Which of the following is/are INCORRECT

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A group of 40 students appeared in an examination of 3 subjects- mathematics, physics and chemistry. It was found that all students passed in at least one of the subjects 20 students passed in mathematics, 25 students passed in physics, 16 students passed in chemistry, at most 11 students passed in both mathematics and physics, at most 15 students passed in both physics and chemistry at most 15 students passed in both mathematics and chemistry. The maximum number of students passed in all the three subjects is 10

Let $a_{1}, a_{2}, \dots a_{10}$ be 10 observations such that $\sum_{k = 1}^{10} a_{k} = 50$ and $\underset{1 \leq k < j \leq 10}{\sum \sum} a_{k} . a_{j} = 1100$ .Then the standard deviation of $a_{1}, a_{2}, \dots ., a_{10}$ is equal to 5

If $T_{n} = \frac{(2 n + 1) (2 n^{2} + 2 n - 1)}{{(n - 1)}^{2} n^{2} {(n + 1)}^{2} {(n + 2)}^{2}}, n \in N, n \geq 2$ . If $K = lim_{n \to \infty}_{r = 2}^{n} T_{r}$ then k is 3

The constant term in the expansion of ${(2 x + \frac{1}{x^{7}} + 3 x^{2})}^{5}$ is 1080

answer is C, D.

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Detailed Solution

A) general term $\frac{5}{n! . n_{2}! . n_{3}!} {(2 x)}^{n_{1}} . {(x^{- 7})}^{n_{2}} {(3 x^{2})}^{n_{3}}$
Here $n_{1} + n_{2} + n_{3} = 5$ and $n_{1} - 7 n_{2} + 2 n_{3} = 0$
Only possibility is $n_{1} = 1, n_{2} = 1, n_{3} = 3$
Constant term = 1080
B) If x = 11 then a = 0.
p + b = 9; q + c = 14
b + c + r = 5 and p + q + r + b + c = 29
$\Rightarrow r + 23 = 29 \Rightarrow r = 6 is not possible$
$∴ x = 11$ is wrong
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C) $T_{r} = \frac{(2 r + 1) (2 r^{2} + 2 r - 1)}{{(r^{2} - 1)}^{2} {(r^{2} + 2 r)}^{2}} = \frac{(2 r + 1) (2 r^{2} + 2 r - 1)}{{(r^{2} - 1)}^{2} {({(r + 1)}^{2} - 1)}^{2}}$

$∴ T_{r} = \frac{1}{{(r^{2} - 1)}^{2}} - \frac{1}{{({(r + 1)}^{2} - 1)}^{2}} ∴ K = lim_{n \to \infty} \sum^{} ((\frac{1}{{(2^{2} - 1)}^{2}} - \frac{1}{{(3^{2} - 1)}^{2}}) + (\frac{1}{{(3^{2} - 1)}^{2}} - \frac{1}{{(4^{2} - 1)}^{2}}) + \dots) ∴ k = \frac{1}{9}$

D) $∴ {(\sum_{k = 1}^{10} a_{k})}^{2} = \sum_{k = 1}^{10} a_{k}^{2} + 2 (\underset{1 \leq k < j \leq 10}{\sum \sum} a_{k} . a_{j})$
$\Rightarrow {(50)}^{2} = \sum_{k = 1}^{10} a_{k}^{2} + 2 (1100) \Rightarrow \sum_{k = 1}^{10} a_{k}^{2} = 2500 - 2200 = 300 ∵ σ^{2} = (\frac{\sum_{k = 1}^{10} a_{k}^{2}}{10}) - {(\frac{\sum_{k = 1}^{10} a_{k}}{10})}^{2} \Rightarrow σ^{2} = \frac{300}{10} - {(\frac{50}{10})}^{2} = 30 - 25 ∴ standard deviation = \sqrt{5}$