यदि (a) और (b) धनात्मक पूर्णांक हैं तथा \(\sqrt{a}+\sqrt{b}\) परिमेय है, जबकि (a) पूर्ण वर्ग नहीं है, तो (b) के बारे में कौन-सा निष्कर्ष निश्चित रूप से सही हो सकता है?
If (a) and (b) are positive integers and \(\sqrt{a}+\sqrt{b}\) is rational, while (a) is not a perfect square, which conclusion about (b) can definitely be true?
#irrational numbers
#square roots
#expert
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A (b) भी (a) जैसा समान अपूर्ण वर्ग भाग रखता है / (b) also has the same non-square part as (a)
B (b) हमेशा पूर्ण वर्ग होगा / (b) will always be a perfect square
C ऐसा होना संभव नहीं है / This is not possible
D (b) अवश्य अभाज्य होगा / (b) must be prime
Explanation opens after your attempt
Correct Answer
C. ऐसा होना संभव नहीं है / This is not possible
Step 1
Concept
(a) पूर्ण वर्ग नहीं है, इसलिए \(\sqrt{a}\) अपरिमेय है। / Since (a) is not a perfect square, \(\sqrt{a}\) is irrational.
Step 2
Why this answer is correct
दो धनात्मक वर्गमूलों का योग परिमेय तभी हो सकता है जब अपरिमेय भाग कटे, पर यहाँ दोनों पद धनात्मक हैं इसलिए कटना संभव नहीं है। / A sum of two positive square roots could become rational only if irrational parts cancel, but both terms are positive here.
Step 3
Exam Tip
धनात्मक मूलों के योग में विपरीत चिह्न न होने पर अपरिमेय भाग बचता है। / Without opposite signs, irrational surd parts remain in the sum.
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कौन-सा विकल्प \(\frac{\sqrt{45}+\sqrt{20}}{\sqrt{5}}\) का सही मान देता है?
Which option gives the correct value of \(\frac{\sqrt{45}+\sqrt{20}}{\sqrt{5}}\)?
#surds
#simplification
#rational result
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A (5)
B (7)
C \(\sqrt{65}\)
D (13)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{20}=2\sqrt{5}\) हैं। / \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\).
Step 2
Why this answer is correct
ऊपर का योग \(5\sqrt{5}\) है, इसलिए \(\frac{5\sqrt{5}}{\sqrt{5}}=5\)। / The numerator becomes \(5\sqrt{5}\), so \(\frac{5\sqrt{5}}{\sqrt{5}}=5\).
Step 3
Exam Tip
भाग से पहले ऊपर के मूलों को समान रूप में बदलें। / Before division, convert the numerator surds into like terms.
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यदि \(x=\sqrt{7}+\sqrt{28}\), तो \(\frac{x^2}{7}\) का मान क्या है?
If \(x=\sqrt{7}+\sqrt{28}\), what is the value of \(\frac{x^2}{7}\)?
#surd square
#real numbers
#expert
#class 10
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A (9)
B (16)
C (25)
D (36)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{28}=2\sqrt{7}\), इसलिए \(x=3\sqrt{7}\)। / \(\sqrt{28}=2\sqrt{7}\), so \(x=3\sqrt{7}\).
Step 2
Why this answer is correct
(x-2 =\(3\sqrt{7}\)2 =63), अतः \(\frac{x^2}{7}=9\)। / (x-2 =\(3\sqrt{7}\)2 =63), hence \(\frac{x^2}{7}=9\).
Step 3
Exam Tip
वर्ग करने से पहले समान मूल वाले पद जोड़ना सरल रहता है। / Combine like surds before squaring.
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कौन-सा कथन \(\sqrt{5}+\sqrt{20}-\sqrt{45}\) के लिए सही है?
Which statement is correct for \(\sqrt{5}+\sqrt{20}-\sqrt{45}\)?
#cancellation of surds
#rational result
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A यह (0) है और परिमेय है / It is (0) and rational
B यह \(\sqrt{5}\) है और अपरिमेय है / It is \(\sqrt{5}\) and irrational
C यह \(6\sqrt{5}\) है और अपरिमेय है / It is \(6\sqrt{5}\) and irrational
D यह (10) है और परिमेय है / It is (10) and rational
Explanation opens after your attempt
Correct Answer
A. यह (0) है और परिमेय है / It is (0) and rational
Step 1
Concept
\(\sqrt{20}=2\sqrt{5}\) और \(\sqrt{45}=3\sqrt{5}\) लिखें। / Write \(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\).
Step 2
Why this answer is correct
\(\sqrt{5}+2\sqrt{5}-3\sqrt{5}=0\), जो परिमेय है। / \(\sqrt{5}+2\sqrt{5}-3\sqrt{5}=0\), which is rational.
Step 3
Exam Tip
अपरिमेय दिखने वाले पद कटकर परिमेय उत्तर दे सकते हैं। / Terms that look irrational may cancel to give a rational result.
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यदि (p) अभाज्य संख्या है, तो \(\sqrt{p}\) की अपरिमेयता सिद्ध करने में कौन-सा मुख्य विचार काम आता है?
If (p) is a prime number, which main idea is used to prove that \(\sqrt{p}\) is irrational?
#irrationality proof
#prime factor
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A यदि \(p\mid a^2\), तो \(p\mid a\) / If \(p\mid a^2\), then \(p\mid a\)
B यदि \(p\mid a\), तो (a=0) / If \(p\mid a\), then (a=0)
C हर अभाज्य संख्या पूर्ण वर्ग होती है / Every prime number is a perfect square
D हर वर्गमूल परिमेय होता है / Every square root is rational
Explanation opens after your attempt
Correct Answer
A. यदि \(p\mid a^2\), तो \(p\mid a\) / If \(p\mid a^2\), then \(p\mid a\)
Step 1
Concept
प्रमाण में \(\sqrt{p}=\frac{a}{b}\) मानकर वर्ग किया जाता है। / In the proof, assume \(\sqrt{p}=\frac{a}{b}\) and square both sides.
Step 2
Why this answer is correct
\(a^2=pb^2\) से \(p\mid a^2\) मिलता है, इसलिए \(p\mid a\) का विचार प्रयोग होता है। / From \(a^2=pb^2\), we get \(p\mid a^2\), so the idea \(p\mid a\) is used.
Step 3
Exam Tip
अभाज्य गुणनखंड वाला तर्क विरोध तक पहुँचाता है। / The prime factor argument leads to a contradiction.
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कौन-सा विकल्प (\(\sqrt{11}+\sqrt{3}\)\(\sqrt{11}-\sqrt{3}\)) की प्रकृति सही बताता है?
Which option correctly describes the nature of (\(\sqrt{11}+\sqrt{3}\)\(\sqrt{11}-\sqrt{3}\))?
#conjugate surds
#difference of squares
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A (8), परिमेय / (8), rational
B (14), परिमेय / (14), rational
C \(2\sqrt{33}\), अपरिमेय / \(2\sqrt{33}\), irrational
D \(\sqrt{8}\), अपरिमेय / \(\sqrt{8}\), irrational
Explanation opens after your attempt
Correct Answer
A. (8), परिमेय / (8), rational
Step 1
Concept
यह ((u+v)(u-v)) के रूप में है। / This is of the form ((u+v)(u-v)).
Step 2
Why this answer is correct
मान (11-3=8) आता है, जो परिमेय है। / The value is (11-3=8), which is rational.
Step 3
Exam Tip
संयुग्मी पदों का गुणन अक्सर अपरिमेय भाग हटा देता है। / Multiplying conjugate surds often removes the irrational part.
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यदि \(x=\frac{1}{\sqrt{6}-\sqrt{5}}\), तो (x) किसके बराबर है?
If \(x=\frac{1}{\sqrt{6}-\sqrt{5}}\), what is (x) equal to?
#rationalization
#conjugates
#expert
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A \(\sqrt{6}+\sqrt{5}\)
B \(\sqrt{6}-\sqrt{5}\)
C \(\frac{\sqrt{6}+\sqrt{5}}{11}\)
D (1)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{6}+\sqrt{5}\)
Step 1
Concept
हर का संयुग्मी \(\sqrt{6}+\sqrt{5}\) है। / The conjugate of the denominator is \(\sqrt{6}+\sqrt{5}\).
Step 2
Why this answer is correct
हर (\(\sqrt{6}\)2 -\(\sqrt{5}\)2 =6-5=1) बनता है। / The denominator becomes (\(\sqrt{6}\)2 -\(\sqrt{5}\)2 =6-5=1).
Step 3
Exam Tip
जब हर में दो मूलों का अंतर हो, तो संयुग्मी से गुणा करें। / When the denominator is a difference of two surds, multiply by its conjugate.
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किस विकल्प में दिया गया दशमलव निश्चित रूप से अपरिमेय है?
Which given decimal is definitely irrational?
#decimal expansion
#non recurring
#irrational
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A \(0.246824682468\ldots\)
B \(0.1357913579\ldots\)
C \(0.120120012000120000\ldots\)
D \(0.777777\ldots\)
Explanation opens after your attempt
Correct Answer
C. \(0.120120012000120000\ldots\)
Step 1
Concept
पहले देखें कि कोई निश्चित अंकों का समूह बार-बार आ रहा है या नहीं। / First check whether a fixed block of digits repeats.
Step 2
Why this answer is correct
\(0.120120012000120000\ldots\) में शून्यों की संख्या बदलती जाती है, इसलिए स्थिर आवर्तन नहीं है। / In \(0.120120012000120000\ldots\), the number of zeros keeps changing, so there is no fixed repetition.
Step 3
Exam Tip
असांत अनावर्ती दशमलव अपरिमेय होता है। / A non-terminating non-recurring decimal is irrational.
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यदि \(x=\sqrt{3}+\sqrt{2}\), तो \(\frac{1}{x}\) का परिमेय हर वाला रूप कौन-सा है?
If \(x=\sqrt{3}+\sqrt{2}\), which is the rationalized form of \(\frac{1}{x}\)?
#rationalization
#surd reciprocal
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A \(\sqrt{3}-\sqrt{2}\)
B \(\sqrt{3}+\sqrt{2}\)
C \(\frac{\sqrt{3}-\sqrt{2}}{5}\)
D \(\frac{1}{\sqrt{3}-\sqrt{2}}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{3}-\sqrt{2}\)
Step 1
Concept
\(\sqrt{3}+\sqrt{2}\) का संयुग्मी \(\sqrt{3}-\sqrt{2}\) है। / The conjugate of \(\sqrt{3}+\sqrt{2}\) is \(\sqrt{3}-\sqrt{2}\).
Step 2
Why this answer is correct
हर (3-2=1) बनता है, इसलिए \(\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}\)। / The denominator becomes (3-2=1), so \(\frac{1}{\sqrt{3}+\sqrt{2}}=\sqrt{3}-\sqrt{2}\).
Step 3
Exam Tip
जिन दो मूलों के वर्गों का अंतर (1) हो, वहाँ उत्तर बहुत सरल आता है। / When the difference of the squared surds is (1), the result becomes very simple.
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कौन-सा विकल्प दो अपरिमेय संख्याओं के योग को परिमेय बनाता है?
Which option makes the sum of two irrational numbers rational?
#sum of irrationals
#counterexample
#class 10
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A (\(2+\sqrt{5}\)+\(\sqrt{5}-2\))
B (\(4+\sqrt{7}\)+\(4-\sqrt{7}\))
C (\(\sqrt{3}+1\)+\(\sqrt{3}-1\))
D (\(\sqrt{2}+\sqrt{3}\)+\(\sqrt{2}-\sqrt{3}\))
Explanation opens after your attempt
Correct Answer
B. (\(4+\sqrt{7}\)+\(4-\sqrt{7}\))
Step 1
Concept
\(4+\sqrt{7}\) और \(4-\sqrt{7}\) दोनों अपरिमेय हैं। / \(4+\sqrt{7}\) and \(4-\sqrt{7}\) are both irrational.
Step 2
Why this answer is correct
उनका योग (8) है, जो परिमेय है। / Their sum is (8), which is rational.
Step 3
Exam Tip
ऐसे उदाहरणों में समान अपरिमेय पद विपरीत चिह्न के साथ कटते हैं। / In such examples, equal irrational parts cancel with opposite signs.
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यदि \(a=\sqrt{8}+\sqrt{18}\) और \(b=\sqrt{8}-\sqrt{18}\), तो (ab) का मान क्या है?
If \(a=\sqrt{8}+\sqrt{18}\) and \(b=\sqrt{8}-\sqrt{18}\), what is the value of (ab)?
#conjugate product
#negative rational
#class 10
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A (-10)
B (10)
C (26)
D \(12\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
(ab=\(\sqrt{8}\)2 -\(\sqrt{18}\)2 ) है। / (ab=\(\sqrt{8}\)2 -\(\sqrt{18}\)2 ).
Step 2
Why this answer is correct
(ab=8-18=-10), जो परिमेय है। / (ab=8-18=-10), which is rational.
Step 3
Exam Tip
संयुग्मी गुणन में मूलों को अलग-अलग सरल करना जरूरी नहीं होता। / In conjugate multiplication, you do not always need to simplify each radical first.
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कौन-सा विकल्प \(\sqrt{50}+\sqrt{72}-\sqrt{98}\) का सही सरल रूप है?
Which option is the correct simplified form of \(\sqrt{50}+\sqrt{72}-\sqrt{98}\)?
#multiple surds
#simplification
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A \(4\sqrt{2}\)
B \(5\sqrt{2}\)
C \(6\sqrt{2}\)
D \(15\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{2}\)
Step 1
Concept
\(\sqrt{50}=5\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), और \(\sqrt{98}=7\sqrt{2}\)। / \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{98}=7\sqrt{2}\).
Step 2
Why this answer is correct
\(5\sqrt{2}+6\sqrt{2}-7\sqrt{2}=4\sqrt{2}\)। / \(5\sqrt{2}+6\sqrt{2}-7\sqrt{2}=4\sqrt{2}\).
Step 3
Exam Tip
सभी पद समान मूल में बदल जाएँ तो केवल गुणांक जोड़ें या घटाएँ। / Once all terms are like surds, add or subtract only the coefficients.
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यदि \(\sqrt{m}+\sqrt{n}=5\) और (m,n) धनात्मक पूर्णांक हैं, तो कौन-सा युग्म संभव है?
If \(\sqrt{m}+\sqrt{n}=5\) and (m,n) are positive integers, which pair is possible?
#perfect squares
#square root sum
#class 10
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A (m=4,n=9)
B (m=2,n=9)
C (m=8,n=1)
D (m=5,n=4)
Explanation opens after your attempt
Correct Answer
A. (m=4,n=9)
Step 1
Concept
\(\sqrt{4}=2\) और \(\sqrt{9}=3\)। / \(\sqrt{4}=2\) and \(\sqrt{9}=3\).
Step 2
Why this answer is correct
इनका योग (2+3=5) है। / Their sum is (2+3=5).
Step 3
Exam Tip
परिमेय पूर्णांक योग पाने के लिए पूर्ण वर्गों को पहले जाँचें। / To get a rational integer sum, check perfect squares first.
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कौन-सी संख्या (1) और (2) के बीच एक अपरिमेय संख्या है, पर \(\sqrt{2}\) से बड़ी है?
Which number is an irrational number between (1) and (2), but greater than \(\sqrt{2}\)?
#number line
#comparison of irrationals
#class 10
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A \(\sqrt{3}\)
B \(\sqrt{2}\)
C \(\frac{3}{2}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{3}\)
Step 1
Concept
\(\sqrt{3}\) लगभग (1.732) है, इसलिए यह (1) और (2) के बीच है। / \(\sqrt{3}\) is about (1.732), so it lies between (1) and (2).
Step 2
Why this answer is correct
(3>2), इसलिए \(\sqrt{3}>\sqrt{2}\)। / Since (3>2), \(\sqrt{3}>\sqrt{2}\).
Step 3
Exam Tip
धनात्मक वर्गमूलों की तुलना में अंदर की संख्याओं की तुलना कर सकते हैं। / For positive square roots, compare the numbers inside the roots.
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यदि \(x=\sqrt{10}-\sqrt{2}\), तो \(x^2\) किसके बराबर है?
If \(x=\sqrt{10}-\sqrt{2}\), what is \(x^2\) equal to?
#square of surd difference
#irrational expression
#class 10
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A \(12-4\sqrt{5}\)
B (8)
C \(12+4\sqrt{5}\)
D \(10-2\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(12-4\sqrt{5}\)
Step 1
Concept
((a-b)2 =a-2 -2ab+b-2 ) लगाएँ। / Use ((a-b)2 =a-2 -2ab+b-2 ).
Step 2
Why this answer is correct
\(x^2=10-2\sqrt{20}+2=12-4\sqrt{5}\)। / \(x^2=10-2\sqrt{20}+2=12-4\sqrt{5}\).
Step 3
Exam Tip
बीच वाले पद (-2ab) में चिह्न और मूल दोनों ध्यान से लिखें। / In the middle term (-2ab), write both the sign and the surd carefully.
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कौन-सा विकल्प \(2\sqrt{3}+3\sqrt{2}\) को एक वर्गमूल के वर्ग के रूप में पहचानने में मदद करता है?
Which option helps identify \(2\sqrt{3}+3\sqrt{2}\) as a square of a surd expression?
#surd identity
#error detection
#class 10
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A (\(\sqrt{3}+\sqrt{2}\)2 -5)
B (\(\sqrt{3}+\sqrt{2}\)2 )
C (\(\sqrt{6}+1\)2 )
D (\(3+\sqrt{2}\)2 )
Explanation opens after your attempt
Correct Answer
A. (\(\sqrt{3}+\sqrt{2}\)2 -5)
Step 1
Concept
(\(\sqrt{3}+\sqrt{2}\)2 =3+2+2\sqrt{6}=5+2\sqrt{6}) होता है, यह दिए गए पद जैसा नहीं है। / (\(\sqrt{3}+\sqrt{2}\)2 =5+2\sqrt{6}), which does not match the given expression.
Step 2
Why this answer is correct
दिए गए \(2\sqrt{3}+3\sqrt{2}\) को सीधे इस रूप में मिलाना संभव नहीं है; इसलिए यह विकल्पों में कोई सीधा वर्ग नहीं बनाता। / The expression \(2\sqrt{3}+3\sqrt{2}\) does not directly match any listed square form.
Step 3
Exam Tip
ऐसे प्रश्न में पहले प्रसार करके मिलान करें, अनुमान से नहीं। / Always expand and match, not guess by appearance.
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कौन-सा विकल्प \(5+2\sqrt{6}\) के बराबर है?
Which option is equal to \(5+2\sqrt{6}\)?
#surd square
#algebraic identity
#class 10
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A (\(\sqrt{3}+\sqrt{2}\)2 )
B (\(\sqrt{6}+1\)2 )
C (\(3+\sqrt{2}\)2 )
D (\(2+\sqrt{6}\)2 )
Explanation opens after your attempt
Correct Answer
A. (\(\sqrt{3}+\sqrt{2}\)2 )
Step 1
Concept
(\(\sqrt{3}+\sqrt{2}\)2 =3+2+2\sqrt{6})। / (\(\sqrt{3}+\sqrt{2}\)2 =3+2+2\sqrt{6}).
Step 2
Why this answer is correct
यह \(5+2\sqrt{6}\) के बराबर है। / This equals \(5+2\sqrt{6}\).
Step 3
Exam Tip
दो मूलों के योग का वर्ग करते समय बीच वाला पद \(2\sqrt{6}\) बनता है। / When squaring a sum of two surds, the middle term becomes \(2\sqrt{6}\).
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यदि \(x=\sqrt{a}\) अपरिमेय है और (a<50) धनात्मक पूर्णांक है, तो कौन-सा (a) उपयुक्त नहीं है?
If \(x=\sqrt{a}\) is irrational and (a<50) is a positive integer, which (a) is not suitable?
#perfect square
#irrational root
#class 10
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A (18)
B (27)
C (36)
D (48)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{a}\) अपरिमेय तभी होगा जब (a) पूर्ण वर्ग न हो। / \(\sqrt{a}\) is irrational only when (a) is not a perfect square.
Step 2
Why this answer is correct
(36) पूर्ण वर्ग है और \(\sqrt{36}=6\), इसलिए यह उपयुक्त नहीं है। / (36) is a perfect square and \(\sqrt{36}=6\), so it is not suitable.
Step 3
Exam Tip
विकल्पों में पूर्ण वर्ग को तुरंत पहचानें। / Quickly identify perfect squares among the options.
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किस विकल्प में संख्या अपरिमेय है, लेकिन उसका व्युत्क्रम भी अपरिमेय है?
In which option is the number irrational and its reciprocal also irrational?
#reciprocal
#irrational numbers
#class 10
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A \(\sqrt{9}\)
B \(\sqrt{12}\)
C \(\frac{1}{4}\)
D (0.25)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{12}\)
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) अपरिमेय है। / \(\sqrt{12}=2\sqrt{3}\) is irrational.
Step 2
Why this answer is correct
इसका व्युत्क्रम \(\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\) भी अपरिमेय है। / Its reciprocal \(\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\) is also irrational.
Step 3
Exam Tip
अशून्य अपरिमेय मूल के व्युत्क्रम को परिमेय मानने की गलती न करें। / Do not assume the reciprocal of a non-zero irrational surd is rational.
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यदि \(x=\sqrt{2}+\sqrt{5}\) और \(y=\sqrt{5}-\sqrt{2}\), तो (xy) का मान क्या है?
If \(x=\sqrt{2}+\sqrt{5}\) and \(y=\sqrt{5}-\sqrt{2}\), what is the value of (xy)?
#conjugate surds
#rational product
#class 10
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A (3)
B (7)
C \(\sqrt{10}\)
D \(2\sqrt{10}\)
Explanation opens after your attempt
Step 1
Concept
गुणन को (\(\sqrt{5}+\sqrt{2}\)\(\sqrt{5}-\sqrt{2}\)) की तरह देखें। / View the product as (\(\sqrt{5}+\sqrt{2}\)\(\sqrt{5}-\sqrt{2}\)).
Step 2
Why this answer is correct
यह (5-2=3) देता है। / It gives (5-2=3).
Step 3
Exam Tip
जोड़ के क्रम को बदलकर संयुग्मी रूप पहचान सकते हैं। / You can rearrange the order of addition to recognize a conjugate form.
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कौन-सा विकल्प \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\) जैसी गलत सोच को खंडित करता है?
Which option disproves the wrong idea \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\)?
#common mistake
#square roots
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A (a=4,b=9)
B (a=0,b=9)
C (a=1,b=0)
D (a=0,b=0)
Explanation opens after your attempt
Correct Answer
A. (a=4,b=9)
Step 1
Concept
(a=4,b=9) रखने पर बायाँ पक्ष (2+3=5) है। / For (a=4,b=9), the left side is (2+3=5).
Step 2
Why this answer is correct
दायाँ पक्ष \(\sqrt{13}\) है, जो (5) नहीं है। / The right side is \(\sqrt{13}\), which is not (5).
Step 3
Exam Tip
वर्गमूलों को जोड़ते समय अंदर की संख्याएँ सीधे नहीं जोड़ी जातीं। / When adding square roots, the numbers inside the roots are not added directly.
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किस विकल्प में \(\frac{\sqrt{a}}{\sqrt{b}}\) अपरिमेय है?
In which option is \(\frac{\sqrt{a}}{\sqrt{b}}\) irrational?
#quotient of radicals
#error detection
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A (a=18,b=2)
B (a=50,b=2)
C (a=12,b=3)
D (a=45,b=5)
Explanation opens after your attempt
Correct Answer
B. (a=50,b=2)
Step 1
Concept
\(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) है। / \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\).
Step 2
Why this answer is correct
(a=50,b=2) पर \(\sqrt{\frac{50}{2}}=\sqrt{25}=5\), यह परिमेय है; इसलिए इसे नहीं चुनना चाहिए। / For (a=50,b=2), it becomes \(\sqrt{25}=5\), which is rational, so it should not be selected.
Step 3
Exam Tip
सही अपरिमेय के लिए भागफल पूर्ण वर्ग न हो, जैसे यहाँ दिए विकल्पों में कोई अपरिमेय परिणाम नहीं बनता। / For an irrational quotient, \(\frac{a}{b}\) should not be a perfect square; none of the listed options gives that.
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कौन-सा विकल्प \(\frac{\sqrt{18}}{\sqrt{5}}\) की प्रकृति सही बताता है?
Which option correctly describes the nature of \(\frac{\sqrt{18}}{\sqrt{5}}\)?
#quotient of surds
#irrational result
#class 10
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A परिमेय, क्योंकि (18) सम है / Rational because (18) is even
B अपरिमेय, क्योंकि \(\frac{18}{5}\) पूर्ण वर्ग नहीं है / Irrational because \(\frac{18}{5}\) is not a perfect square
C परिमेय, क्योंकि (5) अभाज्य है / Rational because (5) is prime
D पूर्णांक, क्योंकि दोनों वर्गमूल हैं / Integer because both are square roots
Explanation opens after your attempt
Correct Answer
B. अपरिमेय, क्योंकि \(\frac{18}{5}\) पूर्ण वर्ग नहीं है / Irrational because \(\frac{18}{5}\) is not a perfect square
Step 1
Concept
\(\frac{\sqrt{18}}{\sqrt{5}}=\sqrt{\frac{18}{5}}\) है। / \(\frac{\sqrt{18}}{\sqrt{5}}=\sqrt{\frac{18}{5}}\).
Step 2
Why this answer is correct
\(\frac{18}{5}\) किसी परिमेय संख्या का पूर्ण वर्ग नहीं है, इसलिए परिणाम अपरिमेय है। / \(\frac{18}{5}\) is not a perfect square of a rational number, so the result is irrational.
Step 3
Exam Tip
भाग वाले मूलों में अंदर के भिन्न को पूर्ण वर्ग है या नहीं, यह देखें। / In quotients of radicals, check whether the fraction inside is a perfect square.
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यदि \(x=2+\sqrt{3}\), तो \(x+\frac{1}{x}\) का मान क्या है?
If \(x=2+\sqrt{3}\), what is the value of \(x+\frac{1}{x}\)?
#rationalization
#reciprocal
#expert
#class 10
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A (4)
B \(2\sqrt{3}\)
C \(4+2\sqrt{3}\)
D (1)
Explanation opens after your attempt
Step 1
Concept
\(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\) होता है। / \(\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\).
Step 2
Why this answer is correct
(x+\frac{1}{x}=\(2+\sqrt{3}\)+\(2-\sqrt{3}\)=4)। / (x+\frac{1}{x}=\(2+\sqrt{3}\)+\(2-\sqrt{3}\)=4).
Step 3
Exam Tip
संयुग्मी व्युत्क्रम को पहचानने से लंबी गणना बचती है। / Recognizing the conjugate reciprocal saves long calculation.
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यदि \(x=3+\sqrt{8}\), तो (x) की प्रकृति और सरल रूप के बारे में सही कथन कौन-सा है?
If \(x=3+\sqrt{8}\), which statement about the nature and simplified form of (x) is correct?
#surd simplification
#irrational expression
#class 10
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A \(x=3+2\sqrt{2}\), अपरिमेय / \(x=3+2\sqrt{2}\), irrational
B \(x=5\sqrt{2}\), अपरिमेय / \(x=5\sqrt{2}\), irrational
C (x=11), परिमेय / (x=11), rational
D \(x=\sqrt{11}\), अपरिमेय / \(x=\sqrt{11}\), irrational
Explanation opens after your attempt
Correct Answer
A. \(x=3+2\sqrt{2}\), अपरिमेय / \(x=3+2\sqrt{2}\), irrational
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\) है। / \(\sqrt{8}=2\sqrt{2}\).
Step 2
Why this answer is correct
इसलिए \(x=3+2\sqrt{2}\), जिसमें अपरिमेय भाग है। / So \(x=3+2\sqrt{2}\), which contains an irrational part.
Step 3
Exam Tip
परिमेय और अपरिमेय पदों को सीधे जोड़कर एक मूल न बनाएं। / Do not combine rational and irrational terms into a single radical.
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कौन-सा विकल्प बताता है कि \(\sqrt{2}+\sqrt{3}\) अपरिमेय है?
Which option explains why \(\sqrt{2}+\sqrt{3}\) is irrational?
#proof idea
#sum of surds
#irrationality
#class 10
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A यदि यह परिमेय हो, तो वर्ग करने पर \(5+2\sqrt{6}\) परिमेय होगा और \(\sqrt{6}\) परिमेय निकल आएगा / If it were rational, squaring would make \(5+2\sqrt{6}\) rational and then \(\sqrt{6}\) would be rational
B क्योंकि हर योग अपरिमेय होता है / Because every sum is irrational
C क्योंकि \(\sqrt{2}\) और \(\sqrt{3}\) दोनों धनात्मक हैं / Because \(\sqrt{2}\) and \(\sqrt{3}\) are both positive
D क्योंकि (2+3=5) है / Because (2+3=5)
Explanation opens after your attempt
Correct Answer
A. यदि यह परिमेय हो, तो वर्ग करने पर \(5+2\sqrt{6}\) परिमेय होगा और \(\sqrt{6}\) परिमेय निकल आएगा / If it were rational, squaring would make \(5+2\sqrt{6}\) rational and then \(\sqrt{6}\) would be rational
Step 1
Concept
मान लें \(\sqrt{2}+\sqrt{3}\) परिमेय है। / Assume \(\sqrt{2}+\sqrt{3}\) is rational.
Step 2
Why this answer is correct
वर्ग करने पर \(5+2\sqrt{6}\) परिमेय होगा, जिससे \(\sqrt{6}\) परिमेय मानना पड़ेगा, जो गलत है। / Squaring gives \(5+2\sqrt{6}\) rational, which would force \(\sqrt{6}\) to be rational, impossible.
Step 3
Exam Tip
दो अलग मूलों के योग में वर्ग विधि उपयोगी होती है। / Squaring is useful for sums of two different surds.
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यदि \(x=\sqrt{5}+\sqrt{3}\), तो \(x-\frac{2}{x}\) का मान क्या है?
If \(x=\sqrt{5}+\sqrt{3}\), what is the value of \(x-\frac{2}{x}\)?
#rationalization
#algebraic surds
#class 10
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A \(2\sqrt{3}\)
B \(2\sqrt{5}\)
C \(\sqrt{5}-\sqrt{3}\)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{3}\)
Step 1
Concept
\(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{2}\) होता है। / \(\frac{1}{\sqrt{5}+\sqrt{3}}=\frac{\sqrt{5}-\sqrt{3}}{2}\).
Step 2
Why this answer is correct
इसलिए \(\frac{2}{x}=\sqrt{5}-\sqrt{3}\)। / Therefore \(\frac{2}{x}=\sqrt{5}-\sqrt{3}\).
Step 3
Exam Tip
(x-\frac{2}{x}=\(\sqrt{5}+\sqrt{3}\)-\(\sqrt{5}-\sqrt{3}\)=2\sqrt{3})। / (x-\frac{2}{x}=\(\sqrt{5}+\sqrt{3}\)-\(\sqrt{5}-\sqrt{3}\)=2\sqrt{3}).
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कौन-सा विकल्प \(\sqrt{a}+\sqrt{b}\) को अपरिमेय बनाता है?
Which option makes \(\sqrt{a}+\sqrt{b}\) irrational?
#sum of roots
#irrational expression
#class 10
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A (a=9,b=16)
B (a=25,b=36)
C (a=4,b=18)
D (a=49,b=64)
Explanation opens after your attempt
Correct Answer
C. (a=4,b=18)
Step 1
Concept
(a=4) पर \(\sqrt{4}=2\) है। / For (a=4), \(\sqrt{4}=2\).
Step 2
Why this answer is correct
(b=18) पर \(\sqrt{18}=3\sqrt{2}\), जो अपरिमेय है; इसलिए योग \(2+3\sqrt{2}\) अपरिमेय है। / For (b=18), \(\sqrt{18}=3\sqrt{2}\), which is irrational; so the sum \(2+3\sqrt{2}\) is irrational.
Step 3
Exam Tip
यदि एक पद परिमेय और दूसरा अपरिमेय हो, तो योग अपरिमेय रहता है। / A rational plus an irrational remains irrational.
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यदि \(x=\sqrt{2}+\sqrt{7}\), तो \(x^2-9\) का मान क्या है?
If \(x=\sqrt{2}+\sqrt{7}\), what is the value of \(x^2-9\)?
#surd square
#irrational result
#class 10
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A \(2\sqrt{14}\)
B \(\sqrt{14}\)
C (9)
D (14)
Explanation opens after your attempt
Correct Answer
A. \(2\sqrt{14}\)
Step 1
Concept
\(x^2=2+7+2\sqrt{14}=9+2\sqrt{14}\)। / \(x^2=2+7+2\sqrt{14}=9+2\sqrt{14}\).
Step 2
Why this answer is correct
इसलिए \(x^2-9=2\sqrt{14}\), जो अपरिमेय है। / Therefore \(x^2-9=2\sqrt{14}\), which is irrational.
Step 3
Exam Tip
पहले वर्ग करें, फिर परिमेय भाग घटाएँ। / Square first, then subtract the rational part.
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कौन-सा विकल्प \(\sqrt{3}\) और \(\sqrt{12}\) के बीच संबंध सही बताता है?
Which option correctly states the relation between \(\sqrt{3}\) and \(\sqrt{12}\)?
#simplifying radicals
#like surds
#class 10
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A \(\sqrt{12}=4\sqrt{3}\)
B \(\sqrt{12}=2\sqrt{3}\)
C \(\sqrt{12}=\sqrt{3}+3\)
D \(\sqrt{12}=6\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{12}=2\sqrt{3}\)
Step 1
Concept
\(12=4\times3\) है। / \(12=4\times3\).
Step 2
Why this answer is correct
\(\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}\)। / \(\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}\).
Step 3
Exam Tip
पूर्ण वर्ग गुणनखंड को मूल से बाहर निकालें। / Take the perfect square factor outside the radical.
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यदि \(x=\sqrt{3}-1\), तो ((x+1)2 ) का मान क्या है?
If \(x=\sqrt{3}-1\), what is the value of ((x+1)2 )?
#algebra with surds
#square
#class 10
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A (3)
B \(\sqrt{3}\)
C (4)
D \(2\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
\(x+1=\sqrt{3}\) है। / \(x+1=\sqrt{3}\).
Step 2
Why this answer is correct
इसलिए ((x+1)2 =\(\sqrt{3}\)2 =3)। / Therefore ((x+1)2 =\(\sqrt{3}\)2 =3).
Step 3
Exam Tip
पहले भीतर के पद को सरल करें, फिर वर्ग करें। / Simplify the inner expression first, then square it.
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कौन-सा विकल्प बताता है कि \(2\sqrt{3}\) और \(3\sqrt{2}\) में कौन बड़ा है?
Which option correctly tells which is greater between \(2\sqrt{3}\) and \(3\sqrt{2}\)?
#comparison of surds
#number sense
#class 10
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A \(2\sqrt{3}\) बड़ा है / \(2\sqrt{3}\) is greater
B \(3\sqrt{2}\) बड़ा है / \(3\sqrt{2}\) is greater
C दोनों बराबर हैं / Both are equal
D तुलना संभव नहीं है / Comparison is not possible
Explanation opens after your attempt
Correct Answer
B. \(3\sqrt{2}\) बड़ा है / \(3\sqrt{2}\) is greater
Step 1
Concept
दोनों संख्याएँ धनात्मक हैं, इसलिए वर्ग करके तुलना करें। / Both numbers are positive, so compare their squares.
Step 2
Why this answer is correct
(\(2\sqrt{3}\)2 =12) और (\(3\sqrt{2}\)2 =18), इसलिए \(3\sqrt{2}\) बड़ा है। / (\(2\sqrt{3}\)2 =12) and (\(3\sqrt{2}\)2 =18), so \(3\sqrt{2}\) is greater.
Step 3
Exam Tip
धनात्मक मूलों की तुलना में वर्ग करना सुरक्षित तरीका है। / Squaring is a safe method for comparing positive surds.
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यदि \(x=2\sqrt{5}\) और \(y=5\sqrt{2}\), तो (xy) की प्रकृति क्या है?
If \(x=2\sqrt{5}\) and \(y=5\sqrt{2}\), what is the nature of (xy)?
#product of surds
#irrational product
#class 10
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A परिमेय / Rational
B अपरिमेय / Irrational
C पूर्णांक / Integer
D शून्य / Zero
Explanation opens after your attempt
Correct Answer
B. अपरिमेय / Irrational
Step 1
Concept
\(xy=2\sqrt{5}\times5\sqrt{2}=10\sqrt{10}\)। / \(xy=2\sqrt{5}\times5\sqrt{2}=10\sqrt{10}\).
Step 2
Why this answer is correct
\(\sqrt{10}\) अपरिमेय है, इसलिए \(10\sqrt{10}\) अपरिमेय है। / \(\sqrt{10}\) is irrational, so \(10\sqrt{10}\) is irrational.
Step 3
Exam Tip
गुणन के बाद अंदर की संख्या पूर्ण वर्ग नहीं बने तो परिणाम अपरिमेय रह सकता है। / If the product inside the root is not a perfect square, the result may remain irrational.
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कौन-सा विकल्प \(0.10110111011110\ldots\) के लिए सही है, जहाँ प्रत्येक चरण में (1) की संख्या बढ़ती जाती है?
Which option is correct for \(0.10110111011110\ldots\), where the number of (1)'s increases at each stage?
#decimal pattern
#irrational decimal
#class 10
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A यह सांत परिमेय है / It is terminating rational
B यह असांत आवर्ती परिमेय है / It is non-terminating recurring rational
C यह असांत अनावर्ती अपरिमेय है / It is non-terminating non-recurring irrational
D यह पूर्णांक है / It is an integer
Explanation opens after your attempt
Correct Answer
C. यह असांत अनावर्ती अपरिमेय है / It is non-terminating non-recurring irrational
Step 1
Concept
यह दशमलव समाप्त नहीं होता। / This decimal does not terminate.
Step 2
Why this answer is correct
(1) की संख्या बदलती रहती है, इसलिए कोई निश्चित आवर्ती समूह नहीं है। / The number of (1)'s keeps changing, so there is no fixed recurring block.
Step 3
Exam Tip
असांत अनावर्ती दशमलव को अपरिमेय माना जाता है। / A non-terminating non-recurring decimal is irrational.
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यदि \(x=\sqrt{13}+2\), तो \(x^2-4x\) का मान क्या है?
If \(x=\sqrt{13}+2\), what is the value of \(x^2-4x\)?
#hidden conjugate
#algebraic surds
#class 10
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A (9)
B (13)
C \(\sqrt{13}\)
D (17)
Explanation opens after your attempt
Step 1
Concept
(x-2 -4x=x(x-4)) लिखें। / Write (x-2 -4x=x(x-4)).
Step 2
Why this answer is correct
\(x=\sqrt{13}+2\) होने पर \(x-4=\sqrt{13}-2\), इसलिए गुणन (13-4=9) है। / With \(x=\sqrt{13}+2\), \(x-4=\sqrt{13}-2\), so the product is (13-4=9).
Step 3
Exam Tip
ऐसे रूप में संयुग्मी छिपा हो सकता है। / A conjugate form may be hidden in such expressions.
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कौन-सा विकल्प \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\) के सही सरल रूप के बराबर है?
Which option is equal to the simplified form of \(\frac{2+\sqrt{3}}{2-\sqrt{3}}\)?
#rationalization
#conjugate fraction
#class 10
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A \(7+4\sqrt{3}\)
B \(7-4\sqrt{3}\)
C (1)
D \(4+\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(7+4\sqrt{3}\)
Step 1
Concept
हर को परिमेय बनाने के लिए \(2+\sqrt{3}\) से गुणा करें। / Multiply by \(2+\sqrt{3}\) to rationalize the denominator.
Step 2
Why this answer is correct
(\frac{\(2+\sqrt{3}\)2 }{4-3}=4+4\sqrt{3}+3=7+4\sqrt{3})। / (\frac{\(2+\sqrt{3}\)2 }{4-3}=4+4\sqrt{3}+3=7+4\sqrt{3}).
Step 3
Exam Tip
संयुग्मी से गुणा करते समय ऊपर भी पूरा वर्ग बनता है। / When multiplying by the conjugate, the numerator may become a full square.
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यदि \(a=7+4\sqrt{3}\), तो कौन-सा विकल्प (a) का वर्गमूल दर्शाता है?
If \(a=7+4\sqrt{3}\), which option represents a square root of (a)?
#surd square root
#identity
#class 10
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A \(2+\sqrt{3}\)
B \(2-\sqrt{3}\)
C \(\sqrt{7}+2\)
D \(\sqrt{3}+1\)
Explanation opens after your attempt
Correct Answer
A. \(2+\sqrt{3}\)
Step 1
Concept
(\(2+\sqrt{3}\)2 =4+4\sqrt{3}+3)। / (\(2+\sqrt{3}\)2 =4+4\sqrt{3}+3).
Step 2
Why this answer is correct
यह \(7+4\sqrt{3}\) के बराबर है। / This equals \(7+4\sqrt{3}\).
Step 3
Exam Tip
ऐसे प्रश्नों में \(m+n+2\sqrt{mn}\) का रूप पहचानें। / In such questions, identify the form \(m+n+2\sqrt{mn}\).
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कौन-सा विकल्प \(\sqrt{80}-\sqrt{45}+\sqrt{20}\) का सही सरल रूप देता है?
Which option gives the correct simplified form of \(\sqrt{80}-\sqrt{45}+\sqrt{20}\)?
#surd simplification
#error detection
#class 10
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A \(4\sqrt{5}\)
B \(5\sqrt{5}\)
C \(6\sqrt{5}\)
D \(\sqrt{55}\)
Explanation opens after your attempt
Correct Answer
B. \(5\sqrt{5}\)
Step 1
Concept
\(\sqrt{80}=4\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\), और \(\sqrt{20}=2\sqrt{5}\)। / \(\sqrt{80}=4\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\), and \(\sqrt{20}=2\sqrt{5}\).
Step 2
Why this answer is correct
\(4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}\), इसलिए दिए विकल्पों में कोई सही नहीं दिखता। / \(4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}\), so none of the listed options is correct.
Step 3
Exam Tip
ऐसे प्रश्न में विकल्प से पहले अपनी गणना पर भरोसा करें। / In such questions, trust your simplification before matching options.
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कौन-सा विकल्प \(\sqrt{80}-\sqrt{45}+\sqrt{20}\) का सही सरल रूप है?
Which option is the correct simplified form of \(\sqrt{80}-\sqrt{45}+\sqrt{20}\)?
#surds
#addition subtraction
#class 10
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A \(2\sqrt{5}\)
B \(3\sqrt{5}\)
C \(4\sqrt{5}\)
D \(5\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \(3\sqrt{5}\)
Step 1
Concept
\(\sqrt{80}=4\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\), और \(\sqrt{20}=2\sqrt{5}\)। / \(\sqrt{80}=4\sqrt{5}\), \(\sqrt{45}=3\sqrt{5}\), and \(\sqrt{20}=2\sqrt{5}\).
Step 2
Why this answer is correct
\(4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}\), जो अपरिमेय है। / \(4\sqrt{5}-3\sqrt{5}+2\sqrt{5}=3\sqrt{5}\), which is irrational.
Step 3
Exam Tip
तीन पदों में चिह्नों को ध्यान से संभालें। / Handle the signs carefully when three terms are involved.
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यदि (x) अपरिमेय है और \(x+ \sqrt{2}\) परिमेय है, तो (x) का संभावित रूप कौन-सा हो सकता है?
If (x) is irrational and \(x+\sqrt{2}\) is rational, which can be a possible form of (x)?
#irrational cancellation
#expression
#class 10
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A \(3-\sqrt{2}\)
B \(3+\sqrt{2}\)
C \(\sqrt{3}\)
D \(2\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(3-\sqrt{2}\)
Step 1
Concept
\(x+\sqrt{2}\) को परिमेय बनाने के लिए (x) में \(-\sqrt{2}\) वाला भाग होना चाहिए। / To make \(x+\sqrt{2}\) rational, (x) should contain a \(-\sqrt{2}\) part.
Step 2
Why this answer is correct
\(x=3-\sqrt{2}\) रखने पर \(x+\sqrt{2}=3\), जो परिमेय है। / If \(x=3-\sqrt{2}\), then \(x+\sqrt{2}=3\), which is rational.
Step 3
Exam Tip
अपरिमेय भाग के कटने की संभावना खोजें। / Look for cancellation of the irrational part.
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कौन-सा विकल्प (\(\sqrt{7}+\sqrt{2}\)2 -\(\sqrt{7}-\sqrt{2}\)2 ) के बराबर है?
Which option is equal to (\(\sqrt{7}+\sqrt{2}\)2 -\(\sqrt{7}-\sqrt{2}\)2 )?
#algebraic identity
#surds
#class 10
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A \(4\sqrt{14}\)
B (9)
C \(2\sqrt{14}\)
D (18)
Explanation opens after your attempt
Correct Answer
A. \(4\sqrt{14}\)
Step 1
Concept
((u+v)2 -(u-v)2 =4uv) होता है। / ((u+v)2 -(u-v)2 =4uv).
Step 2
Why this answer is correct
यहाँ \(u=\sqrt{7}\) और \(v=\sqrt{2}\), इसलिए मान \(4\sqrt{14}\) है। / Here \(u=\sqrt{7}\) and \(v=\sqrt{2}\), so the value is \(4\sqrt{14}\).
Step 3
Exam Tip
पहचान का प्रयोग करने से विस्तार छोटा हो जाता है। / Using the identity makes the expansion shorter.
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यदि \(x=\sqrt{6}+\sqrt{2}\) और \(y=\sqrt{6}-\sqrt{2}\), तो \(\frac{x}{y}\) का सरल रूप क्या है?
If \(x=\sqrt{6}+\sqrt{2}\) and \(y=\sqrt{6}-\sqrt{2}\), what is the simplified form of \(\frac{x}{y}\)?
#rationalization
#quotient of surds
#class 10
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A \(2+\sqrt{3}\)
B \(2-\sqrt{3}\)
C \(3+2\sqrt{2}\)
D \(\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. \(2+\sqrt{3}\)
Step 1
Concept
\(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\) में हर को संयुग्मी से परिमेय करें। / Rationalize the denominator of \(\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}\).
Step 2
Why this answer is correct
ऊपर (\(\sqrt{6}+\sqrt{2}\)2 =8+4\sqrt{3}) और नीचे (6-2=4), इसलिए मान \(2+\sqrt{3}\) है। / The numerator becomes (\(\sqrt{6}+\sqrt{2}\)2 =8+4\sqrt{3}), and the denominator is (6-2=4), so the value is \(2+\sqrt{3}\).
Step 3
Exam Tip
भाग में संयुग्मी से गुणा करना प्रभावी तरीका है। / Multiplying by the conjugate is effective in such quotients.
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किस विकल्प में दी गई संख्या \(2\sqrt{3}\) से छोटी और \(\sqrt{11}\) से बड़ी है?
Which given number is smaller than \(2\sqrt{3}\) and greater than \(\sqrt{11}\)?
#comparison
#irrational numbers
#number line
#class 10
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A (3.2)
B (3.4)
C (3.6)
D (3.8)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{11}\) लगभग (3.316) है और \(2\sqrt{3}\) लगभग (3.464) है। / \(\sqrt{11}\) is about (3.316), and \(2\sqrt{3}\) is about (3.464).
Step 2
Why this answer is correct
(3.4) इन दोनों के बीच है। / (3.4) lies between them.
Step 3
Exam Tip
निकट मानों की तुलना में दो दशमलव तक अनुमान काफी मदद करता है। / For close values, estimating to two decimal places is helpful.
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यदि \(x=\sqrt{2}\), तो \(x^4-4x^2+4\) का मान क्या है?
If \(x=\sqrt{2}\), what is the value of \(x^4-4x^2+4\)?
#powers of surds
#algebra
#class 10
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A (0)
B (2)
C (4)
D \(\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
\(x^2=2\) है। / \(x^2=2\).
Step 2
Why this answer is correct
इसलिए (x-4 =\(x^2\)2 =4), और मान (4-8+4=0) है। / Therefore (x-4 =\(x^2\)2 =4), and the value is (4-8+4=0).
Step 3
Exam Tip
मूल वाली संख्या पर घात लगाते समय पहले \(x^2\) निकालें। / For powers of a surd, first find \(x^2\).
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कौन-सा विकल्प \(\sqrt{2}\) की अपरिमेयता के प्रमाण में गलत कदम है?
Which option is a wrong step in the proof of irrationality of \(\sqrt{2}\)?
#proof of sqrt 2
#logical step
#class 10
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A \(\sqrt{2}=\frac{p}{q}\) मानना / Assuming \(\sqrt{2}=\frac{p}{q}\)
B (p) और (q) को सहअभाज्य मानना / Taking (p) and (q) as coprime
C \(p^2=2q^2\) लिखना / Writing \(p^2=2q^2\)
D \(p^2=2q^2\) से (q) सम है, सीधे मान लेना / Directly assuming from \(p^2=2q^2\) that (q) is even
Explanation opens after your attempt
Correct Answer
D. \(p^2=2q^2\) से (q) सम है, सीधे मान लेना / Directly assuming from \(p^2=2q^2\) that (q) is even
Step 1
Concept
\(p^2=2q^2\) से पहले \(p^2\) सम और फिर (p) सम मिलता है। / From \(p^2=2q^2\), first \(p^2\) is even and hence (p) is even.
Step 2
Why this answer is correct
(p=2k) रखने के बाद \(q^2=2k^2\) से (q) सम निकलता है। / After writing (p=2k), we get \(q^2=2k^2\), so (q) is even.
Step 3
Exam Tip
प्रमाण में क्रम छोड़ने से तर्क अधूरा हो जाता है। / Skipping this order makes the proof incomplete.
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यदि \(a=1+\sqrt{5}\), तो \(a^2-2a\) का मान क्या है?
If \(a=1+\sqrt{5}\), what is the value of \(a^2-2a\)?
#hidden conjugate
#surd algebra
#class 10
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A (4)
B (5)
C \(\sqrt{5}\)
D \(2\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
(a-2 -2a=a(a-2)) है। / (a-2 -2a=a(a-2)).
Step 2
Why this answer is correct
\(a-2=\sqrt{5}-1\), इसलिए (a(a-2)=\(1+\sqrt{5}\)\(\sqrt{5}-1\)=4)। / \(a-2=\sqrt{5}-1\), so (a(a-2)=\(1+\sqrt{5}\)\(\sqrt{5}-1\)=4).
Step 3
Exam Tip
छिपे हुए संयुग्मी रूप को पहचानना तेज तरीका है। / Recognizing the hidden conjugate form is a quick method.
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कौन-सा विकल्प \(\sqrt{48}+\sqrt{75}-\sqrt{27}\) को सरल करके देता है?
Which option gives the simplified form of \(\sqrt{48}+\sqrt{75}-\sqrt{27}\)?
#surd simplification
#like radicals
#class 10
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A \(4\sqrt{3}\)
B \(6\sqrt{3}\)
C \(8\sqrt{3}\)
D \(\sqrt{96}\)
Explanation opens after your attempt
Correct Answer
B. \(6\sqrt{3}\)
Step 1
Concept
\(\sqrt{48}=4\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), और \(\sqrt{27}=3\sqrt{3}\)। / \(\sqrt{48}=4\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{27}=3\sqrt{3}\).
Step 2
Why this answer is correct
\(4\sqrt{3}+5\sqrt{3}-3\sqrt{3}=6\sqrt{3}\)। / \(4\sqrt{3}+5\sqrt{3}-3\sqrt{3}=6\sqrt{3}\).
Step 3
Exam Tip
एक ही मूल वाले पदों में गुणांकों पर काम करें। / For like surds, work with the coefficients.
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यदि \(x=\sqrt{3}+\sqrt{2}\), तो (\(x-\sqrt{3}\)\(x-\sqrt{2}\)) का मान क्या है?
If \(x=\sqrt{3}+\sqrt{2}\), what is the value of (\(x-\sqrt{3}\)\(x-\sqrt{2}\))?
#substitution
#surd product
#class 10
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A \(\sqrt{6}\)
B (1)
C (5)
D (0)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{6}\)
Step 1
Concept
\(x-\sqrt{3}=\sqrt{2}\) और \(x-\sqrt{2}=\sqrt{3}\)। / \(x-\sqrt{3}=\sqrt{2}\) and \(x-\sqrt{2}=\sqrt{3}\).
Step 2
Why this answer is correct
उनका गुणन \(\sqrt{2}\times\sqrt{3}=\sqrt{6}\) है। / Their product is \(\sqrt{2}\times\sqrt{3}=\sqrt{6}\).
Step 3
Exam Tip
पहले छोटे-छोटे कोष्ठकों को सरल करें। / Simplify the small brackets first.
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कौन-सा विकल्प \(\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}\) का सही सरल रूप है?
Which option is the correct simplified form of \(\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}\)?
#series of surds
#simplification
#class 10
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A \(10\sqrt{2}\)
B \(8\sqrt{2}\)
C \(12\sqrt{2}\)
D (60)
Explanation opens after your attempt
Correct Answer
A. \(10\sqrt{2}\)
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), और \(\sqrt{32}=4\sqrt{2}\)। / \(\sqrt{8}=2\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), and \(\sqrt{32}=4\sqrt{2}\).
Step 2
Why this answer is correct
कुल योग \(1\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}=10\sqrt{2}\) है। / The total is \(1\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}=10\sqrt{2}\).
Step 3
Exam Tip
क्रमबद्ध मूलों में गुणांक का पैटर्न पहचानें। / In ordered surds, identify the coefficient pattern.
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यदि \(x=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\), तो (x) का मान और प्रकृति क्या है?
If \(x=\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\), what is the value and nature of (x)?
#conjugate surds
#rationalization
#irrational numbers
#class 10
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A (12), परिमेय / (12), rational
B \(6\sqrt{35}\), अपरिमेय / \(6\sqrt{35}\), irrational
C (24), परिमेय / (24), rational
D \(\sqrt{35}\), अपरिमेय / \(\sqrt{35}\), irrational
Explanation opens after your attempt
Correct Answer
A. (12), परिमेय / (12), rational
Step 1
Concept
पहले दोनों भिन्नों का साझा रूप देखें और \(a=\sqrt{7}+\sqrt{5}\) तथा \(b=\sqrt{7}-\sqrt{5}\) मानें। / First observe the common structure and take \(a=\sqrt{7}+\sqrt{5}\) and \(b=\sqrt{7}-\sqrt{5}\).
Step 2
Why this answer is correct
\(\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\) होगा। यहाँ \(a^2+b^2=24\) और (ab=2) इसलिए (x=12) है। / \(\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\). Here \(a^2+b^2=24\) and (ab=2), so (x=12).
Step 3
Exam Tip
संयुग्मी मूलों वाले भिन्नों में सीधे लंबा प्रसार करने के बजाय (a) और (b) रखकर हल करें। / For fractions with conjugate surds, use substitution instead of expanding everything directly.
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