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(A) For a convex lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Let $u = -x$ (where $x > 0$ is the distance of the object

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(A) For a convex lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Let $u = -x$ (where $x > 0$ is the distance of the object from the lens) and $v = y$ (where $y > 0$ is the distance of the real image).The formula becomes $\frac{1}{y} + \frac{1}{x} = \frac{1}{f}$,which implies $y = \frac{xf}{x-f}$.We are interested in the sum $S = x + y = x + \frac{xf}{x-f} = \frac{x^2 - xf + xf}{x-f} = \frac{x^2}{x-f}$.To find the minimum value of $S$,we differentiate with respect to $x$ and set it to zero: $\frac{dS}{dx} = \frac{(x-f)(2x) - x^2(1)}{(x-f)^2} = 0$.This gives $2x^2 - 2xf - x^2 = 0$,or $x^2 - 2xf = 0$,which means $x = 2f$ (since $x 
eq 0$).At $x = 2f$,$S = \frac{(2f)^2}{2f-f} = \frac{4f^2}{f} = 4f$.For $x > f$,the function $S(x)$ has a minimum at $x = 2f$ with a value of $4f$,and it increases as $x$ moves away from $2f$ in either direction. Thus,the graph is a curve that reaches a minimum at $(2f, 4f)$.

For a convex lens,if a real image is formed,the graph between $(u + v)$ and $u$ or $v$ is as follows:

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