Statement 1 : $a + b + c + d + e = 8$ and $a^{2} + b^{2} + c^{2} + d^{2} + e^{2} = 16,$ where $a, b, c, d, e$ are non-negative real numbers, then $e \in [0, \frac{16}{5}]$
Statement 2: ${(\frac{a + b + c + d}{4})}^{2} \leq \frac{a^{2} + b^{2} + c^{2} + d^{2}}{4}$ where $a, b, c, d \in R$

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Statement -1 is true, statement -2 is false

Both the Statements are true and statement 2 is not the correct explanation for statement 1.

Statement -1 is false, Statement -2 is true

Both the Statements are true and statement 2 is correct explanation for statement 1.

answer is A.

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Detailed Solution

${(\frac{a + b + c + d}{4})}^{2} \leq \frac{a^{2} + b^{2} + c^{2} + d^{2}}{4}$

$\frac{{(8 - e)}^{2}}{16} \leq \frac{16 - e^{2}}{4}$

$64 + e^{2} - 16 e \leq 64 - 4 e^{2}$

$5 e^{2} - 16 e \leq 0$

$0 \leq e \leq \frac{16}{5}$

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