CBSE Class 9 Mathematics Polynomials Worksheet Set 06

Question 1. (-2 - \sqrt{3})(-2 + \sqrt{3}) when simplified is:
(a) positive and irrational
(b) positive and rational
(c) negative and irrational
(d) negative and rational
Answer: (b) positive and rational
Applying the algebraic identity \( (a - b)(a + b) = a^2 - b^2 \) where \( a = -2 \) and \( b = \sqrt{3} \):
\( (-2)^2 - (\sqrt{3})^2 = 4 - 3 = 1 \)
Since 1 is an integer greater than zero, it is both positive and rational.
In simple words: Multiplying the terms eliminates the square root and leaves 1, which is a positive rational number.

Exam Tip: Using the identity \( (a - b)(a + b) = a^2 - b^2 \) avoids long expansion and removes the radical directly.

Question 2. Two rational numbers between \(\frac{1}{2}\) and \(\frac{5}{3}\) are:
(a) 1/6 and 2/6
(b) 1/2 and 2/1
(c) 5/6 and 7/6
(d) 2/3 and 4/3
Answer: (c) 5/6 and 7/6
Expressing both fractions with a common denominator of 6 gives:
\( \frac{1}{2} = \frac{3}{6} \) and \( \frac{5}{3} = \frac{10}{6} \)
Any fraction with denominator 6 having a numerator strictly between 3 and 10 lies in this interval. Both \( \frac{5}{6} \) and \( \frac{7}{6} \) satisfy this condition.
In simple words: Converting to the common denominator 6 shows that 5/6 and 7/6 sit comfortably between 3/6 and 10/6.

Exam Tip: Convert all given fractions and choices to a common denominator to compare values instantly.

Question 3. The sum of the digits of a number is subtracted from the number, the resulting number is always divisible by:
(a) 2
(b) 5
(c) 8
(d) 9
Answer: (d) 9
Let a two-digit number be represented as \( 10a + b \). Subtracting the sum of its digits \( (a + b) \) gives:
\( (10a + b) - (a + b) = 9a \)
Because \( 9a \) is an exact multiple of 9, the resulting difference is always divisible by 9. This property holds generally for numbers of any number of digits.
In simple words: Taking away the digit sum cancels out the units and leaves multiples of 9, making the result always divisible by 9.

Exam Tip: Express numbers in expanded place-value form like \( 10a + b \) or \( 100a + 10b + c \) to prove divisibility rules easily.

Question 4. (6 + \sqrt{27}) - (3 + \sqrt{3}) + (1 - 2\sqrt{3}) when simplified is:
(a) positive and irrational
(b) negative and rational
(c) positive and rational
(d) negative and irrational
Answer: (c) positive and rational
Simplifying the surd \( \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \), the expression becomes:
\( (6 + 3\sqrt{3}) - 3 - \sqrt{3} + 1 - 2\sqrt{3} \)
Grouping the rational terms and irrational terms separately:
\( (6 - 3 + 1) + (3\sqrt{3} - \sqrt{3} - 2\sqrt{3}) = 4 + 0\sqrt{3} = 4 \)
The final simplified value is 4, which is positive and rational.
In simple words: The square root terms completely cancel each other out, leaving just positive 4.

Exam Tip: Always reduce radicals to their simplest form first so you can identify and cancel like terms.

Question 5. Two rational numbers between 1/5 and 4/5 are:
(a) 1 and 3/5
(b) 2/5 and 3/5
(c) 1/2 and 2/1
(d) 3/5 and 6/5
Answer: (b) 2/5 and 3/5
Since both fractions share the common denominator 5, any fraction with denominator 5 and numerator between 1 and 4 lies in the interval. The values \( \frac{2}{5} \) and \( \frac{3}{5} \) both lie strictly between \( \frac{1}{5} \) and \( \frac{4}{5} \).
In simple words: Between 1 fifth and 4 fifths sit 2 fifths and 3 fifths.

Exam Tip: Check whether the numerators of the choices lie strictly inside the boundaries when denominators match.

Question 6. Add 5\sqrt{2} + 3\sqrt{3} and \sqrt{2} - 5\sqrt{3} .
(a) \( 7\sqrt{2} - 2\sqrt{3} \)
(b) \( 6\sqrt{2} - 2\sqrt{3} \)
(c) \( 6\sqrt{2} - 8\sqrt{3} \)
(d) \( 6\sqrt{2} + 8\sqrt{3} \)
Answer: (b) \( 6\sqrt{2} - 2\sqrt{3} \)
Combining like radicals together:
\( (5\sqrt{2} + \sqrt{2}) + (3\sqrt{3} - 5\sqrt{3}) = (5 + 1)\sqrt{2} + (3 - 5)\sqrt{3} = 6\sqrt{2} - 2\sqrt{3} \)
In simple words: Group the root-2 terms together and the root-3 terms together, just like combining like variables in algebra.

Exam Tip: Treat identical square roots like variables: add coefficients of \( \sqrt{2} \) together and coefficients of \( \sqrt{3} \) together.

Question 7. A number is an irrational if and only if its decimal representation is:
(a) non- terminating
(b) non - terminating and repeating
(c) non - terminating and non- repeating
(d) terminating
Answer: (c) non - terminating and non- repeating
Rational numbers have decimals that either stop or repeat periodically. By definition, a real number is irrational if and only if its decimal expansion never terminates and never forms an endlessly repeating pattern.
In simple words: An irrational decimal goes on forever without any repeating pattern.

Exam Tip: Remember: terminating or repeating decimals are rational; non-terminating and non-repeating decimals are irrational.

Question 8. The value of \(\sqrt[4]{(64)^{-2}}\) is:
(a) 1/8
(b) 1/2
(c) 8
(d) 1/64
Answer: (a) 1/8
Writing 64 as a power of 2, we have \( 64 = 2^6 \). Therefore:
\( (64)^{-2} = (2^6)^{-2} = 2^{-12} \)
Taking the fourth root:
\( \sqrt[4]{2^{-12}} = (2^{-12})^{\frac{1}{4}} = 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \)
In simple words: Change 64 to 2 to the power 6, multiply the powers together to get 2 to the power -3, which equals 1/8.

Exam Tip: Break bases down into prime factors before applying exponent rules like \( (a^m)^n = a^{mn} \).

Question 9. (5 + \sqrt{5})(5 - \sqrt{5}) on simplification gives:
(a) 20
(b) \( 2\sqrt{5} \)
(c) 10
(d) 25
Answer: (a) 20
Using the difference of squares identity \( (a + b)(a - b) = a^2 - b^2 \):
\( (5)^2 - (\sqrt{5})^2 = 25 - 5 = 20 \)
In simple words: Square the first number (25), square the second number (5), and subtract them to get 20.

Exam Tip: The product of conjugate surds \( (a + \sqrt{b})(a - \sqrt{b}) \) always simplifies to a rational number \( a^2 - b \).

Question 10. \(\sqrt[3]{\frac{54}{250}}\) is:
(a) 9/25
(b) 3/5
(c) 27/125
(d) \(\sqrt[3]{\frac{2}{5}}\)
Answer: (b) 3/5
Reducing the fraction inside the radical by dividing numerator and denominator by 2 gives:
\( \frac{54}{250} = \frac{27}{125} \)
Taking the cube root of both numerator and denominator:
\( \sqrt[3]{\frac{27}{125}} = \frac{\sqrt[3]{27}}{\sqrt[3]{125}} = \frac{3}{5} \)
In simple words: Simplify the fraction inside to 27/125, then find the cube root of 27 (which is 3) and 125 (which is 5).

Exam Tip: Always reduce the fraction inside a root first to see if perfect squares or cubes appear.

Question 11. The value of \(7^{\frac{1}{2}} 8^{\frac{1}{2}}\) is :
(a) \( 28^{\frac{1}{2}} \)
(b) \( 56^{\frac{1}{2}} \)
(c) \( 14^{\frac{1}{2}} \)
(d) \( 42^{\frac{1}{2}} \)
Answer: (b) \( 56^{\frac{1}{2}} \)
By the power rule \( a^m \cdot b^m = (ab)^m \):
\( 7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}} = (7 \times 8)^{\frac{1}{2}} = 56^{\frac{1}{2}} \)
In simple words: Since both numbers share the same power of 1/2, simply multiply 7 and 8 inside the brackets.

Exam Tip: When the exponents are identical, multiply the base numbers together under that single exponent.

Question 12. The value of \(\sqrt[4]{\sqrt[3]{2^2}}\) equal to :
(a) \( 2^{-\frac{1}{6}} \)
(b) \( 2^{-6} \)
(c) \( 2^{\frac{1}{6}} \)
(d) \( 2^6 \)
Answer: (c) \( 2^{\frac{1}{6}} \)
Expressing nested radical signs in fractional exponent notation:
\( \sqrt[4]{\sqrt[3]{2^2}} = \left(\left(2^2\right)^{\frac{1}{3}}\right)^{\frac{1}{4}} = 2^{2 \times \frac{1}{3} \times \frac{1}{4}} = 2^{\frac{2}{12}} = 2^{\frac{1}{6}} \)
In simple words: Turn the roots into fraction powers and multiply them: 2 times 1/3 times 1/4 gives 1/6.

Exam Tip: Combine successive roots using \( \sqrt[m]{\sqrt[n]{a}} = a^{\frac{1}{mn}} \) to simplify quickly.

Question 13. When \(15\sqrt{15}\) is divided by \(3\sqrt{3}\) the quotient is:
(a) \( 5\sqrt{3} \)
(b) \( 5\sqrt{5} \)
(c) \( 5\sqrt{5} \)
(d) \( 3\sqrt{3} \)
Answer: (b) \( 5\sqrt{5} \)
Dividing the rational parts and the radical parts separately:
\( \frac{15\sqrt{15}}{3\sqrt{3}} = \left(\frac{15}{3}\right) \times \sqrt{\frac{15}{3}} = 5 \times \sqrt{5} = 5\sqrt{5} \)
In simple words: Divide 15 by 3 to get 5, and divide root 15 by root 3 to get root 5.

Exam Tip: Separate terms into \( \frac{a}{c} \cdot \sqrt{\frac{b}{d}} \) to simplify rational and radical components systematically.

Question 14. Which of the following number is irrational?
(a) \( \sqrt{16} \)
(b) \( (3 - \sqrt{3})(3 + \sqrt{3}) \)
(c) \( \sqrt{5} + 3 \)
(d) \( -\sqrt{25} \)
Answer: (c) \( \sqrt{5} + 3 \)
Evaluating each option:
(a) \( \sqrt{16} = 4 \) (rational)
(b) \( (3 - \sqrt{3})(3 + \sqrt{3}) = 3^2 - 3 = 6 \) (rational)
(c) The sum of an irrational number (\(\sqrt{5}\)) and a rational number (3) is always irrational.
(d) \( -\sqrt{25} = -5 \) (rational)
In simple words: Adding a rational number to a non-repeating square root leaves the number irrational.

Exam Tip: The sum or difference of a rational number and an irrational number is always irrational.

Question 15. The value of \(\frac{2^0 + 7^0}{5^0}\) is:
(a) 2
(b) 0
(c) 9/5
(d) 1/5
Answer: (a) 2
Since any non-zero real number raised to the power of 0 equals 1:
\( 2^0 = 1, \quad 7^0 = 1, \quad 5^0 = 1 \)
Substituting these values into the fraction:
\( \frac{1 + 1}{1} = \frac{2}{1} = 2 \)
In simple words: Any non-zero number to power zero is 1, so the expression simplifies to (1 + 1) / 1 = 2.

Exam Tip: Never mistake \( a^0 \) for 0; for any non-zero value, \( a^0 = 1 \).

Question 16. (5 + \sqrt{8}) + (3 - \sqrt{2}) - (\sqrt{2} - 6) when simplified is:
(a) positive and irrational
(b) negative and irrational
(c) positive and rational
(d) negative and rational
Answer: (c) positive and rational
Expressing \( \sqrt{8} \) in simplest radical form: \( \sqrt{8} = 2\sqrt{2} \).
Expanding the brackets and tracking signs:
\( 5 + 2\sqrt{2} + 3 - \sqrt{2} - \sqrt{2} + 6 \)
Combining like terms:
\( (5 + 3 + 6) + (2\sqrt{2} - \sqrt{2} - \sqrt{2}) = 14 + 0 = 14 \)
The result is 14, which is positive and rational.
In simple words: The root terms cancel out completely to zero, leaving a positive whole number 14.

Exam Tip: Take care with the minus sign in front of \( (\sqrt{2} - 6) \); distributing it gives \( -\sqrt{2} + 6 \).

Question 17. An irrational number between \(\frac{5}{7}\) and \(\frac{7}{9}\) is:
(a) 0.75
(b) \( \sqrt{6} \)
(c) 0.750750075000…
(d) 0.7512
Answer: (c) 0.750750075000…
Converting the fractions to decimals:
\( \frac{5}{7} \approx 0.714285\dots \) and \( \frac{7}{9} \approx 0.777777\dots \)
An irrational number must have a non-terminating and non-repeating decimal expansion. The number \( 0.750750075000\dots \) lies within the interval \( (0.714\dots, 0.777\dots) \) and has a non-repeating pattern, confirming it is irrational.
In simple words: The decimal 0.750750075000... lies between 0.714 and 0.777 and never repeats, making it irrational.

Exam Tip: Terminating decimals like 0.75 and 0.7512 are rational; look for non-repeating ellipsis patterns for irrational numbers.

Question 18. Simplified value of \((25)^{\frac{1}{3}} \times (5)^{\frac{1}{3}}\) is:
(a) 25
(b) 3
(c) 1
(d) 5
Answer: (d) 5
Using the exponent law \( a^m \times b^m = (ab)^m \):
\( (25 \times 5)^{\frac{1}{3}} = (125)^{\frac{1}{3}} \)
Since \( 125 = 5^3 \), we find:
\( (5^3)^{\frac{1}{3}} = 5 \)
In simple words: Multiply 25 and 5 to get 125, then find the cube root of 125, which is 5.

Exam Tip: Combining bases under a common fractional power helps reveal perfect powers like \( 125 = 5^3 \).

Question 19. Which of the following is an irrational number?
(a) 2.2
(b) \(\pi\)
(c) 3.763
(d) 3.763
Answer: (b) \(\pi\)
The constant \( \pi \) represents the ratio of a circle's circumference to its diameter and is proven to be non-terminating and non-repeating, making it an irrational number. The other choices are terminating decimals and therefore rational.
In simple words: Pi cannot be written as a simple fraction, so it is an irrational number.

Exam Tip: Note that \( \frac{22}{7} \) is only an approximation of \( \pi \); the true value of \( \pi \) is irrational.

Question 20. Which of the following is an irrational number?
(a) \(\sqrt{23}\)
(b) \(\sqrt{225}\)
(c) 0.3796
(d) 7.478
Answer: (a) \(\sqrt{23}\)
Because 23 is a prime number and not a perfect square, its square root \( \sqrt{23} \) is irrational. By comparison, \( \sqrt{225} = 15 \) is rational, and options (c) and (d) are rational terminating decimals.
In simple words: Since 23 is not a square number, its square root is an irrational number.

Exam Tip: The square root of any positive integer that is not a perfect square is always irrational.

Question 21. Which of the following is the value of \((\sqrt{11} - \sqrt{7})(\sqrt{11} + \sqrt{7})\)
(a) -4
(b) 4
(c) \(\sqrt{11}\)
(d) \(\sqrt{7}\)
Answer: (b) 4
Using the identity \( (a - b)(a + b) = a^2 - b^2 \):
\( (\sqrt{11})^2 - (\sqrt{7})^2 = 11 - 7 = 4 \)
In simple words: The square of root 11 is 11, the square of root 7 is 7, and 11 minus 7 gives 4.

Exam Tip: Apply \( (\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) = a - b \) directly for quick computation.

Question 22. Which of the following is a rational number?
(a) \( 1 + \sqrt{3} \)
(b) \(\pi\)
(c) \( 2\sqrt{3} \)
(d) 0
Answer: (d) 0
The number 0 can be written in the form \( \frac{p}{q} \) with integers \( p = 0 \) and \( q = 1 \) where \( q \neq 0 \), which fits the definition of a rational number. The other three options are irrational numbers.
In simple words: Zero can be written as 0/1, which makes it a rational number.

Exam Tip: Zero is a rational number because it can be expressed as \( \frac{0}{q} \) for any non-zero integer \( q \).

Question 23. Simplified value of \((16)^{-\frac{1}{4}} \times \sqrt[4]{16}\) is:
(a) 16
(b) 4
(c) 1
(d) 0
Answer: (c) 1
Rewriting \( \sqrt[4]{16} \) in exponential notation gives \( (16)^{\frac{1}{4}} \). Combining terms using the product rule \( a^m \times a^n = a^{m+n} \):
\( (16)^{-\frac{1}{4}} \times (16)^{\frac{1}{4}} = (16)^{-\frac{1}{4} + \frac{1}{4}} = (16)^0 = 1 \)
In simple words: Multiplying numbers with opposite powers adds them to zero, and any non-zero base to power zero equals 1.

Exam Tip: Recognizing inverse quantities \( a^{-n} \cdot a^n = a^0 = 1 \) saves calculation time.

Question 24. Value of \(\sqrt{(3^{-2})}\) is:
(a) 1/9
(b) 9
(c) -3
(d) 1/3
Answer: (d) 1/3
Expressing the square root as an exponent of \( \frac{1}{2} \):
\( \left(3^{-2}\right)^{\frac{1}{2}} = 3^{-2 \times \frac{1}{2}} = 3^{-1} = \frac{1}{3} \)
In simple words: The square root cuts the power -2 in half to -1, and 3 to the power -1 equals 1/3.

Exam Tip: A negative exponent indicates a reciprocal: \( a^{-1} = \frac{1}{a} \).

Question 25. Zero of the polynomial p(x) where p(x) = ax, a \neq 0 is:
(a) 1
(b) a
(c) 0
(d) 1/a
Answer: (c) 0
To find the zero of the polynomial, set \( p(x) = 0 \):
\( ax = 0 \)
Since it is given that \( a \neq 0 \), dividing both sides by \( a \) yields:
\( x = \frac{0}{a} = 0 \)
In simple words: Setting ax equal to zero means x must be zero because a is not zero.

Exam Tip: To find the zero of any polynomial \( p(x) \), set \( p(x) = 0 \) and solve for \( x \).

Question 26. If \(\sqrt{3} = 1.732\) and \(\sqrt{2} = 1.414\), the value of \(\frac{1}{\sqrt{3} - \sqrt{2}}\) is:
(a) 0.318
(b) 3.146
(c) \(\frac{1}{3.146}\)
(d) \( \sqrt{1.732} - \sqrt{1.414} \)
Answer: (b) 3.146
Rationalizing the denominator by multiplying top and bottom by the conjugate \( \sqrt{3} + \sqrt{2} \):
\( \frac{1}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{\sqrt{3} + \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2} = \frac{\sqrt{3} + \sqrt{2}}{3 - 2} = \sqrt{3} + \sqrt{2} \)
Substituting the given approximations:
\( 1.732 + 1.414 = 3.146 \)
In simple words: Rationalizing removes the denominator completely, leaving the sum of the two roots: 1.732 + 1.414 = 3.146.

Exam Tip: Never substitute decimal approximations into the denominator directly; always rationalize first.

Question 27. Which one of the following is an irrational number?
(a) 0.14
(b) 0.1416
(c) 0.1416
(d) 0.4014001400014…..
Answer: (d) 0.4014001400014…..
An irrational number exhibits an infinite, non-repeating decimal expansion. The pattern in option (d) introduces an increasing number of zeros between the blocks of 14, meaning no fixed sequence repeats indefinitely, which defines an irrational number.
In simple words: The number in (d) keeps changing its pattern of zeros and never ends, so it is irrational.

Exam Tip: A decimal where the number of intervening zeros grows continuously is a standard model of a non-repeating, irrational number.

Question 28. \(\pi\) is:
(a) a rational number
(b) an integer
(c) an irrational number
(d) a whole number
Answer: (c) an irrational number
The mathematical constant \( \pi \) cannot be written as the ratio of two integers. Its decimal value continues infinitely without periodic recurrence, making it an irrational number.
In simple words: Pi is an irrational number because its decimal digits go on forever without a repeating block.

Exam Tip: While \( \frac{22}{7} \) is a rational approximation used for calculations, \( \pi \) itself is strictly irrational.

Question 29. The decimal form of \(\frac{56}{100}\) is:
(a) 0.56
(b) 0.056
(c) 0.0056
(d) 5.6
Answer: (a) 0.56
Dividing an integer by 100 shifts the decimal point two positions to the left:
\( \frac{56}{100} = 0.56 \)
In simple words: Dividing by 100 moves the decimal point two places left, giving 0.56.

Exam Tip: Count the number of zeros in the power of 10 to shift the decimal point to the left by that exact number of places.

Question 30. The decimal expansion of \(\sqrt{2}\) is:
(a) finite decimal
(b) 1.4121
(c) non- terminating recurring
(d) non-terminating non- recurring
Answer: (d) non-terminating non- recurring
Since 2 is not a square of any rational number, \( \sqrt{2} \) is irrational. By the fundamental properties of real numbers, every irrational number has a non-terminating and non-recurring decimal expansion.
In simple words: Square root of 2 is an irrational number, so its decimal never ends and never repeats.

Exam Tip: The decimal expansion of every irrational number is uniquely characterized as non-terminating and non-recurring.

Question 31. Simplify: \(\frac{13^{\frac{1}{5}}}{13^{\frac{1}{3}}}\)
(a) \( 13^{\frac{2}{15}} \)
(b) \( 13^{\frac{8}{15}} \)
(c) \( 13^{\frac{1}{3}} \)
(d) \( 13^{-\frac{2}{15}} \)
Answer: (d) \( 13^{-\frac{2}{15}} \)
Applying the quotient law of exponents \( \frac{a^m}{a^n} = a^{m - n} \):
\( 13^{\frac{1}{5} - \frac{1}{3}} \)
Finding a common denominator for the exponents:
\( \frac{1}{5} - \frac{1}{3} = \frac{3 - 5}{15} = -\frac{2}{15} \)
Thus, the simplified result is \( 13^{-\frac{2}{15}} \).
In simple words: Subtract the bottom power 1/3 from the top power 1/5 to get -2/15.

Exam Tip: Remember to subtract the denominator's exponent from the numerator's exponent: \( m - n \), not the other way around.

Question 32. \(\frac{p}{q}\) form of the number 0.3 is:
(a) 3/10
(b) 3/100
(c) 1/3
(d) 1/2
Answer: (a) 3/10
The decimal 0.3 has one digit following the decimal point, representing three tenths:
\( 0.3 = \frac{3}{10} \)
In simple words: 0.3 means 3 tenths, which is written as the fraction 3/10.

Exam Tip: Note whether the decimal has a repeating bar or not; terminating 0.3 is \( \frac{3}{10} \), whereas recurring \( 0.\overline{3} \) equals \( \frac{1}{3} \).

Question 33. The simplest rationalization factor of \(\sqrt{50}\) is:
(a) \( 5\sqrt{2} \)
(b) \( \sqrt{2} \)
(c) 50
(d) \( \sqrt{50} \)
Answer: (b) \( \sqrt{2} \)
Factoring out square factors from the radicand:
\( \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \)
To convert \( 5\sqrt{2} \) into a rational number, it needs only to be multiplied by \( \sqrt{2} \), which gives \( 5 \times 2 = 10 \). Hence, the simplest rationalizing factor is \( \sqrt{2} \).
In simple words: Simplifying root 50 gives 5 root 2, so multiplying by just root 2 makes it a whole number.

Exam Tip: Always reduce the surd to its simplest form \( a\sqrt{b} \) before finding the lowest rationalizing factor \( \sqrt{b} \).

Question 34. The value of \((125)^{-\frac{1}{3}}\) is:
(a) 25
(b) 1/5
(c) 5
(d) 1/25
Answer: (b) 1/5
Expressing 125 as a power of 5: \( 125 = 5^3 \).
Applying the power of a power rule:
\( (125)^{-\frac{1}{3}} = \left(5^3\right)^{-\frac{1}{3}} = 5^{3 \times \left(-\frac{1}{3}\right)} = 5^{-1} = \frac{1}{5} \)
In simple words: The cube root of 125 is 5, and the negative power flips it into 1/5.

Exam Tip: Deal with the fraction power first to find the root, then invert the base for the negative sign.

Question 35. The product of Quotient of a non-zero rational number with an irrational number is:
(a) Irrational number
(b) Rational number
(c) Whole number
(d) Natural number
Answer: (a) Irrational number
Multiplying or dividing any non-zero rational number by an irrational number always yields an irrational result.
In simple words: Combining a non-zero rational number and an irrational number by multiplication or division always gives an irrational number.

Exam Tip: Remember that zero is the only rational number whose product with an irrational number produces a rational result (0).

Question 36. The value of \(\sqrt{20} \times \sqrt{5}\) is
(a) 10
(b) \( 2\sqrt{5} \)
(c) \( 20\sqrt{5} \)
(d) \( 4\sqrt{5} \)
Answer: (a) 10
Using the multiplication property of radicals \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \):
\( \sqrt{20 \times 5} = \sqrt{100} = 10 \)
In simple words: Multiply 20 and 5 under the square root to get root 100, which equals 10.

Exam Tip: Combine numbers under a single radical sign before extracting the root.

Question 37. Which of the following is irrational number?
(a) 0.15
(b) 0.1516
(c) 0.1516
(d) 0.501500150001---
Answer: (d) 0.501500150001---
Options (a), (b), and (c) represent terminating decimals, all of which are rational. Option (d) contains an infinite, non-repeating pattern (with an increasing count of zeros), making it an irrational number.
In simple words: A decimal that never terminates and never repeats regular blocks is an irrational number.

Exam Tip: Look for expanding sequences of digits like 0, 00, 000 between numbers to identify irrational decimals.

Question 38. If \(x = 2 + \sqrt{3}\), then \(\left(x + \frac{1}{x}\right)\) equals to:
(a) \( -2\sqrt{3} \)
(b) 2
(c) 4
(d) \( 4 - 2\sqrt{3} \)
Answer: (c) 4
Finding the reciprocal \( \frac{1}{x} \) by rationalizing the denominator:
\( \frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3} \)
Now, computing the sum:
\( x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \)
In simple words: 1/x is the conjugate 2 - root 3; adding x and 1/x cancels the roots and leaves 2 + 2 = 4.

Exam Tip: For expressions of the form \( x = a + \sqrt{b} \) where \( a^2 - b = 1 \), the reciprocal is simply \( \frac{1}{x} = a - \sqrt{b} \).

Question 39. A rational number lying between \(\sqrt{2}\) and \(\sqrt{3}\) is:
(a) \(\frac{\sqrt{2} + \sqrt{3}}{2}\)
(b) \(\sqrt{6}\)
(c) 1.6
(d) -1
Answer: (c) 1.6
Approximating the irrational values gives \( \sqrt{2} \approx 1.414 \) and \( \sqrt{3} \approx 1.732 \).
The number 1.6 can be written as \( \frac{16}{10} = \frac{8}{5} \), which is rational, and it lies strictly between 1.414 and 1.732. Options (a) and (b) are irrational numbers.
In simple words: 1.6 is a terminating fraction that sits right between 1.414 and 1.732.

Exam Tip: Make sure the selected option is not just inside the range, but is also explicitly rational as asked.

Question 40. The value of \(\sqrt[3]{216} - \sqrt[3]{125}\) is:
(a) 1
(b) 0
(c) 2
(d) -1
Answer: (a) 1
Calculating the cube roots of the perfect cubes:
\( \sqrt[3]{216} = 6 \) (since \( 6^3 = 216 \))
\( \sqrt[3]{125} = 5 \) (since \( 5^3 = 125 \))
Subtracting these results:
\( 6 - 5 = 1 \)
In simple words: The cube root of 216 is 6, the cube root of 125 is 5, and 6 minus 5 is 1.

Exam Tip: Memorize cubes of numbers from 1 to 10 to evaluate cube root expressions rapidly.

Question 41. Which of the following is a rational number?
(a) \(\sqrt{5}\)
(b) \(\pi\)
(c) 0.101001000100001…..
(d) 0.853853853…….
Answer: (d) 0.853853853…….
The decimal in option (d) repeats the group of digits 853 endlessly (\( 0.\overline{853} \)). Every non-terminating repeating decimal can be converted into a fraction of integers, making it a rational number. The other three choices are non-repeating and irrational.
In simple words: A decimal that repeats the same pattern over and over can always be written as a fraction.

Exam Tip: Repeating decimal expansions always represent rational numbers, while non-repeating ones represent irrational numbers.

Question 42. A rational number between -3 and 3 is
(a) 0
(b) -4.3
(c) -3.4
(d) 1.101100110001….
Answer: (a) 0
0 is an integer, so it is a rational number (\( \frac{0}{1} \)), and it lies between -3 and 3. The numbers -4.3 and -3.4 fall outside the range, and (d) is an irrational number.
In simple words: 0 is a rational number that lies right in the middle of -3 and 3.

Exam Tip: Check both conditions: the number must be rational, and it must lie strictly inside the specified interval.

Question 43. Which of the following is an irrational number?
(a) 3.3
(b) 3.763
(c) 3.763
(d) 3.101100110001…..
Answer: (d) 3.101100110001…..
Option (d) has a non-terminating and non-repeating decimal pattern, which fits the definition of an irrational number. The other options are terminating rational decimals.
In simple words: The decimal in (d) never stops and never repeats, which makes it an irrational number.

Exam Tip: A decimal that shows dots (...) at the end without a repeating bar must be examined for a changing digit pattern to confirm it is irrational.

Question 44. The factors of (2a – b)3 + (b – 2c)3 + 8(c – a)3 is:
(a) (2a – b)(b – 2c)(c –a)
(b) 3(2a – b)(b – 2c)(c –a)
(c) 6(2a –b)(b – 2c)(c –a)
(d) 2a × b × 2c
Answer: (c) 6(2a –b)(b – 2c)(c –a)
Notice that \( 8(c - a)^3 = [2(c - a)]^3 = (2c - 2a)^3 \).
Let \( x = 2a - b \), \( y = b - 2c \), and \( z = 2c - 2a \).
Summing the three variables:
\( x + y + z = (2a - b) + (b - 2c) + (2c - 2a) = 0 \)
When \( x + y + z = 0 \), the conditional identity gives \( x^3 + y^3 + z^3 = 3xyz \).
Substituting back:
\( 3(2a - b)(b - 2c)(2c - 2a) = 3(2a - b)(b - 2c) \cdot 2(c - a) = 6(2a - b)(b - 2c)(c - a) \)
In simple words: Since the three terms add up to zero, their cubes sum to 3 times their product, which simplifies to 6(2a - b)(b - 2c)(c - a).

Exam Tip: Remember to write \( 8(c - a)^3 \) as \( [2(c - a)]^3 \) before testing the condition \( x + y + z = 0 \).

Question 45. In which of the following (x + 2) is a factor?
(a) 4x3 – 13x + 6
(b) x3 + x2 + x + 4
(c) 4x3 + 13x – 25
(d) -2x3 + x2 – x – 19
Answer: (a) 4x3 – 13x + 6
By the Factor Theorem, \( (x + 2) \) is a factor of a polynomial \( p(x) \) if and only if \( p(-2) = 0 \).
Testing the polynomial in option (a), \( p(x) = 4x^3 - 13x + 6 \):
\( p(-2) = 4(-2)^3 - 13(-2) + 6 = 4(-8) + 26 + 6 = -32 + 32 = 0 \)
Since the remainder is 0, \( (x + 2) \) is a factor of \( 4x^3 - 13x + 6 \).
In simple words: Putting -2 into the first polynomial gives 0, so x + 2 is an exact factor.

Exam Tip: Set the divisor to zero (\( x + 2 = 0 \implies x = -2 \)) and substitute this value into each choice.

Question 46. Which of the following is a binomial in y?
(a) 2y + 3y
(b) 2y + 1
(c) \(\sqrt{y} + \sqrt{2}y\)
(d) \(y\sqrt{y} + 1\)
Answer: (b) 2y + 1
A binomial is a polynomial consisting of exactly two non-zero terms with non-negative integer exponents. In (a), \( 2y + 3y = 5y \) reduces to a single term (monomial). In (c) and (d), the variable has fractional exponents (\( y^{\frac{1}{2}} \) and \( y^{\frac{3}{2}} \)), so they are not polynomials. Option (b) has two terms with whole-number powers of \( y \).
In simple words: A binomial must be a polynomial with two terms, and 2y + 1 is the only one fitting this description.

Exam Tip: Simplify like terms first (e.g. \( 2y + 3y = 5y \)) before classifying an expression by its number of terms.

Question 47. Which of the following polynomials has -3 as a zero?
(a) (x – 3)
(b) x2 – 9
(c) x2 – 3x
(d) x2 + 3
Answer: (b) x2 – 9
A number \( c \) is a zero of \( p(x) \) if \( p(c) = 0 \).
Substituting \( x = -3 \) into \( p(x) = x^2 - 9 \):
\( (-3)^2 - 9 = 9 - 9 = 0 \)
Thus, -3 is a zero of \( x^2 - 9 \).
In simple words: Squaring -3 gives positive 9, and subtracting 9 gives zero.

Exam Tip: Be careful with signs: squaring a negative number always yields a positive value: \( (-3)^2 = 9 \).

Question 48. Which of the following is a polynomial in x?
(a) \( x + \frac{1}{x} \)
(b) \( x^2 + \sqrt{x} \)
(c) \( x + \sqrt{2}x^2 + 1 \)
(d) \( \sqrt{3x} + 1 \)
Answer: (c) \( x + \sqrt{2}x^2 + 1 \)
An algebraic expression is a polynomial in \( x \) if all exponents of the variable \( x \) are non-negative integers. In option (c), the powers of \( x \) are 1 and 2, which are whole numbers (the square root applies only to the coefficient \( \sqrt{2} \)). In the other options, \( x \) has negative or fractional powers (\( x^{-1} \) or \( x^{\frac{1}{2}} \)).
In simple words: A polynomial can have square roots on its numbers, but the variable x must have positive whole-number powers.

Exam Tip: Check the exponent of the variable, not the coefficients; real numbers like \( \sqrt{2} \) are perfectly valid coefficients.

Question 49. The remainder when x2 + 2x + 1 is divided by (x + 1) is
(a) 4
(b) 0
(c) 1
(d) -2
Answer: (b) 0
By the Remainder Theorem, dividing \( p(x) = x^2 + 2x + 1 \) by \( (x + 1) \) yields remainder \( p(-1) \):
\( p(-1) = (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0 \)
Alternatively, notice that \( x^2 + 2x + 1 = (x + 1)^2 \), which is divisible by \( (x + 1) \) with zero remainder.
In simple words: The expression is the exact square of (x + 1), so it divides with no remainder left over.

Exam Tip: Recognizing standard expansions like \( (x + 1)^2 = x^2 + 2x + 1 \) can answer remainder questions immediately.

Question 50. Which of the following is a trinomial in x?
(a) x3 + 1
(b) x3 + x2 + x
(c) \( x\sqrt{x} + \sqrt{x} + 1 \)
(d) x3 + 2x
Answer: (b) x3 + x2 + x
A trinomial is a polynomial with exactly three non-zero terms having whole-number exponents. Option (b) has three terms (\( x^3 \), \( x^2 \), and \( x \)), all with non-negative integer powers. Option (c) is not a polynomial because the variable appears under a square root.
In simple words: Option (b) is a proper polynomial that has exactly three different terms.

Exam Tip: An expression must first be a valid polynomial before it can be classified as a monomial, binomial, or trinomial.

Question 51. The value of the polynomial x2 – x – 1 at x = -1 is:
(a) -3
(b) 1
(c) -1
(d) 0
Answer: (b) 1
Substituting \( x = -1 \) into the polynomial:
\( (-1)^2 - (-1) - 1 = 1 + 1 - 1 = 1 \)
In simple words: Square -1 to get 1, minus -1 turns into plus 1, then subtract 1 to get 1.

Exam Tip: Be careful with the double negative: \( -(-1) = +1 \).

Question 52. If P(x) = 7 – 3x + 2x2 then value of P (-2) is :
(a) 12
(b) 31
(c) 21
(d) 22
Answer: (c) 21
Substituting \( x = -2 \) into \( P(x) \):
\( P(-2) = 7 - 3(-2) + 2(-2)^2 = 7 + 6 + 2(4) = 7 + 6 + 8 = 21 \)
In simple words: Replace x with -2, work out each part (7 + 6 + 8), and add them to get 21.

Exam Tip: Evaluate powers first before multiplying by coefficients: calculate \( (-2)^2 = 4 \) before multiplying by 2.

Question 53. The coefficient of x2 in (3x + x3) (x + \frac{1}{x}) is:
(a) 3
(b) 1
(c) 4
(d) 2
Answer: (c) 4
Expanding the product using the distributive property:
\( (3x + x^3)\left(x + \frac{1}{x}\right) = 3x(x) + 3x\left(\frac{1}{x}\right) + x^3(x) + x^3\left(\frac{1}{x}\right) \)
\( = 3x^2 + 3 + x^4 + x^2 = x^4 + 4x^2 + 3 \)
The total coefficient of \( x^2 \) is \( 3 + 1 = 4 \).
In simple words: Multiplying gives two x-squared terms, 3x^2 and 1x^2, which add together to make 4x^2.

Exam Tip: Expand all products and combine like terms completely before identifying the requested coefficient.

Question 54. What is remainder when x3 – 2x2 + x + 1 is divided by (x – 1)?
(a) 0
(b) -1
(c) 1
(d) 2
Answer: (c) 1
According to the Remainder Theorem, the remainder when dividing \( p(x) \) by \( (x - 1) \) equals \( p(1) \):
\( p(1) = (1)^3 - 2(1)^2 + (1) + 1 = 1 - 2 + 1 + 1 = 1 \)
In simple words: Plug 1 into the polynomial to find the remainder directly, which comes out to 1.

Exam Tip: When dividing by \( (x - a) \), the remainder is simply the value of the polynomial evaluated at \( x = a \).

Question 55. Degree of which of the following polynomial is zero?
(a) x
(b) 15
(c) y
(d) \( x + \frac{1}{x} \)
Answer: (b) 15
Any non-zero constant polynomial has a degree of 0 because it can be written as \( c \cdot x^0 \). Here, \( 15 = 15x^0 \), so its degree is 0. The polynomials \( x \) and \( y \) have degree 1.
In simple words: A constant number without variables has a degree of zero.

Exam Tip: The degree of any non-zero constant polynomial is always 0, whereas the degree of the zero polynomial is undefined.

Question 56. When p(x) is divided by ax – b then the remainder is:
(a) p (a + b)
(b) p (-b /a)
(c) p (a / b)
(d) p (b / a)
Answer: (d) p (b / a)
By the Remainder Theorem, setting the linear divisor to zero gives:
\( ax - b = 0 \implies ax = b \implies x = \frac{b}{a} \)
Therefore, the remainder is \( p\left(\frac{b}{a}\right) \).
In simple words: Set ax - b to zero to get x = b/a, so the remainder is p(b/a).

Exam Tip: Always find the zero of the linear divisor \( ax - b \) to know what value to substitute into \( p(x) \).

Question 57. If x2 + kx + 6 = (x + 2)(x + 3) for all x, the value of k is:
(a) 1
(b) -1
(c) 5
(d) 3
Answer: (c) 5
Expanding the right-hand side:
\( (x + 2)(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6 \)
Comparing this with \( x^2 + kx + 6 \), the coefficient of \( x \) is \( k = 5 \).
In simple words: Multiplying (x + 2) and (x + 3) gives x^2 + 5x + 6, so k must be 5.

Exam Tip: Equate corresponding coefficients of like powers of \( x \) on both sides of an algebraic identity.

Question 58. Zero of the zero polynomial is:
(a) 0
(b) 1
(c) any real number
(d) not defined
Answer: (c) any real number
The zero polynomial is defined as \( p(x) = 0 \). For any real number \( c \), evaluating gives \( p(c) = 0 \). Hence, every real number satisfies the condition to be a zero of the zero polynomial.
In simple words: Since the zero polynomial is always 0, putting in any real number gives zero.

Exam Tip: Do not confuse the zero of the zero polynomial (which is any real number) with the degree of the zero polynomial (which is not defined).

Question 59. Which of the following is cubic polynomial
(a) x3 + 3x2 – 4x + 3
(b) x2 + 4x – 7
(c) 3x2 + 4
(d) 3 (x2 + x + 1)
Answer: (a) x3 + 3x2 – 4x + 3
A cubic polynomial is defined as a polynomial of degree 3 (highest exponent of the variable is 3). In option (a), the highest power of \( x \) is 3, making it a cubic polynomial. The other choices all have degree 2 (quadratic polynomials).
In simple words: A cubic polynomial must have 3 as the highest power of the variable.

Exam Tip: Determine the degree of a polynomial from the term with the highest power of the variable.

Question 60. If x51 + 51 is divided by (x + 1) the remainder is:
(a) 0
(b) 1
(c) 49
(d) 50
Answer: (d) 50
Using the Remainder Theorem, the remainder of \( p(x) = x^{51} + 51 \) when divided by \( (x + 1) \) is \( p(-1) \):
\( p(-1) = (-1)^{51} + 51 \)
Since 51 is an odd number, \( (-1)^{51} = -1 \). Thus:
\( -1 + 51 = 50 \)
In simple words: -1 raised to an odd power is -1, and -1 + 51 gives a remainder of 50.

Exam Tip: An odd power of -1 is always -1: \( (-1)^{\text{odd}} = -1 \), whereas \( (-1)^{\text{even}} = 1 \).