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#### Explain Solution R.D. Sharma Class 12 Chapter 17 Maxima and Minima  Exercise 17.3 Question 7 Maths Textbook Solution.

\begin{aligned} &a=-3 \\ &b=-9 \end{aligned}

Hint:

Put,

\begin{aligned} &\mathrm{f}^{\prime}(-1)=0 \\ &\mathrm{f}^{\prime}(3)=0 \end{aligned}

Given:

$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{C}$

Explanation:

Differentiating f(x) with respect to x

\begin{aligned} \mathrm{f}^{\prime}(\mathrm{x}) &=3 \mathrm{x}^{2}+2 \mathrm{ax}+\mathrm{b}+0 \\ &=3 \mathrm{x}^{2}+2 \mathrm{ax}+\mathrm{b} \end{aligned}

$Calculate \mathrm{f}^{\prime}(\mathrm{x}) at \mathrm{x}=-1\: and\: \mathrm{x}=3$d

\begin{aligned} \mathrm{f}^{\prime}(-1) &=3(-1)^{2}+2 \mathrm{a}(-1)=\mathrm{b} \\ &=3-2 \mathrm{a}+\mathrm{b} \\ \mathrm{f}^{\prime}(3) &=3(3)^{2}+2 \mathrm{a}(3)+\mathrm{b} \\ &=27+6 \mathrm{a}+\mathrm{b} \end{aligned}

Put $f^{\prime}(-1)=0$ ad $\mathrm{f}^{\prime}(3)=0$ as they are the extremum values hence $f(x)=0$

\begin{aligned} &\therefore 3-2 \mathrm{a}+\mathrm{b}=0 \\ &7-2 \mathrm{a}+\mathrm{b}=-3-(1) \\ &27-6 \mathrm{a}+\mathrm{b}=20 \\ &6 \mathrm{a}+\mathrm{b}=-27-(2) \end{aligned}

$(1) -(2), \mathrm{wc} get$

\begin{aligned} &-2 a-6 a=-3+27 \\ &-8 a=24 \\ &a=-3 \end{aligned}
\begin{aligned} &\text { Put } a=3 \text { in }(1) \\ &-2(3)^{2}-b=-3 \\ &b=-9 \end{aligned}