Let [ε0] denote the dimensional formula of the permittivity of the vaccum, and [μ0] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current : [JEE-1998' 2/200]a)[ε0] = M-1L-3T2Ib)[ε0] = M-1L-3T4I2c)[μ0] = MLT-2I-2d)[μ0] = M L2T-1ICorrect answer is option 'B,C'. Can you explain this answer?

Question

Let [&#949;0] denote the dimensional formula of the permittivity of the vaccum, and [&#956;0] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current :          [JEE-1998' 2/200]a)[&#949;0] = M-1L-3T2Ib)[&#949;0] = M-1L-3T4I2c)[&#956;0] = MLT-2I-2d)[&#956;&#8203;0] = M L2T-1ICorrect answer is option 'B,C'. Can you explain this answer?

Answer

X_1, x_2, \ldots, x_n] be a list of n real numbers. We define the range of the list to be the difference between the largest and smallest elements in the list. That is,

range = max(x_1, x_2, \ldots, x_n) - min(x_1, x_2, \ldots, x_n)

For example, if the list is [1, 2, 3, 4, 5], then the range is 5 - 1 = 4.

The mean of the list is defined as the sum of the elements divided by n:

mean = (x_1 + x_2 + \ldots + x_n) / n

For example, if the list is [1, 2, 3, 4, 5], then the mean is (1+2+3+4+5)/5 = 3.

To find the standard deviation of the list, we first calculate the variance, which is the average of the squared differences between each element and the mean:

variance = ((x_1 - mean)^2 + (x_2 - mean)^2 + \ldots + (x_n - mean)^2) / n

The standard deviation is then the square root of the variance:

standard deviation = sqrt(variance)

For example, if the list is [1, 2, 3, 4, 5], then the mean is 3 and the variance is ((1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2)/5 = 2.

The standard deviation is then sqrt(2) = 1.414.

Answer

A_1, a_2, \ldots, a_n] be a sequence of positive real numbers. We define the sequence [b_1, b_2, \ldots, b_n] as follows:

b_1 = a_1

b_i = \frac{a_i}{\sum_{j=1}^{i-1} b_j} for i = 2, 3, \ldots, n.

Prove that b_n \leq \frac{2}{n-1} \sum_{i=1}^{n-1} a_i.

Solution: We will prove the inequality by induction on n. For n = 2, we have b_1 = a_1 and b_2 = \frac{a_2}{b_1}. Since b_1 = a_1 > 0, we have b_2 = \frac{a_2}{b_1} \leq 2a_2, which is equivalent to b_2 \leq \frac{2}{1} \sum_{i=1}^{1} a_i. Therefore, the inequality holds for n = 2.

Assume that the inequality holds for some positive integer k \geq 2, i.e., b_k \leq \frac{2}{k-1} \sum_{i=1}^{k-1} a_i. We need to prove that the inequality also holds for n = k+1.

Using the definition of b_i, we have

b_{k+1} = \frac{a_{k+1}}{\sum_{j=1}^{k} b_j} = \frac{a_{k+1}}{\sum_{j=1}^{k} \frac{a_j}{\sum_{i=1}^{j-1} b_i} } = \frac{a_{k+1}}{\sum_{i=1}^{k-1} b_i + \frac{a_k}{b_k} + \frac{a_{k+1}}{\sum_{i=1}^{k-1} b_i + \frac{a_k}{b_k}} }.

We can rewrite the denominator as

\sum_{i=1}^{k-1} b_i + \frac{a_k}{b_k} + \frac{a_{k+1}}{\sum_{i=1}^{k-1} b_i + \frac{a_k}{b_k}} = \sum_{i=1}^{k-1} b_i + \frac{a_k}{b_k} + \frac{a_{k+1}b_k}{\sum_{i=1}^{k-1} b_i b_k + a_k} = \frac{(\sum_{i=1}^{k-1} b_i)b_k + a_k + a_{k+1}b_k}{\sum_{i=1}^{k-1} b_i b_k + a_k}.

By the induction hypothesis, we have

\sum_{i=1}^{k-1} b_i \leq \frac{2}{k-2} \sum_{i=1}^{k-2} a_i,

which implies

(\sum_{i=1}^{

Answer

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