If Rolle's theorem holds for the function $f(x)=x^3+b x^2+a x+5$ on $[1,3]$ with $c=2+\frac{1}{\sqrt{3}}$, then the values of $a$ and $b$ respectively are

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Aicharya · Accepted Answer

Since $\mathrm{f}(x)$ satisfies the Rolle's theorem, $f(1)=f(3)$ $	herefore \quad 1+b+a+5=27+9 b+3 a+5$ $\Rightarrow 2 a+8 b=-26$ $\Rightarrow a+4 b=-13$ ...(i) $\vec{f}(x)=x^3+b x^2+a x+5$ $	herefore \quad \mathrm{f}^{\prime}(x)=3 x^2+2 \mathrm{~b} x+\mathrm{a}$ Now, $\mathrm{f}^{\prime}(\mathrm{c})=0$ $\begin{aligned} & \Rightarrow \mathrm{f}^{\prime}\left(2+\frac{1}{\sqrt{3}}ight)=0 \ & \Rightarrow 3\left(2+\frac{1}{\sqrt{3}}ight)^2+2 \mathrm{~b}\left(2+\frac{1}{\sqrt{3}}ight)+\mathrm{a}=0 \ & \Rightarrow 3\left(4+\frac{4}{\sqrt{3}}+\frac{1}{3}ight)+4 \mathrm{~b}+\frac{2 \mathrm{~b}}{\sqrt{3}}+\mathrm{a}=0 \ & \Rightarrow \mathrm{a}+4 \mathrm{~b}+\frac{2 \mathrm{~b}+12}{\sqrt{3}}+13=0 \ & \Rightarrow-13+\frac{2 \mathrm{~b}+12}{\sqrt{3}}+13=0 \ & \Rightarrow \frac{2 \mathrm{~b}+12}{\sqrt{3}}=0 \ & \Rightarrow \mathrm{b}=-6 \end{aligned}$ Substituting $b=-6$ in (i), we get $a=11$

If Rolle's theorem holds for the function $f(x)=x^3+b x^2+a x+5$ on $[1,3]$ with $c=2+\frac{1}{\sqrt{3}}$,…

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