For the frequency distribution: Variate $$(x)$$: $$x_1, x_2, x_3, \ldots, x_{15}$$
Frequency $$(f)$$: $$f_1, f_2, f_3, \ldots, f_{15}$$
where $$0 < x_1 < x_2 < x_3 < \ldots < x_{15} = 10$$ and $$\sum_{i=1}^{15} f_i > 0$$, the standard deviation cannot be

We have a grouped (but discrete) frequency distribution whose observed values satisfy $$0<x_1<x_2<\dots<x_{15}=10.$$ Thus every variate value lies strictly between a positive number (call it $$m=x_1$$) and $$10$$. Hence each $$x_i$$ is contained in the closed interval $$[\,m,\,10\,]$$ where $$0<m<10.$$

Let the total frequency be $$N=\displaystyle\sum_{i=1}^{15}f_i,$$ which is given to be positive. The (arithmetic) mean of the distribution is

$$\bar x \;=\;\dfrac{\sum_{i=1}^{15}f_i\,x_i}{\sum_{i=1}^{15}f_i} \;=\;\dfrac{\sum f_i\,x_i}{N}.$$

The population standard deviation (about the mean) is defined by the well-known formula

$$\sigma \;=\;\sqrt{\dfrac{\sum_{i=1}^{15}f_i\,(x_i-\bar x)^2}{\sum_{i=1}^{15}f_i}} \;=\;\sqrt{\dfrac{\sum f_i\,(x_i-\bar x)^2}{N}}.$$

To find an upper bound on $$\sigma$$ we invoke an elementary but very useful inequality for any set of numbers lying in a fixed interval. Let the interval be $$[a,b]$$ and denote the mean by $$\mu.$$ Since every observation satisfies $$a\le x\le b,$$ we can write

$$x-\mu \;\le\; b-\mu\quad\text{and}\quad \mu-x \;\le\; \mu-a.$$

Consequently the squared deviation satisfies

$$0\;\le\;(x-\mu)^2\;\le\;\max\{(b-\mu)^2,\,(\mu-a)^2\}.$$ Because $$\mu$$ itself is inside the interval $$[a,b],$$ both distances $$b-\mu$$ and $$\mu-a$$ are at most the total length $$b-a$$. Furthermore, at least one of them is not larger than half that length; precisely, we always have

$$\max\{\,b-\mu,\,\mu-a\,\}\;\le\;\dfrac{b-a}{2}.$$

Squaring this inequality gives

$$(x-\mu)^2\;\le\;\left(\dfrac{b-a}{2}\right)^2.$$

Averaging the above inequality over all observations can never increase the right-hand side; therefore

$$\sigma^2 =\dfrac{\sum f_i\,(x_i-\mu)^2}{N} \;\le\;\left(\dfrac{b-a}{2}\right)^2,$$ or equivalently

$$\sigma\;\le\;\dfrac{b-a}{2}.$$

This result—which is sometimes quoted as “the variance of any sample contained in an interval of length $$L$$ cannot exceed $$\dfrac{L^2}{4}$$”—tells us that the standard deviation never exceeds half the range.

Now we apply the inequality to the present question. Our lower bound is $$a=m$$ and our upper bound is $$b=10.$$ Therefore

$$\sigma\;\le\;\dfrac{\,10-m\,}{2}.$$

Because $$m>0,$$ we have $$10-m<10,$$ and hence the right-hand side is strictly less than $$\dfrac{10}{2}=5.$$ Putting it plainly,

$$\sigma\;<\;5.$$

Thus no matter how the actual frequencies $$f_1,f_2,\dots,f_{15}$$ are chosen (provided they are all positive), the standard deviation of the distribution can never reach $$5,$$ let alone exceed it.

Let us compare this strict upper bound with the four numerical options offered in the question:

Option A: $$4 \quad(\text{possible, because }4<5)$$
Option B: $$1 \quad(\text{certainly possible})$$
Option C: $$6 \quad(\text{impossible, because }6>5)$$
Option D: $$2 \quad(\text{possible})$$

Only the value $$6$$ violates the theoretical ceiling of $$5$$, so it is the unique value that the standard deviation cannot take.

Hence, the correct answer is Option C.