100 results found for "two-term-plus-constant" in Class 10.
If the constant term of (p(x)=x-2+ax+4) is (4), which term is the constant term?
C. (4)
स्थिर पद में (x) नहीं होता। इसलिए (4) स्थिर पद है। / The constant term does not contain (x). So (4) is the constant term.
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What is the constant term in (p(x)=7x-2)?
C. (0)
कोई अलग नियत पद नहीं है, इसलिए नियत पद (0) है। केवल (x) वाले पद को नियत पद न मानें। / There is no separate constant term, so the constant term is (0). Do not treat a term containing (x) as constant.
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Which option is the constant term in \(6x^2+x-8=0\)?
C. (-8)
स्थिर पद में चर नहीं होता। यहां स्थिर पद (-8) है। / The constant term has no variable. Here the constant term is (-8).
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Which option is the constant term in \(2x^2+3x+1=0\)?
C. (1)
स्थिर पद में (x) नहीं होता। इसलिए (1) स्थिर पद है। / The constant term does not contain (x). Hence (1) is the constant term.
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If the (5)th term of an AP is (27) and the (14)th term is (90), what is the (20)th term?
C. (132)
\(d=\frac{90-27}{14-5}=7\) और \(a_{20}=90+6\times7=132\)। पहले (d) निकालें फिर आगे बढ़ें। / \(d=\frac{90-27}{14-5}=7\) and \(a_{20}=90+6\times7=132\). First find (d) then move forward.
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If the (4)th term of an AP is (15) and the (12)th term is (55), what is the (16)th term?
C. (75)
\(d=\frac{55-15}{12-4}=5\) और \(a_{16}=55+4\times5=75\)। पहले (d) फिर वांछित पद निकालें। / \(d=\frac{55-15}{12-4}=5\) and \(a_{16}=55+4\times5=75\). First find (d), then the required term.
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If the (3)rd term of an AP is (8) and the (8)th term is (3), what is the (11)th term?
A. (0)
\(d=\frac{3-8}{8-3}=-1\), इसलिए (a_{11}=3+3(-1)=0)। ज्ञात पास वाले पद से आगे बढ़ना सरल है। / \(d=\frac{3-8}{8-3}=-1\), so (a_{11}=3+3(-1)=0). Moving from the nearer known term is simple.
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If the (5)th term of an AP is (16) and the (9)th term is (32), what is the (13)th term?
C. (48)
\(d=\frac{32-16}{9-5}=4\), इसलिए \(a_{13}=32+4\times4=48\)। समान स्थान अंतर होने पर पदों का अंतर भी समान होता है। / \(d=\frac{32-16}{9-5}=4\), so \(a_{13}=32+4\times4=48\). Equal position gaps give equal term gaps in an AP.
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C. (32)
मान लें बड़ी संख्या (x) और छोटी (y) है। (x+y=52) और (3x+2y=136) से (x=32)। / Let the larger number be (x) and the smaller be (y). From (x+y=52) and (3x+2y=136), (x=32).
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What is the constant term in ((4x+3)\(x^2-2x+5\))?
D. (15)
अचर पद केवल \(3\cdot5=15\) से आता है। अचर पद के लिए अचर भागों का गुणन देखें। / The constant term comes only from \(3\cdot5=15\). For the constant term, multiply the constant parts.
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What is the constant term in ((3x-2)\(x^2+5x-1\))?
C. (2)
अचर पद केवल ((-2)(-1)=2) से आता है। अचर पद के लिए अचर भागों का गुणन देखें। / The constant term comes only from ((-2)(-1)=2). For the constant term, multiply the constant parts.
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What is the constant term in ((2x-3)\(x^2+x+1\))?
A. (-3)
अचर पद केवल ((-3)\cdot1=-3) से आता है। अचर पद के लिए अचर पदों का गुणन देखें। / The constant term comes only from ((-3)\cdot1=-3). For the constant term, multiply the constant parts.
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In (p(x)=3x-3-7x-2+2x-11), what is the sum of the coefficient of \(x^2\) and the constant term?
A. (-18)
\(x^2\) का गुणांक (-7) और अचर पद (-11) है, इसलिए योग (-18) है। संकेत सहित जोड़ना न भूलें। / The coefficient of \(x^2\) is (-7) and the constant term is (-11), so the sum is (-18). Do not forget to add with signs.
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Which is the constant term in the polynomial (p(t)=4t-5-6t-2+t+15)?
D. (15)
जिस पद में (t) नहीं है वह अचर पद है। यहां अचर पद (15) है। / The term without (t) is the constant term. Here the constant term is (15).
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Which option correctly gives the degree and constant term of \(2x^3-5x^2+x-4\)?
A. घात (3), नियत पद (-4)Degree (3), constant term (-4)
सबसे बड़ी घात (3) है और बिना (x) वाला पद (-4) है। इसलिए सही जोड़ी घात (3), नियत पद (-4) है। / The highest power is (3) and the term without (x) is (-4). So the correct pair is degree (3), constant term (-4).
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If the constant term in (p(x)=x-2+5x+c) is (-6), what will (p(0)) be?
B. (-6)
(p(0)) हमेशा नियत पद के बराबर होता है। यहाँ नियत पद (-6) है। / (p(0)) is always equal to the constant term. Here the constant term is (-6).
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Which is the constant term in \(12x^2-9x+4\)?
C. (4)
जिस पद में (x) नहीं है वह नियत पद होता है। यहाँ नियत पद (4) है। / The term without (x) is the constant term. Here the constant term is (4).
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Which is the constant term in \(5x^4-8x+11\)?
C. (11)
जिस पद में (x) नहीं होता वह अचर पद है। यहां अचर पद (11) है। / The term without (x) is the constant term. Here the constant term is (11).
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Which is the constant term in \(2x^2+3x-11\)?
C. (-11)
जिस पद में (x) नहीं होता वह नियत पद होता है। यहाँ नियत पद (-11) है। / The term without (x) is the constant term. Here the constant term is (-11).
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In \(2x^2+3x+4\), what are the coefficient of (x) and the constant term respectively?
A. (3,4)
(x) का गुणांक (3) और नियत पद (4) है। क्रमशः पूछे गए क्रम का ध्यान रखें। / The coefficient of (x) is (3), and the constant term is (4). Keep the asked order in mind.
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What is the constant term in \(6x^2-3x+10\)?
C. (10)
जिस पद में (x) नहीं है वह नियत पद होता है। यहाँ वह (10) है। / The term without (x) is the constant term. Here it is (10).
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In which option is the constant term absent but the equation is quadratic?
A. \(2x^2+7x=0\)
\(2x^2+7x=0\) में \(x^2\) पद है और स्थिर पद अनुपस्थित है। स्थिर पद न होने पर भी समीकरण द्विघात हो सकता है। / In \(2x^2+7x=0\), the \(x^2\) term is present and the constant term is absent. An equation can be quadratic even without a constant term.
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What is the constant term in \(2x^2+3x-8=0\)?
D. (-8)
जिस पद में (x) नहीं होता वह स्थिर पद होता है। यहाँ स्थिर पद (-8) है। / The term without (x) is the constant term. Here the constant term is (-8).
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If (p(x)=x-2-2x-3\sqrt{2}), what does the constant term tell about the zeroes?
A. शून्यकों का गुणनफल \(-3\sqrt{2}\) हैThe product of zeroes is \(-3\sqrt{2}\)
एकक द्विघात में स्थिर पद शून्यकों का गुणनफल होता है। यहाँ \(\alpha\beta=-3\sqrt{2}\) है। / In a monic quadratic, the constant term is the product of zeroes. Here \(\alpha\beta=-3\sqrt{2}\).
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A. (12)
स्थिर पद शून्यकों के गुणनफल के बराबर है। (\(2\sqrt{2}\)\(3\sqrt{2}\)=12) है। / The constant term equals the product of the zeroes. (\(2\sqrt{2}\)\(3\sqrt{2}\)=12).
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C. (x+82)
(7d=35) से (d=5)। \(a_{20}=a_{12}+8d=x+42+40=x+82\)। / From (7d=35), (d=5). \(a_{20}=a_{12}+8d=x+42+40=x+82\).
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If the (p)th term of an AP is (q) and the (q)th term is (p), what is its ((p+q))th term?
A. (0)
संबंधों को घटाने पर (d=-1) और आगे रखने पर \(a_{p+q}=0\) मिलता है। प्रतीकात्मक AP में भी (a_n=a+(n-1)d) ही लगाएं। / Subtracting the relations gives (d=-1), and substitution gives \(a_{p+q}=0\). Even in symbolic APs, use (a_n=a+(n-1)d).
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A. \(a^2-b\)
एकक बहुपद में स्थिर पद शून्यकों का गुणनफल होता है। यहाँ गुणनफल (\(a+\sqrt{b}\)\(a-\sqrt{b}\)=a-2-b) है। / In a monic polynomial, the constant term is the product of zeroes. Here the product is (\(a+\sqrt{b}\)\(a-\sqrt{b}\)=a-2-b).
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C. (x+181)
(12d=72), इसलिए (d=6)। \(a_{35}=x+91+15\times6=x+181\)। / (12d=72), so (d=6). \(a_{35}=x+91+15\times6=x+181\).
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B. (x+188)
(11d=77) से (d=7)। \(a_{32}=a_{18}+14d=x+90+98=x+188\)। / From (11d=77), (d=7). \(a_{32}=a_{18}+14d=x+90+98=x+188\).
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B. (x+137)
(9d=54) से (d=6)। \(a_{27}=a_{15}+12d=x+65+72=x+137\)। / From (9d=54), (d=6). \(a_{27}=a_{15}+12d=x+65+72=x+137\).
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The (12)th term of an AP is (71) and the common difference is (5). What is the first term?
D. (16)
\(71=a+11\times5\) से (a=16)। ज्ञात पद से पहले पद तक जाने के लिए (11d) घटाएं। / From \(71=a+11\times5\), (a=16). To move from the known term to the first term subtract (11d).
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The (9)th term of an AP is (47) and the common difference is (4). What is the first term?
D. (15)
\(47=a+8\times4\) से (a=15)। ज्ञात पद से पहले पद तक जाने के लिए (8d) घटाएं। / From \(47=a+8\times4\), (a=15). To move from the known term to the first term, subtract (8d).
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The (6)th term of an AP is (23) and the common difference is (5). What is the first term?
D. (-2)
(23=a+5d=a+25), इसलिए (a=-2)। दिए गए पद से पीछे जाते समय (5d) घटाया जाता है। / (23=a+5d=a+25), so (a=-2). When moving backward from a given term, subtract (5d).
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The general term of an AP is \(a_n=7n+2\). Find the (11)th term.
C. (79)
\(a_{11}=7\times11+2=79\)। सामान्य पद में सही पद संख्या रखना ही मुख्य कदम है। / \(a_{11}=7\times11+2=79\). The main step is substituting the correct term number in the general term.
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A. (36)
लड़के (x) और लड़कियां (y) मानें। (x+y=74) और (3x+2y=186) से (x=38), इसलिए (y=36)। / Let boys be (x) and girls be (y). From (x+y=74) and (3x+2y=186), (x=38), so (y=36).
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D. (24)
लड़के (x) और लड़कियां (y) मानें। (x+y=68) और (2x+y=112) से (x=44), इसलिए (y=24)। / Let boys be (x) and girls be (y). From (x+y=68) and (2x+y=112), (x=44), so (y=24).
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B. (1485)
शर्त से (6+13d=3(6+3d)), इसलिए (d=3) और \(S_{30}=1485\) है। पदों की शर्त को पहले समीकरण में बदलें। / The condition gives (6+13d=3(6+3d)), so (d=3) and \(S_{30}=1485\). Convert the term condition into an equation first.
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A. (1050)
शर्त से (5+11d=4(5+2d)), इसलिए (d=5) और \(S_{20}=1050\) है। पदों की शर्त को पहले समीकरण में बदलें। / The condition gives (5+11d=4(5+2d)), so (d=5) and \(S_{20}=1050\). Convert the term condition into an equation first.
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C. (1050)
शर्त से (6+8d=2(6+3d)), इसलिए (d=3) और \(S_{25}=1050\) है। पदों की शर्त को पहले समीकरण में बदलें। / The condition gives (6+8d=2(6+3d)), so (d=3) and \(S_{25}=1050\). Convert the term condition into an equation first.
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C. (576)
शर्त से (4+7d=3(4+2d)), इसलिए (d=8) और \(S_{12}=576\)। पदों की शर्त को पहले समीकरण में बदलें। / The condition gives (4+7d=3(4+2d)), so (d=8) and \(S_{12}=576\). Convert the term condition into an equation first.
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The (n)th-term formula of the AP \(23,31,39,\ldots\) is \(a_n=8n+15\). What is the (36)th term?
C. (303)
\(a_{36}=8\times36+15=303\)। बने हुए सूत्र में (n=36) रखकर सीधे उत्तर पाएं। / \(a_{36}=8\times36+15=303\). Put (n=36) in the formed formula to get the answer directly.
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If the (r)th term of an AP is (3r-2) and (d=4), what is the ((r+6))th term?
C. (3r+22)
((r+6))वां पद (r)वें पद से (6d) आगे है इसलिए (3r-2+24=3r+22)। प्रतीकात्मक प्रश्न में स्थान अंतर देखें। / The ((r+6))th term is (6d) ahead of the (r)th term so (3r-2+24=3r+22). In symbolic questions look at the position gap.
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The (n)th-term formula of the AP \(19,26,33,\ldots\) is \(a_n=7n+12\). What is the (41)st term?
D. (299)
\(a_{41}=7\times41+12=299\)। बने हुए सूत्र में केवल (n=41) रखें। / \(a_{41}=7\times41+12=299\). Put only (n=41) in the formed formula.
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If the (m)th term of an AP is (2m+3) and (d=2), what is the ((m+5))th term?
D. (2m+13)
((m+5))वां पद (m)वें पद से (5d) आगे है इसलिए (2m+3+10=2m+13)। प्रतीकात्मक पदों में भी स्थान अंतर देखें। / The ((m+5))th term is (5d) ahead of the (m)th term, so (2m+3+10=2m+13). In symbolic terms, look at the position gap.
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The first term of an AP is (25) and the (18)th term is (93). What is the common difference?
C. (4)
(93=25+17d) से (68=17d) इसलिए (d=4)। (18)वें पद के लिए गुणक (17) होता है। / From (93=25+17d), (68=17d) so (d=4). For the (18)th term, the multiplier is (17).
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The (n)th term of the AP \(11,17,23,\ldots\) is \(a_n=6n+5\). What is its (32)nd term?
B. (197)
\(a_{32}=6\times32+5=197\)। बनाए गए सूत्र में (n) का सही मान रखें। / \(a_{32}=6\times32+5=197\). Substitute the correct value of (n) in the formed formula.
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The first term of an AP is (18) and the (16)th term is (93). What is the common difference?
A. (5)
(93=18+15d), इसलिए (75=15d) और (d=5)। (16)वें पद के लिए गुणक (15) होगा। / (93=18+15d), so (75=15d) and (d=5). For the (16)th term, the multiplier is (15).
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If the (10)th term of an AP is (41) and the (20)th term is (81), what is its common difference?
C. (4)
दो पदों का अंतर (81-41=40) है और पद संख्या का अंतर (10), इसलिए (d=4)। \(दो ज्ञात पदों में (d=\frac{\)पदों का अंतर}{स्थान का अंतर}) लगाएं। \(/ The difference of terms is (81-41=40) and the difference of positions is (10), so (d=4). Use (d=\frac{\)difference of terms}{difference of positions}).
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If the first term of an AP is (7) and the common difference is (4), what is its (18)th term?
A. (75)
सूत्र (a_n=a+(n-1)d) लगाने पर \(7+17\times4=75\)। परीक्षा में (n-1) को भूलना नहीं चाहिए। / Using (a_n=a+(n-1)d), \(7+17\times4=75\). Exam tip: do not forget (n-1).
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If the (4)th term of an AP is (21) and (d=6), what is the (10)th term?
C. (57)
(10)वाँ पद (4)थे पद से (6d) आगे है इसलिए (21+36=57)। पद संख्या का अंतर सीधे उपयोग करें। / The (10)th term is (6d) after the (4)th term, so (21+36=57). Use the difference in term numbers directly.
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If the first term of an AP is (-12) and the common difference is (5), what is the (16)th term?
B. (63)
\(a_{16}=-12+15\times5=63\)। पहले (15d) निकालें फिर प्रथम पद जोड़ें। / \(a_{16}=-12+15\times5=63\). First calculate (15d), then add the first term.
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If the first term of an AP is (9) and the common difference is (10), what is the (7)th term?
B. (69)
\(a_7=9+6\times10=69\)। (7)वें पद के लिए (6d) जोड़ना होता है। / \(a_7=9+6\times10=69\). For the (7)th term, (6d) is added.
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If the (3)rd term of an AP is (14) and (d=5), what is the (7)th term?
C. (34)
(7)वाँ पद (3)रे पद से (4d) आगे है इसलिए (14+20=34)। बीच के पदों की संख्या सही गिनें। / The (7)th term is (4d) after the (3)rd term, so (14+20=34). Count the gap between terms correctly.
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If the first term of an AP is (-2) and the common difference is (6), what is the (12)th term?
B. (64)
\(a_{12}=-2+11\times6=64\)। पहले गुणा करें फिर (-2) जोड़ें। / \(a_{12}=-2+11\times6=64\). Multiply first and then add (-2).
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If the (5)th term of an AP is (22) and (d=6), what is the (8)th term?
D. (40)
(8)वाँ पद (5)वें पद से (3d) आगे है इसलिए (22+18=40)। पदों का अंतर गिनना आसान तरीका है। / The (8)th term is (3d) after the (5)th term, so (22+18=40). Counting the gap between terms is an easy method.
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If the (7)th term of an AP is (31) and (d=4), what is the (10)th term?
C. (43)
(10)वाँ पद (7)वें पद से (3d) आगे है इसलिए \(31+3\times4=43\)। पास के पदों के लिए अंतर विधि तेज होती है। / The (10)th term is (3d) after the (7)th term, so \(31+3\times4=43\). The difference method is quick for nearby terms.
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If the first term of an AP is (15) and the common difference is (-4), what is the (8)th term?
A. (-13)
(a_8=15+7(-4)=-13)। ऋणात्मक (d) को कोष्ठक में लिखना उपयोगी है। / (a_8=15+7(-4)=-13). Writing negative (d) in brackets is useful.
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If the first term of an AP is (5) and common difference is (3), what is the (8)th term?
C. (26)
सूत्र (a_n=a+(n-1)d) लगाएं तो \(5+7\times3=26\)। परीक्षा में (n-1) लेना न भूलें। / Using (a_n=a+(n-1)d), \(5+7\times3=26\). In exams, do not forget (n-1).
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The (n)th term of an AP is \(a_n=2n+3\). What will be the (18)th term?
A. (39)
(n=18) रखने पर \(a_{18}=2\times18+3=39\)। दिए गए \(a_n\) में सीधे पद संख्या रखें। / Putting (n=18), \(a_{18}=2\times18+3=39\). Substitute the term number directly in the given \(a_n\).
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The first term of an AP is (25) and the common difference is (4). What will be the (21)st term?
B. (105)
\(a_{21}=25+20\times4=105\)। पद संख्या से (1) घटाना न भूलें। / \(a_{21}=25+20\times4=105\). Do not forget to subtract (1) from the term number.
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If the first term of an AP is (6) and the common difference is (7), what is the (11)th term?
D. (76)
\(a_{11}=6+10\times7=76\)। (11)वें पद तक (10) बार अंतर जुड़ता है। / \(a_{11}=6+10\times7=76\). Up to the (11)th term, the difference is added (10) times.
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In an AP, first term (a=4) and common difference (d=3). What is the (10)th term?
A. (31)
सूत्र (a_n=a+(n-1)d) लगाने पर \(a_{10}=4+9\times3=31\)। परीक्षा में (n-1) को ध्यान से लिखें। / Using (a_n=a+(n-1)d), \(a_{10}=4+9\times3=31\). Exam tip: write (n-1) carefully.
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The square of a number (x) plus (3) times the number is (10). Which equation is correct?
B. \(x^2+3x=10\)
वर्ग का अर्थ \(x^2\) और (3) गुना का अर्थ (3x) है। शब्द प्रश्न में वाक्य को धीरे-धीरे समीकरण में बदलें। / Square means \(x^2\) and (3) times means (3x). In word problems, convert the sentence slowly into an equation.
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For what value of (r) can ((r+4)x-3+(r-1)x-2+9) become a constant polynomial?
D. ऐसा कोई मान नहींNo such value
अचर बहुपद के लिए (r+4=0) और (r-1=0) दोनों चाहिए, जो साथ संभव नहीं हैं। सभी चर वाले पद हटने चाहिए। / For a constant polynomial, both (r+4=0) and (r-1=0) are needed, which is impossible together. All variable terms must vanish.
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Which of the following is a constant polynomial?
A. (7)
स्थिर बहुपद में चर का पद नहीं होता, इसलिए (7) स्थिर बहुपद है। अशून्य स्थिर बहुपद की घात (0) होती है। / A constant polynomial has no variable term, so (7) is a constant polynomial. A non-zero constant polynomial has degree (0).
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For what value of (n) can ((n-2)x-2+(n+1)x+5) become a constant polynomial?
D. ऐसा कोई मान नहींNo such value
अचर बहुपद के लिए (n-2=0) और (n+1=0) दोनों चाहिए, जो साथ संभव नहीं हैं। सभी चर वाले पद हटने चाहिए। / For a constant polynomial, both (n-2=0) and (n+1=0) are needed, which is impossible together. All variable terms must vanish.
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For what value of (m) can ((m+1)x-2+(m-2)x+7) become a constant polynomial?
C. ऐसा कोई मान नहींNo such value
अचर बहुपद के लिए (m+1=0) और (m-2=0) दोनों चाहिए, जो साथ संभव नहीं हैं। सभी चर वाले पद हटने चाहिए। / For a constant polynomial, both (m+1=0) and (m-2=0) are needed, which is impossible together. All variable terms must vanish.
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What is the degree of the non-zero constant polynomial (p(x)=-18)?
A. (0)
अशून्य अचर बहुपद की घात (0) होती है। अचर संख्या का अर्थ घात (0) से जुड़ा है। / A non-zero constant polynomial has degree (0). A constant number is linked with degree (0).
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If the degree of a non-zero constant polynomial is asked, what will be the correct answer?
A. (0)
गैर-शून्य नियत बहुपद की घात (0) होती है। इसे शून्य बहुपद से अलग रखें। / The degree of a non-zero constant polynomial is (0). Keep it separate from the zero polynomial.
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Which of the following is not a constant polynomial?
D. (2x+1)
(2x+1) में चर (x) है इसलिए यह नियत बहुपद नहीं है। नियत बहुपद में चर वाला पद नहीं होता। / (2x+1) contains the variable (x), so it is not a constant polynomial. A constant polynomial has no variable term.
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What is the degree of the non-zero constant polynomial (p(x)=12)?
A. (0)
अशून्य अचर बहुपद की घात (0) होती है। शून्य बहुपद से इसे अलग रखें। / A non-zero constant polynomial has degree (0). Keep it separate from the zero polynomial.
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Which is a constant polynomial?
A. (6)
जिस बहुपद में चर नहीं होता वह नियत बहुपद है। (6) में (x) नहीं है। / A polynomial with no variable is a constant polynomial. (6) has no (x).
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How many real zeroes does the constant polynomial (p(x)=5) have?
A. शून्यZero
रेखा (y=5) (x)-अक्ष के समानांतर है और उससे नहीं मिलती। इसलिए कोई वास्तविक शून्यक नहीं है। / The line (y=5) is parallel to the (x)-axis and does not meet it. So it has no real zero.
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What is the relation of the graph of the non-zero constant polynomial (p(x)=5) with the (x)-axis?
A. यह (x)-अक्ष के समांतर है और उसे नहीं काटताIt is parallel to the (x)-axis and does not cut it
(p(x)=5) कभी (0) नहीं होता इसलिए शून्यक नहीं है। टिप: अशून्य स्थिर बहुपद का शून्यक नहीं होता। / The value (p(x)=5) is never (0) so it has no zero. Tip: a non-zero constant polynomial has no zero.
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Why does the graph of the non-zero constant polynomial (p(x)=-3) not cut the (x)-axis?
A. क्योंकि (y) हमेशा (-3) रहता हैBecause (y) always remains (-3)
(p(x)=-3) का (y)-मान कभी (0) नहीं होता। इसलिए ग्राफ (x)-अक्ष को नहीं काटता और कोई शून्यक नहीं है। / For (p(x)=-3), the (y)-value is never (0). So the graph does not cut the (x)-axis and has no zero.
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What does the graph of a non-zero constant polynomial like (p(x)=5) show about its zeroes?
A. कोई शून्यक नहींNo zero
(p(x)=5) का ग्राफ (x)-अक्ष के समानांतर रेखा है जो (x)-अक्ष को नहीं काटती। इसलिए इसका कोई शून्यक नहीं है। / The graph of (p(x)=5) is a line parallel to the (x)-axis and does not cut it. Hence it has no zero.
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A. (-11)
\(x^2\) का गुणांक (-2) और नियत पद (-9) है। उनका योग (-11) है। / The coefficient of \(x^2\) is (-2) and the constant term is (-9). Their sum is (-11).
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What is the constant term in the polynomial \(7x^2+5x-11\)?
C. \(-11\)
स्थिर पद वह होता है जिसमें (x) नहीं होता। यहाँ स्थिर पद (-11) है। / The constant term is the term without (x). Here the constant term is (-11).
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A. (5)
स्थिर पद गुणनफल है और (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5)। संयुग्मी गुणनफल में बीच का अपरिमेय भाग हट जाता है। / The constant term is the product, and (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5). In conjugate products, the irrational middle part cancels.
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If \(9,9,9,\ldots\) is added term by term to \(2,5,8,\ldots\), what will be (d) of the new sequence?
B. (3)
स्थिर अनुक्रम का (d=0) है और दूसरे का (d=3) है, इसलिए योग का (d=3)। स्थिर अनुक्रम जोड़ने से (d) नहीं बदलता। / The constant sequence has (d=0), and the second has (d=3), so the sum has (d=3). Adding a constant sequence does not change (d).
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A. (-3)
स्थिर अनुक्रम का (d=0) और दूसरे का (d=3) है, इसलिए नया (d=0-3=-3)। स्थिर अनुक्रम घटाव में भी (d) बदल सकता है। / The constant sequence has (d=0), and the second has (d=3), so the new (d=0-3=-3). Even with a constant sequence, subtraction can change (d).
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B. (2)
पहले अनुक्रम का (d=0) और दूसरे का (d=2) है, इसलिए नया (d=2)। स्थिर अनुक्रम जोड़ने से (d) नहीं बदलता। / The first sequence has (d=0) and the second has (d=2), so the new (d=2). Adding a constant sequence does not change (d).
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How can divide and rule policy be linked with both short term control and long term crisis?
A. इससे तत्काल विरोध कमजोर हो सकता था लेकिन सामाजिक अविश्वास बचता थाIt could weaken immediate resistance but leave social distrust
यह नीति नियंत्रण के लिए विभाजन का उपयोग करती थी। परीक्षा में इसकी विरासत भी लिखें। / This policy used division for control. For exams also write its legacy.
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C. (3000)
\(a_{21}+a_{40}=a_1+a_{60}=300\), इसलिए (20) पदों का योग (3000) है। सममित पदों का योग बराबर होता है। / \(a_{21}+a_{40}=a_1+a_{60}=300\), so the sum of (20) terms is (3000). Sums of symmetric terms are equal in an AP.
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B. (2100)
\(a_{11}+a_{30}=a_1+a_{40}=210\), इसलिए (20) पदों का योग (2100) है। सममित पदों का योग बराबर होता है। / \(a_{11}+a_{30}=a_1+a_{40}=210\), so the sum of (20) terms is (2100). Sums of symmetric terms are equal in an AP.
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B. (14)
(740=10(a+60)) से (a=14) मिलता है। जब (n)वाँ पद दिया हो तो उसे अंतिम पद की तरह इस्तेमाल करें। / From (740=10(a+60)), (a=14). When the (n)th term is given, use it as the last term for the first (n) terms.
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What is the last term among the positive multiples of (13) less than (600)?
B. (598)
(13n<600) में सबसे बड़ा (n=46) है इसलिए पद \(13\times46=598\)। सीमा से कम सबसे बड़ा पूर्णांक लें। / In (13n<600), the greatest (n=46) so the term is \(13\times46=598\). Take the greatest integer below the limit.
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The (n)th term of the AP \(73,68,63,\ldots\) is (-12). What is (n)?
C. (18)
(-12=73+(n-1)(-5)) से (85=5(n-1)) इसलिए (n=18)। घटती AP में ऋणात्मक लक्ष्य तक चिन्ह सही रखें। / From (-12=73+(n-1)(-5)), (85=5(n-1)) so (n=18). In a decreasing AP keep signs correct up to the negative target.
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If the (19)th term of the AP \(z,z+5,z+10,\ldots\) is (112), what is (z)?
C. (22)
\(112=z+18\times5\) से (z=22)। चर वाले पहले पद को (a) मानकर सूत्र लगाएं। / From \(112=z+18\times5\), (z=22). Treat the variable first term as (a) and apply the formula.
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In the AP \(7,12,17,\ldots\), what is the last term less than (180)?
C. (177)
पद (7+5(n-1)) हैं। (180) से कम अंतिम पद (177) है क्योंकि अगला पद (182) होगा। / The terms are (7+5(n-1)). The last term less than (180) is (177) because the next term will be (182).
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In an AP (a=63) and (d=-4). Which term will be (-1)?
C. (17)वां(17)th
(-1=63+(n-1)(-4)) से (64=4(n-1)) इसलिए (n=17)। ऋणात्मक लक्ष्य पद में चिन्हों को ध्यान से संभालें। / From (-1=63+(n-1)(-4)), (64=4(n-1)) so (n=17). Handle signs carefully with a negative target term.
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Which is the first negative term of the AP \(105,98,91,\ldots\)?
C. (17)वां(17)th
(a_n=105+(n-1)(-7)=112-7n)। \(a_n<0\) से (n>16) इसलिए पहला ऋणात्मक पद (17)वां है। / (a_n=105+(n-1)(-7)=112-7n). From \(a_n<0\), (n>16) so the first negative term is the (17)th.
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Which term of the AP \(26,35,44,\ldots\) is (206)?
C. (21)वां(21)st
(206=26+(n-1)9) से (180=9(n-1)) इसलिए (n=21)। पद और पहले पद का अंतर (d) से भाग दें। / From (206=26+(n-1)9), (180=9(n-1)) so (n=21). Divide the difference between the term and first term by (d).
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The (14)th term of an AP is (92) and (d=7). What is \(a_1\)?
A. (1)
\(92=a+13\times7\) से (a=1)। (14)वें पद से पहले पद तक (13d) घटता है। / From \(92=a+13\times7\), (a=1). From the (14)th term to the first term (13d) is subtracted.
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Which term of the AP \(9,17,25,\ldots\) is (201)?
C. (25)वां(25)th
(201=9+(n-1)8) से (192=8(n-1)) और (n=25)। पद संख्या पूर्णांक आए तो उत्तर सही दिशा में है। / From (201=9+(n-1)8), (192=8(n-1)) and (n=25). If the term number is an integer the answer is on the right track.
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What is the last term among the positive multiples of (11) less than (500)?
B. (495)
(11n<500) में सबसे बड़ा (n=45) है इसलिए पद \(11\times45=495\)। सीमा से कम सबसे बड़ा पूर्णांक लें। / In (11n<500), the greatest (n=45), so the term is \(11\times45=495\). Take the greatest integer below the limit.
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The (n)th term of the AP \(55,51,47,\ldots\) is (-17). What is (n)?
C. (19)
(-17=55+(n-1)(-4)) से (72=4(n-1)) इसलिए (n=19)। ऋणात्मक पद तक जाते समय चिन्ह सही रखें। / From (-17=55+(n-1)(-4)), (72=4(n-1)) so (n=19). Keep signs correct while reaching a negative term.
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If the (12)th term of the AP \(y,y+6,y+12,\ldots\) is (89), what is the value of (y)?
D. (23)
\(89=y+11\times6\) से (y=23)। चर वाले पहले पद को सीधे (a) मानें। / From \(89=y+11\times6\), (y=23). Treat the variable first term directly as (a).
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In the AP \(2,6,10,\ldots\), what is the last term less than (75)?
A. (74)
इस श्रेणी के पद (2+4(n-1)) हैं। (75) से कम अंतिम पद (74) है। / The terms of this sequence are (2+4(n-1)). The last term less than (75) is (74).
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