If real numbers, a, b, c, d, e satisfy a + 1 = b + 2 = c + 3 = d + 4 = e + 5 = a + b + c + d + e + 3.
Find the value of a^2+b^2+c^2+d^2+e^2?
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Answered by
1
Step-by-step explanation:
As we know
(
4
a+b+c+d
)
2
≤
4
a
2
+b
2
+c
2
+d
2
(i)(Using Tchebycheff's Inequality)
Where a+b+c+d+e=8 and a
2
+b
2
+c
2
+d
2
+e
2
=16
∴ Eq. (i), reduces to
(
4
8−e
)
2
≤
4
16−e
2
⇒ 64+e
2
−16e≤4(16−e
2
)
⇒ 5e
2
−16e≤0
⇒ e(5e−16)≤0 (Using number line rule)
⇒ 0≤e≤
5
16
Thus, range of eϵ[0,
5
16
Answered by
0
Answer:
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