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Convex lens: f is positive.  u is negative.    u = - 30 cm

We donot know if real or virtual image is formed.

1) Assume a real image is formed. v = +45 cm
    \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ = \frac{1}{45} + \frac{1}{30} \\ \\ = \frac{5}{90} \\ \\ f = 16 cm \\ \\

  Let us say that the object is at 30cm from the lens on the left side then
    image is formed inverted on the right side of lens at 45cm.

2) Assume a virtual image is formed v = -45 cm

    \frac{1}{f} = \frac{-1}{45} + \frac{1}{30} \\ \\ f = 90 cm \\ \\

     image is formed erect and on the same side as the object wrt to lens

1 5 1
sir but if not mentioned in the question then if we do any one is it right
i do not know what the expectation of examiner could be. if it is said, image is formed on the other side of lens, then there is only one option. what do u think ?
because sir i dint marks for that sum as i got 90 cms