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Sequences & Series: Non Negative Reals Range

JEE Maths question with a full step-by-step solution.

Question

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 reals and the range of

e

\left[\ell,\frac{m}{n}\right]

with

\gcd(m,n)=1

, then

\ell+m+n

21

correct

26

20

31

Solution

Step 1: For

a,b,c,d

, the quadratic mean is at least the arithmetic mean:

\left(\frac{a+b+c+d}{4}\right)^2\le\frac{a^2+b^2+c^2+d^2}{4}.

Step 2: Substitute

a+b+c+d=8-e

and

a^2+b^2+c^2+d^2=16-e^2

\left(\frac{8-e}{4}\right)^2\le\frac{16-e^2}{4}\ \Rightarrow\ 64-16e+e^2\le 64-4e^2.

Step 3:

5e^2-16e\le 0\Rightarrow 0\le e\le\dfrac{16}{5}

. So

\ell=0,\ m=16,\ n=5

and

\ell+m+n=21.

Correct answer: (1)

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