Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(12\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=64\) gives \(r=4\sqrt{2}\), and \(p=3r=12\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=64\) से \(r=4\sqrt{2}\) और \(p=3r=12\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{21\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=49\) और \(p=3r=\frac{21\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।
Let the roots be (-r) and (-2r), then \(2r^2=36\) gives \(r=3\sqrt{2}\), and \(p=3r=9\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(9\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=36\) gives \(r=3\sqrt{2}\), and \(p=3r=9\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=36\) से \(r=3\sqrt{2}\) और \(p=3r=9\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{15\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=25\) और \(p=3r=\frac{15\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।
Let the roots be (-r) and (-2r), then \(2r^2=16\) gives \(r=2\sqrt{2}\), and \(p=3r=6\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=16\) gives \(r=2\sqrt{2}\), and \(p=3r=6\sqrt{2}\). In exams, keep signs of both roots carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=16\) से \(r=2\sqrt{2}\) और \(p=3r=6\sqrt{2}\) है। परीक्षा में दोनों मूलों के चिन्ह ध्यान से रखें।
Let the roots be (-r) and (-2r), then \(2r^2=9\) and the sum is (-3r), so \(p=3r=\frac{9}{\sqrt{2}}\). In exams, assume the roots and form equations carefully.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}\). Let the roots be (-r) and (-2r), then \(2r^2=9\) and the sum is (-3r), so \(p=3r=\frac{9}{\sqrt{2}}\). In exams, assume the roots and form equations carefully.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=9\) और योग (-3r) है, इसलिए \(p=3r=3\sqrt{\frac{9}{2}}\) नहीं बल्कि \(r=\frac{3}{\sqrt{2}}\), अतः \(p=\frac{9}{\sqrt{2}}\) होता है। परीक्षा में ऐसे प्रश्नों में मानकर समीकरण बनाएं।
A. तत्व दृश्य सामग्री हैं और सिद्धांत उनके संगठन के तरीके हैं/Elements are visual materials and principles are ways of organizing them
Step 1
Concept
Elements tell what and principles tell how they are organized. Exam tip: remember what and how difference.
Step 2
Why this answer is correct
The correct answer is A. तत्व दृश्य सामग्री हैं और सिद्धांत उनके संगठन के तरीके हैं / Elements are visual materials and principles are ways of organizing them. Elements tell what and principles tell how they are organized. Exam tip: remember what and how difference.
Step 3
Exam Tip
तत्व क्या है और सिद्धांत कैसे व्यवस्थित है बताता है। परीक्षा में what और how का अंतर याद रखें।
A. रेखा मान रंग और स्थान को संतुलन जोर और लय बनाने के लिए व्यवस्थित किया जाता है/Line value colour and space are arranged to create balance emphasis and rhythm
Step 1
Concept
Elements are visual materials and principles explain their arrangement. Exam tip: keep what and how separate.
Step 2
Why this answer is correct
The correct answer is A. रेखा मान रंग और स्थान को संतुलन जोर और लय बनाने के लिए व्यवस्थित किया जाता है / Line value colour and space are arranged to create balance emphasis and rhythm. Elements are visual materials and principles explain their arrangement. Exam tip: keep what and how separate.
Step 3
Exam Tip
तत्व दृश्य सामग्री हैं और सिद्धांत उनकी व्यवस्था बताते हैं। परीक्षा में what and how अलग रखें।
A. तत्व दृश्य सामग्री हैं और सिद्धांत उनके संगठन के तरीके हैं/Elements are visual materials and principles are ways of organizing them
Step 1
Concept
Elements tell what and principles tell how it is organized. Exam tip: remember the difference of what and how.
Step 2
Why this answer is correct
The correct answer is A. तत्व दृश्य सामग्री हैं और सिद्धांत उनके संगठन के तरीके हैं / Elements are visual materials and principles are ways of organizing them. Elements tell what and principles tell how it is organized. Exam tip: remember the difference of what and how.
Step 3
Exam Tip
तत्व क्या है और सिद्धांत कैसे व्यवस्थित है बताता है। परीक्षा में what और how का अंतर याद रखें।
A. रेखा और रंग तत्व हैं जिन्हें जोर और संतुलन बनाने में व्यवस्थित किया जाता है/Line and colour are elements arranged to create emphasis and balance
Step 1
Concept
Elements are materials and principles are their organization. Exam tip: keep difference between what and how.
Step 2
Why this answer is correct
The correct answer is A. रेखा और रंग तत्व हैं जिन्हें जोर और संतुलन बनाने में व्यवस्थित किया जाता है / Line and colour are elements arranged to create emphasis and balance. Elements are materials and principles are their organization. Exam tip: keep difference between what and how.
Step 3
Exam Tip
तत्व सामग्री हैं और सिद्धांत उनका संगठन हैं। परीक्षा में what और how का अंतर रखें।
A. तत्व दृश्य घटक हैं और सिद्धांत उनके संगठन के तरीके हैं/Elements are visual components and principles are ways of organizing them
Step 1
Concept
Elements are materials and principles tell arrangement. Exam tip: understand elements as what and principles as how.
Step 2
Why this answer is correct
The correct answer is A. तत्व दृश्य घटक हैं और सिद्धांत उनके संगठन के तरीके हैं / Elements are visual components and principles are ways of organizing them. Elements are materials and principles tell arrangement. Exam tip: understand elements as what and principles as how.
Step 3
Exam Tip
तत्व सामग्री हैं और सिद्धांत व्यवस्था बताते हैं। परीक्षा में elements को what और principles को how समझें।
The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 2
Why this answer is correct
The correct answer is C. \(18 / 2=27 / 3 \ne 126 / 16\). The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग है। इसलिए कोई हल नहीं है।
The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 2
Why this answer is correct
The correct answer is A. (16 / 32=(-9) / (-18) \ne 55 / 113). The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं और तीसरा अलग है। इसलिए कोई हल नहीं होगा।
Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is C. (10 / 30=13 / 39=(-71) / (-213)). Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 3
Exam Tip
यहां तीनों अनुपात बराबर हैं। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं।
The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 2
Why this answer is correct
The correct answer is C. \(14 / 2=21 / 3 \ne 98 / 17\). The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग है। इसलिए कोई हल नहीं है।
The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 2
Why this answer is correct
The correct answer is A. (12 / 24=(-5) / (-10) \ne 37 / 79). The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं और तीसरा अलग है। इसलिए कोई हल नहीं होगा।
Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is C. (8 / 24=13 / 39=(-59) / (-177)). Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 3
Exam Tip
यहां तीनों अनुपात बराबर हैं। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं।
The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 2
Why this answer is correct
The correct answer is C. \(12 / 2=18 / 3 \ne 72 / 15\). The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग है। इसलिए कोई हल नहीं है।
The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 2
Why this answer is correct
The correct answer is A. (10 / 20=(-7) / (-14) \ne 31 / 65). The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं और तीसरा अलग है। इसलिए कोई हल नहीं होगा।
Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is C. (6 / 18=11 / 33=(-47) / (-141)). Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 3
Exam Tip
यहां तीनों अनुपात बराबर हैं। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं।
The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 2
Why this answer is correct
The correct answer is C. \(10 / 2=15 / 3 \ne 50 / 13\). The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग है। इसलिए कोई हल नहीं है।
The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 2
Why this answer is correct
The correct answer is A. (8 / 16=(-3) / (-6) \ne 22 / 47). The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं और तीसरा अलग है। इसलिए कोई हल नहीं होगा।
Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is C. (4 / 12=9 / 27=(-31) / (-93)). Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 3
Exam Tip
यहां तीनों अनुपात बराबर हैं। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं।
The first two ratios are equal but the constant ratio is different. Therefore there is no solution.
Step 2
Why this answer is correct
The correct answer is C. \(10 / 2=15 / 3 \ne 50 / 13\). The first two ratios are equal but the constant ratio is different. Therefore there is no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग है। इसलिए कोई हल नहीं है।
The first two ratios are equal and the third is different. Therefore there will be no solution.
Step 2
Why this answer is correct
The correct answer is A. (8 / 16=(-3) / (-6) \ne 22 / 47). The first two ratios are equal and the third is different. Therefore there will be no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं और तीसरा अलग है। इसलिए कोई हल नहीं होगा।
Here all three ratios are equal. Therefore both lines are coincident and have infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is C. (4 / 12=9 / 27=(-31) / (-93)). Here all three ratios are equal. Therefore both lines are coincident and have infinitely many solutions.
Step 3
Exam Tip
यहां तीनों अनुपात बराबर हैं। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं।
The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 2
Why this answer is correct
The correct answer is C. \(8 / 2=12 / 3 \ne 40 / 12\). The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग है। इसलिए कोई हल नहीं है।
The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 2
Why this answer is correct
The correct answer is A. (6 / 12=(-7) / (-14) \ne 9 / 25). The first two ratios are equal and the third is different. Therefore, there will be no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं और तीसरा अलग है। इसलिए कोई हल नहीं होगा।
Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is C. (3 / 9=5 / 15=(-20) / (-60)). Here all three ratios are equal. Therefore, both lines are coincident and have infinitely many solutions.
Step 3
Exam Tip
यहां तीनों अनुपात बराबर हैं। इसलिए दोनों रेखाएं संपाती हैं और अनंत हल हैं।
The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 2
Why this answer is correct
The correct answer is C. \(6 / 2=9 / 3 \ne 30 / 12\). The first two ratios are equal but the constant ratio is different. Therefore, there is no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं लेकिन स्थिर पद का अनुपात अलग है। इसलिए कोई हल नहीं है।
The first two ratios are equal and the third is different. Therefore, the lines are parallel and there is no solution.
Step 2
Why this answer is correct
The correct answer is A. (4 / 8=(-5) / (-10) \ne 7 / 9). The first two ratios are equal and the third is different. Therefore, the lines are parallel and there is no solution.
Step 3
Exam Tip
पहले दो अनुपात बराबर हैं और तीसरा अलग है। इसलिए रेखाएं समानांतर हैं और कोई हल नहीं है।
Here all three ratios are equal to (1/3). Therefore, the equations will have infinitely many solutions.
Step 2
Why this answer is correct
The correct answer is C. (2 / 6=3 / 9=(-11) / (-33)). Here all three ratios are equal to (1/3). Therefore, the equations will have infinitely many solutions.
Step 3
Exam Tip
यहां तीनों अनुपात (1/3) के बराबर हैं। इसलिए समीकरणों के अनंत हल होंगे।
A. उनका लघुत्तम समापवर्त्य उनके गुणनफल के बराबर होता है/Their LCM is equal to their product
Step 1
Concept
Co-prime numbers have HCF 1.
Step 2
Why this answer is correct
For two numbers, product (=) HCF \(\times\) LCM.
Step 3
Exam Tip
Therefore, the LCM of co-prime numbers is equal to their product. चरण 1: सह-अभाज्य संख्याओं का महत्तम समापवर्तक 1 होता है। चरण 2: दो संख्याओं के लिए गुणनफल (=) महत्तम समापवर्तक \(\times\) लघुत्तम समापवर्त्य होता है। चरण 3: इसलिए सह-अभाज्य संख्याओं का लघुत्तम समापवर्त्य उनके गुणनफल के बराबर होता है।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
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।
