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If a, b, c are real numbers, then the intervals in which fx=x+a2abacab      x+b2bcacbc       x+c2  is strictly  decreasing

a
x < 0
b
x > 0
c
0 < x < 1
d
[−23,  0]

detailed solution

Correct option is D

f(x)=1abcax+a2ab2ac2a2bbx+b2bc2a2cb2ccx+c2=abcabcx+a2b2c2a2x+b2c2a2b2x+c2=x+a2+b2+c2b2c2x+a2+b2+c2x+b2c2x+a2+b2+c2b2x+c2=x+a2+b2+c21b2c21x+b2c21b2x+c2=x2x+a2+b2+c2f′(x)=3x2+2xa2+b2+c2f′(x)<0x(3x+2)a2+b2+c2≤0−23a2+b2+c2≤x≤0x(3x+2)≤0−23≤x≤0

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If a, b, c are real numbers, then the intervals in which f(x)  =  x+a2    ab     acab      x+b2  bcac       bc       x+c2  is strictly  decreasing


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