PE 3 - Surds & Indices

u₁ = $\sqrt{3}$, 

u₂ = $\sqrt{3\sqrt{3}}$ = $\sqrt{3^1\times 3^{1/2}}$ = $3^{\frac{1}{2}}\times 3^{\frac{1}{4}}$ = $3^{\frac{1}{2}+\frac{1}{4}}$ = 3^{​​​​​​​3/4}, 

u₃ = $\sqrt{3\sqrt{3\sqrt{3}}}$ = $3^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}$ = 3^{​​​​​​​7/8}

∴ u_n = $3^{\frac{2^n-1}{2^n}}$

⇒ u₁₀ : u₉ = $3^{\frac{2^{10}-1}{2^{10}}}$ : $3^{\frac{2^9-1}{2^9}}$ 

= $3^{\frac{1023}{1024}-\frac{511}{512}}$ 

= $3^{\frac{1}{1024}}$ 

= $3^{\frac{1}{2^{10}}}$

Hence, option (d).

$\frac{2^{\mathrm{n}+4}-2\left(2^{\mathrm{n}}\right)}{2\left(2^{\mathrm{n}+3}\right)}$

= $\frac{2^{\mathrm{n}+4}-2^{\mathrm{n}+1}}{2^{\mathrm{n}+4}}$

= $\frac{2^{\mathrm{n}+1}(2^3-1)}{2^{\mathrm{n}+4}}$

= $\frac{(2^3-1)}{2^3}$

= $\frac{7}{8}$

Hence, option (d).

$2+\sqrt{2}$ + $\frac{1}{2+\sqrt{2}}$ + $\frac{1}{\sqrt{2}-2}$

= $2+\sqrt{2}$ + $\frac{1}{2+\sqrt{2}}$ - $\frac{1}{2-\sqrt{2}}$

= $\frac{\left(2+\sqrt{2}\right)\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)+\left(2-\sqrt{2}\right)-\left(2+\sqrt{2}\right)}{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}$

= $\frac{\left(2+\sqrt{2}\right)(4-2)+2-\sqrt{2}-2-\sqrt{2}}{4-2}$

= $\frac{4+2\sqrt{2}-2\sqrt{2}}{4-2}$

= $\frac{4}{2}$

= 2

Hence, option (a).

Cube-root of 26 + 15√3 will be of the form x + √y

Now, 26 + 15√3 = (x + √y)³

⇒ 26 + 15√3 = x³ + y√y + 3x²√y + 3xy

⇒ 26 + 15√3 = (x³ + 3xy) +  (y + 3x²)√y

By comparing rational and irrational terms on both sides

√y = √3 ⇒ y = 3

and, x³ + 3xy = 26
⇒ x³ + 9x = 26
⇒ x = 2 (by trial and error)

∴ Cube-root of 26 + 15√3 = 2 + √3

Hence, option (c).

Let us multiply and divide the given expression with √x 

The given expression becomes $\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}}}\times \sqrt{\mathrm{x}+\sqrt{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{\frac{\mathrm{x}+\sqrt{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}}{x}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{\frac{\mathrm{x}}{\mathrm{x}}+\frac{\sqrt{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}}{\mathrm{x}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\frac{\sqrt{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}}{\mathrm{x}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{\frac{{\mathrm{x}}^2+\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}{{\mathrm{x}}^2}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{\frac{{\mathrm{x}}^2}{{\mathrm{x}}^2}+\frac{\sqrt{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}}{{\mathrm{x}}^2}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{1+\sqrt{\frac{{\mathrm{x}}^4+\sqrt{{\mathrm{x}}^8+...\infty }}{{\mathrm{x}}^4}}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{1+\sqrt{\frac{{\mathrm{x}}^4}{{\mathrm{x}}^4}+\frac{\sqrt{{\mathrm{x}}^8+...\infty }}{{\mathrm{x}}^4}}}}$

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{\frac{{\mathrm{x}}^8+...\infty }{x^8}}}}}$

Finally, we will get the following expression

= $\frac{\sqrt{\mathrm{x}}}{1}\times \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...\infty }}}}}$   ...(1)

Let y = $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...\infty }}}}}$

⇒ y = $\sqrt{1+y}$

⇒ y² = 1 + y

⇒ y² - y - 1 = 0

⇒ y = $\frac{-(-1)\pm \sqrt{{(-1)}^2-4\times 1\times -1}}{2\times 1}$ = $\frac{1+\sqrt{5}}{2}$ [we will accept only positive value]

From (1), the original expression becomes = $\sqrt{x}\times \left(\frac{1+\sqrt{5}}{2}\right)$

Hence, option (b).