PE 1 - Surds & Indices

Let x = $\sqrt{\sqrt[3]{\sqrt[4]{\sqrt[5]{\sqrt[6]{8^{240}}}}}}$

⇒ x = ${\left({\left({\left({\left({\left(8^{240}\right)}^{\frac{1}{6}}\right)}^{\frac{1}{5}}\right)}^{\frac{1}{4}}\right)}^{\frac{1}{3}}\right)}^{\frac{1}{2}}$

⇒ x = $8^{240\times \frac{1}{6}\times \frac{1}{5}\times \frac{1}{4}\times \frac{1}{3}\times \frac{1}{2}}$

⇒ x = $8^{\frac{240}{720}}$

⇒ x = $8^{\frac{1}{3}}$

⇒ x = ${\left(2^3\right)}^{\frac{1}{3}}$

⇒ x = $2^{3\times \frac{1}{3}}$ = 2

Hence, option (a).

Let x = $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$

Rationalising with $\sqrt{a+b}+\sqrt{a-b}$

⇒ x = $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$ × $\frac{\sqrt{a+b}+\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}}$

⇒ x = $\frac{{\left(\sqrt{a+b}+\sqrt{a-b}\right)}^2}{{\left(\sqrt{a+b}\right)}^2-{\left(\sqrt{a-b}\right)}^2}$

⇒ x = $\frac{2a+2\sqrt{a^2-b^2}}{2b}$

Now, a = 2 and b = 1

∴ x = $\frac{4+2\sqrt{3}}{2}$

Now $\sqrt{x}$ = $\sqrt{\frac{4+2\sqrt{3}}{2}}$

∴ $\sqrt{x}$ = $\frac{\sqrt{3}+1}{\sqrt{2}}$

Hence, option (b).

Given $\sqrt[ab]{x^{c^2}}\times \sqrt[bc]{x^{a^2}}\times \sqrt[ca]{{x^b}^2}$& 

= $x^{\frac{c^2}{ab}}\times x^{\frac{a^2}{bc}}\times x^{\frac{b^2}{ca}}$

= $x^{\frac{c^3}{abc}}\times x^{\frac{a^3}{abc}}\times x^{\frac{b^3}{abc}}$

= $x^{\frac{a^3+b^3+c^3}{abc}}$

Since, a + b + c = 0, it means a³ + b³ + c³ = 3abc

∴ $\sqrt[ab]{x^c}\times \sqrt[bc]{x^a}\times \sqrt[ca]{x^b}$ = $x^{\frac{3abc}{abc}}$ = x³

Hence, option (a).

Given, x = 3 + $\sqrt[3]{3}$ + $\sqrt[3]{9}$

⇒ x - 3 = $\sqrt[3]{3}$ + $\sqrt[3]{9}$

Cubing both sides 

⇒ x³ - 27 - 9x² + 27x = 12 + 3 × $\sqrt[3]{3}$ × $\sqrt[3]{9}$ × ($\sqrt[3]{3}$ + $\sqrt[3]{9}$)

⇒ x³ - 9x² + 27x - 27 = 12 + 3 × 3 × x = 12 + 9x

⇒ x³ - 18x² + 18x = 39

∴ x³ - 18x² + 18x + 27 = 66

Hence, option (d).

Let 3^x = a and 5^y = b

⇒ 3^x + 5^y = 52
⇒ a + b = 52

Also, 3^2x-2 + 5^y+1 = 206
⇒ 3^2x/9 + 5 × 5^y = 206
⇒ 3^2x + 45 × 5^y = 1854
⇒ a² + 45b = 1854   ...(2)

(2) - 45 × (1)

⇒ a² - 45a = 1854 - 2340
⇒ a² - 45a + 486 = 0
⇒ (a - 18)(a - 27) = 0
⇒ a = 18 or 27

⇒ 3^x = 18 or 27​​​​​​​
Since x is an integer, 3^x = 27 and x = 3.

Also, b = 5^y = 52 - 27 = 25, hence, y = 2

∴ x + y = 3 + 2 = 5.

Hence, option (c).

Let a = 11^x + 11^x+1 + 11^x+2​​​​​​​

⇒ a = 11^x × (1 + 11¹ + 11²​​​​​​​)

⇒ a = 11^x × 133

Since x is a positive integer, the least value of 11^x will always be a multiple of 11 and its least value will be 11.

∴ Least possible value of a = 11 × 133 = 1463

Hence, option (d).

Let x = 5 + √6 + √10 + √15

Now, 2x = 10 + 2√6 + 2√10 + 2√15

This can be written as:

2x = (√2)² + (√3)² + (√5)² + 2 × √2 × √3 + 2 × √2 × √5 + 2 × √3 × √5

The RHS is of the form a² + b² + c² + 2ab + 2bc + 2ca which is equal to (a + b + c)²

∴ 2x = (√2 + √3 + √5)²

∴ x = $\frac{{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)}^2}{2}$

⇒ √x = $\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)}{\sqrt{2}}$

∴ √x = 1 + $\sqrt{\frac{3}{2}}$ + $\sqrt{\frac{5}{2}}$

Hence, option (b).

2^x = 8^y+1
∴ 2^x = (2³)^y+1
∴ 2^x = 2^3(y+1)
∴ x = 3y+3   ...(1)

Consider, 
9^y = 3^x–9
∴ (3²)^y = 3^x–9
∴ 3^2y = 3^x–9
∴ 2y = x – 9       …(2) 

Solving (1) and (2), we get, 
x = 21 and y = 6
∴ x + y = 21 + 6 = 27

Hence, option (d).

By the condition given in the question,
r = (2a)^2b
⇒ r = 2^2b × a^2b
⇒ r = (2²)^b × (a^b)²
⇒ r = (4)^b × (a^b)²

Also, r = a^b × x^b

∴ a^b × x^b = 4^b × (a^b)²

∴ x^b = 4^b × a^b
∴ x^b = (4a)^b
∴ x = 4a

Hence, option (b).

$\sqrt{\frac{4}{3}}-\sqrt{\frac{3}{4}}$

= $\frac{\sqrt{4}\times \sqrt{4}-\sqrt{3}\times \sqrt{3}}{\sqrt{3\times 4}}$

= $\frac{4-3}{\sqrt{12}}$

= $\frac{1}{\sqrt{12}}$

= $\frac{1}{2\sqrt{3}}$

= $\frac{\sqrt{3}}{6}$

Hence, option (a).