Practice Questions for Algebra - Surds & Indices

1. PE 1 - Surds & Indices | Algebra - Surds & Indices

Simplify: $\sqrt{\sqrt[3]{\sqrt[4]{\sqrt[5]{\sqrt[6]{8^{240}}}}}}$

A.
2
B.
64
C.
8
D.
None of these

Answer: Option A

Explanation :

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

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

⇒ 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 = ${(2^{3})}^{\frac{1}{3}}$

⇒ x = $2^{3 \times \frac{1}{3}}$ = 2

Hence, option (a).

2. PE 1 - Surds & Indices | Algebra - Surds & Indices

Find the value of $\sqrt{\frac{\sqrt{a + b} + \sqrt{a - b}}{\sqrt{a + b} - \sqrt{a - b}}}$ , where a = 2 and b = 1.

A. $\frac{\sqrt{3} - 1}{\sqrt{2}}$
B. $\frac{\sqrt{3} + 1}{\sqrt{2}}$
C. $\frac{\sqrt{2} + 1}{\sqrt{3}}$
D.
None of these

Answer: Option B

Explanation :

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

⇒ x = $\frac{2 a + 2 \sqrt{a^{2} - b^{2}}}{2 b}$

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).

3. PE 1 - Surds & Indices | Algebra - Surds & Indices

Find the value of $\sqrt[a b]{x^{c^{2}}} \times \sqrt[b c]{x^{a^{2}}} \times \sqrt[c a]{{x^{b}}^{2}}$ & if a + b + c = 0.

A.
x³
B.
x
C.
x²
D.
x^abc

Answer: Option A

Explanation :

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

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

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

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

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

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

Hence, option (a).

4. PE 1 - Surds & Indices | Algebra - Surds & Indices

If x = 3 + $\sqrt[3]{3}$ + $\sqrt[3]{9}$ , then find the value of x³ - 18x² + 18x + 27?

A.
√108
B.
8
C.
√66
D.
66
E.
None of these

Answer: Option D

Explanation :

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).

5. PE 1 - Surds & Indices | Algebra - Surds & Indices

If 3^x + 5^y = 52 and 3^2x-2 + 5^y+1 = 206 where x and y are integers, find x + y.

A.
3
B.
4
C.
5
D.
6
E.
Cannot be determined

6. PE 1 - Surds & Indices | Algebra - Surds & Indices

If x is a positive integer, the greatest integer by which 11^x + 11^x+1 + 11^x+2 is always divisible is

A.
11
B.
121
C.
133
D.
1463

7. PE 1 - Surds & Indices | Algebra - Surds & Indices

Find square root of 5 + √6 + √10 + √15.

A.
√2 + √3 + √5
B.
1 + $\sqrt{\frac{3}{2}}$ + $\sqrt{\frac{5}{2}}$
C.
(√2 + √3 + √5)/2
D.
None of these

Answer: Option B

Explanation :

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 = $\frac{{(\sqrt{2} + \sqrt{3} + \sqrt{5})}^{2}}{2}$

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

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

Hence, option (b).

8. PE 1 - Surds & Indices | Algebra - Surds & Indices

Given 2^x = 8^y+1 and 9^y = 3^x–9; the value of x + y is:

A.
18
B.
21
C.
24
D.
27

9. PE 1 - Surds & Indices | Algebra - Surds & Indices

Let r be the result of doubling both the base and the exponent of a^b, b ≠ 0. If r equals the product of a^b and x^b, then x equals:

A.
2a
B.
4a
C.
2
D.
4

10. PE 1 - Surds & Indices | Algebra - Surds & Indices

The expression $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}$ is equal to:

A.
$\frac{\sqrt{3}}{6}$
B.
- $\frac{\sqrt{3}}{6}$
C.
$\frac{\sqrt{-3}}{6}$
D.
5 $\frac{\sqrt{3}}{6}$

Answer: Option A

Explanation :

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

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

= $\frac{4 - 3}{\sqrt{12}}$

= $\frac{1}{\sqrt{12}}$

= $\frac{1}{2 \sqrt{3}}$

= $\frac{\sqrt{3}}{6}$

Hence, option (a).

PE 1 - Surds & Indices | Algebra - Surds & Indices

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