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SubjectsMathematics
अपरिमेय संख्याएँ और वास्तविक संख्याएँ
In this Class 10 Mathematics topic, students build a clear understanding of real numbers as the collection of rational and irrational numbers. They learn to identify irrational numbers, compare and represent real numbers on the number line, and interpret terminating, recurring, and non-terminating non-recurring decimals. The topic also develops confidence with properties and operations involving real numbers, providing useful foundations for reading polynomial expressions, coefficients, and real zeros in the surrounding Polynomials chapter.
Correct answer: A. (8\sqrt{2})
Explanation: (\sqrt{50}=5\sqrt{2}), (3\sqrt{8}=6\sqrt{2}) and (\sqrt{18}=3\sqrt{2}). Hence the value is (8\sqrt{2}).
Correct answer: A. (2\sqrt{2})
Explanation: (x^2=3+2\sqrt{2}) and (x^3=7+5\sqrt{2}), so (x^3-3x=4+2\sqrt{2}). The correct value is not in the options so calculate carefully.
Correct answer: A. (4+2\sqrt{2})
Explanation: (x^3=7+5\sqrt{2}) and (3x=3+3\sqrt{2}), so the difference is (4+2\sqrt{2}). In exams calculate powers step by step.
Correct answer: A. (9+\sqrt{7}) and (9-\sqrt{7})
Explanation: The sum of (9+\sqrt{7}) and (9-\sqrt{7}) is (18), and the product is (81-7=74). In exams check both sum and product of options.
Correct answer: A. (\frac{15+\sqrt{221}}{2})
Explanation: Multiplying by the conjugate gives (\frac{30+2\sqrt{221}}{4}=\frac{15+\sqrt{221}}{2}). In exams divide by the common factor at the end.
Correct answer: A. Terminating decimal
Explanation: After cancelling (385=5\cdot7\cdot11), only (2^3) remains in the denominator. In exams always check the denominator in lowest form.
Correct answer: A. (8+2\sqrt{21})
Explanation: The first product is (14-6=8), and (\sqrt{84}=2\sqrt{21}). In exams use both conjugate multiplication and radical simplification.
Correct answer: A. (17)
Explanation: The other zero will be (6-\sqrt{19}), and the product is (36-19=17). In exams connect the constant term with the product of zeroes.
