1a+11a+1\frac{1}{{a + 1}} + 1b+1" role="presentation">1b+11b+1\frac{1}{{b + 1}} + 1c+1" role="presentation">1c+11c+1\frac{1}{{c + 1}} ?" /> 1a+11a+1\frac{1}{{a + 1}} + 1b+1" role="presentation">1b+11b+1\frac{1}{{b + 1}} + 1c+1" role="presentation">1c+11c+1\frac{1}{{c + 1}} ?" /> 1a+11a+1\frac{1}{{a + 1}} + 1b+1" role="presentation">1b+11b+1\frac{1}{{b + 1}} + 1c+1" role="presentation">1c+11c+1\frac{1}{{c + 1}} ?" />

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If a² = b + c, b² = a + c, c² = b + a, then what will be the value of $\frac{1}{a + 1}$ $\frac{1}{{a + 1}}$ + $\frac{1}{b + 1}$ $\frac{1}{{b + 1}}$ + $\frac{1}{c + 1}$ $\frac{1}{{c + 1}}$ ?

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a^{2} = b + c, b^{2} = a + c, c^{2} = b + a \\ Taking a = 2, b = 2 and c = 2 \\ So, {(2)}^{2} = 2 + 2 \\ 4 = 4 \\ Now, \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \\ Put a = 2, b = 2 and c = 2 \\ = \frac{1}{2 + 1} + \frac{1}{2 + 1} + \frac{1}{2 + 1} \\ = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \\ = 1 \end{aligned}

$\eqalign{ & {a^2} = b + c,{\text{ }}{b^2} = a + c,{\text{ }}{c^2} = b + a \cr & {\text{Taking }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & {\text{So,}}{\left( 2 \right)^2} = 2 + 2 \cr & \boxed{4 = 4} \cr & {\text{Now,}}\frac{1}{{a + 1}} + {\text{ }}\frac{1}{{b + 1}} + {\text{ }}\frac{1}{{c + 1}} \cr & {\text{Put }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & = \frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} \cr & = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \cr & = 1 \cr}$

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+

\frac{1}{b + 1}

$\frac{1}{{b + 1}}$ +

\frac{1}{c + 1}

$\frac{1}{{c + 1}}$ ?", "text": "If a² = b + c, b² = a + c, c² = b + a, then what will be the value of

\frac{1}{a + 1}

$\frac{1}{{a + 1}}$ +

\frac{1}{b + 1}

$\frac{1}{{b + 1}}$ +

\frac{1}{c + 1}

$\frac{1}{{c + 1}}$ ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a^{2} = b + c, b^{2} = a + c, c^{2} = b + a \\ Taking a = 2, b = 2 and c = 2 \\ So, {(2)}^{2} = 2 + 2 \\ 4 = 4 \\ Now, \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \\ Put a = 2, b = 2 and c = 2 \\ = \frac{1}{2 + 1} + \frac{1}{2 + 1} + \frac{1}{2 + 1} \\ = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \\ = 1 \end{aligned}

$\eqalign{ & {a^2} = b + c,{\text{ }}{b^2} = a + c,{\text{ }}{c^2} = b + a \cr & {\text{Taking }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & {\text{So,}}{\left( 2 \right)^2} = 2 + 2 \cr & \boxed{4 = 4} \cr & {\text{Now,}}\frac{1}{{a + 1}} + {\text{ }}\frac{1}{{b + 1}} + {\text{ }}\frac{1}{{c + 1}} \cr & {\text{Put }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & = \frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} \cr & = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a^{2} = b + c, b^{2} = a + c, c^{2} = b + a \\ Taking a = 2, b = 2 and c = 2 \\ So, {(2)}^{2} = 2 + 2 \\ 4 = 4 \\ Now, \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \\ Put a = 2, b = 2 and c = 2 \\ = \frac{1}{2 + 1} + \frac{1}{2 + 1} + \frac{1}{2 + 1} \\ = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \\ = 1 \end{aligned}

$\eqalign{ & {a^2} = b + c,{\text{ }}{b^2} = a + c,{\text{ }}{c^2} = b + a \cr & {\text{Taking }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & {\text{So,}}{\left( 2 \right)^2} = 2 + 2 \cr & \boxed{4 = 4} \cr & {\text{Now,}}\frac{1}{{a + 1}} + {\text{ }}\frac{1}{{b + 1}} + {\text{ }}\frac{1}{{c + 1}} \cr & {\text{Put }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & = \frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} \cr & = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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