PE 2 - Surds & Indices | Algebra - Surds & Indices

Answer: Option A

Explanation :

Let x = $\frac{2\mathrm{ }}{2+\frac{2\mathrm{ }}{\mathrm{ }2+\frac{2\mathrm{ }}{\mathrm{ }2+\mathrm{ }\frac{2\mathrm{ }}{\dots \dots \dots ..\mathrm{\infty }}\mathrm{ }}}}$

⇒ x = $\frac{2}{2+x}$

⇒ 2x + x² = 2

⇒ x² + 2x – 2 = 0

⇒ x = -1 ± √3 (Using formula for quadratic equations)

Here, since x > 0, we will reject the negative value.

∴ x = -1 + √3

Hence, option (a).

Answer: 33

Explanation :

Given, x = 3 + √8 and x × y = 1

⇒ y = $\frac{1}{3+\sqrt{8}}$ = $\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}$ = 3 - √8

∴ x² – xy + y² = x² – 2xy + y² + xy

= (x - y)² + xy

= (2√8)² + 1

= 32 + 1 = 33

Hence, 33.

Answer: 16

Explanation :

x = 1 + √2 + √3

⇒ x - 1 = √2 + √3

Squaring both sides, we get

⇒ x² – 2x + 1 = 2 + 3 + 2√6

⇒ x² – 2x - 4 = 2√6

Again, squaring both sides, we get

⇒ x⁴ + 4x² + 16 – 4x³ – 8x² + 16x = 24.

⇒ x⁴ – 4x³ - 4x² + 16x + 16 = 24.

⇒ x⁴ – 4x³ - 4x² + 16x = 8.

Now, we have 

x⁴ – 4x³ – 4x² + 16x + 8 = 8 + 8 = 16.

Hence, 16.

Answer: Option A

Explanation :

x = $\frac{7}{\sqrt{11-6\sqrt{2}}}$ = $\frac{7}{3-\sqrt{2}}$ = $\frac{7}{3-\sqrt{2}}$ × $\frac{3+\sqrt{2}}{3+\sqrt{2}}$ = 3 + √2

⇒ x - 3 = √2   ...(1)

⇒ (x - 3)³ = (√2)³

⇒ x³ - 27 - 9x² + 27x = 2√2

⇒ x³ - 27 - 9x² + 27x = 2(x - 3) [from (1)]

⇒ x³ - 21 - 9x² + 25x = 0

⇒ x³ - 9x² + 25x - 20 = 1

Hence, option (b).

Answer: Option A

Explanation :

Given, $xyz$ = ${\left(\mathrm{xy}.\mathrm{z}\right)}^a$ = ${\left(x.y\mathrm{z}\right)}^b$ = ${\left(0.\mathrm{xyz}\right)}^c$

$xyz$ = ${\left(\mathrm{xy}.\mathrm{z}\right)}^a$ 
⇒ $xy.z$ = ${\left(\mathrm{xyz}\right)}^{1/a}$
⇒ $xyz\times 0.1$ = ${\left(\mathrm{xyz}\right)}^{1/a}$    ...(1)
⇒ 0.1 = ${\left(\mathrm{xyz}\right)}^{\frac{1}{a}-1}$    ...(1)

$xyz$ = ${\left(x.y\mathrm{z}\right)}^b$ 
⇒ $x.yz$ = ${\left(\mathrm{xyz}\right)}^{1/b}$
⇒ $xyz\times 0.01$ = ${\left(\mathrm{xyz}\right)}^{1/b}$
⇒ 0.01 = ${\left(\mathrm{xyz}\right)}^{\frac{1}{b}-1}$    ...(2)

$xyz$ = ${\left(0.\mathrm{xyz}\right)}^c$ 
⇒ $0.xyz$ = ${\left(\mathrm{xyz}\right)}^{1/c}$
⇒ $xyz\times 0.001$ = ${\left(\mathrm{xyz}\right)}^{1/c}$
⇒ 0.001 = ${\left(\mathrm{xyz}\right)}^{\frac{1}{c}-1}$    ...(3)

We konw, (1) × (2) = (3)

⇒ 0.1 × 0.01 = 0.001

⇒  ${\left(\mathrm{xyz}\right)}^{\frac{1}{a}-1}$ × ${\left(\mathrm{xyz}\right)}^{\frac{1}{b}-1}$ = ${\left(\mathrm{xyz}\right)}^{\frac{1}{c}-1}$

⇒  ${\left(\mathrm{xyz}\right)}^{\frac{1}{a}+\frac{1}{b}-2}$ = ${\left(\mathrm{xyz}\right)}^{\frac{1}{c}-1}$

⇒ $\frac{1}{a}$ + $\frac{1}{b}$ - 2 = $\frac{1}{c}$ - 1

⇒ $\frac{1}{a}$ + $\frac{1}{b}$ = $\frac{1}{c}$ + 1

Hence, option (a)

Answer: Option C

Explanation :

Given, $\frac{1}{1+\mathrm{a}+{\mathrm{a}}^2}$ + $\frac{1}{{\mathrm{a}}^{-1}+1+\mathrm{a}}$ + $\frac{1}{{\mathrm{a}}^{-2}+{\mathrm{a}}^{-1}+1}$.

= $\frac{1}{1+\mathrm{a}+{\mathrm{a}}^2}$ + $\frac{1}{{\displaystyle \frac{1}{a}}+1+\mathrm{a}}$ + $\frac{1}{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{a}}+1}$.

= $\frac{1}{1+\mathrm{a}+{\mathrm{a}}^2}$ + $\frac{\mathrm{a}}{1+\mathrm{a}+{\mathrm{a}}^2}$ + $\frac{{\mathrm{a}}^2}{1+\mathrm{a}+{\mathrm{a}}^2}$

= $\frac{1+\mathrm{a}+{\mathrm{a}}^2}{1+\mathrm{a}+{\mathrm{a}}^2}$

= 1

Hence, option (c).

Answer: 2

Explanation :

Given, 3^2x+2 + 3^2x-2 = 738

⇒ $9\times 3^{2x}$ + $\frac{3^{2x}}{9}$ = 738

⇒ $3^{2x}\left(9+\frac{1}{9}\right)$ = 738

⇒ $3^{2x}\times \frac{82}{9}$ = 738

⇒ $3^{2x}$ = 738 × $\frac{9}{82}$

⇒ $3^{2x}$ = 9 × 9 = 81

⇒ $3^{2x}$ = 3⁴

⇒ 2x = 4

⇒  x = 2

Hence, 2.

Answer: Option C

Explanation :

Let x = $\frac{1}{\sqrt{\sqrt{6+2\sqrt{5}}+\sqrt{5}(7+4\sqrt{5})}}$

Let us first calculate $\sqrt{6+2\sqrt{5}}$ = $\sqrt{5}+1$

Now, x = $\frac{1}{\sqrt{(\sqrt{5}+1)+\sqrt{5}(7+4\sqrt{5})}}$

⇒ x = $\frac{1}{\sqrt{(\sqrt{5}+1)+7\sqrt{5}+20}}$

⇒ x = $\frac{1}{\sqrt{21+8\sqrt{5}}}$

Now, $\sqrt{21+8\sqrt{5}}$ = $4+\sqrt{5}$

⇒ x = $\frac{1}{4+\sqrt{5}}$

On rationalising with 4 - √5, we get

⇒ x = $\frac{4-\sqrt{5}}{11}$

Hence, option (c).

Answer: Option D

Explanation :

${\left[\sqrt[3]{\sqrt[6]{a^9}}\right]}^4$.${\left[\sqrt[6]{\sqrt[3]{a^9}}\right]}^4$ 

= $a^{9\times \frac{1}{6}\times \frac{1}{3}\times 4}$ × $a^{9\times \frac{1}{3}\times \frac{1}{6}\times 4}$ 

= a² × a² = a⁴

Hence, option (d).

Answer: Option A

Explanation :

1 - $\frac{1}{1+\sqrt{3}}$ + $\frac{1}{1-\sqrt{3}}$

Taking LCM of denominators, we get

= $\frac{(1+\sqrt{3})(1-\sqrt{3})-(1-\sqrt{3})+(1+\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}$

= $\frac{1-3-1+\sqrt{3}+1+\sqrt{3}}{1-3}$

= $\frac{-2+2\sqrt{3}}{-2}$

= $1-\sqrt{3}$

Hence, option (a).