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Question

# If x,y,z are real numbers such that x+y+z=4 and x2+y2+z2=6, then the range of x is

A
[23,2]
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B
(1,1)
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C
[0,2]
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D
[2,3]
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Solution

## The correct option is A [23,2]Given, x2+y2+z2=6 and x+y+z=4 Squaring both the sides, (x+y+z)2=(4)2 ⇒x2+y2+z2+2(xy+yz+zx)=16 ⇒6+2(xy+yz+zx)=16 ⇒xy+yz+zx=5 y+z=4−x ⋯(1) Using (1), we get y(4−x−y)=5−x(4−x) ⇒4y−xy−y2=5−x(4−x) ⇒y2−(4−x)y+5−x(4−x)=0 If y is real, then D≥0 ⇒(4−x)2−4(5−x(4−x))≥0 ⇒16+x2−8x−20+4x(4−x)≥0 ⇒16+x2−8x−20+16x−4x2≥0 ⇒−3x2+8x−4≥0 ⇒3x2−8x+4≤0 ⇒(3x−2)(x−2)≤0 ⇒x∈[23,2]

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