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if $a^2+b^2+c^2=1$ then $ab+bc+ac$ lies in the interval

  1. $[1,2/3]$
  2. $[-1/2,1]$
  3. $[-1,1/2]$
  4. $[2,-4]$ 
in Numerical Ability
edited by
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1 Answer

18 votes
 
Best answer

Ans (B) 
$(a+b+c)^2$ should be $\geq 0.$
$ \therefore a^2 + b^2  +c^2  + 2ab +2bc + 2ca \geq 0$

Given $a^2 + b^2  +c^2  = 1$ therefore $1+ 2(ab+bc+ca)\geq 0$
$\implies ab+bc+ca\geq -\frac{1}{2}$

Now to find the upper limit,
$(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$

Expanding we get, $2(a^2 + b^2 + c^2) - 2(ab+bc+ca) \geq 0$
$\implies 2-2(ab+bc+ca)\geq 0$
$\implies ab+bc+ca\leq 1$ 

Answer B. [-1/2,1]


edited by
0

(a-b)2+(b-c)2+(c-a)2 >=0 

How did you come up with this equation? 

1

@Tuhin Dutta It is the sum of squares. Therefore, it will never be negative. It will either be positive or zero.

Answer:

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