CBSE Class 9 Mathematics Number System Assignment Set 01

Question. Which of the following is not a rational number?
(A) √2
(B) √4
(C) √9
(D) √16
Answer : A

Question. of which of the following is the set of natural numbers a subset?
(A) The set of even numbers.
(B) The set of odd numbers.
(C) The set of composite numbers.
(D) The set of real numbers.
Answer : D

Question. Given X : Every fraction is a rational number. 
and Y : Every rational number is a fraction.
Which of the following is correct?
(A) X is False andY is True.
(B) X is True andY is False.
(C) Both X andY are True.
(D) Both X andY are False.
Answer : B

Question. What is the product of ³√4 and ³√22 ?
(A) 2³√11
(B) (3√11)
(C) 4√11
(D) 8³√11
Answer : A

Question. What is the quotient when ⁶√12 is divided by √3 ³√2?
(A) 1/²√3
(B) 1/³√3
(C) 1/⁴√3
(D) 1/⁵√3
Answer : B

Question. If -4/7 = -32/x, what is the value of x?
(A) - 56
(B) 56
(C) - 65
(D) 65
Answer : B

Question. What is the rationalizing factor of (2³√5) ?
(A) ³√5
(B) ³√5²
(C) 5²
(D) 5³
Answer : B

Question. What is the rationalizing factor of ⁵√a²b³c⁴ ?
(A) ⁵√a³b²c
(B) ⁴√a³b²c
(C) ³√a³b²c
(D) √a³b²c
Answer : A

Question. What is the rational denominator of 2³√/³√9?
(A) 1
(B) 2
(C) 3
(D) 4
Answer : C

Question. Which is an irrational number bet ween √2 and v3?
(A) 2^1/2
(B) 3^1/4
(C) 6^1/4
(D) 6^1/8
Answer : C

Question. What t ype of a number is (6 + √2)(6 - √2)?
(A) Rational number
(B) Irrational number
(C) Prime number
(D) Negative Integer
Answer : A

Question. What type of a number is (√2 + √3)² ?
(A) A rational number
(B) An irrational number
(C) A fraction
(D) A decimal number
Answer : B

Question. If 9^x+2 = 240 + 9^x, find X.
(A) 0.5
(B) 0.2
(C) 0.4
(D) 0.1
Answer : A

Question. Which expression is equal to (x^-1 + y^-1)¹ ?
(A) xy
(B) x + y
(C) xy/x+y
(D) x+y/xy
Answer : C

Question. If 10^x = 64, what is the value of 10^x/2+1 ?
(A) 18
(B) 42
(C) 80
(D) 81
Answer : C

Question. If x = 2/3-√7 find the value of (x - 3)²
(A) 1
(B) 3
(C) 6
(D) 7
Answer : D

Question. Find the difference between 1.4¯ and 0.2¯.
(A) 0.2¯
(B) 1.3¯
(C) 11/9
(D) 12/9
Answer : C

Question. If √a is an irrational number, what is a?
(A) Rational
(B) Irrational
(C) 0
(D) Real
Answer : A

Question. Which of the following is an irrational number?
(A) √9
(B) π
(C) 22/7
(D) 2.5151515 ....
Answer : B

Question. If x = 7 + 4√3 and xy = 1, what is the value of 1/x² + 1/y²
(A) 64
(B) 134
(C) 194
(D) 1/49
Answer : C

Question. If √5 = 2.236 and √3 = 1.732, find the value of 1/√5 - √3 .
(A) 3.968
(B) 1/3.968
(C) 1.984
(D) √0.504
Answer : C

Question. Find the value of (⁶√27 - √6(3/4))²
(A) √5/2
(B) 3/2
(C) √3/4
(D) 3/4
Answer : D

Question. Find the value of x - y^x-y when x = 2 and y= -2.
(A) 18
(B) - 18
(C) 14
(D) - 14
Answer : D

Question. If x/x¹⁵ = 8x^-1 and x > 0, find x.
(A) √2/4
(B) 2√2
(C) 4
(D) 64 
Answer : D

Question. If 4^x - 4^x-1 = 24, what is the value of (2x)^x ?
(A) 5√5
(B) √5
(C) 25√5
(D) 125
Answer : C

Question. If x = √S + 2, find the value of x - 1/x.
(A) 2√S
(B) 4
(C) 2
(D) √5
Answer : B

Question. Identify a rational number between -5 and 5.
(A) 0
(B) - 7.3
(C) -5.7
(D) 1.101100110001..
Answer : A

Question. The sum of the digits of a number is subtracted from the number, the resulting number is always divisible by which of the following numbers?
(A) 2
(B) 5
(C) 8
(D) 9
Answer : D

Question. Which of the following numbers is an irrational number?
(A) √16 - 4
(B) (3 - √3)(3 + √3)
(C) √5 + 3
(D) -√25
Answer : C

Question. m and n are integers and √mn = 10. Which of the following cannot be a value of m + n?
(A) 25
(B) 52
(C) 101
(D) 50
Answer : C

Question. If 2.5252525 ..... = p/q (in the lowest form) what is the value of q/p?
(A) 0.4
(B) 0.42525
(C) 0.0396
(D) 0.396
Answer : D

Question. If x^x√x = (x√x)^x find the value of x.
(A) 3/2
(B) 2/9
(C) 9/4
(D) 4/9
Answer : C

Question. If x = 2 + 2^1/3 + 2^2/3, find the value of (x³ - 6x² + 6x).
(A) 2
(B) 4
(C) 8
(D) 6
Answer : A

Question. If x = 1/2 - √3, find the value of x³ - 2x² - 7x + 10
(A) 14
(B) 8
(C) 4
(D) 16
Answer : B

Question. If a = 3 and b = 5, find the value of a^a + b^b.
(A) 512
(B) 251
(C) 513
(D) 152
Answer : D

Question. If p = 3 and q = 5, find the value of (1/p + 1/q)^p
(A) 3375/512
(B) 512/3375
(C) 512/225
(D) 54/225
Answer : B

Question. Fmd the value of 1/√5 - √4
(A) √5 - √4
(B) √5 + √4
(C) √5 + √4 / 9
(D) √4 -√S
Answer : B

Question. Find the rational number for 0.5¯.
(A) 9/5
(B) 5/10
(C) 5/9
(D) 10/5
Answer : C

Question. Identify the simplest form of 5√3 + 3√5 / 5√3 - 3√5
(A) 6 + √15
(B) 4 + √15
(C) 4 + √3
(D) 6 + √5
|Answer : B

Question. Find the value of (0.6)⁶ - (0.1)^-1/(3/8)^-1(3/2)³ + (-1/3)^-1
(A) -2/3
(B) 3/2
(C) - (3/2)
(D) 2/3
Answer : C
 

SECTION A: (1 MARK)

Question 1. Given that \(\sqrt{10} = 3.162\), find the value of \(\frac{1}{\sqrt{10}}\)
Answer: Rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{10}\):
\(\frac{1}{\sqrt{10}} = \frac{1 \times \sqrt{10}}{\sqrt{10} \times \sqrt{10}}\)
\(= \frac{\sqrt{10}}{10}\)
Substitute the given value \(\sqrt{10} = 3.162\):
\(= \frac{3.162}{10} = 0.3162\)
In simple words: Multiply top and bottom by the square root to remove it from the denominator, then divide by 10.

Exam Tip: Rationalization is a standard technique used to simplify fractions containing square roots in the denominator.

Question 2. Simplify: \(\frac{28}{\sqrt{28} - \sqrt{7}}\)
Answer: Rationalize the denominator by multiplying the numerator and denominator by the conjugate \(\sqrt{28} + \sqrt{7}\):
\(= \frac{28(\sqrt{28} + \sqrt{7})}{(\sqrt{28} - \sqrt{7})(\sqrt{28} + \sqrt{7})}\)
Using the identity \((a-b)(a+b) = a^2 - b^2\) in the denominator:
Denominator = \((\sqrt{28})^2 - (\sqrt{7})^2 = 28 - 7 = 21\)
Now, substitute back into the fraction:
\(= \frac{28(\sqrt{28} + \sqrt{7})}{21}\)
Simplify the coefficients by dividing numerator and denominator by 7:
\(= \frac{4(\sqrt{28} + \sqrt{7})}{3}\)
Simplify \(\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}\):
\(= \frac{4(2\sqrt{7} + \sqrt{7})}{3} = \frac{4(3\sqrt{7})}{3} = 4\sqrt{7}\)
In simple words: Use the conjugate to clear the root from the denominator, simplify the numbers, and combine like terms.

Exam Tip: Always look for ways to simplify surds like \(\sqrt{28}\) into \(2\sqrt{7}\) before performing final multiplications.

Question 3. Simplify: \(16^{-\frac{1}{4}} \times \sqrt[4]{16}\)
Answer: Rewrite both terms using exponents:
\(16^{-\frac{1}{4}} \times 16^{\frac{1}{4}}\)
Using the laws of exponents \(a^m \times a^n = a^{m+n}\):
\(= 16^{-\frac{1}{4} + \frac{1}{4}}\)
\(= 16^0\)
Any non-zero number raised to the power of 0 is 1:
\(= 1\)
In simple words: When multiplying identical bases with opposite powers, add the exponents to get zero, making the result 1.
Exam Tip: Recall that \(\sqrt[4]{16}\) can be written in exponential form as \(16^{\frac{1}{4}}\).

Question 4. Simplify: \(\sqrt{10} \times \sqrt{15}\)
Answer: Using the multiplication property of square roots \(\sqrt{a} \times \sqrt{b} = \sqrt{a \times b}\):
\(= \sqrt{10 \times 15}\)
\(= \sqrt{150}\)
Factorize 150 to extract perfect squares:
\(= \sqrt{25 \times 6}\)
\(= 5\sqrt{6}\)
In simple words: Combine the numbers under a single square root, then factor out any perfect square numbers.

Exam Tip: Always check if the resulting product under the root has a factor that is a perfect square (like 4, 9, 16, 25, etc.).

SECTION B: (2 MARKS)

Question 5. If \(x = 3 + 2\sqrt{2}\), then find whether \(x + \frac{1}{x}\) is rational or irrational
Answer: First, find \(\frac{1}{x}\):
\(\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}}\)
Rationalize the denominator by multiplying by the conjugate \(3 - 2\sqrt{2}\):
\(\frac{1}{x} = \frac{3 - 2\sqrt{2}}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})}\)
\(= \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2}\)
\(= \frac{3 - 2\sqrt{2}}{9 - 8} = \frac{3 - 2\sqrt{2}}{1} = 3 - 2\sqrt{2}\)
Now, add \(x\) and \(\frac{1}{x}\):
\(x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2})\)
\(= 3 + 3 + 2\sqrt{2} - 2\sqrt{2} = 6\)
Since 6 is a rational number, \(x + \frac{1}{x}\) is rational.
In simple words: Find the reciprocal of \(x\) by rationalizing, then add it to \(x\); the irrational parts cancel out leaving a rational number.

Exam Tip: Whenever \(x = a + \sqrt{b}\) and \(a^2 - b = 1\), its reciprocal \(\frac{1}{x}\) is simply \(a - \sqrt{b}\).

Question 6. Simplify: \(\left(\frac{125}{27}\right)^{-\frac{1}{3}} \times \left(\frac{5}{2}\right)^2 \div \left(\frac{2}{3}\right)^3\)
Answer: Simplify each term step by step:
First term: \(\left(\frac{125}{27}\right)^{-\frac{1}{3}} = \left(\frac{5^3}{3^3}\right)^{-\frac{1}{3}} = \left(\frac{5}{3}\right)^{-1} = \frac{3}{5}\)
Second term: \(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\)
Third term: \(\left(\frac{2}{3}\right)^3 = \frac{8}{27}\)
Now, substitute these back into the expression: 
\(= \frac{3}{5} \times \frac{25}{4} \div \frac{8}{27}\)
Change division to multiplication by taking the reciprocal:
\(= \frac{3}{5} \times \frac{25}{4} \times \frac{27}{8}\)
Cancel out common factors (5 in denominator and 25 in numerator):
\(= \frac{3}{1} \times \frac{5}{4} \times \frac{27}{8}\)
Multiply numerators and denominators:
Numerator = \(3 \times 5 \times 27 = 405\)
Denominator = \(1 \times 4 \times 8 = 32\)
\(= \frac{405}{32}\)
In simple words: Convert negative exponents to reciprocals, evaluate powers, and change division into multiplication by flipping the fraction.

Exam Tip: Always deal with negative exponents first by inverting the base fraction before applying fractional powers.

Question 7. Express \(1.4191919\dots\) in the form of \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\)
Answer: Let \(x = 1.4191919\dots\) ---- (1)
Here, two digits (19) are repeating. Multiply both sides by 100:
\(100x = 141.919191\dots\) ---- (2)
Subtract equation (1) from equation (2):
\(100x - x = 141.919191\dots - 1.4191919\dots\)
\(99x = 140.5000\dots\)
\(99x = \frac{1405}{10}\)
\(x = \frac{1405}{990}\)
Simplify by dividing numerator and denominator by their greatest common divisor (5):
\(x = \frac{1405 \div 5}{990 \div 5} = \frac{281}{198}\)
In simple words: Set the repeating decimal equal to \(x\), multiply by 10^k (where k is the number of repeating digits) to align the decimals, and subtract.

Exam Tip: Count the number of repeating digits carefully to decide whether to multiply by 10, 100, or 1000.

Question 8. Find the product: \(\sqrt[3]{2} \times \sqrt{2} \times \sqrt[12]{32}\)
Answer: Convert all radical terms into fractional exponents with a common base (2):
\(\sqrt[3]{2} = 2^{\frac{1}{3}}\)
\(\sqrt{2} = 2^{\frac{1}{2}}\)
\(\sqrt[12]{32} = (2^5)^{\frac{1}{12}} = 2^{\frac{5}{12}}\)
Now, multiply the terms by adding their exponents:
\(2^{\frac{1}{3}} \times 2^{\frac{1}{2}} \times 2^{\frac{5}{12}} = 2^{\left(\frac{1}{3} + \frac{1}{2} + \frac{5}{12}\right)}\)
Find the common denominator for the fractions, which is 12:
\(\frac{1}{3} = \frac{4}{12}, \quad \frac{1}{2} = \frac{6}{12}, \quad \frac{5}{12} = \frac{5}{12}\)
Add the fractions in the exponent:
\(\frac{4}{12} + \frac{6}{12} + \frac{5}{12} = \frac{15}{12} = \frac{5}{4}\)
Thus, the product is \(2^{\frac{5}{4}} \dots\) Wait, let us check: \(2^{\frac{15}{12}} = 2^{\frac{5}{4}}\). If the intended answer is 2, let us re-verify: \(2^{1/3} \times 2^{1/2} \times 2^{5/12} = 2^{(4+6+5)/12} = 2^{15/12} = 2^{5/4} = \sqrt[4]{32} = 2\sqrt[4]{2}\). Re-reading standard textbook problems, if the question asks for product evaluation, exponential form or simplified surd form \(2^{\frac{5}{4}}\) is standard.
In simple words: Convert radicals to exponents, find a common denominator for the fractions, add them, and simplify.

Exam Tip: Converting radical expressions to exponential form makes multiplication and division of surds much easier.

Question 9. Insert a rational number and an irrational number between 0.17 and 0.18
Answer: Rational number between 0.17 and 0.18:
Any terminating decimal between them works, such as \(0.175\) (or \(\frac{175}{1000} = \frac{7}{40}\)).
Irrational number between 0.17 and 0.18:
Any non-terminating, non-recurring decimal between them works, such as \(0.171010010001\dots\)
In simple words: A rational number can be a simple terminating decimal like 0.175, while an irrational number must have a non-repeating pattern like 0.1710100...

Exam Tip: There are infinitely many rational and irrational numbers between any two given numbers.

SECTION C: (3 MARKS)

Question 10. If \(x = 2 + \sqrt{3}\), find the value of \(x^2 + \frac{1}{x^2}\)
Answer: Given \(x = 2 + \sqrt{3}\).
Find \(\frac{1}{x}\):
\(\frac{1}{x} = \frac{1}{2 + \sqrt{3}}\)
Rationalize the denominator:
\(\frac{1}{x} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}\)
Now, find \(x + \frac{1}{x}\):
\(x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4\)
Square both sides to find \(x^2 + \frac{1}{x^2}\):
\(\left(x + \frac{1}{x}\right)^2 = 4^2\)
\(x^2 + \frac{1}{x^2} + 2(x)\left(\frac{1}{x}\right) = 16\)
\(x^2 + \frac{1}{x^2} + 2 = 16\)
\(x^2 + \frac{1}{x^2} = 16 - 2 = 14\)
In simple words: Find \(\frac{1}{x}\), add it to \(x\) to get a simple integer, and then square that sum using algebraic identities.

Exam Tip: Remember the algebraic identity: \(\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2\).

Question 11. Simplify: \(3\sqrt{45} - \sqrt{125} + \sqrt{200} - \sqrt{50}\)
Answer: Simplify each surd by finding perfect square factors:
\(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\)
So, \(3\sqrt{45} = 3 \times 3\sqrt{5} = 9\sqrt{5}\)
\(\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}\)
\(\sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2}\)
\(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\)
Substitute these simplified forms back into the expression:
\(= 9\sqrt{5} - 5\sqrt{5} + 10\sqrt{2} - 5\sqrt{2}\)
Group like surds together:
\(= (9\sqrt{5} - 5\sqrt{5}) + (10\sqrt{2} - 5\sqrt{2})\)
\(= 4\sqrt{5} + 5\sqrt{2}\)
In simple words: Break down each square root into simpler surds, then combine terms that have the same square root part.

Exam Tip: Only surds with identical numbers under the root can be added or subtracted directly.

Question 12. Represent \(\sqrt{8.5}\) on the number line.
Answer: Steps of construction:
1. Draw a line segment \(AB = 8.5\) units on a number line.
2. Extend the line from point \(B\) by 1 unit to a new point \(C\), such that \(BC = 1\). The total length \(AC = 9.5\).
3. Find the midpoint \(O\) of the line segment \(AC\).
4. Taking \(O\) as the center and \(OC\) (or \(OA\)) as the radius, draw a semi-circle passing through \(A\) and \(C\).
5. At point \(B\), draw a perpendicular line intersecting the semi-circle at point \(D\).
6. The length \(BD\) is equal to \(\sqrt{8.5}\).
7. Taking \(B\) as the center and radius \(BD\), draw an arc intersecting the number line to mark the position of \(\sqrt{8.5}\).
In simple words: Draw a line of 8.5 units, add 1 unit, find the middle to draw a semicircle, draw a vertical line from the 8.5 mark to the semicircle, and transfer that length to the number line.

Exam Tip: Use a sharp pencil and a compass for precise geometric constructions on number lines.

Question 13. Classify the following numbers as rational or irrational with justification:
(a) \(\frac{\sqrt{12}}{\sqrt{3}}\)
(b) \(0.1416\)
(c) \(-\sqrt{0.4}\)
(d) \(0.401400140001\dots\)
(e) \(\frac{\sqrt{9}}{\sqrt{27}}\)
(f) \((1 + \sqrt{5}) - (4 + \sqrt{5})\)
Answer:
(a) \(\frac{\sqrt{12}}{\sqrt{3}} = \sqrt{\frac{12}{3}} = \sqrt{4} = 2\). Since 2 can be written as \(\frac{2}{1}\), it is rational.
(b) \(0.1416\) is a terminating decimal, so it is rational.
(c) \(-\sqrt{0.4} = -\frac{2}{\sqrt{10}}\). Since the square root cannot be fully resolved, it is irrational.
(d) \(0.401400140001\dots\) is a non-terminating and non-recurring decimal, so it is irrational.
(e) \(\frac{\sqrt{9}}{\sqrt{27}} = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}\). Since the denominator contains an unresolved square root, it is irrational.
(f) \((1 + \sqrt{5}) - (4 + \sqrt{5}) = 1 + \sqrt{5} - 4 - \sqrt{5} = -3\). Since -3 is an integer, it is rational.
In simple words: Simplify each expression first; if roots disappear and decimals terminate or repeat, it is rational. Otherwise, it is irrational.

Exam Tip: Always simplify radical expressions completely before deciding whether a number is rational or irrational.

Question 14. Write \(\sqrt[3]{4}, \sqrt{3}, \sqrt[4]{6}\) in ascending order
Answer: Convert each radical to fractional exponents:
\(\sqrt[3]{4} = 4^{\frac{1}{3}}\)
\(\sqrt{3} = 3^{\frac{1}{2}}\)
\(\sqrt[4]{6} = 6^{\frac{1}{4}}\)
The denominators of the fractional exponents are 3, 2, and 4. The least common multiple (LCM) of 3, 2, and 4 is 12.
Convert each exponent to have a denominator of 12:
\(4^{\frac{1}{3}} = 4^{\frac{4}{12}} = (4^4)^{\frac{1}{12}} = (256)^{\frac{1}{12}}\)
\(3^{\frac{1}{2}} = 3^{\frac{6}{12}} = (3^6)^{\frac{1}{12}} = (729)^{\frac{1}{12}}\)
\(6^{\frac{1}{4}} = 6^{\frac{3}{12}} = (6^3)^{\frac{1}{12}} = (216)^{\frac{1}{12}}\)
Comparing the base numbers under the common 12th root: \(216 < 256 < 729\)
Therefore, in ascending order: \(\sqrt[4]{6}, \sqrt[3]{4}, \sqrt{3}\)
In simple words: Make the root indices equal by finding the LCM, then compare the numbers inside the roots.

Exam Tip: Equalizing the indices of surds is the standard method to compare their magnitudes.

Question 15. Locate \(\sqrt{3}\) on the number line
Answer: Steps of construction:
1. Draw a number line and mark point \(O\) as 0 and point \(A\) as 1 unit.
2. At point \(A\), draw a perpendicular line segment \(AB\) of length 1 unit.
3. Join \(O\) to \(B\). By Pythagoras theorem, \(OB = \sqrt{1^2 + 1^2} = \sqrt{2}\).
4. At point \(B\), draw a perpendicular line segment \(BC\) of length 1 unit.
5. Join \(O\) to \(C\). By Pythagoras theorem, \(OC = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}\).
6. Taking \(O\) as the center and radius \(OC = \sqrt{3}\), draw an arc that intersects the number line to mark \(\sqrt{3}\).
In simple words: First construct \(\sqrt{2}\), then build a perpendicular of 1 unit on it to construct \(\sqrt{3}\), and use a compass to transfer it to the number line.

Exam Tip: This spiral method uses the Pythagorean theorem successively to construct square roots of consecutive integers.

Question 16. Visualize on the number line \(2.463\)
Answer: Steps of successive magnification:
1. Observe that \(2.463\) lies between 2 and 3. Divide the segment between 2 and 3 into 10 equal parts (\(2.1, 2.2, \dots, 2.9\)).
2. Locate the interval between \(2.4\) and \(2.5\), and divide it into 10 equal parts (\(2.41, 2.42, \dots, 2.49\)).
3. Locate the interval between \(2.46\) and \(2.47\), and divide it into 10 equal parts (\(2.461, 2.462, 2.463, \dots, 2.469\)).
4. Point \(2.463\) is clearly located on the third division between \(2.46\) and \(2.47\).
In simple words: Zoom in progressively from one decimal place to two, and then to three decimal places on the number line.

Exam Tip: Mention the principle of successive magnification clearly when explaining decimal visualization.

SECTION D: (4 MARKS)

Question 17. Simplify: \(\frac{1}{3 - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2}\)
Answer: Rationalize each term individually:
1st term: \(\frac{1}{3 - \sqrt{8}} = \frac{3 + \sqrt{8}}{9 - 8} = 3 + \sqrt{8}\)
2nd term: \(\frac{1}{\sqrt{8} - \sqrt{7}} = \frac{\sqrt{8} + \sqrt{7}}{8 - 7} = \sqrt{8} + \sqrt{7}\)
3rd term: \(\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}\)
4th term: \(\frac{1}{\sqrt{6} - \sqrt{5}} = \frac{\sqrt{6} + \sqrt{5}}{6 - 5} = \sqrt{6} + \sqrt{5}\)
5th term: \(\frac{1}{\sqrt{5} - 2} = \frac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2\)
Substitute these back into the expression with their respective signs:
\(= (3 + \sqrt{8}) - (\sqrt{8} + \sqrt{7}) + (\sqrt{7} + \sqrt{6}) - (\sqrt{6} + \sqrt{5}) + (\sqrt{5} + 2)\)
Expand and cancel out all intermediate terms:
\(= 3 + \sqrt{8} - \sqrt{8} - \sqrt{7} + \sqrt{7} + \sqrt{6} - \sqrt{6} - \sqrt{5} + \sqrt{5} + 2\)
\(= 3 + 2 = 5\)
In simple words: Rationalize each denominator, notice that the denominators become 1, and all middle terms cancel out in pairs.

Exam Tip: This is a telescoping series where almost all terms cancel out neatly. Be very careful with minus signs between fractions.

Question 18. Find the value of \(p\) and \(q\) if \(\frac{7 + \sqrt{5}}{7 - \sqrt{5}} - \frac{7 - \sqrt{5}}{7 + \sqrt{5}} = p - 7\sqrt{5}q\)
Answer: Simplify the left-hand side (LHS) by taking the LCM of denominators:
LCM = \((7 - \sqrt{5})(7 + \sqrt{5}) = 7^2 - (\sqrt{5})^2 = 49 - 5 = 44\)
LHS numerator:
\(= (7 + \sqrt{5})^2 - (7 - \sqrt{5})^2\)
Using the algebraic identity \((a+b)^2 - (a-b)^2 = 4ab\):
Numerator = \(4(7)(\sqrt{5}) = 28\sqrt{5}\)
So, LHS = \(\frac{28\sqrt{5}}{44}\)
Simplify by dividing numerator and denominator by 4:
LHS = \(\frac{7\sqrt{5}}{11} = 0 + \frac{7}{11}\sqrt{5}\)
Compare LHS with the given RHS expression \(p - 7\sqrt{5}q\):
\(p = 0\)
\(-7\sqrt{5}q = \frac{7\sqrt{5}}{11} \implies -7q = \frac{7}{11} \implies q = -\frac{1}{11}\)
In simple words: Combine fractions using a common denominator, apply algebraic expansion identities, simplify the surd, and compare coefficients with the right side.

Exam Tip: Recognizing algebraic identities like \((a+b)^2 - (a-b)^2 = 4ab\) saves time during expansion.

Question 19. If \(x = \frac{\sqrt{a+1} + \sqrt{a-1}}{\sqrt{a+1} - \sqrt{a-1}}\), then show that \(x^2 - 2ax + 1 = 0\)
Answer: Rationalize the expression for \(x\):
\(x = \frac{(\sqrt{a+1} + \sqrt{a-1})^2}{(\sqrt{a+1} - \sqrt{a-1})(\sqrt{a+1} + \sqrt{a-1})}\)
Denominator = \((a+1) - (a-1) = a + 1 - a + 1 = 2\)
Numerator = \((\sqrt{a+1})^2 + (\sqrt{a-1})^2 + 2\sqrt{(a+1)(a-1)}\)
\(= (a + 1) + (a - 1) + 2\sqrt{a^2 - 1} = 2a + 2\sqrt{a^2 - 1}\)
Thus:
\(x = \frac{2a + 2\sqrt{a^2 - 1}}{2} = a + \sqrt{a^2 - 1}\)
Rearranging to isolate the radical:
\(x - a = \sqrt{a^2 - 1}\)
Square both sides:
\((x - a)^2 = a^2 - 1\)
\(x^2 - 2ax + a^2 = a^2 - 1\)
Subtract \(a^2\) from both sides:
\(x^2 - 2ax = -1\)
\(x^2 - 2ax + 1 = 0\)
In simple words: Rationalize \(x\), isolate the square root term, square both sides to eliminate the root, and rearrange to get the desired quadratic equation.

Exam Tip: Isolating the radical term before squaring is a key strategy for eliminating square roots in algebraic proofs.

Question 20. If \(x = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\) and \(y = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\), find the value of \(x^2 + y^2 + xy\)
Answer: First, observe that \(y = \frac{1}{x}\).
Calculate \(x\):
\(x = \frac{(\sqrt{3} - \sqrt{2})(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} = \frac{(\sqrt{3} - \sqrt{2})^2}{3 - 2} = (\sqrt{3} - \sqrt{2})^2 = 3 + 2 - 2\sqrt{6} = 5 - 2\sqrt{6}\)
Since \(y = \frac{1}{x}\), \(y = 5 + 2\sqrt{6}\).
Now find \(x + y\) and \(xy\):
\(x + y = (5 - 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10\)
\(xy = x \times \frac{1}{x} = 1\)
We need to find the value of \(x^2 + y^2 + xy\).
Using the identity \((x+y)^2 = x^2 + y^2 + 2xy\):
\(x^2 + y^2 = (x+y)^2 - 2xy\)
Substitute \(x+y = 10\) and \(xy = 1\):
\(x^2 + y^2 = 10^2 - 2(1) = 100 - 2 = 98\)
Now add \(xy\):
\(x^2 + y^2 + xy = 98 + 1 = 99\)
In simple words: Notice that \(x\) and \(y\) are reciprocals, find their sum and product, and use algebraic identities to evaluate the final expression.

Exam Tip: Recognizing reciprocals simplifies calculations of sums and products significantly.