Question

For the function $f\left(x\right)=x^3-6x^2+ax+b.$ If Rolle’s theorem holds in $[1,3]$ with $c=2+\frac{1}{\sqrt{3}}$ then $(a,b)$

• A. $(11,12)$
• B. $(11,11)$
• C. $(11,anyrealvalue)$
• D. $(anyrealvalue,0)$

Answer 1

The correct option is C $(11,anyrealvalue)$

Given, $f\left(x\right)=x^3-6x^2+ax+b$

Rolle's theorem states that if $f\left(x\right)$ be continuous on $[a,b]$, differentiable on $(a,b)$ and $f\left(a\right)=f\left(b\right)$ then there exists some $c$ between $a$ and $b$ such that $f^{{}^{\prime }}\left(c\right)=0$

Given $(a,b)=(1,3)$ and $c=2+\frac{1}{\sqrt{3}}$

$f\left(1\right)=1-6+a+b=a+b-5$ and

$f\left(3\right)=27-54+3a+b=3a+b-27$

According to Rolle's Theorem, $f\left(1\right)=f\left(3\right)$

$⟹a+b-5=3a+b-27$

$⟹2a=22$

$⟹a=11$

$f^{{}^{\prime }}\left(x\right)=3x^2-12x+a$

According to Rolle's Theorem, $f^{{}^{\prime }}\left(c\right)=0$

$⟹f^{{}^{\prime }}\left(c\right)=3c^2-12c+a=0$

$⟹3(2+\frac{1}{\sqrt{3}}{)}^2-12(2+\frac{1}{\sqrt{3}})+a=0$

$⟹12+1+4\sqrt{3}-24-4\sqrt{3}+a=0$

$⟹a=11$

Therefore, $(a,b)=(11,\text{any real value})$