CBSE Class 9 Mathematics Revision Assignments Set 06

Number Systems and Polynomials

Question 1. Which of the following is an irrational number : (a) \(\sqrt{9}\) (b) \(3.026026026\dots\) (c) \(4.020020002\dots\) (d) \(\frac{22}{7}\)
(a) \(\sqrt{9} = 3\) (rational)
(b) \(3.026026026\dots\) (non-terminating repeating, so rational)
(c) \(4.020020002\dots\) (non-terminating non-repeating, so irrational)
(d) \(\frac{22}{7}\) (rational)
Answer: (c) \(4.020020002\dots\)
An irrational number has a decimal representation that is neither terminating nor repeating.
In simple words: A decimal that goes on forever without a repeating pattern is irrational.

Exam Tip: Remember \(\sqrt{9} = 3\) is a whole number, hence rational. Non-terminating and non-recurring decimals are always irrational.

Question 2. If \(x = 2 + \sqrt{3}\), then \(\left( x + \frac{1}{x} \right) =\) (a) \(-2\sqrt{3}\) (b) \(2\) (c) \(4\) (d) \(4 - 2\sqrt{3}\)
Given \(x = 2 + \sqrt{3}\)
\(\frac{1}{x} = \frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}\)
Now, \(x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4\)
Answer: (c) \(4\)
Rationalize the denominator of \(\frac{1}{x}\) and add it to \(x\).
In simple words: Find the reciprocal of x by rationalizing, then add it to x.

Exam Tip: For \(x = a + \sqrt{b}\), the reciprocal \(\frac{1}{x} = a - \sqrt{b}\) if \(a^2 - b = 1\).

Question 3. Which of the following numbers has terminating decimal form: (a) \(\frac{37}{45}\) (b) \(\frac{21}{2^3 \times 5^6}\) (c) \(\frac{17}{49}\) (d) \(\frac{89}{2^2 \times 3^2}\)
A rational number in its simplest form has a terminating decimal expansion if the prime factorization of its denominator is of the form \(2^m \times 5^n\), where m and n are non-negative integers.
(a) Denominator \(45 = 3^2 \times 5\) (has factor 3, non-terminating repeating)
(b) Denominator \(2^3 \times 5^6\) (only powers of 2 and 5, terminating)
(c) Denominator \(49 = 7^2\) (has factor 7, non-terminating repeating)
(d) Denominator \(2^2 \times 3^2\) (has factor 3, non-terminating repeating)
Answer: (b) \(\frac{21}{2^3 \times 5^6}\)
Only option (b) has prime factors of the denominator restricted to 2 and 5.
In simple words: A fraction gives a terminating decimal only if its denominator's prime factors are strictly 2 and 5.

Exam Tip: Always check if the fraction is in lowest terms before inspecting the denominator's prime factors.

Question 4. What will be the value of a if \(x + a\) is a factor of \(x^3 + ax^2 - 2x + a + 6\) : (a) \(3\) (b) \(4\) (c) \(-2\) (d) \(-3\)
Let \(p(x) = x^3 + ax^2 - 2x + a + 6\)
Since \(x + a\) is a factor, by the Factor Theorem, \(p(-a) = 0\)
\((-a)^3 + a(-a)^2 - 2(-a) + a + 6 = 0\)
\(-a^3 + a^3 + 2a + a + 6 = 0\)
\(3a + 6 = 0\)
\(3a = -6\)
\(a = -2\)
Answer: (c) \(-2\)
Substitute \(x = -a\) into the polynomial and equate it to zero to solve for a.
In simple words: Put \(x = -a\) in the expression and solve the resulting linear equation for a.

Exam Tip: Factor Theorem states that if \(x - c\) is a factor of \(p(x)\), then \(p(c) = 0\).

Question 5. The degree of the polynomial \(8x^4 - 36x + 5x^7 + 1\) is : (a) \(4\) (b) \(1\) (c) \(5\) (d) \(7\)
Given polynomial: \(p(x) = 8x^4 - 36x + 5x^7 + 1\)
The powers of x present are 4, 1, 7, and 0.
The highest power of x is 7.
Answer: (d) \(7\)
The degree of a polynomial is the highest power of the variable in the polynomial.
In simple words: The degree is simply the highest exponent of x in the expression.

Exam Tip: Rearrange the polynomial in descending order of powers to easily spot the highest degree term.

Question 6. Find two rational numbers lying between \(\frac{1}{2}\) and \(\frac{2}{5}\).
First, make the denominators equal:
\(\frac{1}{2} = \frac{5}{10}\)
\(\frac{2}{5} = \frac{4}{10}\)
To find two rational numbers, multiply numerator and denominator by \(2 + 1 = 3\) (or any larger number like 10):
\(\frac{4}{10} = \frac{40}{100}\)
\(\frac{5}{10} = \frac{50}{100}\)
Two rational numbers between them are \(\frac{41}{100}\) and \(\frac{42}{100}\) (or \(\frac{41}{100}, \frac{43}{100}\)).
Answer: \(\frac{41}{100}\) and \(\frac{42}{100}\)
Convert fractions to equivalent fractions with larger denominators.
In simple words: Expand the fractions by multiplying top and bottom by 10 to easily pick numbers in between.

Exam Tip: There are infinitely many rational numbers between any two numbers; any correct values in the range will receive full credit.

Question 7. Express \(0.25626262\dots\) in the form of \(\frac{p}{q}\).
Let \(x = 0.25626262\dots\) ---- (1)
Here, two digits (62) repeat after one decimal place (2). Multiply equation (1) by 10:
\(10x = 2.5626262\dots\) ---- (2)
Now multiply equation (2) by 100 (since two digits repeat):
\(1000x = 256.262626\dots\) ---- (3)
Subtract equation (2) from equation (3):
\(1000x - 10x = 256.262626\dots - 2.5626262\dots\)
\(990x = 253.7 = \frac{2537}{10}\)
\(x = \frac{2537}{9900}\)
Answer: \(\frac{2537}{9900}\)
Set up algebraic equations by shifting decimals to eliminate the repeating part.
In simple words: Multiply to shift the decimal past the non-repeating part and then past the repeating block, then subtract.

Exam Tip: Always verify that the repeating decimal subtraction cancels out all infinite tails correctly.

Question 8. Give an example of two irrational numbers, whose (i) Product is a rational number. (ii) Difference is a rational number.
(i) Let \(a = 3 + \sqrt{2}\) and \(b = 3 - \sqrt{2}\) (or simply \(\sqrt{2}\) and \(\sqrt{2}\)).
Product: \(\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4\) (which is rational).
(ii) Let \(a = 5 + \sqrt{3}\) and \(b = \sqrt{3}\).
Difference: \((5 + \sqrt{3}) - \sqrt{3} = 5\) (which is rational).
Answer: (i) \(\sqrt{2}\) and \(\sqrt{8}\) (ii) \(5 + \sqrt{3}\) and \(\sqrt{3}\)
Choose irrational numbers such that their product or difference eliminates the radical.
In simple words: Pick surds that cancel each other out when multiplied or subtracted.

Exam Tip: Conjugate surds like \(a + \sqrt{b}\) and \(a - \sqrt{b}\) are very useful for these examples.

Question 9. Simplify: (i) \(\left( \frac{15^{1/3}}{9^{1/4}} \right)^{-6}\) (ii) \(\frac{1}{(27)^{-1/3}} + \frac{1}{(625)^{-1/4}}\)
(i) \(\left( \frac{15^{1/3}}{9^{1/4}} \right)^{-6} = \left( \frac{9^{1/4}}{15^{1/3}} \right)^6 = \frac{(9^{1/4})^6}{(15^{1/3})^6} = \frac{9^{6/4}}{15^{6/3}} = \frac{9^{3/2}}{15^2}\)
\(9^{3/2} = (3^2)^{3/2} = 3^3 = 27\)
\(15^2 = 225\)
Result = \(\frac{27}{225} = \frac{3}{25}\)

(ii) \(\frac{1}{(27)^{-1/3}} + \frac{1}{(625)^{-1/4}} = (27)^{1/3} + (625)^{1/4}\)
\((27)^{1/3} = (3^3)^{1/3} = 3\)
\((625)^{1/4} = (5^4)^{1/4} = 5\)
Result = \(3 + 5 = 8\)
Answer: (i) \(\frac{3}{25}\) (ii) \(8\)
Apply laws of exponents such as \((a^m)^n = a^{m \times n}\) and negative exponent rule \(a^{-m} = \frac{1}{a^m}\).
In simple words: Use exponent rules to simplify fractional powers into whole numbers.

Exam Tip: Convert bases into prime exponent form before applying fractional powers.

Question 10. Find the value of a and b if \(\frac{5 + \sqrt{6}}{5 - \sqrt{6}} = a + b\sqrt{6}\).
Rationalize the LHS:
\(\frac{5 + \sqrt{6}}{5 - \sqrt{6}} \times \frac{5 + \sqrt{6}}{5 + \sqrt{6}} = \frac{(5 + \sqrt{6})^2}{5^2 - (\sqrt{6})^2}\)
Numerator = \(25 + 6 + 10\sqrt{6} = 31 + 10\sqrt{6}\)
Denominator = \(25 - 6 = 19\)
So, LHS = \(\frac{31}{19} + \frac{10}{19}\sqrt{6}\)
Comparing with \(a + b\sqrt{6}\):
\(a = \frac{31}{19}\), \(b = \frac{10}{19}\)
Answer: \(a = \frac{31}{19}\), \(b = \frac{10}{19}\)
Rationalize denominator and compare rational and irrational parts on both sides.
In simple words: Simplify the fraction by rationalizing, then match the whole number and root parts to find a and b.

Exam Tip: Equating coefficients of rational and irrational parts is a standard technique in surds.

Question 11. If the polynomials \(px^3 + 4x^2 + 3x - 4\) and \(x^3 - 4x + p\) are divided by \(x - 3\) then the remainder in each case is the same. Find the value of p.
Let \(f(x) = px^3 + 4x^2 + 3x - 4\) and \(g(x) = x^3 - 4x + p\)
By Remainder Theorem, the remainders when divided by \(x - 3\) are \(f(3)\) and \(g(3)\).
\(f(3) = p(3)^3 + 4(3)^2 + 3(3) - 4 = 27p + 36 + 9 - 4 = 27p + 41\)
\(g(3) = (3)^3 - 4(3) + p = 27 - 12 + p = p + 15\)
Given \(f(3) = g(3)\):
\(27p + 41 = p + 15\)
\(26p = 15 - 41 = -26\)
\(p = -1\)
Answer: \(p = -1\)
Evaluate both polynomials at \(x = 3\) using the Remainder Theorem and equate the expressions.
In simple words: Substitute 3 for x in both polynomials, set them equal to each other, and solve for p.

Exam Tip: Remainder Theorem states that remainder of \(p(x)\) divided by \(x - c\) is \(p(c)\).

Question 12. Factorize: (i) \(9x^2 - 22x + 8\). (ii) \(x^3 - x^2 - 14x + 24\) (by using factor theorem)
(i) Using splitting the middle term:
\(9x^2 - 18x - 4x + 8\)
\(= 9x(x - 2) - 4(x - 2)\)
\(= (9x - 4)(x - 2)\)

(ii) Let \(p(x) = x^3 - x^2 - 14x + 24\)
Factors of constant term 24 are \(\pm 1, \pm 2, \pm 3, \pm 4 \dots\)
Test \(x = 2\):
\(p(2) = (2)^3 - (2)^2 - 14(2) + 24 = 8 - 4 - 28 + 24 = 0\)
So \(x - 2\) is a factor.
Divide \(p(x)\) by \(x - 2\) or use synthetic division to get the quadratic quotient: \(x^2 + x - 12\).
Factorize \(x^2 + x - 12 = (x + 4)(x - 3)\).
Thus, factors are \((x - 2)(x + 4)(x - 3)\).
Answer: (i) \((9x - 4)(x - 2)\) (ii) \((x - 2)(x + 4)(x - 3)\)
Use middle-term splitting for quadratics and Factor Theorem for cubics.
In simple words: Find one linear factor by trial for cubics, then divide to find the remaining quadratic factor.

Exam Tip: Always check constant term factors first when applying the Factor Theorem to cubic polynomials.

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