If a2 = b + c, b2 = a + c, c2 = b + a, then what will be the value of $$\frac{1}{{a + 1}}$$ + $$\frac{1}{{b + 1}}$$ + $$\frac{1}{{c + 1}}$$ ?
A. -1
B. 2
C. 1
D. 0
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {a^2} = b + c,{\text{ }}{b^2} = a + c,{\text{ }}{c^2} = b + a \cr & {\text{Taking }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & {\text{So,}}{\left( 2 \right)^2} = 2 + 2 \cr & \boxed{4 = 4} \cr & {\text{Now,}}\frac{1}{{a + 1}} + {\text{ }}\frac{1}{{b + 1}} + {\text{ }}\frac{1}{{c + 1}} \cr & {\text{Put }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & = \frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} \cr & = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \cr & = 1 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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