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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