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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

\frac{1}{1+a} + \frac{1}{b+1} + \frac{1}{1+c}

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

Solution:

Concept:

Use the given relations to write each fraction in terms of

a+b+c

.

Explanation:

For

a+b+c\neq0

, let

S=a+b+c

.

From

a^2=b+c

, add

a

to both sides:

a^2+a=a+b+c

a(a+1)=S

So,

\frac{1}{a+1}=\frac{a}{S}

.

Similarly, from

b^2=c+a

:

b(b+1)=S \Rightarrow \frac{1}{b+1}=\frac{b}{S}

And from

c^2=a+b

:

c(c+1)=S \Rightarrow \frac{1}{c+1}=\frac{c}{S}

Now add the three fractions:

\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=\frac{a}{S}+\frac{b}{S}+\frac{c}{S}

=\frac{a+b+c}{S}=\frac{S}{S}=1

Answer:

The required value is

1

.

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