RRB NTPC Mains or Stage 2 Model Paper 8 with solutions

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Question : 15 of 120

Marks: +1, -0

If

a^2 = b + c, b^2 = c + a, c^2 = a + b

, then the value of

1/{1+a}+ 1/ {b+1} +1 /{1+c}

abc
$a^2 b^2 c^2$
1
0

Solution:

(c)1/{1+ a} + 1/ {1+ b} +1/{1 +c}...(i)
Given that,

a^2 = b + c

a + a^2 = a + b + c

a (a + 1) = a + b + c

a + 1 ={a+ b+ c}/a

1/{a+1}=a/ {a+ b+ c}

Similarly,

1/{b+1}=b/ {a+ b+ c}

1/{c+1}=c/ {a+ b+ c}

Put in eq. (i)

∴a/ {a+ b +c}+ b/ {a+ b+ c}+c/{ a+ b+ c}

={a+ b+ c}/{a +b+ c}=1

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