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If a₁, a₂, a₃, ........., a_n are positive real numbers whose product is a fixed number c, then the minimum value of a₁ + a₂ + a₃ + .... + a_{n – 1} + 2a_n is

$n (2 c)^{1 / n}$
$(n + 1) c^{1 / n}$
$2 {nc}^{1 / n}$
$(n + 1) (2 c)^{1 / n}$

hint Hint:

In this question, we have to find the minimum value of a1 + a2 + a3 + …… + an-1 + 2an. For solving this question, remember the inequality of A.M and G. M.

The correct answer is: $n left parenthesis 2 c right parenthesis to the power of 1 divided by n end exponent$

Here we have to find the minimum value of a1 + a2 + a3 + …… + an-1 + 2an.
Firstly, given is
a1. a2. a3….an =c ------(1)
We know that,
Always A.M ≥ G.M
So, A.M = $fraction numerator left parenthesis a 1 space plus space a 2 space plus space a 3 space plus space horizontal ellipsis plus space a n minus 1 space plus space 2 a n right parenthesis over denominator n end fraction$ And
G.M = (a1.a2.a3……an-1 (2an)) ^^{$1 over n$}
From, A.M ≥ G.M, we can write
=> (a1 + a2 + a3 + …+ an-1 + 2an) / n ≥ (a1.a2.a3……an-1 (2an)) ^^{$1 over n$}Cross multiply,
=> a1 + a2 + a3 + …+ an-1 + 2an ≥ n (a1.a2.a3……an-1 (2an)) ^ $1 over n$
=> a1 + a2 + a3 + …+ an-1 + 2an ≥ n 2 (c)^ $1 over n$ [since, c = a1. a2. a3…...an]
=> a1 + a2 + a3 + …+ an-1 + 2an ≥ 2nc^ $1 over n$
Therefore, the minimum value of a1 + a2 + a3 + …… + an-1 + 2an is 2nc^ $1 over n$ .
The correct answer is 2nc^ $1 over n$ .

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If a1, a2, a3, ........., an are positive real numbers whose product is a fixed number c, then the minimum value of a1 + a2 + a3 + .... + an – 1 + 2an is

In this question, we have to find the minimum value of a1 + a2 + a3 + …… + an-1 + 2an. For solving this question, remember the inequality of A.M and G. M.

The correct answer is:

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The correct answer is: $n left parenthesis 2 c right parenthesis to the power of 1 divided by n end exponent$

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Range of the function $f open parentheses x close parentheses equals fraction numerator x to the power of 2 end exponent over denominator x to the power of 2 end exponent plus 1 end fraction$ is

Range of the function $f open parentheses x close parentheses equals fraction numerator x to the power of 2 end exponent over denominator x to the power of 2 end exponent plus 1 end fraction$ is

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