(a)
Hint: Use mirror equation.

Step 1: Find the mirror equation for a concave mirror.
For a concave mirror, the focal length, f<0.
When the object is placed on the left side of the mirror, the object distance, u<0.
For image distance v,

$\begin{array}{l}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \cdots \left(1\right)\end{array}$

Step 2: Find position of the image.
The object lies between f and 2f.
$\begin{array}{l}\therefore 2f<u<f \left(\because u and f are negative\right)\\ \frac{1}{2f}>\frac{1}{u}>\frac{1}{f}\\ -\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}\\ \frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<0 \cdots \left(2\right)\end{array}$
Using equation (1),

$\frac{1}{2f}<\frac{1}{v}<0$
$\Rightarrow \frac{1}{v}=-ve\Rightarrow v=-ve$

$\frac{1}{2f}<\frac{1}{v}\Rightarrow 2f>v\Rightarrow -v>-2f$.

Therefore, the image lies beyond 2f.

(b)
Hint: Focal length of the convex mirror is positive.

Step: Find the position of the image.
When the object is placed on the left side of the mirror, the object distance (u) is negative.
For image distance v,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

As u is always negative and f is always positive for a convex mirror,

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=+ve$
$v=+ve$

Thus, the image is formed on the backside of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object's distance.

(c)
Hint: $m=-\frac{v}{u}$

Step 1: Find the image distance.
For image distance v,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$As u=-ve$
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\Rightarrow \frac{1}{v}>\frac{1}{f}$
$\Rightarrow v<f$
$So the image always lies between the focus and the pole.$
$As \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{uf}\Rightarrow v=\frac{uf}{u-f}$

So the image formed by a convex mirror is always diminished.

Hence, the image formed is diminished and is located between the focus (f) and the pole.

(d)
Hint: Use mirror equation.

Step 1: Find the nature of the image
When it is placed between the focus (f) and the pole:

$\therefore f<u \left(Both u and f are -ve\right)$
$\frac{1}{f}>\frac{1}{u}$
$\frac{1}{f}-\frac{1}{u}>0$

For image distance v,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}>0\Rightarrow \frac{1}{v}=+ve\Rightarrow v=+ve$

The image is formed on the right side of the mirror. Hence, it is a virtual image.
Step 2: Find the magnification.

$Magnification=-\frac{v}{u}=-\frac{f}{u-f}$
$As u and f are -ve for a concave mirror,$
$m=\frac{f}{f-u}>1$

Hence, the formed image is enlarged.