(a) If y =√(a x+ b), prove that y ((d^2 y)/(d x^2))+(dy/dx)^2=0.

$\displaystyle \mathbf{y}=\sqrt{\mathbf{a x}+\mathbf{b}} \Rightarrow \mathbf{y}^{2}=\mathbf{a x}+\mathbf{b}$

Differentiate with respect to ' $\displaystyle x$ ', $\displaystyle 2 \mathrm{y} \frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}=\mathrm{a}$

Differentiate with respect to ' $\displaystyle x, 2 \mathrm{y} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}+2\left(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\right)^{2}=0 \Rightarrow \mathrm{y} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}+\left(\frac{\mathrm{d} \mathrm{y}}{\mathrm{d} x}\right)^{2}=0$

Or (b) $\displaystyle \mathbf{f}(\mathbf{x})$ is differentiable in $\displaystyle (\mathbf{0 , 2}) \Rightarrow \mathbf{f}(\mathbf{x})$ is continuous on $\displaystyle (\mathbf{0 , 2}) \Rightarrow \mathrm{f}(x)$ is continuous at $\displaystyle x=1 \therefore \lim _{x \rightarrow 1^{-}}(\mathrm{a} x+\mathrm{b})=\lim _{x \rightarrow 1^{+}}\left(2 x^{2}-x\right) \Rightarrow \mathrm{a}+\mathrm{b}=1$

Also, $\displaystyle \mathrm{f}(x)$ is differentiable at $\displaystyle x=1, \therefore$ L.H.D. $\displaystyle (x=1)=$ R.H.D. $\displaystyle (x=1)$

$$\Rightarrow \mathrm{a}=$\displaystyle 4$($\displaystyle 1$)-$\displaystyle 1$ \therefore \mathrm{a}=$\displaystyle 3$ & \mathrm{b}=$\displaystyle 1$-\mathrm{a}=-$\displaystyle 2$