#### Need solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise 17.3  question 3

$a = -\frac{2}{3}$

$\mathrm{b}=-\frac{1}{6}$

Hint:

Put?  $\frac{d y}{d x}=0 \text { at } x=1,2$

Given:

$y=\operatorname{alog} x+b x^{2}+x$

Explanation:

Differentiating y with respect to x

\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{~b} \mathrm{x}+1 \\ &\quad=\frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{bx}+1 \\ &\text { At } \mathrm{x}=1 \mid \\ &\begin{aligned} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) &=\frac{\mathrm{a}}{1}+2 \mathrm{~b}(1)+1 \\ &=\mathrm{a}+2 \mathrm{~b}+1-(1) \end{aligned} \end{aligned}

\begin{aligned} &\text { At } x=2 \\ &\frac{d y}{d x}=\frac{a}{2}+2(b) \cdot 2+1 \\ &=\frac{a}{2}+4 b+1-(2) \end{aligned}

Since x=1 and x=2 are the extreme values,f’(x)=0 at x=1 & 2

Then equation (1) and (2) become

\begin{aligned} &a+2 b+1=0-(3)\\ &\frac{a}{2}+4 b+1=0\\ &a+8 b+2=0 \quad-(4)\\ &\text { [Multiplying both sides by 2] } \end{aligned}

Solving , (3) and (4) , we get

$\begin{array}{r} 6 b-1=0 \\ b=-\frac{1}{6} \end{array}$

Putting $B = -1/6 in (3) , we get$

\begin{aligned} &a-\frac{1}{3}+120 \\ &a=-\frac{2}{3} \end{aligned}