]$$, differentiable on $$(a,b)$$ and $$f(a)=f(b)$$ then there exists some $$c \in (a,b)$$ such that $$f'(c)=0$$

Given that the function $$f(x)$$ satisfies the conditions of Rolle's theorem with $$c=\dfrac 32$$

$$f(1)=1^4-2(1)^3+a(1)^2+b(1)=1-2+a+b=a+b-1$$

$$f(3)=3^4-2(3)^3+a(3)^2+b(3)=81-54+9a+3b=9a+3b-27$$

from Rolles's theorem, $$f(1)=f(3)$$

$$\implies a+b-1=9a+3b-27$$

$$\implies 8a+2b=26$$

$$\implies 4a+b=13$$ .......(1)

from Rolles's theorem, $$f'(c)=4c^3-6c^2+2ac+b=0$$ 

$$\implies f'(\dfrac 32)=4(\dfrac 32)^3-6(\dfrac 32)^2+2a(\dfrac 32)+b=0$$

$$\implies \dfrac{27}{2}-\dfrac{27}{2}+3a+b=0$$

$$\implies 3a+b=0$$ .......(2)

(1) - (2) $$\implies a=13$$

substituting $$a=13$$ in (1) 

$$\implies 4(13)+b=13$$

$$\implies b=-39$$

Therefore, $$a=13,b=-39$$

Mathematics

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