A diminished image of an object is to be obtained on a screen $1.0m$ from it. This can be achieved by appropriately placing.

The correct option is C A convex lens of focal length less than $0.25m$
A convex mirror and a concave lens always produce semi image for the objects. Therefore, option (b) and (d) are not correct. The image by a convex lens is diminished when the object is placed beyond $2f$.
Let $u=2f+x$
Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$,
$\Rightarrow \frac{1}{v}-\frac{1}{-(2f+x)}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{2f+x}=\frac{2f+x-f}{f(2f+x)}=\frac{(f+x)}{f(2f+x)}$
But $u=v=1$ (given) $(2f+x)+\frac{f(2f+x)}{f+x}\le 1$
$\Rightarrow 2f=x[1+\frac{f}{f+x}]\le 1\Rightarrow \frac{(2f+x{)}^2}{f+x}\le 1$
$\Rightarrow (2f+x{)}^2\le f+x$
The above is true for $f<0.25m$
c. is the correct answer.