If for some $$x \in$$ R, the frequency distribution of the marks obtained by 20 students in a test is:

Then the mean of the marks is:

We are given four distinct marks $$2,\;3,\;5,\;7$$ with corresponding frequencies $$f_1=(x+1)^2,\;f_2=(2x-5),\;f_3=(x^2-3x),\;f_4=x$$. The total number of students is stated to be $$20$$, so by definition of a frequency distribution we must have

$$f_1+f_2+f_3+f_4 \;=\;20.$$

Substituting the four expressions we get

$$ (x+1)^2 \;+\;(2x-5)\;+\;(x^2-3x)\;+\;x \;=\;20. $$

Now we expand and collect like terms step by step. First, expand the square: $$ (x+1)^2 = x^2 + 2x + 1.$$ Placing this into the equation,

$$ x^2 + 2x + 1 \;+\;2x - 5 \;+\;x^2 - 3x \;+\;x \;=\;20. $$

Combine the $$x^2$$ terms: $$x^2 + x^2 = 2x^2.$$

Combine the $$x$$ terms: $$2x + 2x - 3x + x = 2x.$$

Combine the constant terms: $$1 - 5 = -4.$$

So the left-hand side simplifies to

$$ 2x^2 + 2x - 4. $$

Equating this to $$20$$ and bringing every term to one side gives

$$ 2x^2 + 2x - 4 - 20 = 0, $$

which is

$$ 2x^2 + 2x - 24 = 0. $$

Dividing every term by $$2$$ for convenience,

$$ x^2 + x - 12 = 0. $$

This is a quadratic equation. The quadratic formula $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = 1,\,b = 1,\,c = -12$$ gives

$$x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-12)}}{2(1)} = \dfrac{-1 \pm \sqrt{1 + 48}}{2} = \dfrac{-1 \pm 7}{2}.$$

Thus

$$x = \dfrac{-1 + 7}{2} = 3$$ or $$x = \dfrac{-1 - 7}{2} = -4.$$

Because a frequency cannot be negative, $$x=-4$$ is inadmissible. Therefore we take

$$ x = 3. $$

We now compute each actual frequency:

$$\begin{aligned} f_1 &= (x+1)^2 = (3+1)^2 = 16,\\ f_2 &= 2x-5 = 2(3)-5 = 1,\\ f_3 &= x^2 - 3x = 3^2 - 3(3) = 9 - 9 = 0,\\ f_4 &= x = 3. \end{aligned}$$

A quick check: $$16 + 1 + 0 + 3 = 20,$$ so the frequencies are consistent.

The mean (average) mark is defined as

$$\bar{M} = \dfrac{\sum (\text{mark} \times \text{frequency})}{\sum (\text{frequency})}.$$

Substituting the calculated values,

$$\begin{aligned} \sum (\text{mark} \times \text{frequency}) &= 2 \times 16 \;+\; 3 \times 1 \;+\; 5 \times 0 \;+\; 7 \times 3\\ &= 32 + 3 + 0 + 21\\ &= 56. \end{aligned}$$

Therefore

$$ \bar{M} = \dfrac{56}{20} = 2.8. $$

Hence, the correct answer is Option D.