The point on the curve y²=4x which is nearest to the point (2, 1) is 

Given:

$y^2=4x\Rightarrow x=\frac{y^2}{4}$

$\Rightarrow \mathrm{d}=\sqrt{(x-2{)}^2+(y-1{)}^2}$

$\Rightarrow {\mathrm{d}}^2=(\mathrm{x}-2{)}^2+(\mathrm{y}-1{)}^2$

$\Rightarrow d^2={\left(\frac{y^2}{4}-2\right)}^2+(y-1{)}^2$

Let $\mathrm{u}={\left(\frac{y^2}{4}-2\right)}^2+(\mathrm{y}-1{)}^2$

$\Rightarrow \frac{du}{dy}=2\left(\frac{y^2}{4}-2\right)\frac{y}{2}+2(\mathrm{y}-1)$

To find minima

$\frac{du}{dy}=0$

$2\left(\frac{y^2}{4}-2\right)\frac{y}{2}+2(\mathrm{y}-1)=0$

$\Rightarrow \mathrm{y}=2\Rightarrow \mathrm{x}=1\left(\because x=\frac{y^2}{4}\right)$

$\frac{d^{2}u}{d{y}^2}=\frac{3y^2}{4}$

$\Rightarrow {\left(\frac{d^{2}u}{d{y}^2}\right)}_{\left(\mathrm{1,2}\right)}=3>0$

Therefore, nearest point is $\left(\mathrm{1,2}\right)$

Hence, correct option is (d).