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

If Rolle's theorem holds for the function $\mathrm{f}(x)=x^3+\mathrm{b} x^2+\mathrm{ax}+5$ on $[1,3]$ with $\mathrm{c}=2+\frac{1}{\sqrt{3}}$, then the values of $a$ and $b$ respectively are
  1. $-11,6$
  2. 11,6
  3. $-11,-6$
  4. $11,-6$

Solution

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

Asked in: MHT CET 2024 (02 May Shift 1)

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