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Question

If the function $$f(x) = x^3 +bx^2 + ax + 5$$ on $$[1, 3]$$ satisfies the conditions of Rolle's Theorem with $$c = 2+\dfrac{1}{\sqrt{3}}$$, then find $$a$$ + $$b$$.


Solution

$$f(x)=x^3+bx^2+ax+5$$
$$f(1)=f(3)$$
$$1+b+a+5=27+9b+3a+5$$
$$a+4b=-13$$
$$f'(x)=3x^2-2bx+a$$
$$f'(c)=0$$
Now $$c=2+\frac{1}{\sqrt{3}}$$
$$3(2+\frac{1}{\sqrt3})^2+b(2+\frac{1}{\sqrt{3}})+a=0$$
$$13+4\sqrt{3}+2b\frac{(6+\sqrt{3})}{3}+a=0$$
solving the above equation
$$x=11$$
$$y=-6$$
$$x+y=5$$

Mathematics

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