CBSE Class 9 Mathematics Revision Worksheet Set 02

Question 1. Classify the following as rational or irrational:
(a) \(2 - \sqrt{5}\)
(b) \(3 + \sqrt{23} - \sqrt{23}\)
(c) \(\frac{2\sqrt{7}}{7\sqrt{7}}\)
(d) \(\frac{1}{\sqrt{2}}\)
(e) \(2\pi\)
Answer:
(a) Irrational (since the difference of a rational number 2 and an irrational number \(\sqrt{5}\) is irrational).
(b) Rational (since \(3 + \sqrt{23} - \sqrt{23} = 3\), which is a rational number).
(c) Rational (since \(\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}\), which is a rational number).
(d) Irrational (since the quotient of a non-zero rational number 1 and an irrational number \(\sqrt{2}\) is irrational).
(e) Irrational (since the product of a non-zero rational number 2 and an irrational number \(\pi\) is irrational).
In simple words: Numbers that cannot be written as simple fractions of integers are irrational.

Exam Tip: Simplify expressions fully before checking if they are rational or irrational.

Question 2. Simplify each of the following:
(a) \((3 + \sqrt{3})(3 - \sqrt{3})\)
(b) \((3 + \sqrt{2})(2 + \sqrt{2})\)
(c) \((\sqrt{5} + \sqrt{2})^2\)
Answer:
(a) Using the identity \((a+b)(a-b) = a^2 - b^2\):
\(= 3^2 - (\sqrt{3})^2\)
\(= 9 - 3 = 6\)
(b) Using distributive multiplication:
\(= 3(2) + 3\sqrt{2} + \sqrt{2}(2) + (\sqrt{2})(\sqrt{2})\)
\(= 6 + 3\sqrt{2} + 2\sqrt{2} + 2\)
\(= 8 + 5\sqrt{2}\)
(c) Using the identity \((a+b)^2 = a^2 + 2ab + b^2\):
\(= (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{2}) + (\sqrt{2})^2\)
\(= 5 + 2\sqrt{10} + 2\)
\(= 7 + 2\sqrt{10}\)
In simple words: Apply standard algebraic expansion formulas and combine like terms.

Exam Tip: Memorize algebraic identities like \((a+b)(a-b) = a^2 - b^2\) to save time.

Question 3. Rationalise the following:
(a) \(\frac{1}{\sqrt{7}}\)
(b) \(\frac{1}{\sqrt{7} - \sqrt{6}}\)
(c) \(\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}}\)
(d) \(\frac{5}{4\sqrt{3} - 3\sqrt{2}}\)
Answer:
(a) Multiply numerator and denominator by \(\sqrt{7}\):
\(= \frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{7}}{7}\)
(b) Multiply numerator and denominator by the conjugate \(\sqrt{7} + \sqrt{6}\):
\(= \frac{1(\sqrt{7} + \sqrt{6})}{(\sqrt{7} - \sqrt{6})(\sqrt{7} + \sqrt{6})}\)
\(= \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}\)
(c) Multiply numerator and denominator by the conjugate \(\sqrt{10} - \sqrt{3}\):
\(= \frac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{(\sqrt{10} + \sqrt{3})(\sqrt{10} - \sqrt{3})}\)
\(= \frac{7\sqrt{30} - 7(3)}{10 - 3} = \frac{7\sqrt{30} - 21}{7} = \sqrt{30} - 3\)
(d) Multiply numerator and denominator by the conjugate \(4\sqrt{3} + 3\sqrt{2}\):
\(= \frac{5(4\sqrt{3} + 3\sqrt{2})}{(4\sqrt{3} - 3\sqrt{2})(4\sqrt{3} + 3\sqrt{2})}\)
\(= \frac{20\sqrt{3} + 15\sqrt{2}}{(4\sqrt{3})^2 - (3\sqrt{2})^2}\)
\(= \frac{20\sqrt{3} + 15\sqrt{2}}{48 - 18} = \frac{20\sqrt{3} + 15\sqrt{2}}{30} = \frac{4\sqrt{3} + 3\sqrt{2}}{6}\)
In simple words: To remove square roots from the denominator, multiply by its conjugate pair.

Exam Tip: The conjugate of \(a - b\) is \(a + b\), and their product is \(a^2 - b^2\).

Question 4. If \(x = 9 - 4\sqrt{5}\), find the value of \(x - \frac{1}{x}\).
Answer:
Given \(x = 9 - 4\sqrt{5}\).
First, find \(\frac{1}{x}\):
\(\frac{1}{x} = \frac{1}{9 - 4\sqrt{5}}\)
Rationalising the denominator:
\(\frac{1}{x} = \frac{9 + 4\sqrt{5}}{(9 - 4\sqrt{5})(9 + 4\sqrt{5})}\)
\(\frac{1}{x} = \frac{9 + 4\sqrt{5}}{9^2 - (4\sqrt{5})^2}\)
\(\frac{1}{x} = \frac{9 + 4\sqrt{5}}{81 - 80} = 9 + 4\sqrt{5}\)
Now, find \(x - \frac{1}{x}\):
\(x - \frac{1}{x} = (9 - 4\sqrt{5}) - (9 + 4\sqrt{5})\)
\(x - \frac{1}{x} = 9 - 4\sqrt{5} - 9 - 4\sqrt{5} = -8\sqrt{5}\)
In simple words: Find the reciprocal of x by rationalisation, then subtract the two values.

Exam Tip: Check if \(a^2 - b^2 = 1\) for terms like \(9^2 - (4\sqrt{5})^2 = 81 - 80 = 1\), which simplifies reciprocals instantly.

Question 5. If \(x = 3 + \sqrt{8}\), find the value of \(x^2 + \frac{1}{x^2}\).
Answer:
Given \(x = 3 + \sqrt{8}\).
First, find \(\frac{1}{x}\):
\(\frac{1}{x} = \frac{1}{3 + \sqrt{8}}\)
Rationalising the denominator:
\(\frac{1}{x} = \frac{3 - \sqrt{8}}{(3 + \sqrt{8})(3 - \sqrt{8})}\)
\(\frac{1}{x} = \frac{3 - \sqrt{8}}{9 - 8} = 3 - \sqrt{8}\)
Now, find \(x + \frac{1}{x}\):
\(x + \frac{1}{x} = (3 + \sqrt{8}) + (3 - \sqrt{8}) = 6\)
Squaring both sides:
\(\left(x + \frac{1}{x}\right)^2 = 6^2\)
\(x^2 + \frac{1}{x^2} + 2 = 36\)
\(x^2 + \frac{1}{x^2} = 36 - 2 = 34\)
In simple words: First find \(x + \frac{1}{x}\), then square the result to get \(x^2 + \frac{1}{x^2}\).

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

Question 6. Find the value of \(a\) and \(b\) if \(\frac{3 - \sqrt{5}}{3 + 2\sqrt{5}} = a\sqrt{5} - b\).
Answer:
Rationalise the L.H.S denominator \(3 + 2\sqrt{5}\) by multiplying numerator and denominator by its conjugate \(3 - 2\sqrt{5}\):
\(\text{L.H.S} = \frac{(3 - \sqrt{5})(3 - 2\sqrt{5})}{(3 + 2\sqrt{5})(3 - 2\sqrt{5})}\)
Numerator: \(3(3) - 3(2\sqrt{5}) - \sqrt{5}(3) + (\sqrt{5})(2\sqrt{5})\)
\(= 9 - 6\sqrt{5} - 3\sqrt{5} + 10 = 19 - 9\sqrt{5}\)
Denominator: \(3^2 - (2\sqrt{5})^2 = 9 - 20 = -11\)
So, L.H.S becomes:
\(= \frac{19 - 9\sqrt{5}}{-11} = -\frac{19}{11} + \frac{9}{11}\sqrt{5}\)
Comparing with R.H.S \(a\sqrt{5} - b\), which can be written as \(a\sqrt{5} + (-b)\):
\(a = \frac{9}{11}\)
\(-b = -\frac{19}{11} \implies b = \frac{19}{11}\)
In simple words: Rationalise the denominator, simplify, and match the rational and irrational parts on both sides.

Exam Tip: Equate the coefficient of \(\sqrt{5}\) and the constant term separately on both sides.

Question 7. Simplify \((32)^{\frac{1}{5}} \times (125)^{-\frac{1}{3}}\).
Answer:
Express 32 and 125 in exponential form:
\(32 = 2^5\)
\(125 = 5^3\)
Substitute into the expression:
\(= (2^5)^{\frac{1}{5}} \times (5^3)^{-\frac{1}{3}}\)
Using the law of exponents \((a^m)^n = a^{m \times n}\):
\(= 2^{5 \times \frac{1}{5}} \times 5^{3 \times (-\frac{1}{3})}\)
\(= 2^1 \times 5^{-1}\)
\(= 2 \times \frac{1}{5} = \frac{2}{5}\)
In simple words: Convert the base numbers into powers of prime numbers before applying exponent rules.

Exam Tip: Remember \(a^{-m} = \frac{1}{a^m}\).

Question 8. Find three rational numbers between -1 and 4.
Answer:
We can write -1 and 4 with a common denominator, or simply choose any three terminating decimals or fractions between them.
Let us choose fractions with denominator 2:
\(-1 = -\frac{2}{2}\)
\(4 = \frac{8}{2}\)
Three rational numbers between \(-\frac{2}{2}\) and \(\frac{8}{2}\) are:
\(0, 1, 2\) (or \(-\frac{1}{2}, 0, \frac{1}{2}\), etc.)
In simple words: Any numbers that can be written as fractions between -1 and 4 are valid answers.

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

Question 9. Find the value of \((729)^{-\frac{1}{6}}\).
Answer:
Find the prime factorization of 729:
\(729 = 3^6\)
Substitute into the expression:
\(= (3^6)^{-\frac{1}{6}}\)
Using the law of exponents \((a^m)^n = a^{m \times n}\):
\(= 3^{6 \times (-\frac{1}{6})}\)
\(= 3^{-1} = \frac{1}{3}\)
In simple words: Write 729 as \(3^6\) so the power of \(\frac{1}{6}\) cancels out the exponent 6.

Exam Tip: Always prime factorize large base numbers to simplify fractional exponents easily.

Question 10. Factorise \(2y^3 + y^2 - 2y - 1\).
Answer:
Let \(p(y) = 2y^3 + y^2 - 2y - 1\).
Group the terms in pairs to factor by grouping:
\(= (2y^3 + y^2) - (2y + 1)\)
Factor out \(y^2\) from the first group:
\(= y^2(2y + 1) - 1(2y + 1)\)
Now, factor out the common binomial \((2y + 1)\):
\(= (2y + 1)(y^2 - 1)\)
Since \(y^2 - 1 = (y - 1)(y + 1)\), we get:
\(= (2y + 1)(y - 1)(y + 1)\)
In simple words: Group terms together to find common factors, then use the difference of squares identity.

Exam Tip: Factorization by grouping is often the fastest method for cubic polynomials with four terms.

Question 11. Use suitable identity to evaluate \((103)^3\).
Answer:
Express 103 as \(100 + 3\):
\((103)^3 = (100 + 3)^3\)
Use the algebraic identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\):
Here \(a = 100\) and \(b = 3\):
\(= (100)^3 + (3)^3 + 3(100)(3)(100 + 3)\)
\(= 1000000 + 27 + 900(103)\)
\(= 1000000 + 27 + 92700\)
\(= 1092727\)
In simple words: Break 103 into \(100 + 3\) and apply the cube of a sum formula.

Exam Tip: Memorize cube identities: \((a \pm b)^3 = a^3 \pm b^3 \pm 3ab(a \pm b)\).

Question 12. Factorise: \(a^6 - b^6\).
Answer:
Write the expression as a difference of squares or cubes:
\(= (a^2)^3 - (b^2)^3\)
Use the difference of cubes identity \(X^3 - Y^3 = (X - Y)(X^2 + XY + Y^2)\) with \(X = a^2\) and \(Y = b^2\):
\(= (a^2 - b^2)((a^2)^2 + (a^2)(b^2) + (b^2)^2)\)
\(= (a - b)(a + b)(a^4 + a^2b^2 + b^4)\)
Further, \(a^4 + a^2b^2 + b^4\) can be factored by completing the square:
\(a^4 + 2a^2b^2 + b^4 - a^2b^2 = (a^2 + b^2)^2 - (ab)^2 = (a^2 + ab + b^2)(a^2 - ab + b^2)\)
Thus, the complete factorization is:
\(= (a - b)(a + b)(a^2 + ab + b^2)(a^2 - ab + b^2)\)
In simple words: Treat \(a^6 - b^6\) as a difference of cubes first, then apply difference of squares.

Exam Tip: Treating \(a^6 - b^6\) as \((a^3)^2 - (b^3)^2\) is an alternative method yielding the same result.

Question 13. If \(x = 3 - 2\sqrt{2}\), find \(x^3 - \frac{1}{x^3}\).
Answer:
Given \(x = 3 - 2\sqrt{2}\).
Find \(\frac{1}{x}\):
\(\frac{1}{x} = \frac{1}{3 - 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} = \frac{3 + 2\sqrt{2}}{9 - 8} = 3 + 2\sqrt{2}\)
Now, find \(x - \frac{1}{x}\):
\(x - \frac{1}{x} = (3 - 2\sqrt{2}) - (3 + 2\sqrt{2}) = -4\sqrt{2}\)
Now use the identity \(\left(x - \frac{1}{x}\right)^3 = x^3 - \frac{1}{x^3} - 3\left(x - \frac{1}{x}\right)\):
\((-4\sqrt{2})^3 = x^3 - \frac{1}{x^3} - 3(-4\sqrt{2})\)
\(-128\sqrt{2} = x^3 - \frac{1}{x^3} + 12\sqrt{2}\)
\(x^3 - \frac{1}{x^3} = -128\sqrt{2} - 12\sqrt{2} = -140\sqrt{2}\)
In simple words: Find \(x - \frac{1}{x}\) first, then cube both sides using the algebraic identity.

Exam Tip: Memorize the cubic expansion: \(x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right)\).

Question 14. Simplify and factorise \((a+b+c)^2 - (a-b-c)^2 + 4b^2 - 4c^2\).
Answer:
Let us use the difference of squares identity \(X^2 - Y^2 = (X - Y)(X + Y)\) for the first two terms:
\(X = a + b + c\)
\(Y = a - b - c\)
\((a+b+c)^2 - (a-b-c)^2 = [(a+b+c) - (a-b-c)][(a+b+c) + (a-b-c)]\)
\(= [a + b + c - a + b + c][a + b + c + a - b - c]\)
\(= [2b + 2c][2a] = 4a(b + c)\)
Now consider the remaining terms: \(4b^2 - 4c^2 = 4(b^2 - c^2) = 4(b - c)(b + c)\)
Combining all parts:
\(= 4a(b + c) + 4(b - c)(b + c)\)
Factor out the common term \(4(b + c)\):
\(= 4(b + c)[a + (b - c)]\)
\(= 4(b + c)(a + b - c)\)
In simple words: Apply the difference of squares formula to the first two terms, then factor out common terms.

Exam Tip: Grouping terms and looking for common binomial factors simplifies complex expressions quickly.

Question 15. If \(a + b + c = 6\) and \(ab + bc + ca = 11\), find the value of \(a^3 + b^3 + c^3 - 3abc\).
Answer:
We know the algebraic identity:
\(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\)
First, find \(a^2 + b^2 + c^2\) using the formula:
\((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
Substitute the given values:
\(6^2 = a^2 + b^2 + c^2 + 2(11)\)
\(36 = a^2 + b^2 + c^2 + 22\)
\(a^2 + b^2 + c^2 = 36 - 22 = 14\)
Now, substitute all values into the main identity:
\(a^3 + b^3 + c^3 - 3abc = (6)[(14) - (11)]\)
\(= 6 \times 3 = 18\)
In simple words: Use the formula for \(a^2 + b^2 + c^2\) first, then plug everything into the cube sum identity.

Exam Tip: This is a standard identity-based question frequently asked in board exams.

Question 16. The polynomials \(bx^3 + 3x^2 - 3\) and \(2x^3 - 5x + b\), when divided by \((x - 4)\), leave the remainder \(R_1\) and \(R_2\) respectively. Find the value of \(b\), if \(2R_1 - R_2 = 0\).
Answer:
By the Remainder Theorem, when a polynomial \(p(x)\) is divided by \(x - 4\), the remainder is \(p(4)\).
Let \(p_1(x) = bx^3 + 3x^2 - 3\).
\(R_1 = p_1(4) = b(4)^3 + 3(4)^2 - 3\)
\(R_1 = 64b + 3(16) - 3 = 64b + 48 - 3 = 64b + 45\)
Let \(p_2(x) = 2x^3 - 5x + b\).
\(R_2 = p_2(4) = 2(4)^3 - 5(4) + b\)
\(R_2 = 2(64) - 20 + b = 128 - 20 + b = 108 + b\)
Given condition: \(2R_1 - R_2 = 0\)
\(2(64b + 45) - (108 + b) = 0\)
\(128b + 90 - 108 - b = 0\)
\(127b - 18 = 0\)
\(127b = 18 \implies b = \frac{18}{127}\)
In simple words: Substitute \(x = 4\) into both polynomials to find remainders \(R_1\) and \(R_2\), then solve the given linear equation for b.

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

Question 17. Plot the points \(A(-2, -2)\), \(B(6, 0)\), \(C(0, 4)\), and \(D(-3, 2)\) on the graph paper. Draw figure ABCD and in which quadrant A and D lie.
Answer:
1. Quadrant identification:
- Point \(A(-2, -2)\) has both coordinates negative, so it lies in the Third Quadrant.
- Point \(D(-3, 2)\) has negative x-coordinate and positive y-coordinate, so it lies in the Second Quadrant.

2. Geometric Construction / Plotting Representation:
(The points \(A(-2,-2), B(6,0), C(0,4), D(-3,2)\) form quadrilateral ABCD on the Cartesian plane).

In simple words: Point A is in Quadrant III and Point D is in Quadrant II. Connect the plotted points to form quadrilateral ABCD.

Exam Tip: Remember the four quadrants: Q1(+,+), Q2(-,+), Q3(-,-), Q4(+,-).

Question 18. If two parallel lines are intersected by a transversal, then prove that bisectors of the interior angles form a rectangle.
Answer:
1. Given:
Line \(l \parallel m\), and transversal \(t\) intersects them at points \(P\) and \(Q\).
Let interior angles on the same side of the transversal be \(\angle BPQ\) and \(\angle PQC\) (where \(A-P-B\) and \(C-Q-D\) are line segments).
Let angle bisectors be \(PS\) and \(QS\) intersecting at point \(S\), forming a quadrilateral with interior angle bisectors.

2. Proof:
- Since \(l \parallel m\), the sum of interior angles on the same side of the transversal is supplementary: \(\angle BPQ + \angle PQC = 180^\circ\).
- Dividing by 2: \(\frac{1}{2}\angle BPQ + \frac{1}{2}\angle PQC = 90^\circ\).
- In triangle \(PSQ\), the sum of angles is \(180^\circ\):
\(\angle SPQ + \angle SQR + \angle PSQ = 180^\circ\)
\(90^\circ + \angle PSQ = 180^\circ \implies \angle PSQ = 90^\circ\).
- Similarly, all interior angles of the quadrilateral formed by angle bisectors are equal to \(90^\circ\).
- Therefore, a quadrilateral with all interior angles equal to \(90^\circ\) is a rectangle.

3. Geometric Diagram:

In simple words: The angle bisectors of interior angles between parallel lines always intersect at right angles, forming a rectangle.

Exam Tip: Use the property that co-interior angles sum to \(180^\circ\) when lines are parallel.