#### Explain Solution R.D.Sharma Class 12 Chapter 17  Maxima and Minima  Exercise 17.3 Question 3 Maths Textbook Solution.

\begin{aligned} &a=-\frac{2}{3} \\ &b=-\frac{1}{6} \end{aligned}

Hint:

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

Given:

$y=a \log x+b x^{2}+x$

Explanation:

Differentiating y with respect to x

\begin{aligned} \frac{d y}{d x} &=\frac{a}{x}+2 b x+1 \\ &=\frac{a}{x}+2 b x+1 \end{aligned}

\begin{aligned} &\text { At } x=1 \\ &\left(\frac{d y}{d x}\right)=\frac{a}{1}+2 b(1)+1 \\ &=a+2 b+1-(1) \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 \\ &\frac{a}{2}+4 b+1=0 \\ &a+8 b+2=0 \end{aligned}                [Multiplying both sides by 2]

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

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

Putting $b=-\frac{1}{6}$  in  (3) , we get

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