CBSE Class 9 Mathematics Number System Assignment Set 07

Question. Rationalize the denominator of
(i) \( \frac{1}{3+\sqrt{2}} \)
(ii) \( \frac{1}{2+\sqrt{3}} \)
Answer:
(i) Multiply numerator and denominator by \( 3-\sqrt{2} \): \[ \frac{1}{3+\sqrt{2}} \times \frac{3-\sqrt{2}}{3-\sqrt{2}} = \frac{3-\sqrt{2}}{3^2 - (\sqrt{2})^2} = \frac{3-\sqrt{2}}{9 - 2} = \frac{3-\sqrt{2}}{7} \]
(ii) Multiply numerator and denominator by \( 2-\sqrt{3} \): \[ \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = \frac{2-\sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2-\sqrt{3}}{4 - 3} = 2-\sqrt{3} \]
In simple words: Multiply top and bottom by the same numbers but with a minus sign in between to clear the root.
Exam Tip: Use the identity \( (a+b)(a-b) = a^2 - b^2 \) in the denominator to remove the radical.

Question. Simplify
(i) \( (\sqrt{5}+\sqrt{2})^2 \)
(ii) \( (\sqrt{11}-\sqrt{5})^2 \)
Answer:
(i) Using the identity \( (a+b)^2 = a^2 + 2ab + b^2 \): \[ (\sqrt{5}+\sqrt{2})^2 = (\sqrt{5})^2 + 2\sqrt{5}\sqrt{2} + (\sqrt{2})^2 \] \[ = 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10} \]
(ii) Using the identity \( (a-b)^2 = a^2 - 2ab + b^2 \): \[ (\sqrt{11}-\sqrt{5})^2 = (\sqrt{11})^2 - 2\sqrt{11}\sqrt{5} + (\sqrt{5})^2 \] \[ = 11 - 2\sqrt{55} + 5 = 16 - 2\sqrt{55} \]
In simple words: Use the algebraic identities for squaring a sum or difference to expand and simplify.

Exam Tip: Ensure you multiply the terms inside the radical when evaluating the middle term \( 2ab \).

Question. Simplify:
(i) \( (5+\sqrt{5})(5-\sqrt{5}) \)
(ii) \( (3+2\sqrt{2})(3-2\sqrt{2}) \)
Answer:
(i) Using \( (a+b)(a-b) = a^2 - b^2 \): \[ (5+\sqrt{5})(5-\sqrt{5}) = 5^2 - (\sqrt{5})^2 = 25 - 5 = 20 \]
(ii) Using \( (a+b)(a-b) = a^2 - b^2 \): \[ (3+2\sqrt{2})(3-2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1 \]
In simple words: This follows the difference of squares rule, so we square the first term and subtract the square of the second term.

Exam Tip: The difference of squares is a highly efficient shortcut for multiplying conjugates.

Question. Simplify
(i) \( (3+\sqrt{3})(2+\sqrt{2}) \)
(ii) \( (5+\sqrt{7})(2+\sqrt{5}) \)
Answer:
(i) Multiply each term in the first bracket by each term in the second: \[ 3(2 + \sqrt{2}) + \sqrt{3}(2 + \sqrt{2}) \] \[ = 6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6} \]
(ii) Multiply terms in both brackets: \[ 5(2 + \sqrt{5}) + \sqrt{7}(2 + \sqrt{5}) \] \[ = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35} \]
In simple words: Use the FOIL method (First, Outer, Inner, Last) to multiply the expressions term by term.

Exam Tip: Check if any of the final radicals can be simplified further. In these cases, they are in simplest form.

Question. Represent \( \sqrt{3} \) on a number line.
Answer: To represent \( \sqrt{3} \) on the number line: 1. Draw a number line with origin \( O \) representing 0 and point \( A \) representing 1 unit. 2. Construct a perpendicular \( AB \) of length 1 unit at \( A \). Join \( OB \). By Pythagoras theorem, \( OB = \sqrt{1^2 + 1^2} = \sqrt{2} \). 3. Now construct a perpendicular \( BC \) of length 1 unit at point \( B \). Join \( OC \). 4. By Pythagoras theorem, \( OC = \sqrt{OB^2 + BC^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3} \). 5. Using a compass with center \( O \) and radius \( OC \), draw an arc that cuts the number line at point \( P \). Point \( P \) represents \( \sqrt{3} \) on the number line.
In simple words: First build a length of root 2, then build a 1-unit perpendicular line on top of it to get root 3, and swing it down with a compass.

Exam Tip: Clearly show the perpendicular construction steps at each vertex.

Question. Represent \( \sqrt{2} \) on a number line.
Answer: To represent \( \sqrt{2} \) on the number line: 1. Draw a horizontal line and mark a point \( O \) as 0 and point \( A \) as 1 unit. 2. Draw a perpendicular line segment \( AB \) of length 1 unit at \( A \). 3. Join \( OB \). By Pythagoras theorem: \[ OB = \sqrt{OA^2 + AB^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ units.} \] 4. With \( O \) as center and \( OB \) as radius, draw an arc that intersects the number line at point \( P \). Point \( P \) represents \( \sqrt{2} \) on the number line.
In simple words: Draw a 1 by 1 square on the number line. The diagonal line is root 2. Swing this diagonal line down to the axis with a compass.

Exam Tip: Use a ruler and compass to ensure your 1-unit measurements are perfectly equal.

Question. You know that \( \frac{1}{7} = 0.\overline{142857} \). Can you predict the decimal expansions of \( \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7} \) without actually doing the long division? If so how?
Answer: Yes, we can predict the decimal expansions of these fractions by multiplying the decimal expansion of \( \frac{1}{7} \) by the respective numerators: \[ \frac{2}{7} = 2 \times 0.\overline{142857} = 0.\overline{285714} \] \[ \frac{3}{7} = 3 \times 0.\overline{142857} = 0.\overline{428571} \] \[ \frac{4}{7} = 4 \times 0.\overline{142857} = 0.\overline{571428} \] \[ \frac{5}{7} = 5 \times 0.\overline{142857} = 0.\overline{714285} \] \[ \frac{6}{7} = 6 \times 0.\overline{142857} = 0.\overline{857142} \]
In simple words: Yes, we can just multiply the repeating decimal of 1/7 by 2, 3, 4, 5, and 6 to find the other values.

Exam Tip: Notice how the digits in the repeating block stay in the same cyclical order; they just start from a different digit.

Question. Represent \( \sqrt{5} \) on the number line.
Answer: To represent \( \sqrt{5} \) on the number line: 1. Draw a horizontal number line. Mark origin \( O \) as 0 and point \( A \) at 2 units from \( O \). 2. Construct a perpendicular line segment \( AB \) of length 1 unit at \( A \). 3. Join \( OB \). By Pythagoras theorem in right-angled triangle \( OAB \): \[ OB = \sqrt{OA^2 + AB^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \text{ units.} \] 4. Using a compass with center \( O \) and radius \( OB \), draw an arc cutting the number line at point \( P \). Point \( P \) represents \( \sqrt{5} \) on the number line.
In simple words: Draw a right triangle with a base of 2 units and a height of 1 unit. The diagonal is root 5. Use a compass to swing it down to the axis.

Exam Tip: Constructing root 5 is simpler than root 2 or root 3 because you can start with a base of 2 units directly.

Question. Construct the square root spiral.
Answer: To construct the square root spiral: 1. Start with a point \( O \) as origin. Draw a horizontal segment \( OP_1 \) of length 1 unit. 2. Draw a perpendicular segment \( P_1P_2 \) of length 1 unit. Join \( OP_2 \). This segment has length \( \sqrt{2} \). 3. Draw a perpendicular segment \( P_2P_3 \) of length 1 unit to \( OP_2 \). Join \( OP_3 \). This segment has length \( \sqrt{3} \). 4. Repeat this step-by-step. Each time, construct a perpendicular segment of length 1 unit to the hypotenuse of the previous triangle. 5. This sequence creates segments \( OP_4, OP_5, OP_6\dots \) of lengths \( \sqrt{4}, \sqrt{5}, \sqrt{6}\dots \) respectively, forming a spiral pattern.
In simple words: Keep adding 1-unit tall perpendicular steps to each new diagonal line. This builds a spiral of square root lines.

Exam Tip: Keep your pencil sharp and your protractor accurate, as small errors quickly accumulate in a spiral.

Question. Find:
(i) \( 9^{\frac{3}{2}} \)
(ii) \( 32^{\frac{2}{5}} \)
(iii) \( 16^{\frac{3}{4}} \)
Answer:
(i) \( 9^{\frac{3}{2}} = (3^2)^{\frac{3}{2}} = 3^{2 \times \frac{3}{2}} = 3^3 = 27 \)
(ii) \( 32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^{5 \times \frac{2}{5}} = 2^2 = 4 \)
(iii) \( 16^{\frac{3}{4}} = (2^4)^{\frac{3}{4}} = 2^{4 \times \frac{3}{4}} = 2^3 = 8 \)
In simple words: Write each base as a power of a smaller prime number, then multiply the power by the fraction exponent to find the final value.

Exam Tip: Expressing composite base numbers as powers of their prime factors is the most reliable way to solve index questions.

Question. Simplify
(i) \( (64)^{\frac{1}{3}} \)
(ii) \( (125)^{-\frac{1}{3}} \)
(iii) \( (27)^{-\frac{2}{3}} \)
(iv) \( \left(\frac{64}{25}\right)^{-\frac{3}{2}} \)
Answer:
(i) \( 64^{\frac{1}{3}} = (4^3)^{\frac{1}{3}} = 4 \)
(ii) \( 125^{-\frac{1}{3}} = (5^3)^{-\frac{1}{3}} = 5^{-1} = \frac{1}{5} \)
(iii) \( 27^{-\frac{2}{3}} = (3^3)^{-\frac{2}{3}} = 3^{-2} = \frac{1}{9} \)
(iv) \( \left(\frac{64}{25}\right)^{-\frac{3}{2}} = \left(\frac{25}{64}\right)^{\frac{3}{2}} = \left( \left(\frac{5}{8}\right)^2 \right)^{\frac{3}{2}} = \left(\frac{5}{8}\right)^3 = \frac{125}{512} \)
In simple words: (i) Cube root of 64 is 4. (ii) Negative power means 1 over the cube root of 125, which is 1/5. (iii) Cube root of 27 is 3, squared is 9, so 1/9. (iv) Flip the fraction to make the power positive, take the square root, and cube the result.

Exam Tip: A negative exponent flips the base fraction: \( (a/b)^{-n} = (b/a)^n \). Apply this rule first.

Question. Simplify
(i) \( \frac{3}{\sqrt{3}+1} + \frac{5}{\sqrt{3}-1} \)
(ii) \( \frac{\sqrt{7}-1}{\sqrt{7}+1} - \frac{\sqrt{7}+1}{\sqrt{7}-1} \)
Answer:
(i) Combine the fractions over a common denominator: \[ \frac{3(\sqrt{3}-1) + 5(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)} \] \[ = \frac{3\sqrt{3} - 3 + 5\sqrt{3} + 5}{3 - 1} \] \[ = \frac{8\sqrt{3} + 2}{2} \] \[ = 4\sqrt{3} + 1 \]
(ii) Combine the fractions over a common denominator: \[ \frac{(\sqrt{7}-1)^2 - (\sqrt{7}+1)^2}{(\sqrt{7}+1)(\sqrt{7}-1)} \] Using identity \( a^2 - b^2 = (a-b)(a+b) \): \[ = \frac{[(\sqrt{7}-1) - (\sqrt{7}+1)][(\sqrt{7}-1) + (\sqrt{7}+1)]}{7 - 1} \] \[ = \frac{[-2][2\sqrt{7}]}{6} \] \[ = \frac{-4\sqrt{7}}{6} \] \[ = -\frac{2\sqrt{7}}{3} \]
In simple words: Multiply crosswise to find a common denominator, expand the terms on top, simplify, and divide by the common denominator.

Exam Tip: Using algebraic identities like \( (x-y)^2 - (x+y)^2 = -4xy \) makes simplifying numerators much faster and less error-prone.

Question. Simplify
(i) \( \sqrt{5} \times \sqrt{45} \)
(ii) \( \sqrt{2x} \times \sqrt{8x} \)
(iii) \( \frac{\sqrt{2}}{\sqrt{50}} \)
Answer:
(i) \( \sqrt{5} \times \sqrt{45} = \sqrt{5 \times 45} = \sqrt{225} = 15 \)
(ii) \( \sqrt{2x} \times \sqrt{8x} = \sqrt{16x^2} = 4x \) (assuming \( x \ge 0 \))
(iii) \( \frac{\sqrt{2}}{\sqrt{50}} = \sqrt{\frac{2}{50}} = \sqrt{\frac{1}{25}} = \frac{1}{5} \)
In simple words: (i) Multiply 5 and 45 to get root 225, which is 15. (ii) Multiply inside to get root 16x^2, which is 4x. (iii) Simplify inside to 1/25, then take the root.

Exam Tip: Group terms together under a single radical sign before simplifying or taking square roots.

Question. Examine, whether the following numbers are rational or irrational.
(i) \( (\sqrt{2}+2)^2 \)
(ii) \( (5+\sqrt{5})(5-\sqrt{5}) \)
(iii) \( \frac{6}{2\sqrt{3}} \)
Answer:
(i) \( (\sqrt{2}+2)^2 = (\sqrt{2})^2 + 2(2)(\sqrt{2}) + 2^2 = 2 + 4\sqrt{2} + 4 = 6 + 4\sqrt{2} \) Since \( 6 + 4\sqrt{2} \) contains an irrational term \( \sqrt{2} \), it is irrational.
(ii) \( (5+\sqrt{5})(5-\sqrt{5}) = 5^2 - (\sqrt{5})^2 = 25 - 5 = 20 \) Since 20 is an integer, it is rational.
(iii) \( \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \) Since 3 is not a perfect square, its square root \( \sqrt{3} \) is irrational.
In simple words: (i) Squaring this gives \( 6 + 4\sqrt{2} \), which is irrational. (ii) Multiplying these gives 20, which is rational. (iii) Simplifying this leaves \( \sqrt{3} \), which is irrational.

Exam Tip: Simplify each expression completely before concluding whether it is rational or irrational.

Question. Write the following in decimal form and find the type of decimal expansion.
(i) \( \frac{36}{100} \)
(ii) \( \frac{1}{11} \)
(iii) \( 4\frac{1}{8} \)
Answer:
(i) Dividing 36 by 100 gives \( 0.36 \), which is a terminating decimal expansion.
(ii) Dividing 1 by 11 gives \( 0.090909... = 0.\overline{09} \), which is a non-terminating repeating decimal expansion.
(iii) Changing \( 4\frac{1}{8} \) to an improper fraction gives \( \frac{33}{8} \). Dividing 33 by 8 results in \( 4.125 \), which is a terminating decimal expansion.
In simple words: A decimal that ends is terminating, while one that keeps repeating the same digits forever is non-terminating repeating.

Exam Tip: If the denominator of a simplified fraction has only 2 and 5 as prime factors, its decimal expansion will terminate; otherwise, it will be non-terminating repeating.

Question. Express the following in the form \( \frac{p}{q} \), where p and q are integers and q \( \neq \) 0.
(i) \( 0.\overline{6} \)
(ii) \( x = 0.4\overline{7} \)
(iii) \( 0.\overline{001} \)
Answer:
(i) Let \( x = 0.\overline{6} = 0.666... \) (Equation 1).
Multiply both sides by 10 to get \( 10x = 6.666... \) (Equation 2).
Subtract Equation 1 from Equation 2:
\( 9x = 6 \)

\( \implies x = \frac{6}{9} = \frac{2}{3} \).

(ii) Let \( x = 0.4\overline{7} = 0.4777... \) (Equation 1).
Multiply both sides by 10 to get \( 10x = 4.777... \) (Equation 2).
Multiply Equation 1 by 100 to get \( 100x = 47.777... \) (Equation 3).
Subtract Equation 2 from Equation 3:
\( 90x = 43 \)

\( \implies x = \frac{43}{90} \).

(iii) Let \( x = 0.\overline{001} = 0.001001... \) (Equation 1).
Multiply both sides by 1000 to get \( 1000x = 1.001001... \) (Equation 2).
Subtract Equation 1 from Equation 2:
\( 999x = 1 \)

\( \implies x = \frac{1}{999} \).
In simple words: To convert repeating decimals to fractions, multiply by 10, 100, or 1000 depending on how many digits repeat, subtract the equations to remove the repeating decimals, and solve for x.

Exam Tip: When converting repeating decimals to a fraction, the number of repeating digits under the bar tells you what power of 10 to multiply by (1 digit = multiply by 10, 3 digits = multiply by 1000).

Question. Identify the following as rational or irrational number.
(i) \( \sqrt{4} \)
(ii) \( \sqrt{100} \)
(iii) \( \sqrt{3} \)
(iv) \( 0.10110111011110... \)
(v) \( 0.375 \)
(vi) \( 22.3333333... \)
Answer:
(i) \( \sqrt{4} = 2 \), which is a whole number, so it is a rational number.
(ii) \( \sqrt{100} = 10 \), which is a whole number, so it is a rational number.
(iii) \( \sqrt{3} \approx 1.732... \) is a non-terminating and non-repeating decimal, so it is an irrational number.
(iv) \( 0.10110111011110... \) is non-terminating and non-repeating, so it is an irrational number.
(v) \( 0.375 \) is a terminating decimal, so it is a rational number.
(vi) \( 22.3333333... = 22.\overline{3} \) is a non-terminating repeating decimal, so it is a rational number.
In simple words: Rational numbers can be written as simple fractions or decimals that either stop or repeat. Irrational numbers are decimals that go on forever without any repeating pattern.

Exam Tip: Square roots of perfect squares are always rational, whereas square roots of non-perfect squares are always irrational.

Most Important Questions

Question. Are all-rational numbers real numbers?
Answer: Yes, all rational numbers are real numbers. Real numbers include both rational and irrational numbers.
In simple words: Yes, every rational number is a part of the real number system.

Exam Tip: Remember the classification of numbers: real numbers are the largest set you study in Class 9, containing all rational and irrational numbers as subsets.

Question. Is it possible to find a natural number between 1 and 2?
Answer: No, it is not possible. Natural numbers are positive counting integers, and there is no integer between the consecutive counting numbers 1 and 2.
In simple words: No, because 1 and 2 are next to each other, so there are no whole counting numbers in between them.

Exam Tip: Do not confuse natural numbers with rational or real numbers; while infinitely many rational numbers exist between 1 and 2, no natural numbers do.

Question. Is each point on the number line of the form \( \sqrt{m} \),where m is a natural numbers?
Answer: No, this statement is false. The number line contains negative real numbers, but the square root of any natural number is always positive. Therefore, negative numbers cannot be written in the form \( \sqrt{m} \).
In simple words: No, because negative numbers are on the number line, but square roots of natural numbers can never be negative.

Exam Tip: Always use negative numbers as a counterexample when dealing with statements about all points on the number line being of the form \( \sqrt{m} \).

Question. Find one rational number between 5 and 6.
Answer: A rational number between 5 and 6 can be found by calculating their arithmetic mean:
\( \frac{5 + 6}{2} = \frac{11}{2} = 5.5 \)
In simple words: You can find a number between 5 and 6 by averaging them, which gives 5.5.

Exam Tip: Finding the mean of two numbers is the easiest way to insert a rational number between them.

Question. Without actual division, find whether the following rational numbers are terminating or non-terminating repeating: 11/50 and 27/56.
Answer:
For \( \frac{11}{50} \), the denominator is 50. The prime factorization of 50 is \( 2 \times 5^2 \). Since the prime factors consist only of 2 and 5, this rational number has a terminating decimal expansion.

For \( \frac{27}{56} \), the denominator is 56. The prime factorization of 56 is \( 2^3 \times 7 \). Since the denominator has a prime factor other than 2 or 5 (which is 7), this rational number has a non-terminating repeating decimal expansion.
In simple words: If the denominator's prime factors only have 2s and 5s, the decimal stops. If there are other prime factors, the decimal repeats forever.

Exam Tip: Ensure the fraction is in its simplest form before analyzing the prime factorization of its denominator.

Question. Name the following:
(a) The outer layer of the cell
(b) The fluid like substance present outside the nucleus
Answer:
(a) Cell membrane (also called Plasma membrane) is the outer layer in animal cells, while Cell wall is the outer layer in plant cells.
(b) Cytoplasm is the fluid-like substance found outside the nucleus.
In simple words: The outer covering of a cell is the cell membrane, and the jelly-like liquid surrounding the nucleus is the cytoplasm.

Exam Tip: Be specific about plant cells versus animal cells when discussing the outermost layer of a cell.

Question. Can photosynthesis take place outside the leaves? If yes, then where this process takes place?
Answer: Yes, photosynthesis can occur outside the leaves. It takes place in other green parts of the plant that contain chlorophyll, such as green stems (like in cacti) and green branches.
In simple words: Yes, any green part of a plant, like a green stem, can perform photosynthesis because it contains chlorophyll.

Exam Tip: In desert plants like cacti, leaves are reduced to spines to prevent water loss, and the thick green stem is the primary site of photosynthesis.

Question. Express the decimal expression \( 0.4\overline{73} \) as a rational numbers.
Answer: Let \( x = 0.4\overline{73} = 0.4737373... \) (Equation 1).
Multiply both sides of Equation 1 by 10 to shift the non-repeating part:
\( 10x = 4.\overline{73} = 4.737373... \) (Equation 2).
Multiply both sides of Equation 1 by 1000:
\( 1000x = 473.\overline{73} = 473.737373... \) (Equation 3).
Subtract Equation 2 from Equation 3:
\( 1000x - 10x = 473.\overline{73} - 4.\overline{73} \)

\( \implies 990x = 469 \)

\( \implies x = \frac{469}{990} \).
In simple words: To write \( 0.4\overline{73} \) as a fraction, we multiply it to isolate the repeating part, subtract to cancel the decimals, and get \( \frac{469}{990} \).

Exam Tip: Always double check that the fraction cannot be simplified further by looking for common factors in the numerator and denominator.

Question. Insert four rational numbers between (1/3) and (1/4).
Answer: First, convert both fractions to have a common denominator. Let's use 60:
\( \frac{1}{4} = \frac{15}{60} \) and \( \frac{1}{3} = \frac{20}{60} \).
Now, we can choose four fractions between \( \frac{15}{60} \) and \( \frac{20}{60} \):
\( \frac{16}{60}, \frac{17}{60}, \frac{18}{60}, \frac{19}{60} \).
Simplifying these gives:
\( \frac{4}{15}, \frac{17}{60}, \frac{3}{10}, \frac{19}{60} \).
In simple words: Change the fractions so they have the same bottom number, then pick four numbers that fall in between them.

Exam Tip: If you need to insert n rational numbers, multiplying the numerator and denominator by \( n+1 \) after finding a common denominator will always give you enough integers in between.

Question. Give one example of each:
(a) a parasitic plant
(b) a parasitic animal
(c) a saprophyte
Answer:
(a) Parasitic plant: Cuscuta (commonly known as Dodder or Amarbel).
(b) Parasitic animal: Tapeworm (or Leech).
(c) Saprophyte: Mushroom (or Yeast).
In simple words: Cuscuta is a plant parasite, tapeworms are animal parasites, and mushrooms are decomposers that live on dead matter.

Exam Tip: Use standard examples from your biology textbook like Cuscuta and Fungi to ensure full marks.

Question. Find two irrational numbers between 1.5 and 1.7.
Answer: Two irrational numbers between 1.5 and 1.7 must be non-terminating and non-repeating decimals. For example:
First number: \( 1.51010010001... \)
Second number: \( 1.62020020002... \)
In simple words: We can write down any two decimal numbers between 1.5 and 1.7 that go on forever without repeating a pattern.

Exam Tip: Create an obvious non-repeating pattern (like adding one more zero in each step) to show that the decimal is truly irrational.

Question. (a) What is a parasite?
(b) State a difference between total parasite and partial parasite?
Answer:
(a) A parasite is an organism that lives on or inside another living organism (the host) to obtain its nutrition, often causing harm to the host in the process.
(b) A total parasite relies entirely on the host for all its nutrients and water (e.g., Cuscuta), whereas a partial parasite can manufacture its own food via photosynthesis but relies on the host for water and essential minerals (e.g., Mistletoe).
In simple words: A parasite feeds off a host. A total parasite takes everything from the host, but a partial parasite only takes water and minerals while making its own food.

Exam Tip: Mentioning specific examples like Cuscuta and Mistletoe is critical to clearly explaining the difference between total and partial parasites.

Question. Find an irrational number between (1/7) and (1/5).
Answer: First, convert the fractions to decimal values:
\( \frac{1}{7} \approx 0.142857... \)
\( \frac{1}{5} = 0.2 \).
Any non-terminating, non-repeating decimal between \( 0.142857... \) and \( 0.2 \) is an irrational number. For example:
\( 0.1501500150001... \)
In simple words: We convert the fractions to decimals and write a non-repeating decimal that sits between them, like \( 0.1501500150001... \).

Exam Tip: Convert fractions to decimals first so you can easily choose a decimal value that sits safely between them.

Question. (a) What is an insectivorous plant?
(b) Give one example of insectivorous plant.
(c) Give the structure and mode of nutrition of one insectivorous plant.
Answer:
(a) An insectivorous plant is a green plant that traps and digests small insects to meet its nitrogen requirements, typically growing in nitrogen-deficient soils.
(b) An example of an insectivorous plant is the Pitcher plant (Nepenthes).
(c) Structure and nutrition of the Pitcher Plant:
- Structure: The leaf blade is modified into a pitcher-like vessel with a lid at the top that can open or close. The interior of the pitcher is lined with slippery walls and downward-pointing hairs.
- Mode of nutrition: When an insect lands on the rim, it slips inside. The lid closes, and the insect is trapped by the downward-pointing hairs. The plant then secretes digestive enzymes to break down the insect and absorb the nutrients.
In simple words: An insectivorous plant eats bugs for nitrogen. The pitcher plant has a leaf shaped like a cup with a lid; when a bug falls in, the plant digests it.

Exam Tip: Clearly state that insectivorous plants are green and perform photosynthesis, but eat insects specifically to acquire nitrogen from poor soil.

Question. Rationalize the expression [1/{(2 \sqrt{3}) + \sqrt{7}}].
Answer: Multiply the numerator and the denominator by the conjugate of the denominator, which is \( 2\sqrt{3} - \sqrt{7} \):
\( \frac{1}{2\sqrt{3} + \sqrt{7}} \times \frac{2\sqrt{3} - \sqrt{7}}{2\sqrt{3} - \sqrt{7}} \)

\( = \frac{2\sqrt{3} - \sqrt{7}}{(2\sqrt{3})^2 - (\sqrt{7})^2} \)

\( = \frac{2\sqrt{3} - \sqrt{7}}{12 - 7} \)

\( = \frac{2\sqrt{3} - \sqrt{7}}{5} \)
In simple words: To remove the roots from the bottom, multiply both the top and bottom by \( 2\sqrt{3} - \sqrt{7} \) and simplify.

Exam Tip: Always use the identity \( (a-b)(a+b) = a^2 - b^2 \) to simplify the denominator when rationalizing.

Question. Find the value of the expression \( \frac{5 - \sqrt{3}}{\sqrt{3} - 3\sqrt{2}} \)
Answer: To evaluate the expression, we rationalize the denominator by multiplying both the numerator and denominator by \( \sqrt{3} + 3\sqrt{2} \):
\( \frac{5 - \sqrt{3}}{\sqrt{3} - 3\sqrt{2}} \times \frac{\sqrt{3} + 3\sqrt{2}}{\sqrt{3} + 3\sqrt{2}} \)

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

\( = \frac{5\sqrt{3} + 15\sqrt{2} - 3 - 3\sqrt{6}}{3 - 18} \)

\( = \frac{5\sqrt{3} + 15\sqrt{2} - 3 - 3\sqrt{6}}{-15} \)

\( = \frac{3 + 3\sqrt{6} - 5\sqrt{3} - 15\sqrt{2}}{15} \)

\( = \frac{3(1 + \sqrt{6})}{15} - \frac{5\sqrt{3} + 15\sqrt{2}}{15} = \frac{1 + \sqrt{6}}{5} - \frac{\sqrt{3}}{3} - \sqrt{2} \)
In simple words: Multiply the top and bottom by \( \sqrt{3} + 3\sqrt{2} \), expand the numerator, and divide by the simplified bottom integer.

Exam Tip: Be careful with the signs when expanding the terms in the numerator and when dividing by a negative denominator.

Question. What is the role of fungi in daily life?
Answer: Fungi play several crucial roles in our daily lives:
1. Food source: Edible mushrooms are consumed directly, and yeast is used extensively in baking bread and brewing beverages.
2. Medicine: Fungi like Penicillium are used to produce antibiotics such as penicillin.
3. Decomposition: Fungi act as decomposers, recycling nutrients in nature by breaking down dead organic matter.
4. Harmful roles: Some fungi cause diseases in plants and animals (like ringworm) or spoil food (like bread mold).
In simple words: Fungi are used to make bread, medicine, and clean up dead waste, though some can cause infections or spoil food.

Exam Tip: Always structure questions about "roles" by listing both beneficial and harmful aspects to provide a complete answer.

Question. If \( \frac{\sqrt{7}-1}{\sqrt{7}+1} - \frac{\sqrt{7}+1}{\sqrt{7}-1} = a + b\sqrt{7} \), find the value of a and b.
Answer: Simplify the left-hand side of the equation by taking the common denominator:
\( \frac{(\sqrt{7}-1)^2 - (\sqrt{7}+1)^2}{(\sqrt{7}+1)(\sqrt{7}-1)} \)

\( = \frac{(7 - 2\sqrt{7} + 1) - (7 + 2\sqrt{7} + 1)}{7 - 1} \)

\( = \frac{(8 - 2\sqrt{7}) - (8 + 2\sqrt{7})}{6} \)

\( = \frac{-4\sqrt{7}}{6} = -\frac{2}{3}\sqrt{7} \)

Now, equate this result to the right-hand side:
\( 0 - \frac{2}{3}\sqrt{7} = a + b\sqrt{7} \)

By comparing the rational and irrational parts on both sides, we get:
\( a = 0 \) and \( b = -\frac{2}{3} \).
In simple words: Combine the fractions on the left, simplify the expression to \( -\frac{2}{3}\sqrt{7} \), and compare it to \( a + b\sqrt{7} \) to find the values of a and b.

Exam Tip: When equating terms, the part without square roots gives the value of a, and the coefficient of the square root term gives the value of b.

Question. Give an example of two irrational numbers whose
a) Sum is a rational number
b) Difference is a rational number.
c) Product is a rational number.
Answer:
a) Let the two irrational numbers be \( 3 + \sqrt{5} \) and \( 3 - \sqrt{5} \).
Sum = \( (3 + \sqrt{5}) + (3 - \sqrt{5}) = 6 \), which is a rational number.
b) Let the two irrational numbers be \( 4 + \sqrt{3} \) and \( 1 + \sqrt{3} \).
Difference = \( (4 + \sqrt{3}) - (1 + \sqrt{3}) = 3 \), which is a rational number.
c) Let the two irrational numbers be \( \sqrt{12} \) and \( \sqrt{3} \).
Product = \( \sqrt{12} \times \sqrt{3} = \sqrt{36} = 6 \), which is a rational number.
In simple words: Even though two numbers are irrational, adding, subtracting, or multiplying them can cancel out the root parts and leave you with a plain rational number.

Exam Tip: Use conjugate pairs like \( x + \sqrt{y} \) and \( x - \sqrt{y} \) to easily show how sum or product can result in a rational number.

Question. Classify the following expressions as rational or irrational.
a) \( (6 - \sqrt{2})^2 \)
b) \( (2 - \sqrt{2})(2 + \sqrt{2}) \)
c) \( (2 + \sqrt{3})(2 + 3\sqrt{3}) \)
Answer:
a) Expand the expression:
\( (6 - \sqrt{2})^2 = 6^2 - 2(6)(\sqrt{2}) + (\sqrt{2})^2 = 36 - 12\sqrt{2} + 2 = 38 - 12\sqrt{2} \)
Since it contains \( \sqrt{2} \), this is an irrational number.
b) Expand using the identity \( (a-b)(a+b) = a^2 - b^2 \):
\( (2 - \sqrt{2})(2 + \sqrt{2}) = 2^2 - (\sqrt{2})^2 = 4 - 2 = 2 \)
Since 2 is an integer, this is a rational number.
c) Multiply out the terms:
\( (2 + \sqrt{3})(2 + 3\sqrt{3}) = 2(2) + 2(3\sqrt{3}) + \sqrt{3}(2) + \sqrt{3}(3\sqrt{3}) = 4 + 6\sqrt{3} + 2\sqrt{3} + 9 = 13 + 8\sqrt{3} \)
Since it contains \( \sqrt{3} \), this is an irrational number.
In simple words: If simplifying an expression completely removes all square roots of non-perfect squares, it is rational; otherwise, it is irrational.

Exam Tip: Never classify an expression based on its initial look; always expand and simplify it completely first.

Question. Prove that the following expression
\( \frac{1}{3 - \sqrt{8}} - \frac{1}{\sqrt{8} - \sqrt{7}} + \frac{1}{\sqrt{7} - \sqrt{6}} - \frac{1}{\sqrt{6} - \sqrt{5}} + \frac{1}{\sqrt{5} - 2} = 5 \)
Answer: Rationalize the denominator of each term individually:
1st term: \( \frac{1}{3 - \sqrt{8}} \times \frac{3 + \sqrt{8}}{3 + \sqrt{8}} = \frac{3 + \sqrt{8}}{9 - 8} = 3 + \sqrt{8} \)
2nd term: \( \frac{1}{\sqrt{8} - \sqrt{7}} \times \frac{\sqrt{8} + \sqrt{7}}{\sqrt{8} + \sqrt{7}} = \frac{\sqrt{8} + \sqrt{7}}{8 - 7} = \sqrt{8} + \sqrt{7} \)
3rd term: \( \frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6} \)
4th term: \( \frac{1}{\sqrt{6} - \sqrt{5}} \times \frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} + \sqrt{5}} = \frac{\sqrt{6} + \sqrt{5}}{6 - 5} = \sqrt{6} + \sqrt{5} \)
5th term: \( \frac{1}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2 \)

Now, substitute these rationalized terms back into the original expression:
L.H.S. = \( (3 + \sqrt{8}) - (\sqrt{8} + \sqrt{7}) + (\sqrt{7} + \sqrt{6}) - (\sqrt{6} + \sqrt{5}) + (\sqrt{5} + 2) \)

\( = 3 + \sqrt{8} - \sqrt{8} - \sqrt{7} + \sqrt{7} + \sqrt{6} - \sqrt{6} - \sqrt{5} + \sqrt{5} + 2 \)

\( = 3 + 2 = 5 \)

L.H.S. = R.H.S. Hence, proved.
In simple words: When we rationalize each fraction, all the intermediate roots cancel each other out, leaving only \( 3 + 2 = 5 \).

Exam Tip: Be exceptionally careful with the minus signs outside the parentheses when expanding the terms, as they distribute to both parts inside.

Question. 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, rationalize both \( x \) and \( y \):
\( x = \frac{(\sqrt{3} - \sqrt{2})^2}{3 - 2} = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6} \)
\( y = \frac{(\sqrt{3} + \sqrt{2})^2}{3 - 2} = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6} \)

Now, find the sum and the product of \( x \) and \( y \):
\( x + y = (5 - 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10 \)
\( xy = (5 - 2\sqrt{6})(5 + 2\sqrt{6}) = 25 - 24 = 1 \)

We can rewrite the expression \( x^2 + y^2 + xy \) using the algebraic identity:
\( x^2 + y^2 + xy = (x + y)^2 - xy \)

Substitute the values of \( x+y \) and \( xy \):
\( x^2 + y^2 + xy = 10^2 - 1 = 100 - 1 = 99 \).
In simple words: Simplify x and y first, then find their sum and product. Use the identity \( (x+y)^2 - xy \) to find the answer quickly as 99.

Exam Tip: Rewriting \( x^2 + y^2 + xy \) as \( (x+y)^2 - xy \) is a massive time-saver compared to squaring \( 5-2\sqrt{6} \) and \( 5+2\sqrt{6} \) individually.

Page 10

Question. Simplify \( \frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} \)
Answer: Let's rationalize each term individually:
First term:
\( \frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} \times \frac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}} = \frac{7\sqrt{30} - 21}{10 - 3} = \frac{7(\sqrt{30} - 3)}{7} = \sqrt{30} - 3 \)

Second term:
\( \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} \times \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} - \sqrt{5}} = \frac{2\sqrt{30} - 10}{6 - 5} = 2\sqrt{30} - 10 \)

Third term:
\( \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} \times \frac{\sqrt{15} - 3\sqrt{2}}{\sqrt{15} - 3\sqrt{2}} = \frac{3\sqrt{30} - 18}{15 - 18} = \frac{3(\sqrt{30} - 6)}{-3} = -\sqrt{30} + 6 \)

Now, substitute these back into the expression:
\( (\sqrt{30} - 3) - (2\sqrt{30} - 10) - (-\sqrt{30} + 6) \)

\( = \sqrt{30} - 3 - 2\sqrt{30} + 10 + \sqrt{30} - 6 \)

\( = (1 - 2 + 1)\sqrt{30} + (-3 + 10 - 6) \)

\( = 0\sqrt{30} + 1 = 1 \).
In simple words: Rationalize each fraction separately, substitute them back into the subtraction, and combine like terms to get the final answer 1.

Exam Tip: Be extremely careful with the negative sign on the denominator of the third term, as it changes the signs of the rationalized numerator.

Question. Simplify \( (\sqrt{x}^{-3})^5 \)
Answer: Write the square root as an exponent of \( \frac{1}{2} \):
\( \sqrt{x} = x^{1/2} \)

So, the expression becomes:
\( \left( (x^{1/2})^{-3} \right)^5 \)

Using the exponent rule \( (a^m)^n = a^{m \times n} \):
\( (x^{-3/2})^5 = x^{-15/2} \)

This can also be written in radical form:
\( \frac{1}{x^{15/2}} = \frac{1}{\sqrt{x^{15}}} \)
In simple words: Convert the square root to a fractional power of 1/2, multiply all the exponents together, and simplify to get \( x^{-15/2} \).

Exam Tip: Always remember the rule \( (x^a)^b = x^{ab} \), which allows you to simply multiply the inner power by the outer power.

Question. Is \( x^a + x^b = x^{a+b} \)
Answer: No, this statement is false. According to the laws of exponents, \( x^a \times x^b = x^{a+b} \). There is no such rule for the addition of exponents with the same base.
In simple words: No, because you only add the powers when you are multiplying the bases, not when you are adding them.

Exam Tip: Be ready to explain exponent rules; addition of powers applies exclusively during multiplication of identical bases.

Question. Simplify \( (16^{-1/5})^{5/2} \)
Answer: Use the power rule of exponents \( (a^m)^n = a^{m \times n} \):
\( 16^{-1/5 \times 5/2} = 16^{-1/2} \)

Since \( 16 = 4^2 \), we can substitute it in:
\( (4^2)^{-1/2} = 4^{2 \times -1/2} = 4^{-1} = \frac{1}{4} \)
In simple words: Multiply the exponents to get \( 16^{-1/2} \), which means finding the square root of 16 on the bottom of a fraction, giving \( \frac{1}{4} \).

Exam Tip: Simplify the outer exponents first before rewriting the base as a power of a smaller prime number to make calculations easier.

Question. Identify as rational or irrational number.
a) \( \sqrt{12} \times \sqrt{12} \)
b) \( \sqrt{4} \times \sqrt{18} \)
Answer:
a) \( \sqrt{12} \times \sqrt{12} = 12 \), which is an integer. Thus, it is a rational number.
b) \( \sqrt{4} \times \sqrt{18} = 2 \times 3\sqrt{2} = 6\sqrt{2} \). Since it contains \( \sqrt{2} \), which is irrational, the product is an irrational number.
In simple words: The first one simplifies to a whole number so it is rational, while the second one still has a square root that cannot be removed, so it is irrational.

Exam Tip: A product of two irrational numbers can sometimes be rational (as in part a), so always simplify the product completely before drawing a conclusion.

Question. Simplify \( (0.008)^{1/3} \)
Answer: Convert the decimal to a fraction:
\( 0.008 = \frac{8}{1000} \)

Now, express the fraction as a perfect cube:
\( \frac{8}{1000} = \left(\frac{2}{10}\right)^3 = (0.2)^3 \)

Substitute this back into the expression:
\( ((0.2)^3)^{1/3} = 0.2 \)
In simple words: Write 0.008 as a fraction \( \frac{8}{1000} \), take the cube root of both 8 and 1000 to get \( \frac{2}{10} \), which is 0.2.

Exam Tip: Converting decimals with three decimal places to fractions of base 1000 makes finding cube roots much more straightforward.

Question. Find the value of x if \( 2^{5x} \div 2^x = \sqrt[5]{2^{20}} \)
Answer: Simplify both sides using exponent rules:
L.H.S.: \( 2^{5x} \div 2^x = 2^{5x - x} = 2^{4x} \)
R.H.S.: \( \sqrt[5]{2^{20}} = (2^{20})^{1/5} = 2^4 \)

Equating L.H.S. and R.H.S.:
\( 2^{4x} = 2^4 \)

Since the bases are the same, equate the exponents:
\( 4x = 4 \)

\( \implies x = 1 \).
In simple words: Subtract the powers on the left to get \( 2^{4x} \) and simplify the root on the right to get \( 2^4 \). Comparing the powers shows that x must be 1.

Exam Tip: Always rewrite roots as fractional exponents so you can use the power of a power rule to simplify them.

Question. Find the value of \( (\sqrt{4})^{-7} \times (\sqrt{2})^{-5} \)
Answer: First, write each term as a power of 2:
\( \sqrt{4} = 2 \), so \( (\sqrt{4})^{-7} = 2^{-7} \)
\( \sqrt{2} = 2^{1/2} \), so \( (\sqrt{2})^{-5} = (2^{1/2})^{-5} = 2^{-5/2} \)

Now, multiply the two terms by adding their exponents:
\( 2^{-7} \times 2^{-5/2} = 2^{-7 - 5/2} = 2^{-19/2} \)

This can also be written as:
\( \frac{1}{2^{19/2}} = \frac{1}{\sqrt{2^{19}}} = \frac{1}{512\sqrt{2}} \)
In simple words: Change both terms into powers of 2, add the powers together during multiplication, and simplify to get \( 2^{-19/2} \).

Exam Tip: Be careful with fractional exponent additions; finding a common denominator is required to add -7 and -5/2.

Question. Find the value of \( \left( \frac{x^{-4}}{x^{-10}} \right)^{5/4} \)
Answer: First, simplify the quotient inside the parentheses by subtracting exponents:
\( \frac{x^{-4}}{x^{-10}} = x^{-4 - (-10)} = x^{6} \)

Now, raise the result to the power of \( 5/4 \):
\( (x^6)^{5/4} = x^{6 \times 5/4} = x^{30/4} = x^{15/2} \)

This can also be expressed in radical form as \( \sqrt{x^{15}} \).
In simple words: Divide the terms inside by subtracting the powers to get \( x^6 \), then multiply 6 by \( \frac{5}{4} \) to get \( x^{15/2} \).

Exam Tip: Remember that \( a^{-n} \) in the denominator is equivalent to \( a^n \) in the numerator, so \( \frac{x^{-4}}{x^{-10}} \) simplifies directly to \( x^{10-4} = x^6 \).

Question. Find the value of x when \( \left(\frac{3}{5}\right)^x \times \left(\frac{5}{3}\right)^{2x} = \frac{125}{27} \)
Answer: Rewrite the terms to have a common base of \( \frac{3}{5} \):
\( \left(\frac{5}{3}\right)^{2x} = \left(\left(\frac{3}{5}\right)^{-1}\right)^{2x} = \left(\frac{3}{5}\right)^{-2x} \)

Also, rewrite the right-hand side with base \( \frac{3}{5} \):
\( \frac{125}{27} = \frac{5^3}{3^3} = \left(\frac{5}{3}\right)^3 = \left(\frac{3}{5}\right)^{-3} \)

Substitute these back into the equation:
\( \left(\frac{3}{5}\right)^x \times \left(\frac{3}{5}\right)^{-2x} = \left(\frac{3}{5}\right)^{-3} \)

Combine the exponents on the left-hand side:
\( \left(\frac{3}{5}\right)^{x - 2x} = \left(\frac{3}{5}\right)^{-3} \)

\( \left(\frac{3}{5}\right)^{-x} = \left(\frac{3}{5}\right)^{-3} \)

Comparing the exponents on both sides:
\( -x = -3 \)

\( \implies x = 3 \).
In simple words: Write all parts of the equation with the base \( \frac{3}{5} \), combine the powers on the left, and match them with the right side to get \( x = 3 \).

Exam Tip: Recognizing that one base is the reciprocal of the other is the key to converting the equation into a single common base.

Question. Simplify the following: \( \frac{(25)^{3/2} \times (243)^{3/5}}{(16)^{5/4} \times (8)^{4/3}} \)
Answer: Express each base number as a prime power:
\( 25 = 5^2 \implies (25)^{3/2} = (5^2)^{3/2} = 5^3 = 125 \)
\( 243 = 3^5 \implies (243)^{3/5} = (3^5)^{3/5} = 3^3 = 27 \)
\( 16 = 2^4 \implies (16)^{5/4} = (2^4)^{5/4} = 2^5 = 32 \)
\( 8 = 2^3 \implies (8)^{4/3} = (2^3)^{4/3} = 2^4 = 16 \)

Substitute these simplified values back into the fraction:
\( \frac{125 \times 27}{32 \times 16} = \frac{3375}{512} \)
In simple words: Find the roots of each number first (like the square root of 25 is 5, then cube it to get 125), and do the same for the others to get \( \frac{3375}{512} \).

Exam Tip: Simplifying the fractional powers individually before doing any multiplication prevents very large numbers and reduces mistakes.

Question. Show that \( \frac{x^{a(b-c)}}{x^{b(a-c)}} \div \left(\frac{x^b}{x^a}\right)^c = 1 \)
Answer: Simplify the left-hand side of the expression:
Numerator: \( x^{a(b-c)} = x^{ab - ac} \)
Denominator: \( x^{b(a-c)} = x^{ab - bc} \)

Dividing these two terms gives:
\( \frac{x^{ab-ac}}{x^{ab-bc}} = x^{(ab-ac) - (ab-bc)} = x^{ab - ac - ab + bc} = x^{bc - ac} \)

Now simplify the second part of the division:
\( \left(\frac{x^b}{x^a}\right)^c = (x^{b-a})^c = x^{c(b-a)} = x^{bc - ac} \)

Now, divide the first simplified part by the second part:
\( x^{bc-ac} \div x^{bc-ac} = 1 \)

L.H.S. = R.H.S. Hence, proved.
In simple words: Expand the exponents, subtract the division powers, and notice that both parts of the division are exactly equal, so dividing them yields 1.

Exam Tip: Keep your work organized by expanding and simplifying the numerator and denominator separately before combining them.

Question. Express the following expression in the form of a rational number
\( \frac{(0.6)^0 - (0.1)^{-1}}{\left(\frac{3}{8}\right)^{-1} \left(\frac{3}{2}\right)^3 + \left(-\frac{1}{3}\right)^{-1}} \)
Answer: Calculate the numerator first:
\( (0.6)^0 = 1 \)
\( (0.1)^{-1} = \left(\frac{1}{10}\right)^{-1} = 10 \)
Numerator = \( 1 - 10 = -9 \)

Now, calculate the terms in the denominator:
\( \left(\frac{3}{8}\right)^{-1} = \frac{8}{3} \)
\( \left(\frac{3}{2}\right)^3 = \frac{27}{8} \)
Their product is \( \frac{8}{3} \times \frac{27}{8} = 9 \)
\( \left(-\frac{1}{3}\right)^{-1} = -3 \)
Denominator = \( 9 + (-3) = 6 \)

Combine the numerator and denominator:
\( \frac{\text{Numerator}}{\text{Denominator}} = \frac{-9}{6} = -\frac{3}{2} = -1.5 \)
In simple words: Solve the top part to get -9, solve the bottom part to get 6, and divide them to get \( -\frac{3}{2} \).

Exam Tip: Remember that any non-zero number raised to the power of 0 is always 1, and a negative exponent simply flips the fraction.

Question. Simplify \( \left( \frac{5^{-1} \times 7^2}{5^2 \times 7^{-4}} \right)^{7/2} \times \left( \frac{5^{-2} \times 7^3}{5^3 \times 7^{-5}} \right)^{-5/2} \)
Answer: Simplify the expression inside the first set of parentheses:
\( \frac{5^{-1} \times 7^2}{5^2 \times 7^{-4}} = 5^{-1-2} \times 7^{2-(-4)} = 5^{-3} \times 7^6 \)

Now, raise this to the power of \( 7/2 \):
\( (5^{-3} \times 7^6)^{7/2} = 5^{-3 \times 7/2} \times 7^{6 \times 7/2} = 5^{-21/2} \times 7^{21} \)

Next, simplify the expression inside the second set of parentheses:
\( \frac{5^{-2} \times 7^3}{5^3 \times 7^{-5}} = 5^{-2-3} \times 7^{3-(-5)} = 5^{-5} \times 7^8 \)

Now, raise this to the power of \( -5/2 \):
\( (5^{-5} \times 7^8)^{-5/2} = 5^{-5 \times -5/2} \times 7^{8 \times -5/2} = 5^{25/2} \times 7^{-20} \)

Finally, multiply both simplified terms together:
\( (5^{-21/2} \times 7^{21}) \times (5^{25/2} \times 7^{-20}) \)

\( = 5^{-21/2 + 25/2} \times 7^{21 - 20} \)

\( = 5^{4/2} \times 7^1 \)

\( = 5^2 \times 7^1 = 25 \times 7 = 175 \).
In simple words: Simplify the fraction inside each group, multiply by the outer power, and then combine the 5s and 7s to get the final answer 175.

Exam Tip: Keep bases grouped (all 5s together and all 7s together) throughout the simplification steps to avoid algebraic confusion.