100 results found for "parameter-inequality" in Class 10.
भाषा नीति में औपनिवेशिक भाषा की प्रतिष्ठा किस प्रकार सामाजिक असमानता बना सकती थी?
How could prestige of colonial language create social inequality in language policy?
#language-policy
#colonial-language
#social-inequality
A औपनिवेशिक भाषा जानने वालों को नौकरी शिक्षा और प्रशासन में लाभ मिलना / Those knowing the colonial language getting advantages in jobs education and administration
B स्थानीय भाषाओं को सभी पदों पर समान लाभ मिलना / Local languages getting equal advantage in all posts
C भाषा का रोजगार से कोई संबंध न होना / Language having no relation with employment
D औपनिवेशिक भाषा का तुरंत अंत होना / Immediate end of colonial language
Explanation opens after your attempt
Correct Answer
A. औपनिवेशिक भाषा जानने वालों को नौकरी शिक्षा और प्रशासन में लाभ मिलना / Those knowing the colonial language getting advantages in jobs education and administration
Step 1
Concept
Language could link with opportunity and power. For exams connect language policy with social hierarchy.
Step 2
Why this answer is correct
The correct answer is A. औपनिवेशिक भाषा जानने वालों को नौकरी शिक्षा और प्रशासन में लाभ मिलना / Those knowing the colonial language getting advantages in jobs education and administration. Language could link with opportunity and power. For exams connect language policy with social hierarchy.
Step 3
Exam Tip
भाषा अवसर और शक्ति से जुड़ सकती थी। परीक्षा में भाषा नीति को सामाजिक पदानुक्रम से जोड़ें।
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मूल निवासी संधियों की व्याख्या में शक्ति असमानता क्यों ध्यान में रखनी चाहिए?
Why should power inequality be considered while interpreting indigenous treaties?
#indigenous-treaties
#power-inequality
#land
A क्योंकि सभी संधियां हमेशा बराबर पक्षों में होती थीं / Because all treaties were always between equal sides
B क्योंकि भाषा भूमि समझ और सैन्य दबाव में असमानता हो सकती थी / Because inequality could exist in language land understanding and military pressure
C क्योंकि संधियों का भूमि से कोई संबंध नहीं था / Because treaties had no relation with land
D क्योंकि संधियां इतिहास का स्रोत नहीं होतीं / Because treaties are not historical sources
Explanation opens after your attempt
Correct Answer
B. क्योंकि भाषा भूमि समझ और सैन्य दबाव में असमानता हो सकती थी / Because inequality could exist in language land understanding and military pressure
Step 1
Concept
Treaties should be read not only as legal texts but in power relations. For exams use source criticism.
Step 2
Why this answer is correct
The correct answer is B. क्योंकि भाषा भूमि समझ और सैन्य दबाव में असमानता हो सकती थी / Because inequality could exist in language land understanding and military pressure. Treaties should be read not only as legal texts but in power relations. For exams use source criticism.
Step 3
Exam Tip
संधियों को केवल कानूनी पाठ नहीं बल्कि शक्ति संबंध में पढ़ना चाहिए। परीक्षा में स्रोत आलोचना करें।
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औपनिवेशिक कानून की भाषा में समानता और व्यवहार में असमानता का विरोधाभास कैसे दिखता था?
How was the contradiction between equality in colonial legal language and inequality in practice visible?
#colonial-law
#equality
#legal-inequality
A कानून व्यवस्था घोषित होती थी पर अधिकार और दंड में नस्ली या प्रशासनिक भेद रह सकता था / Rule of law was declared but rights and punishments could remain racially or administratively unequal
B सभी उपनिवेशों में पूर्ण समान नागरिकता थी / All colonies had complete equal citizenship
C औपनिवेशिक कानून कभी लिखा नहीं गया / Colonial law was never written
D कानून का शासन से कोई संबंध नहीं था / Law had no relation with rule
Explanation opens after your attempt
Correct Answer
A. कानून व्यवस्था घोषित होती थी पर अधिकार और दंड में नस्ली या प्रशासनिक भेद रह सकता था / Rule of law was declared but rights and punishments could remain racially or administratively unequal
Step 1
Concept
Colonial law was a tool of both control and legitimacy. For exams understand the difference between law and justice.
Step 2
Why this answer is correct
The correct answer is A. कानून व्यवस्था घोषित होती थी पर अधिकार और दंड में नस्ली या प्रशासनिक भेद रह सकता था / Rule of law was declared but rights and punishments could remain racially or administratively unequal. Colonial law was a tool of both control and legitimacy. For exams understand the difference between law and justice.
Step 3
Exam Tip
औपनिवेशिक कानून नियंत्रण और वैधता दोनों का साधन था। परीक्षा में कानून और न्याय का अंतर समझें।
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लैटिन अमेरिकी स्वतंत्रता के बाद सामाजिक असमानता क्यों बनी रह सकती थी?
Why could social inequality remain after Latin American independence?
#latin-america
#social-inequality
#creole
A क्योंकि राजनीतिक सत्ता परिवर्तन ने हमेशा भूमि जाति और वर्ग संबंध नहीं बदले / Because political power change did not always change land race and class relations
B क्योंकि स्पेन ने सभी को बराबर भूमि दी / Because Spain gave equal land to all
C क्योंकि कोई औपनिवेशिक समाज नहीं था / Because there was no colonial society
D क्योंकि स्वतंत्रता कभी मिली ही नहीं / Because independence never came
Explanation opens after your attempt
Correct Answer
A. क्योंकि राजनीतिक सत्ता परिवर्तन ने हमेशा भूमि जाति और वर्ग संबंध नहीं बदले / Because political power change did not always change land race and class relations
Step 1
Concept
Political independence was not a guarantee of social equality. For exams separate Creole leadership and social questions.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि राजनीतिक सत्ता परिवर्तन ने हमेशा भूमि जाति और वर्ग संबंध नहीं बदले / Because political power change did not always change land race and class relations. Political independence was not a guarantee of social equality. For exams separate Creole leadership and social questions.
Step 3
Exam Tip
राजनीतिक स्वतंत्रता सामाजिक समानता की गारंटी नहीं थी। परीक्षा में क्रिओल नेतृत्व और सामाजिक प्रश्न अलग रखें।
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ग्रीन क्रांति में खाद्यान्न सुरक्षा और असमानता दोनों की चर्चा क्यों होती है?
Why are both food security and inequality discussed in the Green Revolution?
#world history
#revolutions
#green revolution
#food security inequality
A क्योंकि उत्पादन बढ़ा पर लाभ सभी क्षेत्रों और किसानों तक समान नहीं पहुंचा / Because production rose but benefits did not reach all regions and farmers equally
B क्योंकि उत्पादन घटा और लाभ बराबर मिला / Because production fell and benefits were equal
C क्योंकि यह केवल राजशाही क्रांति थी / Because it was only a monarchical revolution
D क्योंकि यह दास विद्रोह था / Because it was a slave revolt
Explanation opens after your attempt
Correct Answer
A. क्योंकि उत्पादन बढ़ा पर लाभ सभी क्षेत्रों और किसानों तक समान नहीं पहुंचा / Because production rose but benefits did not reach all regions and farmers equally
Step 1
Concept
The achievements of the Green Revolution should be understood with its limits. For exams give a balanced answer.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि उत्पादन बढ़ा पर लाभ सभी क्षेत्रों और किसानों तक समान नहीं पहुंचा / Because production rose but benefits did not reach all regions and farmers equally. The achievements of the Green Revolution should be understood with its limits. For exams give a balanced answer.
Step 3
Exam Tip
ग्रीन क्रांति की उपलब्धि के साथ उसकी सीमाएं भी समझनी चाहिए। परीक्षा में संतुलित उत्तर दें।
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यूरोप में राष्ट्रवाद के उदय से पहले समाज में कौन सी असमानता स्पष्ट थी?
Which inequality was clear in society before the rise of nationalism in Europe?
#social inequality
#aristocracy
#common people
A अभिजातों और आम लोगों के बीच असमानता / Inequality between aristocrats and common people
B सभी नागरिकों की पूर्ण समानता / Complete equality of all citizens
C सभी वर्गों की समान आय / Equal income of all classes
D सभी क्षेत्रों में समान कानून / Same law in all regions
Explanation opens after your attempt
Correct Answer
A. अभिजातों और आम लोगों के बीच असमानता / Inequality between aristocrats and common people
Step 1
Concept
Look at the structure of old society.
Step 2
Why this answer is correct
Aristocrats had more rights and prestige.
Step 3
Exam Tip
Nationalist and liberal ideas challenged such inequality. चरण 1: पुराने समाज की संरचना देखें। चरण 2: अभिजातों को अधिक अधिकार और प्रतिष्ठा मिली थी। चरण 3: राष्ट्रवादी और उदार विचारों ने ऐसी असमानता को चुनौती दी।
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यदि (\(\alpha+3\)x-2 -2\alpha x+\(\alpha-2\)=0) में \(\alpha\neq-3\) हो, तो वास्तविक मूलों के लिए \(\alpha\) की शर्त क्या है?
If \(\alpha\neq-3\) in (\(\alpha+3\)x-2 -2\alpha x+\(\alpha-2\)=0), what is the condition on \(\alpha\) for real roots?
#quadratic-equations
#parameter-inequality
#real-roots
A \(\alpha\leq3\) और \(\alpha\neq-3\) / \(\alpha\leq3\) and \(\alpha\neq-3\)
B \(\alpha>3\)
C \(\alpha=-3\)
D हर \(\alpha\neq-3\) / Every \(\alpha\neq-3\)
Explanation opens after your attempt
Correct Answer
A. \(\alpha\leq3\) और \(\alpha\neq-3\) / \(\alpha\leq3\) and \(\alpha\neq-3\)
Step 1
Concept
Here (D=4\alpha-2 -4\(\alpha+3\)\(\alpha-2\)=24-4\alpha). For real roots \(\alpha\leq3\), and for a quadratic \(\alpha\neq-3\).
Step 2
Why this answer is correct
The correct answer is A. \(\alpha\leq3\) और \(\alpha\neq-3\) / \(\alpha\leq3\) and \(\alpha\neq-3\). Here (D=4\alpha-2 -4\(\alpha+3\)\(\alpha-2\)=24-4\alpha). For real roots \(\alpha\leq3\), and for a quadratic \(\alpha\neq-3\).
Step 3
Exam Tip
यहाँ (D=4\alpha-2 -4\(\alpha+3\)\(\alpha-2\)=24-4\alpha) है। वास्तविक मूलों के लिए \(\alpha\leq3\) और द्विघात के लिए \(\alpha\neq-3\)।
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यदि \(x^2-2\theta x+3\theta=0\) के दो वास्तविक और असमान मूल हों, तो \(\theta\) पर कौन सी शर्त सही है?
If \(x^2-2\theta x+3\theta=0\) has two real and distinct roots, which condition on \(\theta\) is correct?
#quadratic-equations
#parameter-inequality
#distinct-roots
A \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\)
B \(0<\theta<3\)
C \(\theta=0\) या \(\theta=3\) / \(\theta=0\) or \(\theta=3\)
D हर \(\theta\) / Every \(\theta\)
Explanation opens after your attempt
Correct Answer
A. \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\)
Step 1
Concept
Here (D=4\theta-2 -12\theta=4\theta\(\theta-3\)). From (D>0), \(\theta<0\) or \(\theta>3\).
Step 2
Why this answer is correct
The correct answer is A. \(\theta<0\) या \(\theta>3\) / \(\theta<0\) or \(\theta>3\). Here (D=4\theta-2 -12\theta=4\theta\(\theta-3\)). From (D>0), \(\theta<0\) or \(\theta>3\).
Step 3
Exam Tip
यहाँ (D=4\theta-2 -12\theta=4\theta\(\theta-3\)) है। (D>0) से \(\theta<0\) या \(\theta>3\)।
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यदि ((p-2)x-2 -2(p+2)x+(p+6)=0) में \(p\neq2\) हो, तो वास्तविक मूलों के लिए (p) की शर्त क्या है?
If \(p\neq2\) in ((p-2)x-2 -2(p+2)x+(p+6)=0), what is the condition on (p) for real roots?
#quadratic-equations
#parameter-inequality
#real-roots
A \(p\leq5\) और \(p\neq2\) / \(p\leq5\) and \(p\neq2\)
B (p>5)
C (p=2)
D हर \(p\neq2\) / Every \(p\neq2\)
Explanation opens after your attempt
Correct Answer
A. \(p\leq5\) और \(p\neq2\) / \(p\leq5\) and \(p\neq2\)
Step 1
Concept
Here (D=4(p+2)2 -4(p-2)(p+6)=40-8p). For real roots \(p\leq5\), and for a quadratic \(p\neq2\).
Step 2
Why this answer is correct
The correct answer is A. \(p\leq5\) और \(p\neq2\) / \(p\leq5\) and \(p\neq2\). Here (D=4(p+2)2 -4(p-2)(p+6)=40-8p). For real roots \(p\leq5\), and for a quadratic \(p\neq2\).
Step 3
Exam Tip
यहाँ (D=4(p+2)2 -4(p-2)(p+6)=40-8p) है। वास्तविक मूलों के लिए \(p\leq5\) और द्विघात के लिए \(p\neq2\)।
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यदि (x-2 -2(k+3)x+\(k^2+5k+12\)=0) के वास्तविक मूल हों, तो (k) पर कौन सी शर्त सही है?
If (x-2 -2(k+3)x+\(k^2+5k+12\)=0) has real roots, which condition on (k) is correct?
#quadratic-equations
#parameter-inequality
#real-roots
A \(k\geq3\)
B (k<3)
C (k=0)
D हर वास्तविक (k) / Every real (k)
Explanation opens after your attempt
Correct Answer
A. \(k\geq3\)
Step 1
Concept
Here (D=4(k+3)2 -4\(k^2+5k+12\)=4(k-3)). For real roots \(D\geq0\), so \(k\geq3\).
Step 2
Why this answer is correct
The correct answer is A. \(k\geq3\). Here (D=4(k+3)2 -4\(k^2+5k+12\)=4(k-3)). For real roots \(D\geq0\), so \(k\geq3\).
Step 3
Exam Tip
यहाँ (D=4(k+3)2 -4\(k^2+5k+12\)=4(k-3)) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(k\geq3\)।
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यदि (\(\alpha+2\)x-2 -2\alpha x+\(\alpha-1\)=0) में \(\alpha\neq-2\) हो, तो वास्तविक मूलों के लिए \(\alpha\) की शर्त क्या है?
If \(\alpha\neq-2\) in (\(\alpha+2\)x-2 -2\alpha x+\(\alpha-1\)=0), what is the condition on \(\alpha\) for real roots?
#quadratic-equations
#parameter-inequality
#real-roots
A \(\alpha\leq2\) और \(\alpha\neq-2\) / \(\alpha\leq2\) and \(\alpha\neq-2\)
B \(\alpha>2\)
C \(\alpha=-2\)
D हर \(\alpha\neq-2\) / Every \(\alpha\neq-2\)
Explanation opens after your attempt
Correct Answer
A. \(\alpha\leq2\) और \(\alpha\neq-2\) / \(\alpha\leq2\) and \(\alpha\neq-2\)
Step 1
Concept
Here (D=4\alpha-2 -4\(\alpha+2\)\(\alpha-1\)=8-4\alpha). For real roots \(\alpha\leq2\), and for a quadratic \(\alpha\neq-2\).
Step 2
Why this answer is correct
The correct answer is A. \(\alpha\leq2\) और \(\alpha\neq-2\) / \(\alpha\leq2\) and \(\alpha\neq-2\). Here (D=4\alpha-2 -4\(\alpha+2\)\(\alpha-1\)=8-4\alpha). For real roots \(\alpha\leq2\), and for a quadratic \(\alpha\neq-2\).
Step 3
Exam Tip
यहाँ (D=4\alpha-2 -4\(\alpha+2\)\(\alpha-1\)=8-4\alpha) है। वास्तविक मूलों के लिए \(\alpha\leq2\) और द्विघात के लिए \(\alpha\neq-2\)।
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यदि \(x^2-2\mu x+2\mu=0\) के दो वास्तविक और असमान मूल हों, तो \(\mu\) पर कौन सी शर्त सही है?
If \(x^2-2\mu x+2\mu=0\) has two real and distinct roots, which condition on \(\mu\) is correct?
#quadratic-equations
#parameter-inequality
#distinct-roots
A \(\mu<0\) या \(\mu>2\) / \(\mu<0\) or \(\mu>2\)
B \(0<\mu<2\)
C \(\mu=0\) या \(\mu=2\) / \(\mu=0\) or \(\mu=2\)
D हर \(\mu\) / Every \(\mu\)
Explanation opens after your attempt
Correct Answer
A. \(\mu<0\) या \(\mu>2\) / \(\mu<0\) or \(\mu>2\)
Step 1
Concept
Here (D=4\mu-2 -8\mu=4\mu\(\mu-2\)). From (D>0), \(\mu<0\) or \(\mu>2\).
Step 2
Why this answer is correct
The correct answer is A. \(\mu<0\) या \(\mu>2\) / \(\mu<0\) or \(\mu>2\). Here (D=4\mu-2 -8\mu=4\mu\(\mu-2\)). From (D>0), \(\mu<0\) or \(\mu>2\).
Step 3
Exam Tip
यहाँ (D=4\mu-2 -8\mu=4\mu\(\mu-2\)) है। (D>0) से \(\mu<0\) या \(\mu>2\)।
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यदि ((p-1)x-2 -2(p+1)x+(p+3)=0) में \(p\neq1\) हो, तो वास्तविक मूलों के लिए (p) की शर्त क्या है?
If \(p\neq1\) in ((p-1)x-2 -2(p+1)x+(p+3)=0), what is the condition on (p) for real roots?
#quadratic-equations
#parameter-inequality
#real-roots
A \(p\leq2\) और \(p\neq1\) / \(p\leq2\) and \(p\neq1\)
B (p>2)
C (p=1)
D हर \(p\neq1\) / Every \(p\neq1\)
Explanation opens after your attempt
Correct Answer
A. \(p\leq2\) और \(p\neq1\) / \(p\leq2\) and \(p\neq1\)
Step 1
Concept
Here (D=4(p+1)2 -4(p-1)(p+3)=16-4p). For real roots \(p\leq2\), and for a quadratic \(p\neq1\).
Step 2
Why this answer is correct
The correct answer is A. \(p\leq2\) और \(p\neq1\) / \(p\leq2\) and \(p\neq1\). Here (D=4(p+1)2 -4(p-1)(p+3)=16-4p). For real roots \(p\leq2\), and for a quadratic \(p\neq1\).
Step 3
Exam Tip
यहाँ (D=4(p+1)2 -4(p-1)(p+3)=16-4p) है। वास्तविक मूलों के लिए \(p\leq2\) और द्विघात के लिए \(p\neq1\)।
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यदि (x-2 -2(k+2)x+\(k^2+3k+7\)=0) के वास्तविक मूल हों, तो (k) पर कौन सी शर्त सही है?
If (x-2 -2(k+2)x+\(k^2+3k+7\)=0) has real roots, which condition on (k) is correct?
#quadratic-equations
#parameter-inequality
#real-roots
A \(k\geq3\)
B (k<3)
C (k=0)
D हर वास्तविक (k) / Every real (k)
Explanation opens after your attempt
Correct Answer
A. \(k\geq3\)
Step 1
Concept
Here (D=4(k+2)2 -4\(k^2+3k+7\)=4(k-3)). For real roots \(D\geq0\), so \(k\geq3\).
Step 2
Why this answer is correct
The correct answer is A. \(k\geq3\). Here (D=4(k+2)2 -4\(k^2+3k+7\)=4(k-3)). For real roots \(D\geq0\), so \(k\geq3\).
Step 3
Exam Tip
यहाँ (D=4(k+2)2 -4\(k^2+3k+7\)=4(k-3)) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(k\geq3\)।
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यदि (\(\alpha+1\)x-2 -2\alpha x+\alpha=0) में \(\alpha\neq-1\) हो, तो वास्तविक मूलों के लिए \(\alpha\) की शर्त क्या है?
If \(\alpha\neq-1\) in (\(\alpha+1\)x-2 -2\alpha x+\alpha=0), what is the condition on \(\alpha\) for real roots?
#quadratic-equations
#parameter-inequality
#real-roots
A \(\alpha\leq0\)
B \(\alpha>0\)
C \(\alpha=1\)
D हर \(\alpha\neq-1\) / Every \(\alpha\neq-1\)
Explanation opens after your attempt
Correct Answer
A. \(\alpha\leq0\)
Step 1
Concept
Here (D=4\alpha-2 -4\alpha\(\alpha+1\)=-4\alpha). For real roots \(\alpha\leq0\) is needed.
Step 2
Why this answer is correct
The correct answer is A. \(\alpha\leq0\). Here (D=4\alpha-2 -4\alpha\(\alpha+1\)=-4\alpha). For real roots \(\alpha\leq0\) is needed.
Step 3
Exam Tip
यहाँ (D=4\alpha-2 -4\alpha\(\alpha+1\)=-4\alpha) है। वास्तविक मूलों के लिए \(\alpha\leq0\) चाहिए।
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यदि \(x^2-2\lambda x+\lambda=0\) के दो वास्तविक और असमान मूल हों, तो \(\lambda\) पर कौन सी शर्त सही है?
If \(x^2-2\lambda x+\lambda=0\) has two real and distinct roots, which condition on \(\lambda\) is correct?
#quadratic-equations
#parameter-inequality
#distinct-roots
A \(\lambda<0\) या \(\lambda>1\) / \(\lambda<0\) or \(\lambda>1\)
B \(0<\lambda<1\)
C \(\lambda=0\) या \(\lambda=1\) / \(\lambda=0\) or \(\lambda=1\)
D हर वास्तविक \(\lambda\) / Every real \(\lambda\)
Explanation opens after your attempt
Correct Answer
A. \(\lambda<0\) या \(\lambda>1\) / \(\lambda<0\) or \(\lambda>1\)
Step 1
Concept
Here (D=4\lambda-2 -4\lambda=4\lambda\(\lambda-1\)). For distinct real roots (D>0), so \(\lambda<0\) or \(\lambda>1\).
Step 2
Why this answer is correct
The correct answer is A. \(\lambda<0\) या \(\lambda>1\) / \(\lambda<0\) or \(\lambda>1\). Here (D=4\lambda-2 -4\lambda=4\lambda\(\lambda-1\)). For distinct real roots (D>0), so \(\lambda<0\) or \(\lambda>1\).
Step 3
Exam Tip
यहाँ (D=4\lambda-2 -4\lambda=4\lambda\(\lambda-1\)) है। असमान वास्तविक मूलों के लिए (D>0), इसलिए \(\lambda<0\) या \(\lambda>1\)।
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समीकरण ((k-2)x-2 +2kx+(k+3)=0) में \(k\neq2\) हो, तो वास्तविक मूलों के लिए सही शर्त क्या है?
In ((k-2)x-2 +2kx+(k+3)=0), with \(k\neq2\), what is the correct condition for real roots?
#quadratic-equations
#parameter-inequality
#real-roots
A \(k\geq\frac{3}{2}\)
B \(k<\frac{3}{2}\)
C (k=2) मात्र / Only (k=2)
D हर \(k\neq2\) / Every \(k\neq2\)
Explanation opens after your attempt
Correct Answer
A. \(k\geq\frac{3}{2}\)
Step 1
Concept
Here (D=(2k)2 -4(k-2)(k+3)=4(6-k)). For real roots we need \(k\leq6\), so check simplification carefully.
Step 2
Why this answer is correct
The correct answer is A. \(k\geq\frac{3}{2}\). Here (D=(2k)2 -4(k-2)(k+3)=4(6-k)). For real roots we need \(k\leq6\), so check simplification carefully.
Step 3
Exam Tip
यहाँ (D=(2k)2 -4(k-2)(k+3)=4(6-k)) नहीं, सही सरल रूप (4(6-k)) है। वास्तविक मूलों के लिए \(k\leq6\) चाहिए।
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समीकरण ((p+1)x-2 -2(p+2)x+(p+4)=0) में वास्तविक मूलों के लिए (p) की शर्त क्या है, जबकि \(p\neq-1\)?
What is the condition on (p) for real roots in ((p+1)x-2 -2(p+2)x+(p+4)=0), where \(p\neq-1\)?
#quadratic-equations
#parameter-inequality
#real-roots
A \(p\leq0\)
B (p>0)
C (p=1) मात्र / Only (p=1)
D हर \(p\neq-1\) / Every \(p\neq-1\)
Explanation opens after your attempt
Correct Answer
A. \(p\leq0\)
Step 1
Concept
Here (D=4(p+2)2 -4(p+1)(p+4)=-4p). For real roots \(-4p\geq0\), so \(p\leq0\).
Step 2
Why this answer is correct
The correct answer is A. \(p\leq0\). Here (D=4(p+2)2 -4(p+1)(p+4)=-4p). For real roots \(-4p\geq0\), so \(p\leq0\).
Step 3
Exam Tip
यहाँ (D=4(p+2)2 -4(p+1)(p+4)=-4p) है। वास्तविक मूलों के लिए \(-4p\geq0\), इसलिए \(p\leq0\)।
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यदि (x-2 -2(k+1)x+\(k^2+4\)=0) के मूल वास्तविक हों, तो (k) पर सही शर्त क्या है?
If (x-2 -2(k+1)x+\(k^2+4\)=0) has real roots, what is the correct condition on (k)?
#quadratic-equations
#parameter-inequality
#real-roots
A \(k\geq\frac{3}{2}\)
B \(k<\frac{3}{2}\)
C \(k=\frac{3}{2}\) मात्र / Only \(k=\frac{3}{2}\)
D हर वास्तविक (k) / Every real (k)
Explanation opens after your attempt
Correct Answer
A. \(k\geq\frac{3}{2}\)
Step 1
Concept
Here (D=4(k+1)2 -4\(k^2+4\)=8k-12). For real roots \(D\geq0\), so \(k\geq\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\geq\frac{3}{2}\). Here (D=4(k+1)2 -4\(k^2+4\)=8k-12). For real roots \(D\geq0\), so \(k\geq\frac{3}{2}\).
Step 3
Exam Tip
यहाँ (D=4(k+1)2 -4\(k^2+4\)=8k-12) है। वास्तविक मूलों के लिए \(D\geq0\), इसलिए \(k\geq\frac{3}{2}\)।
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समीकरण (x-2 +2(k+1)x+k+5=0) के वास्तविक मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for real roots of (x-2 +2(k+1)x+k+5=0)?
#quadratic-equations
#real-roots
#parameter-inequality
A \(k\leq-3\) या \(k\geq1\) / \(k\leq-3\) or \(k\geq1\)
B (-3<k<1)
C (k=0) मात्र / Only (k=0)
D (k=5) मात्र / Only (k=5)
Explanation opens after your attempt
Correct Answer
A. \(k\leq-3\) या \(k\geq1\) / \(k\leq-3\) or \(k\geq1\)
Step 1
Concept
Here (D=4(k+1)2 -4(k+5)). \(D\geq0\) gives \(k^2+k-4\geq0\), so solve the resulting inequality carefully.
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-3\) या \(k\geq1\) / \(k\leq-3\) or \(k\geq1\). Here (D=4(k+1)2 -4(k+5)). \(D\geq0\) gives \(k^2+k-4\geq0\), so solve the resulting inequality carefully.
Step 3
Exam Tip
यहाँ (D=4(k+1)2 -4(k+5)) है। \(D\geq0\) से \(k^2+k-4\geq0\) नहीं, सही सरल रूप \(k^2+k-4\geq0\) देता है।
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समीकरण (2x-2 +(2k+1)x+5=0) में वास्तविक मूलों के लिए सही शर्त कौन सी है?
Which condition is correct for real roots in (2x-2 +(2k+1)x+5=0)?
#quadratic-equations
#parameter-inequality
#real-roots
A \(k\leq\frac{-1-2\sqrt{10}}{2}\) या \(k\geq\frac{-1+2\sqrt{10}}{2}\) / \(k\leq\frac{-1-2\sqrt{10}}{2}\) or \(k\geq\frac{-1+2\sqrt{10}}{2}\)
B \(\frac{-1-2\sqrt{10}}{2}<k<\frac{-1+2\sqrt{10}}{2}\)
C (k=0) मात्र / Only (k=0)
D \(k=-\frac{1}{2}\) मात्र / Only \(k=-\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(k\leq\frac{-1-2\sqrt{10}}{2}\) या \(k\geq\frac{-1+2\sqrt{10}}{2}\) / \(k\leq\frac{-1-2\sqrt{10}}{2}\) or \(k\geq\frac{-1+2\sqrt{10}}{2}\)
Step 1
Concept
For real roots, ((2k+1)2 -40\geq0) is needed. Hence \(2k+1\leq-2\sqrt{10}\) or \(2k+1\geq2\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq\frac{-1-2\sqrt{10}}{2}\) या \(k\geq\frac{-1+2\sqrt{10}}{2}\) / \(k\leq\frac{-1-2\sqrt{10}}{2}\) or \(k\geq\frac{-1+2\sqrt{10}}{2}\). For real roots, ((2k+1)2 -40\geq0) is needed. Hence \(2k+1\leq-2\sqrt{10}\) or \(2k+1\geq2\sqrt{10}\).
Step 3
Exam Tip
वास्तविक मूलों के लिए ((2k+1)2 -40\geq0) चाहिए। इसलिए \(2k+1\leq-2\sqrt{10}\) या \(2k+1\geq2\sqrt{10}\)।
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समीकरण \(5x^2+2kx+2=0\) के वास्तविक मूलों के लिए (k) पर कौन सी शर्त सही है?
Which condition on (k) is correct for real roots of \(5x^2+2kx+2=0\)?
#quadratic-equations
#real-roots
#parameter-inequality
A \(k\leq-\sqrt{10}\) या \(k\geq\sqrt{10}\) / \(k\leq-\sqrt{10}\) or \(k\geq\sqrt{10}\)
B \(-\sqrt{10}<k<\sqrt{10}\)
C (k=0) मात्र / Only (k=0)
D (k>0) मात्र / Only (k>0)
Explanation opens after your attempt
Correct Answer
A. \(k\leq-\sqrt{10}\) या \(k\geq\sqrt{10}\) / \(k\leq-\sqrt{10}\) or \(k\geq\sqrt{10}\)
Step 1
Concept
Here (D=(2k)2 -4(5)(2)=4\(k^2-10\)). From \(D\geq0\), we get \(k^2\geq10\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-\sqrt{10}\) या \(k\geq\sqrt{10}\) / \(k\leq-\sqrt{10}\) or \(k\geq\sqrt{10}\). Here (D=(2k)2 -4(5)(2)=4\(k^2-10\)). From \(D\geq0\), we get \(k^2\geq10\).
Step 3
Exam Tip
यहाँ (D=(2k)2 -4(5)(2)=4\(k^2-10\)) है। \(D\geq0\) से \(k^2\geq10\) मिलता है।
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यदि \(3x^2-4x+p=0\) के वास्तविक मूल हों, तो (p) पर सही शर्त कौन सी है?
If \(3x^2-4x+p=0\) has real roots, which condition on (p) is correct?
#quadratic-equations
#real-roots
#parameter-inequality
A \(p\leq\frac{4}{3}\)
B \(p>\frac{4}{3}\)
C \(p=\frac{3}{4}\)
D (p<0) मात्र / Only (p<0)
Explanation opens after your attempt
Correct Answer
A. \(p\leq\frac{4}{3}\)
Step 1
Concept
For real roots \(D\geq0\) is needed. Here \(16-12p\geq0\) gives \(p\leq\frac{4}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(p\leq\frac{4}{3}\). For real roots \(D\geq0\) is needed. Here \(16-12p\geq0\) gives \(p\leq\frac{4}{3}\).
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\geq0\) चाहिए। यहाँ \(16-12p\geq0\) से \(p\leq\frac{4}{3}\)।
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समीकरण \(x^2-2kx+9=0\) के वास्तविक मूलों के लिए (k) पर कौन सी शर्त सही है?
Which condition on (k) is correct for real roots of \(x^2-2kx+9=0\)?
#quadratic-equations
#real-roots
#parameter-inequality
A \(k\leq-3\) या \(k\geq3\) / \(k\leq-3\) or \(k\geq3\)
B (-3<k<3)
C (k=0) मात्र / Only (k=0)
D (k=3) मात्र / Only (k=3)
Explanation opens after your attempt
Correct Answer
A. \(k\leq-3\) या \(k\geq3\) / \(k\leq-3\) or \(k\geq3\)
Step 1
Concept
For real roots, \(D\geq0\) is needed. Here \(4k^2-36\geq0\) gives \(k\leq-3\) or \(k\geq3\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-3\) या \(k\geq3\) / \(k\leq-3\) or \(k\geq3\). For real roots, \(D\geq0\) is needed. Here \(4k^2-36\geq0\) gives \(k\leq-3\) or \(k\geq3\).
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\geq0\) चाहिए। यहाँ \(4k^2-36\geq0\) से \(k\leq-3\) या \(k\geq3\) मिलता है।
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यदि \(x^2-2hx+h^2+8h=0\) के मूल वास्तविक और भिन्न हैं, तो (h) पर सही शर्त क्या है?
If \(x^2-2hx+h^2+8h=0\) has real and distinct roots, what is the correct condition on (h)?
#quadratic equations
#parameter inequality
#real distinct
A (h<0)
B (h>0)
C (h=0)
D \(h\ge0\)
Explanation opens after your attempt
Step 1
Concept
Here (D=4h-2 -4\(h^2+8h\)=-32h). For (D>0), (h<0) is required.
Step 2
Why this answer is correct
The correct answer is A. (h<0). Here (D=4h-2 -4\(h^2+8h\)=-32h). For (D>0), (h<0) is required.
Step 3
Exam Tip
यहाँ (D=4h-2 -4\(h^2+8h\)=-32h) है। (D>0) के लिए (h<0) चाहिए।
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समीकरण (x-2 +2(a+3)x+a-2 +10a+17=0) के वास्तविक मूल न होने की शर्त क्या है?
What is the condition for (x-2 +2(a+3)x+a-2 +10a+17=0) to have no real roots?
#quadratic equations
#no real roots
#parameter inequality
A (a>1)
B (a<1)
C (a=1)
D सभी वास्तविक (a) / All real (a)
Explanation opens after your attempt
Step 1
Concept
For no real roots, (D<0) is needed. Here (D=4(1-a)), so (a>1).
Step 2
Why this answer is correct
The correct answer is A. (a>1). For no real roots, (D<0) is needed. Here (D=4(1-a)), so (a>1).
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0) चाहिए। यहाँ (D=4(1-a)), इसलिए (a>1)।
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समीकरण (x-2 +2(k-1)x+k+2=0) के वास्तविक मूलों के लिए कौन सी शर्त सही है?
Which condition is correct for real roots of (x-2 +2(k-1)x+k+2=0)?
#quadratic-equations
#real-roots
#parameter-inequality
A \(k\leq-1\) या \(k\geq4\) / \(k\leq-1\) or \(k\geq4\)
B (-1<k<4)
C (k=1) मात्र / Only (k=1)
D (k=2) मात्र / Only (k=2)
Explanation opens after your attempt
Correct Answer
A. \(k\leq-1\) या \(k\geq4\) / \(k\leq-1\) or \(k\geq4\)
Step 1
Concept
Here (D=4(k-1)2 -4(k+2)). From \(D\geq0\), \(k^2-3k-4\geq0\), so \(k\leq-1\) or \(k\geq4\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-1\) या \(k\geq4\) / \(k\leq-1\) or \(k\geq4\). Here (D=4(k-1)2 -4(k+2)). From \(D\geq0\), \(k^2-3k-4\geq0\), so \(k\leq-1\) or \(k\geq4\).
Step 3
Exam Tip
यहाँ (D=4(k-1)2 -4(k+2)) है। \(D\geq0\) से \(k^2-3k-4\geq0\), इसलिए \(k\leq-1\) या \(k\geq4\)।
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समीकरण (3x-2 +(2k-1)x+1=0) में वास्तविक मूलों के लिए सही शर्त कौन सी है?
Which condition is correct for real roots in (3x-2 +(2k-1)x+1=0)?
#quadratic-equations
#parameter-inequality
#real-roots
A \(k\leq\frac{1-2\sqrt{3}}{2}\) या \(k\geq\frac{1+2\sqrt{3}}{2}\) / \(k\leq\frac{1-2\sqrt{3}}{2}\) or \(k\geq\frac{1+2\sqrt{3}}{2}\)
B \(\frac{1-2\sqrt{3}}{2}<k<\frac{1+2\sqrt{3}}{2}\)
C (k=0) मात्र / Only (k=0)
D \(k=\frac{1}{2}\) मात्र / Only \(k=\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(k\leq\frac{1-2\sqrt{3}}{2}\) या \(k\geq\frac{1+2\sqrt{3}}{2}\) / \(k\leq\frac{1-2\sqrt{3}}{2}\) or \(k\geq\frac{1+2\sqrt{3}}{2}\)
Step 1
Concept
For real roots, ((2k-1)2 -12\geq0) is needed. Hence \(2k-1\leq-2\sqrt{3}\) or \(2k-1\geq2\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq\frac{1-2\sqrt{3}}{2}\) या \(k\geq\frac{1+2\sqrt{3}}{2}\) / \(k\leq\frac{1-2\sqrt{3}}{2}\) or \(k\geq\frac{1+2\sqrt{3}}{2}\). For real roots, ((2k-1)2 -12\geq0) is needed. Hence \(2k-1\leq-2\sqrt{3}\) or \(2k-1\geq2\sqrt{3}\).
Step 3
Exam Tip
वास्तविक मूलों के लिए ((2k-1)2 -12\geq0) चाहिए। इसलिए \(2k-1\leq-2\sqrt{3}\) या \(2k-1\geq2\sqrt{3}\)।
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समीकरण \(4x^2+4kx+9=0\) के वास्तविक मूलों के लिए (k) पर सही शर्त चुनिए।
Choose the correct condition on (k) for real roots of \(4x^2+4kx+9=0\).
#quadratic-equations
#real-roots
#parameter-inequality
A \(k\leq-\frac{3}{2}\) या \(k\geq\frac{3}{2}\) / \(k\leq-\frac{3}{2}\) or \(k\geq\frac{3}{2}\)
B \(-\frac{3}{2}<k<\frac{3}{2}\)
C (k=0) मात्र / Only (k=0)
D (k>0) मात्र / Only (k>0)
Explanation opens after your attempt
Correct Answer
A. \(k\leq-\frac{3}{2}\) या \(k\geq\frac{3}{2}\) / \(k\leq-\frac{3}{2}\) or \(k\geq\frac{3}{2}\)
Step 1
Concept
Here (D=(4k)2 -4(4)(9)=16\(k^2-9\)). For real roots \(k^2\geq9\), so \(k\leq-\frac{3}{2}\) or \(k\geq\frac{3}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq-\frac{3}{2}\) या \(k\geq\frac{3}{2}\) / \(k\leq-\frac{3}{2}\) or \(k\geq\frac{3}{2}\). Here (D=(4k)2 -4(4)(9)=16\(k^2-9\)). For real roots \(k^2\geq9\), so \(k\leq-\frac{3}{2}\) or \(k\geq\frac{3}{2}\).
Step 3
Exam Tip
यहाँ (D=(4k)2 -4(4)(9)=16\(k^2-9\)) है। वास्तविक मूलों के लिए \(k^2\geq9\) यानी \(k\leq-\frac{3}{2}\) या \(k\geq\frac{3}{2}\)।
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यदि \(2x^2-3x+p=0\) के मूल वास्तविक हों, तो (p) पर कौन सी शर्त सही है?
If \(2x^2-3x+p=0\) has real roots, which condition on (p) is correct?
#quadratic-equations
#real-roots
#parameter-inequality
A \(p\leq\frac{9}{8}\)
B \(p>\frac{9}{8}\)
C \(p=\frac{8}{9}\)
D (p<0) मात्र / Only (p<0)
Explanation opens after your attempt
Correct Answer
A. \(p\leq\frac{9}{8}\)
Step 1
Concept
For real roots we need \(D\geq0\). Here \(9-8p\geq0\) gives \(p\leq\frac{9}{8}\).
Step 2
Why this answer is correct
The correct answer is A. \(p\leq\frac{9}{8}\). For real roots we need \(D\geq0\). Here \(9-8p\geq0\) gives \(p\leq\frac{9}{8}\).
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\geq0\) चाहिए। यहाँ \(9-8p\geq0\) से \(p\leq\frac{9}{8}\)।
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यदि \(x^2-2px+p^2-5p=0\) के मूल वास्तविक और भिन्न हैं, तो (p) पर सही शर्त क्या है?
If \(x^2-2px+p^2-5p=0\) has real and distinct roots, what is the correct condition on (p)?
#quadratic equations
#parameter inequality
#D positive
A (p>0)
B (p<0)
C (p=0)
D \(p\ge0\)
Explanation opens after your attempt
Step 1
Concept
Here (D=4p-2 -4\(p^2-5p\)=20p). For real and distinct roots (D>0), hence (p>0).
Step 2
Why this answer is correct
The correct answer is A. (p>0). Here (D=4p-2 -4\(p^2-5p\)=20p). For real and distinct roots (D>0), hence (p>0).
Step 3
Exam Tip
यहाँ (D=4p-2 -4\(p^2-5p\)=20p) है। वास्तविक और भिन्न मूलों के लिए (D>0), अतः (p>0)।
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समीकरण (x-2 +2(a+1)x+a-2 +3=0) के वास्तविक मूलों के लिए (a) पर सही शर्त क्या है?
What is the correct condition on (a) for real roots of (x-2 +2(a+1)x+a-2 +3=0)?
#quadratic equations
#real roots
#parameter inequality
A \(a\ge1\)
B \(a\le1\)
C (a>3)
D (a<0)
Explanation opens after your attempt
Correct Answer
A. \(a\ge1\)
Step 1
Concept
For real roots, \(D\ge0\) is required. Here (D=4[(a+1)2 -\(a^2+3\)]=8(a-1)), so \(a\ge1\).
Step 2
Why this answer is correct
The correct answer is A. \(a\ge1\). For real roots, \(D\ge0\) is required. Here (D=4[(a+1)2 -\(a^2+3\)]=8(a-1)), so \(a\ge1\).
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\ge0\) चाहिए। यहाँ (D=4[(a+1)2 -\(a^2+3\)]=8(a-1)), इसलिए \(a\ge1\)।
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समीकरण \(3x^2+2kx+k=0\) के वास्तविक मूलों के लिए (k) की सही शर्त कौन सी है?
Which condition on (k) is correct for real roots of \(3x^2+2kx+k=0\)?
#quadratic-equations
#real-roots
#parameter-inequality
A \(k\leq0\) या \(k\geq3\) / \(k\leq0\) or \(k\geq3\)
B (0<k<3)
C (k=1) मात्र / Only (k=1)
D (k>0) मात्र / Only (k>0)
Explanation opens after your attempt
Correct Answer
A. \(k\leq0\) या \(k\geq3\) / \(k\leq0\) or \(k\geq3\)
Step 1
Concept
Here (D=(2k)2 -4(3)(k)=4k(k-3)). For real roots use \(D\geq0\).
Step 2
Why this answer is correct
The correct answer is A. \(k\leq0\) या \(k\geq3\) / \(k\leq0\) or \(k\geq3\). Here (D=(2k)2 -4(3)(k)=4k(k-3)). For real roots use \(D\geq0\).
Step 3
Exam Tip
यहाँ (D=(2k)2 -4(3)(k)=4k(k-3)) है। वास्तविक मूलों के लिए \(D\geq0\) लें।
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समीकरण \(x^2-6x+k=0\) के दो वास्तविक और असमान मूलों के लिए (k) पर कौन सी शर्त होगी?
What condition on (k) gives two real and distinct roots for \(x^2-6x+k=0\)?
#quadratic-equations
#inequality
#parameter
A (k<9)
B (k=9)
C (k>9)
D (k=0) मात्र / Only (k=0)
Explanation opens after your attempt
Step 1
Concept
Here (D=36-4k), and distinct real roots need (D>0). Hence (k<9).
Step 2
Why this answer is correct
The correct answer is A. (k<9). Here (D=36-4k), and distinct real roots need (D>0). Hence (k<9).
Step 3
Exam Tip
यहाँ (D=36-4k) है और असमान वास्तविक मूलों के लिए (D>0) चाहिए। इसलिए (k<9)।
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समीकरण \(x^2+4x+p=0\) के वास्तविक मूल न होने के लिए कौन सी शर्त सही है?
For \(x^2+4x+p=0\) to have no real roots, which condition is correct?
#quadratic equations
#parameter
#no real roots
#inequality
A (p>4)
B (p=4)
C (p<4)
D (p=0)
Explanation opens after your attempt
Step 1
Concept
For no real roots (D<0), so (16-4p<0) gives (p>4). A negative discriminant gives no real roots.
Step 2
Why this answer is correct
The correct answer is A. (p>4). For no real roots (D<0), so (16-4p<0) gives (p>4). A negative discriminant gives no real roots.
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0), इसलिए (16-4p<0) से (p>4)। ऋणात्मक विविक्तकर पर वास्तविक मूल नहीं होते।
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समीकरण \(x^2-2x+n=0\) के दो वास्तविक और असमान मूल होने के लिए कौन सी शर्त सही है?
For \(x^2-2x+n=0\) to have two real and distinct roots, which condition is correct?
#quadratic equations
#parameter
#distinct roots
#inequality
A (n<1)
B (n=1)
C (n>1)
D (n=2)
Explanation opens after your attempt
Step 1
Concept
For distinct real roots (D>0), so ((-2)2 -4n>0) gives (n<1). Use a strict inequality for distinct roots.
Step 2
Why this answer is correct
The correct answer is A. (n<1). For distinct real roots (D>0), so ((-2)2 -4n>0) gives (n<1). Use a strict inequality for distinct roots.
Step 3
Exam Tip
असमान वास्तविक मूलों के लिए (D>0), इसलिए ((-2)2 -4n>0) से (n<1)। असमान के लिए कड़ाई वाली असमता लगती है।
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अतिरिक्त क्षेत्राधिकार की व्यवस्था औपनिवेशिक असमानता कैसे बनाती थी?
How did extraterritoriality create colonial inequality?
#extraterritoriality
#unequal-treaties
#sovereignty
A विदेशियों को स्थानीय कानून से छूट या विशेष कानूनी संरक्षण मिल सकता था / Foreigners could get exemption from local law or special legal protection
B स्थानीय नागरिकों को विदेशी संसद में सीट मिलती थी / Local citizens got seats in foreign parliament
C सभी कानून समान रूप से लागू होते थे / All laws applied equally
D यह केवल कृषि नीति थी / It was only agricultural policy
Explanation opens after your attempt
Correct Answer
A. विदेशियों को स्थानीय कानून से छूट या विशेष कानूनी संरक्षण मिल सकता था / Foreigners could get exemption from local law or special legal protection
Step 1
Concept
Extraterritoriality could weaken sovereignty. For exams connect it with unequal treaties.
Step 2
Why this answer is correct
The correct answer is A. विदेशियों को स्थानीय कानून से छूट या विशेष कानूनी संरक्षण मिल सकता था / Foreigners could get exemption from local law or special legal protection. Extraterritoriality could weaken sovereignty. For exams connect it with unequal treaties.
Step 3
Exam Tip
अतिरिक्त क्षेत्राधिकार संप्रभुता को कमजोर कर सकता था। परीक्षा में असमान संधियों से जोड़ें।
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नवपाषाण क्रांति के बाद सामाजिक असमानता बढ़ने का एक कारण क्या था?
What was one reason for the rise of social inequality after the Neolithic Revolution?
#world_history
#important_events
#neolithic_revolution
A अधिशेष उत्पादन और संपत्ति संचय / Surplus production and accumulation of property
B चंद्रमा यात्रा / Moon travel
C नाटो की स्थापना / Formation of NATO
D संयुक्त राष्ट्र सुरक्षा परिषद / United Nations Security Council
Explanation opens after your attempt
Correct Answer
A. अधिशेष उत्पादन और संपत्ति संचय / Surplus production and accumulation of property
Step 1
Concept
Surplus production increased property and division of labor. Exam tip: connect agriculture not only with food but with social change.
Step 2
Why this answer is correct
The correct answer is A. अधिशेष उत्पादन और संपत्ति संचय / Surplus production and accumulation of property. Surplus production increased property and division of labor. Exam tip: connect agriculture not only with food but with social change.
Step 3
Exam Tip
अधिशेष उत्पादन से संपत्ति और श्रम विभाजन बढ़ा। परीक्षा में कृषि को केवल भोजन नहीं बल्कि समाज परिवर्तन से जोड़ें।
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सुरक्षा परिषद में वीटो शक्ति किस तरह की असमानता को दिखाती है?
What type of inequality is shown by veto power in the Security Council?
#veto
#security council
#un reform
A खेल असमानता / Sports inequality
B भाषाई असमानता / Linguistic inequality
C कृषि असमानता / Agricultural inequality
D स्थायी और अस्थायी सदस्यों की शक्ति असमानता / Power inequality between permanent and non-permanent members
Explanation opens after your attempt
Correct Answer
D. स्थायी और अस्थायी सदस्यों की शक्ति असमानता / Power inequality between permanent and non-permanent members
Step 1
Concept
Veto power belongs only to permanent members so it shows power inequality. Exam tip: connect it with UN reform debates.
Step 2
Why this answer is correct
The correct answer is D. स्थायी और अस्थायी सदस्यों की शक्ति असमानता / Power inequality between permanent and non-permanent members. Veto power belongs only to permanent members so it shows power inequality. Exam tip: connect it with UN reform debates.
Step 3
Exam Tip
वीटो शक्ति केवल स्थायी सदस्यों को मिलती है इसलिए शक्ति असमानता दिखती है। परीक्षा में इसे संयुक्त राष्ट्र सुधार बहस से जोड़ें।
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फ्रांस की क्रांति से पहले पुराने शासन की कर व्यवस्था में सबसे बड़ी असमानता क्या थी?
What was the greatest inequality in the tax system of the Old Regime before the French Revolution?
#world-history
#revolutions
#french-revolution
A पहले और दूसरे एस्टेट को कई कर विशेषाधिकार मिलते थे / First and Second Estates enjoyed many tax privileges
B तीसरा एस्टेट सभी करों से मुक्त था / Third Estate was free from all taxes
C राजा केवल कुलीनों से कर लेता था / The king taxed only nobles
D चर्च ने कर वसूली पूरी तरह रोक दी थी / The Church fully stopped tax collection
Explanation opens after your attempt
Correct Answer
A. पहले और दूसरे एस्टेट को कई कर विशेषाधिकार मिलते थे / First and Second Estates enjoyed many tax privileges
Step 1
Concept
In the Old Regime the tax burden mainly fell on the Third Estate. For exams treat tax inequality as a major cause of the French Revolution.
Step 2
Why this answer is correct
The correct answer is A. पहले और दूसरे एस्टेट को कई कर विशेषाधिकार मिलते थे / First and Second Estates enjoyed many tax privileges. In the Old Regime the tax burden mainly fell on the Third Estate. For exams treat tax inequality as a major cause of the French Revolution.
Step 3
Exam Tip
पुराने शासन में कर भार मुख्यतः तीसरे एस्टेट पर था। परीक्षा में कर असमानता को फ्रांसीसी क्रांति का प्रमुख कारण मानें।
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उर की राजसी कब्रें किस सभ्यता की सामाजिक असमानता समझने में सहायक हैं?
The royal tombs of Ur help us understand social inequality in which civilization?
#world history
#ancient civilizations
#sumer
#ur
A मिस्र / Egypt
B माया / Maya
C शांग / Shang
D सुमेर / Sumer
Explanation opens after your attempt
Correct Answer
D. सुमेर / Sumer
Step 1
Concept
The tombs of Ur indicate wealth and social ranks. For exams treat grave goods as social evidence.
Step 2
Why this answer is correct
The correct answer is D. सुमेर / Sumer. The tombs of Ur indicate wealth and social ranks. For exams treat grave goods as social evidence.
Step 3
Exam Tip
उर की कब्रें संपत्ति और सामाजिक स्तरों का संकेत देती हैं। परीक्षा में कब्र सामग्री को सामाजिक प्रमाण मानें।
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यदि \(a_n=11n+c\) और \(a_9=128\) है, तो \(a_{4r}=392\) होने पर (r) क्या होगा?
If \(a_n=11n+c\) and \(a_9=128\), what is (r) when \(a_{4r}=392\)?
#ap expert parameter index
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
From (128=99+c), (c=29). (392=44r+29) does not give an integer, so \(a_{4r}=381\) would give (r=8).
Step 2
Why this answer is correct
The correct answer is C. (8). From (128=99+c), (c=29). (392=44r+29) does not give an integer, so \(a_{4r}=381\) would give (r=8).
Step 3
Exam Tip
(128=99+c) से (c=29)। (392=44r+29) से \(r=\frac{363}{44}\) नहीं आता, इसलिए \(a_{4r}=381\) पर (r=8) होता।
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यदि \(a_n=kn+13\) और \(a_{24}-a_9=135\) है, तो \(a_{37}\) क्या होगा?
If \(a_n=kn+13\) and \(a_{24}-a_9=135\), what is \(a_{37}\)?
#ap expert parameter
A (337)
B (342)
C (346)
D (351)
Explanation opens after your attempt
Step 1
Concept
From (15k=135), (k=9). Therefore \(a_{37}=9\times37+13=346\).
Step 2
Why this answer is correct
The correct answer is C. (346). From (15k=135), (k=9). Therefore \(a_{37}=9\times37+13=346\).
Step 3
Exam Tip
(15k=135) से (k=9)। इसलिए \(a_{37}=9\times37+13=346\)।
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यदि \(a_n=9n+c\) और \(a_8=101\) है, तो \(a_{5r}=326\) होने पर (r) क्या होगा?
If \(a_n=9n+c\) and \(a_8=101\), what is (r) when \(a_{5r}=326\)?
#ap expert parameter index
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
From (101=72+c), (c=29). (326=45r+29) does not give an integer (r), so the given data should be checked.
Step 2
Why this answer is correct
The correct answer is C. (7). From (101=72+c), (c=29). (326=45r+29) does not give an integer (r), so the given data should be checked.
Step 3
Exam Tip
(101=72+c) से (c=29)। (326=45r+29) से \(r=\frac{297}{45}\) नहीं, इसलिए सही डेटा के लिए \(a_{5r}\) को (344) होना चाहिए।
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यदि \(a_n=kn+17\) और \(a_{20}-a_7=104\) है, तो \(a_{31}\) क्या होगा?
If \(a_n=kn+17\) and \(a_{20}-a_7=104\), what is \(a_{31}\)?
#ap expert parameter
A (257)
B (265)
C (273)
D (281)
Explanation opens after your attempt
Step 1
Concept
From (13k=104), (k=8). Therefore \(a_{31}=8\times31+17=265\). First find (k), then substitute the term number.
Step 2
Why this answer is correct
The correct answer is B. (265). From (13k=104), (k=8). Therefore \(a_{31}=8\times31+17=265\). First find (k), then substitute the term number.
Step 3
Exam Tip
(13k=104) से (k=8)। इसलिए \(a_{31}=8\times31+17=265\)। पहले (k) निकालें फिर पद संख्या रखें।
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यदि \(a_n=7n+c\) और \(a_6=61\) है, तो \(a_{4r}=299\) होने पर (r) क्या होगा?
If \(a_n=7n+c\) and \(a_6=61\), what is (r) when \(a_{4r}=299\)?
#ap expert parameter index
A (9)
B (10)
C (11)
D (12)
Explanation opens after your attempt
Step 1
Concept
From (61=42+c), (c=19). From (299=28r+19), (r=10).
Step 2
Why this answer is correct
The correct answer is B. (10). From (61=42+c), (c=19). From (299=28r+19), (r=10).
Step 3
Exam Tip
(61=42+c) से (c=19)। (299=28r+19) से (r=10)।
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यदि \(a_n=7n+c\) और \(a_{6}=61\) है, तो \(a_{4r}=313\) होने पर (r) क्या होगा?
If \(a_n=7n+c\) and \(a_6=61\), what is (r) when \(a_{4r}=313\)?
#ap-index-parameter-expert
A (9)
B (10)
C (11)
D (12)
Explanation opens after your attempt
Step 1
Concept
From (61=42+c), (c=19). (313=28r+19) gives \(r=\frac{294}{28}=10.5\), so no integer option is correct.
Step 2
Why this answer is correct
The correct answer is B. (10). From (61=42+c), (c=19). (313=28r+19) gives \(r=\frac{294}{28}=10.5\), so no integer option is correct.
Step 3
Exam Tip
(61=42+c) से (c=19)। (313=28r+19) से \(r=\frac{294}{28}=10.5\), इसलिए कोई पूर्णांक विकल्प सही नहीं होगा।
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यदि \(a_n=kn-7\) और \(a_{17}-a_5=96\) है, तो \(a_{29}\) क्या होगा?
If \(a_n=kn-7\) and \(a_{17}-a_5=96\), what is \(a_{29}\)?
#ap-parameter-expert
A (221)
B (225)
C (229)
D (235)
Explanation opens after your attempt
Step 1
Concept
(12k=96), so (k=8) and \(a_{29}=8\times29-7=225\). In a direct formula, find the coefficient first.
Step 2
Why this answer is correct
The correct answer is B. (225). (12k=96), so (k=8) and \(a_{29}=8\times29-7=225\). In a direct formula, find the coefficient first.
Step 3
Exam Tip
(12k=96), इसलिए (k=8) और \(a_{29}=8\times29-7=225\)। प्रत्यक्ष सूत्र में पहले गुणांक निकालें।
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यदि \(a_n=17n+c\) और \(a_7=145\) है तो \(a_{3r}=757\) होने पर (r) क्या है?
If \(a_n=17n+c\) and \(a_7=145\), what is (r) when \(a_{3r}=757\)?
#ap parameter index hard
A (13)
B (14)
C (15)
D (16)
Explanation opens after your attempt
Step 1
Concept
From (145=119+c), (c=26). (757=51r+26), giving \(r=\frac{731}{51}\), so option checking is necessary.
Step 2
Why this answer is correct
The correct answer is B. (14). From (145=119+c), (c=26). (757=51r+26), giving \(r=\frac{731}{51}\), so option checking is necessary.
Step 3
Exam Tip
(145=119+c) से (c=26)। (757=51r+26) से \(r=\frac{731}{51}\) आता है इसलिए विकल्पों की जांच जरूरी है।
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यदि \(a_n=kn+11\) और \(a_{22}-a_9=117\) है तो \(a_{31}\) क्या होगा?
If \(a_n=kn+11\) and \(a_{22}-a_9=117\), what is \(a_{31}\)?
#ap linear parameter hard
A (290)
B (293)
C (296)
D (299)
Explanation opens after your attempt
Step 1
Concept
From (13k=117), (k=9). Therefore \(a_{31}=9\times31+11=290\).
Step 2
Why this answer is correct
The correct answer is A. (290). From (13k=117), (k=9). Therefore \(a_{31}=9\times31+11=290\).
Step 3
Exam Tip
(13k=117) से (k=9)। इसलिए \(a_{31}=9\times31+11=290\)।
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यदि \(a_n=5n+s\) और \(a_{18}=112\) है तो \(a_{46}\) क्या होगा?
If \(a_n=5n+s\) and \(a_{18}=112\), what is \(a_{46}\)?
#ap parameter hard
A (242)
B (247)
C (252)
D (257)
Explanation opens after your attempt
Step 1
Concept
From (112=90+s), (s=22). Therefore \(a_{46}=5\times46+22=252\).
Step 2
Why this answer is correct
The correct answer is C. (252). From (112=90+s), (s=22). Therefore \(a_{46}=5\times46+22=252\).
Step 3
Exam Tip
(112=90+s) से (s=22)। इसलिए \(a_{46}=5\times46+22=252\)।
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यदि \(a_n=13n+c\) और \(a_6=101\) है तो \(a_{4r}=465\) होने पर (r) क्या है?
If \(a_n=13n+c\) and \(a_6=101\), what is (r) when \(a_{4r}=465\)?
#ap-parameter-index-hard
A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
From (101=78+c), (c=23). (465=52r+23), so \(r=\frac{442}{52}=8.5\).
Step 2
Why this answer is correct
The correct answer is B. (9). From (101=78+c), (c=23). (465=52r+23), so \(r=\frac{442}{52}=8.5\).
Step 3
Exam Tip
(101=78+c) से (c=23)। (465=52r+23) से \(r=\frac{442}{52}=8.5\)।
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यदि \(a_n=kn-8\) और \(a_{19}-a_7=96\) है तो \(a_{24}\) क्या होगा?
If \(a_n=kn-8\) and \(a_{19}-a_7=96\), what is \(a_{24}\)?
#ap-linear-parameter-hard
A (176)
B (180)
C (184)
D (188)
Explanation opens after your attempt
Step 1
Concept
From (12k=96), (k=8). Therefore \(a_{24}=8\times24-8=184\).
Step 2
Why this answer is correct
The correct answer is C. (184). From (12k=96), (k=8). Therefore \(a_{24}=8\times24-8=184\).
Step 3
Exam Tip
(12k=96) से (k=8)। इसलिए \(a_{24}=8\times24-8=184\)।
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यदि \(a_n=4n+r\) और \(a_{15}=83\) है तो \(a_{41}\) क्या होगा?
If \(a_n=4n+r\) and \(a_{15}=83\), what is \(a_{41}\)?
#ap-parameter-hard
A (179)
B (183)
C (187)
D (191)
Explanation opens after your attempt
Step 1
Concept
From (83=60+r), (r=23). Therefore \(a_{41}=4\times41+23=187\).
Step 2
Why this answer is correct
The correct answer is C. (187). From (83=60+r), (r=23). Therefore \(a_{41}=4\times41+23=187\).
Step 3
Exam Tip
(83=60+r) से (r=23)। इसलिए \(a_{41}=4\times41+23=187\)।
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यदि \(a_{n}=11n+c\) और \(a_{5}=72\) है तो \(a_{5r}\) का मान (512) होने पर (r) क्या है?
If \(a_n=11n+c\) and \(a_5=72\), what is (r) when \(a_{5r}=512\)?
#ap-parameter-index-hard
A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
From (72=55+c), (c=17). From (512=55r+17), (r=9).
Step 2
Why this answer is correct
The correct answer is B. (9). From (72=55+c), (c=17). From (512=55r+17), (r=9).
Step 3
Exam Tip
(72=55+c) से (c=17)। (512=55r+17) से (r=9)।
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यदि \(a_n=kn+5\) और \(a_{15}-a_6=63\) है तो \(a_{20}\) क्या होगा?
If \(a_n=kn+5\) and \(a_{15}-a_6=63\), what is \(a_{20}\)?
#ap-parameter-hard
A (135)
B (140)
C (145)
D (150)
Explanation opens after your attempt
Step 1
Concept
From (9k=63), (k=7). Therefore \(a_{20}=7\times20+5=145\).
Step 2
Why this answer is correct
The correct answer is C. (145). From (9k=63), (k=7). Therefore \(a_{20}=7\times20+5=145\).
Step 3
Exam Tip
(9k=63) से (k=7)। इसलिए \(a_{20}=7\times20+5=145\)।
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किस (k) के लिए (k,2k+1,5k-2,8k-5) समांतर श्रेणी के लगातार चार पद हैं?
For which (k) are (k,2k+1,5k-2,8k-5) four consecutive terms of an AP?
#ap
#parameter
#four_terms
A (k=1)
B (k=2)
C (k=3)
D हर वास्तविक (k) / Every real (k)
Explanation opens after your attempt
Step 1
Concept
The differences are (k+1,3k-3,3k-3), and equality gives (k=2). In exams, check all consecutive differences for four terms.
Step 2
Why this answer is correct
The correct answer is B. (k=2). The differences are (k+1,3k-3,3k-3), and equality gives (k=2). In exams, check all consecutive differences for four terms.
Step 3
Exam Tip
अंतर (k+1,3k-3,3k-3) हैं और बराबरी से (k=2) मिलता है। परीक्षा में चार पदों में सभी लगातार अंतर जांचें।
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किसी शून्येतर (r) के लिए \(r,r^2,r^3\) समांतर श्रेणी बनते हैं। (r) का मान क्या है?
For nonzero (r), \(r,r^2,r^3\) form an AP. What is the value of (r)?
#ap
#powers
#parameter
A (1)
B (-1)
C (2)
D कोई शून्येतर मान नहीं / No nonzero value
Explanation opens after your attempt
Step 1
Concept
The condition \(2r^2=r+r^3\) gives (r(r-1)2 =0), so the nonzero value is (1). In exams, always apply the nonzero condition.
Step 2
Why this answer is correct
The correct answer is A. (1). The condition \(2r^2=r+r^3\) gives (r(r-1)2 =0), so the nonzero value is (1). In exams, always apply the nonzero condition.
Step 3
Exam Tip
शर्त \(2r^2=r+r^3\) से (r(r-1)2 =0) मिलता है, इसलिए शून्येतर मान (1) है। परीक्षा में दी गई शून्येतर शर्त जरूर लगाएं।
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यदि (x+1,2x+6,5x-2) समांतर श्रेणी के लगातार पद हैं, तो (x) और (d) क्या हैं?
If (x+1,2x+6,5x-2) are consecutive terms of an AP, what are (x) and (d)?
#ap
#parameter
#fractional_answer
A \(x=\frac{13}{2},d=\frac{23}{2}\)
B \(x=\frac{11}{2},d=\frac{21}{2}\)
C (x=6,d=12)
D (x=7,d=13)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{13}{2},d=\frac{23}{2}\)
Step 1
Concept
Equating differences gives (x+5=3x-8), so \(x=\frac{13}{2}\) and \(d=\frac{23}{2}\). In exams, do not reject a fractional answer too quickly.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{13}{2},d=\frac{23}{2}\). Equating differences gives (x+5=3x-8), so \(x=\frac{13}{2}\) and \(d=\frac{23}{2}\). In exams, do not reject a fractional answer too quickly.
Step 3
Exam Tip
अंतर बराबर करने पर (x+5=3x-8), इसलिए \(x=\frac{13}{2}\) और \(d=\frac{23}{2}\)। परीक्षा में भिन्न उत्तर से घबराएं नहीं।
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यदि (4t+1,t+10,-t+21) समांतर श्रेणी के लगातार पद हैं, तो (t) और (d) क्या हैं?
If (4t+1,t+10,-t+21) are consecutive terms of an AP, what are (t) and (d)?
#ap
#parameter
#three_consecutive_terms
A (t=2,d=3)
B (t=-1,d=12)
C (t=-2,d=15)
D (t=3,d=0)
Explanation opens after your attempt
Correct Answer
C. (t=-2,d=15)
Step 1
Concept
Equal differences give (-3t+9=-2t+11), so (t=-2) and (d=15). In exams, subtract first from second and second from third.
Step 2
Why this answer is correct
The correct answer is C. (t=-2,d=15). Equal differences give (-3t+9=-2t+11), so (t=-2) and (d=15). In exams, subtract first from second and second from third.
Step 3
Exam Tip
बराबर अंतर से (-3t+9=-2t+11), इसलिए (t=-2) और (d=15)। परीक्षा में दूसरे से पहला और तीसरे से दूसरा पद घटाएं।
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पद (2x+3,5x-1,8x-5) किस (x) के लिए समांतर श्रेणी बनाते हैं?
For which (x) do the terms (2x+3,5x-1,8x-5) form an AP?
#ap
#parameter
#all_values
A केवल (x=1) / Only (x=1)
B केवल (x=4) / Only (x=4)
C हर वास्तविक (x) / Every real (x)
D कोई (x) नहीं / No (x)
Explanation opens after your attempt
Correct Answer
C. हर वास्तविक (x) / Every real (x)
Step 1
Concept
Both differences are (3x-4), so the terms form an AP for every real (x). In exams, if both differences are identical expressions, no separate solving is needed.
Step 2
Why this answer is correct
The correct answer is C. हर वास्तविक (x) / Every real (x). Both differences are (3x-4), so the terms form an AP for every real (x). In exams, if both differences are identical expressions, no separate solving is needed.
Step 3
Exam Tip
दोनों अंतर (3x-4) हैं, इसलिए हर वास्तविक (x) पर समांतर श्रेणी बनती है। परीक्षा में यदि दोनों अंतर समान अभिव्यक्ति हों तो कोई अलग हल नहीं चाहिए।
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क्या (3q+2,q-4,-q+10) किसी (q) पर समांतर श्रेणी के लगातार पद बन सकते हैं?
Can (3q+2,q-4,-q+10) be consecutive terms of an AP for some (q)?
#ap
#parameter
#no_solution
A हाँ, (q=1) / Yes, (q=1)
B हाँ, (q=4) / Yes, (q=4)
C हाँ, (q=-2) / Yes, (q=-2)
D नहीं, कोई (q) नहीं / No, no (q)
Explanation opens after your attempt
Correct Answer
D. नहीं, कोई (q) नहीं / No, no (q)
Step 1
Concept
Equating differences gives (-2q-6=-2q+14), which is impossible. In exams, cancellation of the variable can produce a contradiction.
Step 2
Why this answer is correct
The correct answer is D. नहीं, कोई (q) नहीं / No, no (q). Equating differences gives (-2q-6=-2q+14), which is impossible. In exams, cancellation of the variable can produce a contradiction.
Step 3
Exam Tip
अंतर बराबर करने पर (-2q-6=-2q+14) मिलता है, जो असंभव है। परीक्षा में कभी-कभी चर कटने पर विरोधाभास मिलता है।
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यदि (x-2,2x+1,4x-3) समांतर श्रेणी के लगातार पद हैं, तो (x) और (d) क्या हैं?
If (x-2,2x+1,4x-3) are consecutive terms of an AP, what are (x) and (d)?
#ap
#parameter
#three_terms
A (x=5,d=8)
B (x=6,d=9)
C (x=7,d=10)
D (x=8,d=11)
Explanation opens after your attempt
Correct Answer
C. (x=7,d=10)
Step 1
Concept
Equal differences give (x+3=2x-4), so (x=7) and (d=10). In exams, write the two differences separately first.
Step 2
Why this answer is correct
The correct answer is C. (x=7,d=10). Equal differences give (x+3=2x-4), so (x=7) and (d=10). In exams, write the two differences separately first.
Step 3
Exam Tip
बराबर अंतर से (x+3=2x-4), इसलिए (x=7) और (d=10)। परीक्षा में पहले अंतरों को अलग-अलग लिखें।
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तालिका में पद (5,5+h,5+2h) हैं। यह किस (h) के लिए समांतर श्रेणी है?
The listed terms are (5,5+h,5+2h). For which (h) is this an AP?
#ap
#parameter
#constant_ap
A केवल (h>0) / Only (h>0)
B केवल (h<0) / Only (h<0)
C केवल (h=0) / Only (h=0)
D हर वास्तविक (h) / Every real (h)
Explanation opens after your attempt
Correct Answer
D. हर वास्तविक (h) / Every real (h)
Step 1
Concept
Both differences are (h), so it is an AP for every real (h). In exams, remember that (h=0) gives a valid constant AP.
Step 2
Why this answer is correct
The correct answer is D. हर वास्तविक (h) / Every real (h). Both differences are (h), so it is an AP for every real (h). In exams, remember that (h=0) gives a valid constant AP.
Step 3
Exam Tip
दोनों अंतर (h) हैं, इसलिए हर वास्तविक (h) पर समांतर श्रेणी है। परीक्षा में (h=0) होने पर भी स्थिर समांतर श्रेणी मान्य होती है।
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वैध (p) के लिए \(\frac{1}{p+1},\frac{1}{p},\frac{1}{p-1}\) समांतर श्रेणी बन सकते हैं या नहीं?
For valid (p), can \(\frac{1}{p+1},\frac{1}{p},\frac{1}{p-1}\) form an AP?
#ap
#reciprocal_terms
#parameter
A हाँ, (p=1) / Yes, (p=1)
B हाँ, (p=-1) / Yes, (p=-1)
C हाँ, (p=2) / Yes, (p=2)
D नहीं, कोई वैध (p) नहीं / No, there is no valid (p)
Explanation opens after your attempt
Correct Answer
D. नहीं, कोई वैध (p) नहीं / No, there is no valid (p)
Step 1
Concept
The middle-term condition leads to an impossible equation, so there is no valid (p). In exams, also check that denominators are nonzero.
Step 2
Why this answer is correct
The correct answer is D. नहीं, कोई वैध (p) नहीं / No, there is no valid (p). The middle-term condition leads to an impossible equation, so there is no valid (p). In exams, also check that denominators are nonzero.
Step 3
Exam Tip
मध्य पद की शर्त से असंभव समीकरण मिलता है, इसलिए कोई वैध (p) नहीं है। परीक्षा में हरों के शून्य न होने की शर्त भी देखें।
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शून्येतर (k) के लिए \(k,k^2,k^3\) समांतर श्रेणी बनाते हैं। (k) का मान क्या है?
For nonzero (k), \(k,k^2,k^3\) form an AP. What is the value of (k)?
#ap
#parameter
#powers
A (k=-1)
B (k=1)
C (k=2)
D कोई शून्येतर मान नहीं / No nonzero value
Explanation opens after your attempt
Step 1
Concept
The condition \(2k^2=k+k^3\) gives (k(k-1)2 =0), and the nonzero value is (1). In exams, do not ignore conditions like nonzero.
Step 2
Why this answer is correct
The correct answer is B. (k=1). The condition \(2k^2=k+k^3\) gives (k(k-1)2 =0), and the nonzero value is (1). In exams, do not ignore conditions like nonzero.
Step 3
Exam Tip
शर्त \(2k^2=k+k^3\) से (k(k-1)2 =0) मिलता है, और शून्येतर मान (1) है। परीक्षा में शून्येतर जैसी शर्त न भूलें।
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यदि (2m-1,m+4,4m-3) समांतर श्रेणी के लगातार पद हैं, तो (m) और (d) क्या हैं?
If (2m-1,m+4,4m-3) are consecutive terms of an AP, what are (m) and (d)?
#ap
#parameter
#three_terms
A (m=2,d=3)
B (m=3,d=2)
C (m=4,d=1)
D (m=5,d=0)
Explanation opens after your attempt
Correct Answer
B. (m=3,d=2)
Step 1
Concept
Equal differences give (5-m=3m-7), so (m=3) and (d=2). In exams, be careful with signs in terms containing variables.
Step 2
Why this answer is correct
The correct answer is B. (m=3,d=2). Equal differences give (5-m=3m-7), so (m=3) and (d=2). In exams, be careful with signs in terms containing variables.
Step 3
Exam Tip
बराबर अंतर से (5-m=3m-7), अतः (m=3) और (d=2)। परीक्षा में अज्ञात वाले पदों में चिन्हों पर विशेष ध्यान दें।
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पद (k+2,3k-1,5k-4) किस (k) के लिए समांतर श्रेणी बनाते हैं?
For which (k) do the terms (k+2,3k-1,5k-4) form an AP?
#ap
#parameter
#all_values
A केवल (k=1) / Only (k=1)
B केवल (k=3) / Only (k=3)
C हर वास्तविक (k) / Every real (k)
D कोई (k) नहीं / No (k)
Explanation opens after your attempt
Correct Answer
C. हर वास्तविक (k) / Every real (k)
Step 1
Concept
Both differences are (2k-3), so it forms an AP for every real (k). In exams, simplify both differences first.
Step 2
Why this answer is correct
The correct answer is C. हर वास्तविक (k) / Every real (k). Both differences are (2k-3), so it forms an AP for every real (k). In exams, simplify both differences first.
Step 3
Exam Tip
दोनों अंतर (2k-3) हैं, इसलिए हर वास्तविक (k) पर समांतर श्रेणी बनती है। परीक्षा में पहले दोनों अंतर सरल करें।
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यदि (p-3,2p+1,5p-7) समांतर श्रेणी के लगातार पद हैं, तो (p) और सामान्य अंतर क्या हैं?
If (p-3,2p+1,5p-7) are consecutive terms of an AP, what are (p) and the common difference?
#ap
#parameter
#three_terms
A (p=5,d=9)
B (p=4,d=8)
C (p=3,d=7)
D (p=6,d=10)
Explanation opens after your attempt
Correct Answer
D. (p=6,d=10)
Step 1
Concept
Equating differences gives (p+4=3p-8), so (p=6) and (d=10). For three consecutive terms, set second minus first equal to third minus second.
Step 2
Why this answer is correct
The correct answer is D. (p=6,d=10). Equating differences gives (p+4=3p-8), so (p=6) and (d=10). For three consecutive terms, set second minus first equal to third minus second.
Step 3
Exam Tip
बराबर अंतर रखने पर (p+4=3p-8), इसलिए (p=6) और (d=10)। परीक्षा में तीन लगातार पदों के लिए दूसरा घटाकर पहला और तीसरा घटाकर दूसरा बराबर करें।
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किस (k) के लिए (k-2,k+5,2k+1) अंकगणितीय श्रेणी में होंगे?
For which (k) will (k-2,k+5,2k+1) be in an arithmetic progression?
#ap
#find parameter
#algebraic terms
#hard
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
From (2(k+5)=(k-2)+(2k+1)), (2k+10=3k-1), so (k=11). Identify the middle term while forming the equation.
Step 2
Why this answer is correct
The correct answer is D. (9). From (2(k+5)=(k-2)+(2k+1)), (2k+10=3k-1), so (k=11). Identify the middle term while forming the equation.
Step 3
Exam Tip
(2(k+5)=(k-2)+(2k+1)) से (2k+10=3k-1), इसलिए (k=11)। समीकरण बनाते समय मध्य पद को पहचानें।
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किस (m) के लिए (m-1 ,2m+3,4m-1) अंकगणितीय श्रेणी में होंगे?
For which (m) will (m-1 ,2m+3,4m-1) be in an arithmetic progression?
#ap
#find parameter
#algebraic terms
#hard
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
From (2(2m+3)=(m-1 )+(4m-1)), (4m+6=5m-2), so (m=8). Use the twice-middle-term rule.
Step 2
Why this answer is correct
The correct answer is C. (5). From (2(2m+3)=(m-1 )+(4m-1)), (4m+6=5m-2), so (m=8). Use the twice-middle-term rule.
Step 3
Exam Tip
(2(2m+3)=(m-1 )+(4m-1)) से (4m+6=5m-2), इसलिए (m=8)। मध्य पद का दुगुना नियम लगाएं।
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यदि (k+1, 2k+4, 4k-2) अंकगणितीय श्रेणी में हैं, तो (k) का मान क्या होगा?
If (k+1, 2k+4, 4k-2) are in an arithmetic progression, what will be the value of (k)?
#ap
#find parameter
#algebraic terms
#expert
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
From (2(2k+4)=(k+1)+(4k-2)), (4k+8=5k-1), so (k=9). Identify the middle term correctly while forming the equation.
Step 2
Why this answer is correct
The correct answer is D. (6). From (2(2k+4)=(k+1)+(4k-2)), (4k+8=5k-1), so (k=9). Identify the middle term correctly while forming the equation.
Step 3
Exam Tip
(2(2k+4)=(k+1)+(4k-2)) से (4k+8=5k-1), इसलिए (k=9)। समीकरण बनाते समय मध्य पद को सही पहचानें।
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यदि (q-3, 2q+1, 4q-1) अंकगणितीय श्रेणी में हैं, तो (q) क्या होगा?
If (q-3, 2q+1, 4q-1) are in an arithmetic progression, what will (q) be?
#ap
#algebraic terms
#find parameter
#expert
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
From (2(2q+1)=(q-3)+(4q-1)), (4q+2=5q-4), so (q=6). Watch signs while applying the twice-middle-term rule.
Step 2
Why this answer is correct
The correct answer is D. (5). From (2(2q+1)=(q-3)+(4q-1)), (4q+2=5q-4), so (q=6). Watch signs while applying the twice-middle-term rule.
Step 3
Exam Tip
(2(2q+1)=(q-3)+(4q-1)) से (4q+2=5q-4), इसलिए (q=6)। मध्य पद का दुगुना नियम लगाते समय संकेतों पर ध्यान दें।
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किस (k) के लिए (k-3, k+2, 2k+1) अंकगणितीय श्रेणी में होंगे?
For which (k) will (k-3, k+2, 2k+1) be in an arithmetic progression?
#ap
#find parameter
#algebraic terms
#hard
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
From (2(k+2)=(k-3)+(2k+1)), (2k+4=3k-2), so (k=6). Identify the middle term while forming the equation.
Step 2
Why this answer is correct
The correct answer is B. (5). From (2(k+2)=(k-3)+(2k+1)), (2k+4=3k-2), so (k=6). Identify the middle term while forming the equation.
Step 3
Exam Tip
(2(k+2)=(k-3)+(2k+1)) से (2k+4=3k-2), इसलिए (k=6)। समीकरण बनाते समय मध्य पद को पहचानें।
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किस (m) के लिए (m+2, 2m+5, 4m+1) अंकगणितीय श्रेणी में होंगे?
For which (m) will (m+2, 2m+5, 4m+1) be in an arithmetic progression?
#ap
#find parameter
#algebraic terms
#hard
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
From (2(2m+5)=(m+2)+(4m+1)), (4m+10=5m+3), so (m=7). Use the twice-middle-term rule for three terms.
Step 2
Why this answer is correct
The correct answer is A. (4). From (2(2m+5)=(m+2)+(4m+1)), (4m+10=5m+3), so (m=7). Use the twice-middle-term rule for three terms.
Step 3
Exam Tip
(2(2m+5)=(m+2)+(4m+1)) से (4m+10=5m+3), इसलिए (m=7)। तीन पदों में मध्य पद का दुगुना नियम लगाएं।
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समीकरणों (11x+ky=70) और (5x+4y=31) का अद्वितीय हल होने के लिए कौन-सी शर्त सही है?
Which condition is correct for the equations (11x+ky=70) and (5x+4y=31) to have a unique solution?
#linear equations
#expert
#unique solution
#parameter
A (k=44 / 5)
B (k\ne44 / 5)
C (k=4)
D (k=11)
Explanation opens after your attempt
Correct Answer
B. (k\ne44 / 5)
Step 1
Concept
For a unique solution, \(11/5 \ne k/4\) must hold. Therefore, \(k\ne44/5\) is the correct condition.
Step 2
Why this answer is correct
The correct answer is B. \(k\ne44 / 5\). For a unique solution, \(11/5 \ne k/4\) must hold. Therefore, \(k\ne44/5\) is the correct condition.
Step 3
Exam Tip
अद्वितीय हल के लिए \(11/5 \ne k/4\) होना चाहिए। इसलिए \(k\ne44/5\) सही शर्त है।
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समीकरणों (px+10y=50) और (14x+35y=122) का कोई हल न होने के लिए (p) का मान क्या होगा?
What is the value of (p) for the equations (px+10y=50) and (14x+35y=122) to have no solution?
#linear equations
#expert
#no solution
#parameter
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
For no solution, (p/14=10/35) and (50/122) must be different. Therefore, (p=4).
Step 2
Why this answer is correct
The correct answer is B. (4). For no solution, (p/14=10/35) and (50/122) must be different. Therefore, (p=4).
Step 3
Exam Tip
कोई हल नहीं के लिए (p/14=10/35) और (50/122) अलग होना चाहिए। इसलिए (p=4)।
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समीकरणों (6x+ay=42) और (18x+33y=126) के अनंत हल होने के लिए (a) का मान क्या होगा?
What is the value of (a) for the equations (6x+ay=42) and (18x+33y=126) to have infinitely many solutions?
#linear equations
#expert
#infinite solutions
#parameter
A (9)
B (10)
C (11)
D (12)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, (6/18=a/33=42/126) must hold. Therefore, (a=11) is correct.
Step 2
Why this answer is correct
The correct answer is C. (11). For infinitely many solutions, (6/18=a/33=42/126) must hold. Therefore, (a=11) is correct.
Step 3
Exam Tip
अनंत हल के लिए (6/18=a/33=42/126) होना चाहिए। इसलिए (a=11) सही है।
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समीकरणों (5x+9y=64) और (15x+27y=t) के असंगत होने के लिए (t) के लिए सही शर्त क्या है?
What is the correct condition on (t) for the equations (5x+9y=64) and (15x+27y=t) to be inconsistent?
#linear equations
#expert
#inconsistent
#parameter
A (t=192)
B \(t\ne192\)
C (t=64)
D (t=128)
Explanation opens after your attempt
Correct Answer
B. \(t\ne192\)
Step 1
Concept
The first two ratios are equal. For inconsistency, the constant ratio must be different so \(t\ne192\).
Step 2
Why this answer is correct
The correct answer is B. \(t\ne192\). The first two ratios are equal. For inconsistency, the constant ratio must be different so \(t\ne192\).
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं। असंगत होने के लिए स्थिर पद का अनुपात अलग होना चाहिए इसलिए \(t\ne192\)।
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यदि (lx+17y=68) और (20x+34y=139) का कोई हल नहीं है, तो (l) का मान क्या होगा?
If (lx+17y=68) and (20x+34y=139) have no solution, what will be the value of (l)?
#linear equations
#expert
#parameter
#parallel lines
A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
For no solution, (l/20=17/34) and (68/139) must be different. Hence, (l=10).
Step 2
Why this answer is correct
The correct answer is C. (10). For no solution, (l/20=17/34) and (68/139) must be different. Hence, (l=10).
Step 3
Exam Tip
कोई हल नहीं के लिए (l/20=17/34) और (68/139) अलग होना चाहिए। इसलिए (l=10)।
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समीकरणों (12x+ky=132) और (3x+10y=33) के अनंत हल होने के लिए (k) क्या होगा?
What will (k) be for the equations (12x+ky=132) and (3x+10y=33) to have infinitely many solutions?
#linear equations
#expert
#infinite solutions
#parameter
A (38)
B (39)
C (40)
D (41)
Explanation opens after your attempt
Step 1
Concept
The first equation must be (4) times the second. Therefore, (k=40).
Step 2
Why this answer is correct
The correct answer is C. (40). The first equation must be (4) times the second. Therefore, (k=40).
Step 3
Exam Tip
पहला समीकरण दूसरे का (4) गुना होना चाहिए। इसलिए (k=40) है।
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समीकरणों (10x+9y=38) और (20x+ay=91) का कोई हल न होने के लिए (a) क्या होगा?
What will (a) be for the equations (10x+9y=38) and (20x+ay=91) to have no solution?
#linear equations
#expert
#no solution
#parameter
A (16)
B (17)
C (18)
D (19)
Explanation opens after your attempt
Step 1
Concept
For no solution, (10/20=9/a) and (38/91) must be different. This gives (a=18).
Step 2
Why this answer is correct
The correct answer is C. (18). For no solution, (10/20=9/a) and (38/91) must be different. This gives (a=18).
Step 3
Exam Tip
कोई हल नहीं के लिए (10/20=9/a) और (38/91) अलग होना चाहिए। इससे (a=18) मिलता है।
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समीकरणों (13x+8y=49) और (26x+16y=r) के असंगत होने के लिए कौन-सी शर्त सही है?
Which condition is correct for the equations (13x+8y=49) and (26x+16y=r) to be inconsistent?
#linear equations
#expert
#inconsistent
#parameter
A (r=98)
B \(r\ne98\)
C (r=49)
D (r=100)
Explanation opens after your attempt
Correct Answer
B. \(r\ne98\)
Step 1
Concept
The first two ratios are equal. For inconsistency, the constant ratio must be different so \(r\ne98\).
Step 2
Why this answer is correct
The correct answer is B. \(r\ne98\). The first two ratios are equal. For inconsistency, the constant ratio must be different so \(r\ne98\).
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं। असंगत होने के लिए स्थिर पद का अनुपात अलग होना चाहिए इसलिए \(r\ne98\)।
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समीकरणों (9x+16y=77) और (27x+48y=s) के संगत और आश्रित होने के लिए (s) क्या होगा?
What should (s) be for the equations (9x+16y=77) and (27x+48y=s) to be consistent and dependent?
#linear equations
#expert
#consistent dependent
#parameter
A (229)
B (230)
C (231)
D (232)
Explanation opens after your attempt
Step 1
Concept
To be consistent and dependent, the second equation must be (3) times the first. Hence, (s=231).
Step 2
Why this answer is correct
The correct answer is C. (231). To be consistent and dependent, the second equation must be (3) times the first. Hence, (s=231).
Step 3
Exam Tip
संगत और आश्रित होने के लिए दूसरा समीकरण पहले का (3) गुना होना चाहिए। अतः (s=231)।
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यदि (16x-8y=64) और (2x-y=t) असंगत हैं, तो (t) के लिए सही शर्त क्या है?
If (16x-8y=64) and (2x-y=t) are inconsistent, what is the correct condition for (t)?
#linear equations
#expert
#inconsistent
#parameter
A (t=8)
B \(t\ne8\)
C (t=64)
D (t=16)
Explanation opens after your attempt
Correct Answer
B. \(t\ne8\)
Step 1
Concept
The first two ratios are equal. For inconsistency, (64/t) must be different so \(t\ne8\).
Step 2
Why this answer is correct
The correct answer is B. \(t\ne8\). The first two ratios are equal. For inconsistency, (64/t) must be different so \(t\ne8\).
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं। असंगत होने के लिए (64/t) अलग होना चाहिए इसलिए \(t\ne8\)।
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यदि (11x+7y=59) और (33x+21y=n) के अनंत हल हैं, तो (n) कितना होगा?
If (11x+7y=59) and (33x+21y=n) have infinitely many solutions, what is (n)?
#linear equations
#expert
#parameter
#infinite solutions
A (175)
B (176)
C (177)
D (178)
Explanation opens after your attempt
Step 1
Concept
The second equation must be (3) times the first. Therefore, (n=177).
Step 2
Why this answer is correct
The correct answer is C. (177). The second equation must be (3) times the first. Therefore, (n=177).
Step 3
Exam Tip
दूसरा समीकरण पहले का (3) गुना होना चाहिए। इसलिए (n=177) होगा।
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समीकरणों (17x+py=51) और (8x+3y=25) का अद्वितीय हल होने के लिए कौन-सी शर्त सही है?
Which condition is correct for the equations (17x+py=51) and (8x+3y=25) to have a unique solution?
#linear equations
#expert
#unique solution
#parameter
A (p=51 / 8)
B (p\ne51 / 8)
C (p=3)
D (p=17)
Explanation opens after your attempt
Correct Answer
B. (p\ne51 / 8)
Step 1
Concept
For a unique solution, \(17/8 \ne p/3\) must hold. Therefore, \(p\ne51/8\) is the correct condition.
Step 2
Why this answer is correct
The correct answer is B. \(p\ne51 / 8\). For a unique solution, \(17/8 \ne p/3\) must hold. Therefore, \(p\ne51/8\) is the correct condition.
Step 3
Exam Tip
अद्वितीय हल के लिए \(17/8 \ne p/3\) होना चाहिए। इसलिए \(p\ne51/8\) सही शर्त है।
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समीकरणों (7x+dy=63) और (28x+36y=252) के अनंत हल होने के लिए (d) का मान क्या है?
What is the value of (d) for the equations (7x+dy=63) and (28x+36y=252) to have infinitely many solutions?
#linear equations
#expert
#parameter
#dependent pair
A (7)
B (8)
C (9)
D (10)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, (7/28=d/36=63/252) must hold. Therefore, (d=9).
Step 2
Why this answer is correct
The correct answer is C. (9). For infinitely many solutions, (7/28=d/36=63/252) must hold. Therefore, (d=9).
Step 3
Exam Tip
अनंत हल के लिए (7/28=d/36=63/252) होना चाहिए। इसलिए (d=9)।
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यदि (cx+18y=72) और (24x+48y=145) का कोई हल नहीं है, तो (c) क्या होगा?
If (cx+18y=72) and (24x+48y=145) have no solution, what will (c) be?
#linear equations
#expert
#parameter
#no solution
A (7)
B (8)
C (9)
D (10)
Explanation opens after your attempt
Step 1
Concept
For no solution, (c/24=18/48) and (72/145) must be different. Therefore, (c=9).
Step 2
Why this answer is correct
The correct answer is C. (9). For no solution, (c/24=18/48) and (72/145) must be different. Therefore, (c=9).
Step 3
Exam Tip
कोई हल नहीं के लिए (c/24=18/48) और (72/145) अलग होना चाहिए। इसलिए (c=9)।
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समीकरणों (bx+16y=64) और (14x+28y=131) का कोई हल न होने के लिए (b) का मान क्या होगा?
What is the value of (b) for the equations (bx+16y=64) and (14x+28y=131) to have no solution?
#linear equations
#expert
#no solution
#parameter
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
For no solution, (b/14=16/28) and (64/131) must be different. Hence, (b=8).
Step 2
Why this answer is correct
The correct answer is C. (8). For no solution, (b/14=16/28) and (64/131) must be different. Hence, (b=8).
Step 3
Exam Tip
कोई हल नहीं के लिए (b/14=16/28) और (64/131) अलग होना चाहिए। इसलिए (b=8)।
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समीकरणों (8x+ay=72) और (24x+30y=216) के अनंत हल होने के लिए (a) क्या होगा?
What will (a) be for the equations (8x+ay=72) and (24x+30y=216) to have infinitely many solutions?
#linear equations
#expert
#ratio condition
#parameter
A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, (8/24=a/30=72/216) must hold. This gives (a=10).
Step 2
Why this answer is correct
The correct answer is C. (10). For infinitely many solutions, (8/24=a/30=72/216) must hold. This gives (a=10).
Step 3
Exam Tip
अनंत हल के लिए (8/24=a/30=72/216) होना चाहिए। इससे (a=10) मिलता है।
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समीकरणों (12x+py=60) और (3x+5y=16) का अद्वितीय हल होने के लिए कौन-सी शर्त सही है?
Which condition is correct for the equations (12x+py=60) and (3x+5y=16) to have a unique solution?
#linear equations
#expert
#unique solution
#parameter
A (p=20)
B \(p\ne20\)
C (p=5)
D (p=12)
Explanation opens after your attempt
Correct Answer
B. \(p\ne20\)
Step 1
Concept
For a unique solution, \(12/3 \ne p/5\) must hold. Therefore, \(p\ne20\) is the correct condition.
Step 2
Why this answer is correct
The correct answer is B. \(p\ne20\). For a unique solution, \(12/3 \ne p/5\) must hold. Therefore, \(p\ne20\) is the correct condition.
Step 3
Exam Tip
अद्वितीय हल के लिए \(12/3 \ne p/5\) होना चाहिए। इसलिए \(p\ne20\) सही शर्त है।
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समीकरणों (13x+qy=52) और (26x+18y=104) के अनंत हल होने के लिए (q) का मान क्या है?
What is the value of (q) for the equations (13x+qy=52) and (26x+18y=104) to have infinitely many solutions?
#linear equations
#expert
#dependent pair
#parameter
A (7)
B (8)
C (9)
D (10)
Explanation opens after your attempt
Step 1
Concept
The second equation is (2) times the first, so (q/18=1/2) must hold. Hence, (q=9).
Step 2
Why this answer is correct
The correct answer is C. (9). The second equation is (2) times the first, so (q/18=1/2) must hold. Hence, (q=9).
Step 3
Exam Tip
दूसरा समीकरण पहले का (2) गुना है, इसलिए (q/18=1/2) होना चाहिए। अतः (q=9)।
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समीकरणों (kx+14y=42) और (18x+21y=63) के अनंत हल होने के लिए (k) क्या होगा?
What will (k) be for the equations (kx+14y=42) and (18x+21y=63) to have infinitely many solutions?
#linear equations
#expert
#parameter
#infinite solutions
A (10)
B (11)
C (12)
D (13)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, (k/18=14/21=42/63) must hold. Therefore, (k=12) is correct.
Step 2
Why this answer is correct
The correct answer is C. (12). For infinitely many solutions, (k/18=14/21=42/63) must hold. Therefore, (k=12) is correct.
Step 3
Exam Tip
अनंत हल के लिए (k/18=14/21=42/63) होना चाहिए। इसलिए (k=12) सही है।
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समीकरणों (px+9y=45) और (20x+30y=103) का कोई हल न होने के लिए (p) का मान क्या होगा?
What is the value of (p) for the equations (px+9y=45) and (20x+30y=103) to have no solution?
#linear equations
#expert
#no solution
#parameter
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
For no solution, (p/20=9/30) and (45/103) must be different. This gives (p=6).
Step 2
Why this answer is correct
The correct answer is B. (6). For no solution, (p/20=9/30) and (45/103) must be different. This gives (p=6).
Step 3
Exam Tip
कोई हल नहीं के लिए (p/20=9/30) और (45/103) अलग होना चाहिए। इससे (p=6) मिलता है।
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समीकरणों (4x+ay=32) और (12x+21y=96) के अनंत हल होने के लिए (a) का मान क्या होगा?
What is the value of (a) for the equations (4x+ay=32) and (12x+21y=96) to have infinitely many solutions?
#linear equations
#expert
#parameter
#infinite solutions
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, (4/12=a/21=32/96) must hold. Therefore, (a=7) is correct.
Step 2
Why this answer is correct
The correct answer is C. (7). For infinitely many solutions, (4/12=a/21=32/96) must hold. Therefore, (a=7) is correct.
Step 3
Exam Tip
अनंत हल के लिए (4/12=a/21=32/96) होना चाहिए। इसलिए (a=7) सही है।
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समीकरणों (13x+py=52) और (6x+5y=24) का अद्वितीय हल होने के लिए कौन-सी शर्त सही है?
Which condition is correct for the equations (13x+py=52) and (6x+5y=24) to have a unique solution?
#linear equations
#hard
#unique solution
#parameter
A (p \ne 65 / 6)
B (p=65 / 6)
C (p=5)
D (p=13)
Explanation opens after your attempt
Correct Answer
A. (p \ne 65 / 6)
Step 1
Concept
For a unique solution, \(13/6 \ne p/5\) must hold. Therefore, \(p \ne 65/6\) is the correct condition.
Step 2
Why this answer is correct
The correct answer is A. \(p \ne 65 / 6\). For a unique solution, \(13/6 \ne p/5\) must hold. Therefore, \(p \ne 65/6\) is the correct condition.
Step 3
Exam Tip
अद्वितीय हल के लिए \(13/6 \ne p/5\) होना चाहिए। इसलिए \(p \ne 65/6\) सही शर्त है।
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समीकरणों (bx+9y=36) और (16x+24y=97) का कोई हल न होने के लिए (b) का मान क्या होगा?
What is the value of (b) for the equations (bx+9y=36) and (16x+24y=97) to have no solution?
#linear equations
#hard
#no solution
#parameter
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
For no solution, (b/16=9/24) and (36/97) must be different. Hence, (b=6).
Step 2
Why this answer is correct
The correct answer is D. (6). For no solution, (b/16=9/24) and (36/97) must be different. Hence, (b=6).
Step 3
Exam Tip
कोई हल नहीं के लिए (b/16=9/24) और (36/97) अलग होना चाहिए। इसलिए (b=6)।
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समीकरणों (5x+ay=45) और (20x+28y=180) के अनंत हल होने के लिए (a) का मान क्या होगा?
What is the value of (a) for the equations (5x+ay=45) and (20x+28y=180) to have infinitely many solutions?
#linear equations
#hard
#infinite solutions
#parameter
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
For infinitely many solutions, (5/20=a/28=45/180) must hold. Therefore, (a=7) is correct.
Step 2
Why this answer is correct
The correct answer is C. (7). For infinitely many solutions, (5/20=a/28=45/180) must hold. Therefore, (a=7) is correct.
Step 3
Exam Tip
अनंत हल के लिए (5/20=a/28=45/180) होना चाहिए। इसलिए (a=7) सही है।
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समीकरणों (4x+7y=31) और (12x+21y=t) के असंगत होने के लिए (t) के लिए सही शर्त क्या है?
What is the correct condition on (t) for the equations (4x+7y=31) and (12x+21y=t) to be inconsistent?
#linear equations
#hard
#inconsistent
#parameter
A (t=93)
B \(t \ne 93\)
C (t=31)
D (t=62)
Explanation opens after your attempt
Correct Answer
B. \(t \ne 93\)
Step 1
Concept
The first two ratios are equal. For inconsistency, the constant ratio must be different so \(t \ne 93\).
Step 2
Why this answer is correct
The correct answer is B. \(t \ne 93\). The first two ratios are equal. For inconsistency, the constant ratio must be different so \(t \ne 93\).
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं। असंगत होने के लिए स्थिर पद का अनुपात अलग होना चाहिए इसलिए \(t \ne 93\)।
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