A. \(8+2\sqrt{15}\), अपरिमेय/\(8+2\sqrt{15}\), irrational
Step 1
Concept
\(x^2=5+3+2\sqrt{15}=8+2\sqrt{15}\), so it is irrational. In exams use ((a+b)2) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(8+2\sqrt{15}\), अपरिमेय / \(8+2\sqrt{15}\), irrational. \(x^2=5+3+2\sqrt{15}=8+2\sqrt{15}\), so it is irrational. In exams use ((a+b)2) carefully.
Step 3
Exam Tip
\(x^2=5+3+2\sqrt{15}=8+2\sqrt{15}\), इसलिए यह अपरिमेय है। परीक्षा में ((a+b)2) का प्रयोग सावधानी से करें।
\(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), so the sum is \(2\sqrt{2}\). Rationalising the denominator is a quick exam method.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{2}\). \(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), so the sum is \(2\sqrt{2}\). Rationalising the denominator is a quick exam method.
Step 3
Exam Tip
\(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), इसलिए योग \(2\sqrt{2}\) है। परीक्षा में हर का परिमेयकरण तेज तरीका है।
(p\(\sqrt{2}\)=2-5\sqrt{2}+6=8-5\sqrt{2}), which is irrational. In exams substitute first and then simplify the radical.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. (p\(\sqrt{2}\)=2-5\sqrt{2}+6=8-5\sqrt{2}), which is irrational. In exams substitute first and then simplify the radical.
Step 3
Exam Tip
(p\(\sqrt{2}\)=2-5\sqrt{2}+6=8-5\sqrt{2}), जो अपरिमेय है। परीक्षा में पहले प्रतिस्थापन करें फिर वर्गमूल को सरल करें।
If (x) were rational then \(\sqrt{2}+x\) would be irrational. So (x) must be irrational; remember the sum rule for rational and irrational numbers.
Step 2
Why this answer is correct
The correct answer is B. (x) अपरिमेय है / (x) is irrational. If (x) were rational then \(\sqrt{2}+x\) would be irrational. So (x) must be irrational; remember the sum rule for rational and irrational numbers.
Step 3
Exam Tip
यदि (x) परिमेय होता तो \(\sqrt{2}+x\) अपरिमेय होता। इसलिए (x) अपरिमेय होना चाहिए; परीक्षा में परिमेय और अपरिमेय के योग का नियम याद रखें।
\(\alpha+\beta=10\) and \(\alpha\beta=25-6=19\), so \(\alpha^2+\beta^2=100-38=62\). Use (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta).
Step 2
Why this answer is correct
The correct answer is A. (62). \(\alpha+\beta=10\) and \(\alpha\beta=25-6=19\), so \(\alpha^2+\beta^2=100-38=62\). Use (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta).
Step 3
Exam Tip
\(\alpha+\beta=10\) और \(\alpha\beta=25-6=19\), इसलिए \(\alpha^2+\beta^2=100-38=62\)। पहचान (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) उपयोग करें।
The zeroes are \(x=\pm\sqrt{2}\), and \(\sqrt{2}\) is irrational. In exams, simplify square-root zeroes before deciding the type.
Step 2
Why this answer is correct
The correct answer is B. दोनों अपरिमेय हैं / Both are irrational. The zeroes are \(x=\pm\sqrt{2}\), and \(\sqrt{2}\) is irrational. In exams, simplify square-root zeroes before deciding the type.
Step 3
Exam Tip
शून्यक \(x=\pm\sqrt{2}\) हैं और \(\sqrt{2}\) अपरिमेय है। परीक्षा में वर्गमूल वाले शून्यकों को सरल करके जाँचें।
The average of two numbers lies between them. Hence \(\frac{\sqrt{2}+\sqrt{3}}{2}\) lies between them.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{2}+\sqrt{3}}{2}\). The average of two numbers lies between them. Hence \(\frac{\sqrt{2}+\sqrt{3}}{2}\) lies between them.
Step 3
Exam Tip
दो संख्याओं का औसत उनके बीच होता है। इसलिए \(\frac{\sqrt{2}+\sqrt{3}}{2}\) इनके बीच है।
\(\sqrt{19}\) is irrational because (19) is not a perfect square. Subtracting an irrational from a rational gives an irrational result.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय है / It is irrational. \(\sqrt{19}\) is irrational because (19) is not a perfect square. Subtracting an irrational from a rational gives an irrational result.
Step 3
Exam Tip
\(\sqrt{19}\) अपरिमेय है क्योंकि (19) पूर्ण वर्ग नहीं है। परिमेय से अपरिमेय घटाने पर परिणाम अपरिमेय होता है।
A. गुणनफल परिमेय या अपरिमेय दोनों हो सकता है/The product can be rational or irrational
Step 1
Concept
\(\sqrt{5}\times\sqrt{5}=5\) is rational but \(\sqrt{5}\times\sqrt{2}=\sqrt{10}\) is irrational. So it depends on the case.
Step 2
Why this answer is correct
The correct answer is A. गुणनफल परिमेय या अपरिमेय दोनों हो सकता है / The product can be rational or irrational. \(\sqrt{5}\times\sqrt{5}=5\) is rational but \(\sqrt{5}\times\sqrt{2}=\sqrt{10}\) is irrational. So it depends on the case.
Step 3
Exam Tip
\(\sqrt{5}\times\sqrt{5}=5\) परिमेय है पर \(\sqrt{5}\times\sqrt{2}=\sqrt{10}\) अपरिमेय है। इसलिए स्थिति पर निर्भर करता है।
A. योग परिमेय या अपरिमेय दोनों हो सकता है/The sum can be rational or irrational
Step 1
Concept
(\sqrt{2}+\(-\sqrt{2}\)=0) is rational but \(\sqrt{2}+\sqrt{3}\) is irrational. So there is no single fixed rule.
Step 2
Why this answer is correct
The correct answer is A. योग परिमेय या अपरिमेय दोनों हो सकता है / The sum can be rational or irrational. (\sqrt{2}+\(-\sqrt{2}\)=0) is rational but \(\sqrt{2}+\sqrt{3}\) is irrational. So there is no single fixed rule.
Step 3
Exam Tip
(\sqrt{2}+\(-\sqrt{2}\)=0) परिमेय है पर \(\sqrt{2}+\sqrt{3}\) अपरिमेय है। इसलिए एक ही स्थायी नियम नहीं है।
\(\sqrt{12}\times\sqrt{3}=\sqrt{36}=6\). The product of two irrational numbers is not always irrational.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{12}\times\sqrt{3}\). \(\sqrt{12}\times\sqrt{3}=\sqrt{36}=6\). The product of two irrational numbers is not always irrational.
Step 3
Exam Tip
\(\sqrt{12}\times\sqrt{3}=\sqrt{36}=6\) है। दो अपरिमेय संख्याओं का गुणनफल हमेशा अपरिमेय नहीं होता।
\(\sqrt{3}\), \(\sqrt{5}\), and \(\sqrt{12}\) are all irrational. Remove options with perfect squares and rational numbers.
Step 2
Why this answer is correct
The correct answer is A. \({\sqrt{3},\sqrt{5},\sqrt{12}}\). \(\sqrt{3}\), \(\sqrt{5}\), and \(\sqrt{12}\) are all irrational. Remove options with perfect squares and rational numbers.
Step 3
Exam Tip
\(\sqrt{3}\), \(\sqrt{5}\) और \(\sqrt{12}\) सभी अपरिमेय हैं। पूर्ण वर्ग और परिमेय संख्या वाले विकल्प हटाएँ।
A. यह अनंत और अनावर्ती है/It is non terminating and non repeating
Step 1
Concept
This decimal has no fixed repetition. So it is considered non terminating and non repeating.
Step 2
Why this answer is correct
The correct answer is A. यह अनंत और अनावर्ती है / It is non terminating and non repeating. This decimal has no fixed repetition. So it is considered non terminating and non repeating.
Step 3
Exam Tip
इस दशमलव में निश्चित दोहराव नहीं है। इसलिए यह अनंत और अनावर्ती माना जाता है।
A non zero rational multiplier keeps an irrational number irrational. The condition \(r\neq0\) is important.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय संख्या / Irrational number. A non zero rational multiplier keeps an irrational number irrational. The condition \(r\neq0\) is important.
Step 3
Exam Tip
गैर शून्य परिमेय गुणक अपरिमेय संख्या को अपरिमेय ही रखता है। शर्त \(r\neq0\) महत्वपूर्ण है।
(9) is not a perfect cube so \(\sqrt[3]{9}\) is irrational. Identifying perfect cubes is important in cube roots.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय है / It is irrational. (9) is not a perfect cube so \(\sqrt[3]{9}\) is irrational. Identifying perfect cubes is important in cube roots.
Step 3
Exam Tip
(9) पूर्ण घन नहीं है इसलिए \(\sqrt[3]{9}\) अपरिमेय है। घनमूल में पूर्ण घन पहचानना जरूरी है।
\(\frac{5}{\sqrt{5}}=\sqrt{5}\) because \(5=\sqrt{5}\times\sqrt{5}\). Learn to simplify denominators with roots.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}\). \(\frac{5}{\sqrt{5}}=\sqrt{5}\) because \(5=\sqrt{5}\times\sqrt{5}\). Learn to simplify denominators with roots.
Step 3
Exam Tip
\(\frac{5}{\sqrt{5}}=\sqrt{5}\) है क्योंकि \(5=\sqrt{5}\times\sqrt{5}\)। जड़ वाले हर को सरल करना सीखें।
\(\sqrt{7}\times\sqrt{11}=\sqrt{77}\) and (77) is not a perfect square. So the result is irrational.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय संख्या / Irrational number. \(\sqrt{7}\times\sqrt{11}=\sqrt{77}\) and (77) is not a perfect square. So the result is irrational.
Step 3
Exam Tip
\(\sqrt{7}\times\sqrt{11}=\sqrt{77}\) है और (77) पूर्ण वर्ग नहीं है। इसलिए परिणाम अपरिमेय है।
(4) is not a perfect cube so \(\sqrt[3]{4}\) is irrational. Identify perfect cubes in cube roots.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय है / It is irrational. (4) is not a perfect cube so \(\sqrt[3]{4}\) is irrational. Identify perfect cubes in cube roots.
Step 3
Exam Tip
(4) पूर्ण घन नहीं है इसलिए \(\sqrt[3]{4}\) अपरिमेय है। घनमूल में पूर्ण घन पहचानें।
