If a-3-2-3 and b-3-2-3- then the value of a-1-1-b-1-1 is

The correct option is B 1 Given, a=1(√(3)+√(2))^3,b=1(√(3) √(2))^3 ⇒ #160; (a+1)^ 1+(b+a)^ 1=1a+1+1b+1 =b+1+a+1(a+1)(b+1) =a+b+2ab+a ...

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Standard XII

Mathematics

Theorems for Differentiability

If a=√3+√2-...

Question

If $a = (\sqrt{3} + \sqrt{2})^{- 3}$ and $b = (\sqrt{3} - \sqrt{2})^{- 3}$ , then the value of $(a + 1)^{- 1} + (b + 1)^{- 1}$ is

50 \sqrt{3}

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48 \sqrt{2}

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1

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5

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Solution

The correct option is B $1$
Given, $a = \frac{1}{(\sqrt{3} + \sqrt{2})^{3}}, b = \frac{1}{(\sqrt{3} - \sqrt{2})^{3}}$

$\Rightarrow (a + 1)^{- 1} + (b + a)^{- 1} = \frac{1}{a + 1} + \frac{1}{b + 1}$
$= \frac{b + 1 + a + 1}{(a + 1) (b + 1)}$
$= \frac{a + b + 2}{a b + a + b + 1}$

Now, $a b = \frac{1}{(\sqrt{3} + \sqrt{2})^{3}} \times \frac{1}{(\sqrt{3} - \sqrt{2})^{3}}$ $= \frac{1}{{[(\sqrt{3} + 2 \sqrt{2}) (\sqrt{3} - \sqrt{2})]}^{3}}$
$= \frac{1}{(3 - 2)^{2}} = \frac{1}{1^{1}} = 1$

$∴ (a + 1)^{- 1} + (b + 1)^{- 1} = \frac{a + b + 2}{1 + a + b + 1} = \frac{a + b + 2}{a + b + 2} = 1$

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If a=(√3+√2)−3 and b=(√3−√2)−3, then the value of (a+1)−1+(b+1)−1 is

The correct option is B 1Given, a=1(√3+√2)3,b=1(√3−√2)3⇒(a+1)−1+(b+a)−1=1a+1+1b+1=b+1+a+1(a+1)(b+1)=a+b+2ab+a+b+1Now, ab=1(√3+√2)3×1(√3−√2)3 =1[(√3+2√2)(√3−√2)]3 =1(3−2)2=111=1∴(a+1)−1+(b+1)−1=a+b+21+a+b+1=a+b+2a+b+2=1

If $a = (\sqrt{3} + \sqrt{2})^{- 3}$ and $b = (\sqrt{3} - \sqrt{2})^{- 3}$ , then the value of $(a + 1)^{- 1} + (b + 1)^{- 1}$ is