CRE 2 - Surds | Indices & Surds

Answer: Option C

Explanation :

Rationalizing factor for 5 - 2√6 = 5 + 2√6

∴ Rationalizing factor for 5 - 2√6 = 5 + 2√6 or any rational multiple of it.

Hence, option (c).

Answer: Option B

Explanation :

Given, $\frac{2}{\sqrt{5}-\sqrt{3}}$

Here, the denominator is √5 - √3, its conjugate is √5 + √3.

We multiply and divide the given expression with √5 + √3.

∴ $\frac{2}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ = $\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2}$ = $\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{5-3}$ = √5 + √3.

Hence, option (b).

Answer: Option B

Explanation :

Rationalizing the first term here, $\frac{2}{\sqrt{3}-\sqrt{2}}$ = $\frac{2}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ = 2√3 + 2√2.

Rationalizing the second term here, $\frac{1}{3-2\sqrt{2}}$ = $\frac{1}{3-2\sqrt{2}}\times \frac{3+2\sqrt{2}}{3+2\sqrt{2}}$ = 3 + 2√2

∴ $\frac{2}{\sqrt{3}+\sqrt{2}}-\frac{1}{3-2\sqrt{2}}$ = 2√3 + 2√2 – (3 + 2√2) = 2√3 – 3 = √3(2 - √3)

Hence, option (b).

Answer: Option D

Explanation :

Let X = $\left[1+\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+\mathrm{ }\dots \dots \dots .\mathrm{ }\frac{1}{\sqrt{100}+\sqrt{99}}\right]$

Consider the 2nd term here i.e., $\frac{1}{\sqrt{2}+\sqrt{1}}$.

Upon rationalization, $\frac{1}{\sqrt{2}+\sqrt{1}}$ = √2 - √1.

Similarly, when we rationalize all the terms of the given expression it becomes

X = [1 + (√2 - √1) + (√3 - √2) + (√4 - √3) + … + (√100 - √99)]

⇒ X = [√100] = 10

We next find √X = √10.

Hence, option (d).

Answer: Option C

Explanation :

x³ – y³ = (x - y)(x² + xy + y²)
x³ + y³ = (x + y)(x² - xy + y²)

Consider, a – b

It can be written as (∛a)³ - (∛b)³

∴ a – b = (∛a)³ - (∛b)³ = (∛a - ∛b)((∛a)² + ∛a × ∛b + (∛b)²)

Similarly, 

∴ a + b = (∛a)³ + (∛b)³ = (∛a + ∛b)((∛a)² - ∛a × ∛b + (∛b)²)

We need to find,

$\frac{\mathrm{a}-\mathrm{b}}{\sqrt[3]{{\mathrm{a}}^2}+\sqrt[3]{\mathrm{ab}}+\sqrt[3]{{\mathrm{b}}^2}}$ - $\frac{\mathrm{a}+\mathrm{b}}{\sqrt[3]{{\mathrm{a}}^2}-\sqrt[3]{\mathrm{ab}}+\sqrt[3]{{\mathrm{b}}^2}}$

= $\frac{\left(\sqrt[3]{\mathrm{a}}\mathrm{ }-\mathrm{ }\sqrt[3]{\mathrm{b}}\right)\left({\left(\sqrt[3]{\mathrm{a}}\right)}^2\mathrm{ }+\mathrm{ }\sqrt[3]{\mathrm{ab}}\mathrm{ }+\mathrm{ }{\left(\sqrt[3]{\mathrm{b}}\right)}^2\right)}{\sqrt[3]{{\mathrm{a}}^2}+\sqrt[3]{\mathrm{ab}}+\sqrt[3]{{\mathrm{b}}^2}}$ - $\frac{\left(\sqrt[3]{\mathrm{a}}+\mathrm{ }\sqrt[3]{\mathrm{b}}\right)\left({\left(\sqrt[3]{\mathrm{a}}\right)}^2-\mathrm{ }\sqrt[3]{\mathrm{ab}}\mathrm{ }+\mathrm{ }{\left(\sqrt[3]{\mathrm{b}}\right)}^2\right)}{\sqrt[3]{{\mathrm{a}}^2}-\sqrt[3]{\mathrm{ab}}+\sqrt[3]{{\mathrm{b}}^2}}$

= (∛a - ∛b) – (∛a + ∛b)

= -2 × ∛b

Hence, option (c).

Answer: Option D

Explanation :

(a + b) can be written as ${\left({\mathrm{a}}^{1/3}\right)}^3+{\left({\mathrm{b}}^{1/3}\right)}^3$

Now, ${\left({\mathrm{a}}^{1/3}\right)}^3+{\left({\mathrm{b}}^{1/3}\right)}^3$ = $\left({\mathrm{a}}^{1/3}+{\mathrm{b}}^{1/3}\right)$ × $\left({\mathrm{a}}^{2/3}-{\left(\mathrm{ab}\right)}^{1/3}+{\mathrm{b}}^{2/3}\right)$

∴ a + b =  $\left({\mathrm{a}}^{1/3}+{\mathrm{b}}^{1/3}\right)$ × $\left({\mathrm{a}}^{2/3}-{\left(\mathrm{ab}\right)}^{1/3}+{\mathrm{b}}^{2/3}\right)$

⇒ Product of $\left({\mathrm{a}}^{1/3}+{\mathrm{b}}^{1/3}\right)$ and $\left({\mathrm{a}}^{2/3}-{\left(\mathrm{ab}\right)}^{1/3}+{\mathrm{b}}^{2/3}\right)$ is rational.

∴ Rationalizing factor of $\left({\mathrm{a}}^{1/3}+{\mathrm{b}}^{1/3}\right)$ is $\left({\mathrm{a}}^{2/3}-{\left(\mathrm{ab}\right)}^{1/3}+{\mathrm{b}}^{2/3}\right)$

⇒ Rationalizing factor of $\left(2^{1/3}+3^{1/3}\right)$ is $\left(2^{2/3}-{(2\times 3)}^{1/3}+3^{2/3}\right)$ i.e., $\left(2^{2/3}-6^{1/3}+3^{2/3}\right)$

Hence, option (d). 

Answer: Option C

Explanation :

Given, $\frac{4}{1+\sqrt{2}+\sqrt{3}}$

We first multiply and divide by 1 - (√2 + √3)

∴$\frac{4}{1+\sqrt{2}+\sqrt{3}}$ = $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ × $\frac{1-(\sqrt{2}+\sqrt{3})}{1-(\sqrt{2}+\sqrt{3})}$

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ = $\frac{4\left[1-(\sqrt{2}+\sqrt{3})\right]}{1^2-{(\sqrt{2}+\sqrt{3})}^2}$

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ = $\frac{4\left[1-(\sqrt{2}+\sqrt{3})\right]}{1-(5+2\sqrt{6})}$

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ = $\frac{4\left[1-(\sqrt{2}+\sqrt{3})\right]}{-(4+2\sqrt{6})}$

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ = $\frac{2\left[(\sqrt{2}+\sqrt{3})-1\right]}{2+\sqrt{6}}$

Now, we multiply and divide by 2 - √6

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ = $\frac{2\left[(\sqrt{2}+\sqrt{3})-1\right]}{2+\sqrt{6}}$ × $\frac{2-\sqrt{6}}{2-\sqrt{6}}$

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ = $\frac{2\left[(\sqrt{2}+\sqrt{3})-1\right]\times (2-\sqrt{6})}{2^2-{\left(\sqrt{6}\right)}^2}$

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ = $\frac{2\left[(\sqrt{2}+\sqrt{3})-1\right]\times (2-\sqrt{6})}{-2}$

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ = [(√2 + √3) - 1] × (√6 - 2)

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ =  √12 - 2√2 + √18 - 2√3 - √6 + 2

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ =  2√3 - 2√2 + 3√2 - 2√3 - √6 + 2

⇒ $\frac{4}{1+\sqrt{2}+\sqrt{3}}$ =  √2 - √6 + 2

Hence, option (c).