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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 11251. |
If tan theta+sec theta =x then find the value of sec theta |
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| 11252. |
May I get sample question paper for 10th class 2018 examination |
| Answer» Check sample papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 11253. |
Tignometric identities |
| Answer» {tex}sin\\ ^ 2\\theta +cos\\ ^ 2\\theta=1{/tex}{tex}sec\\ ^2\\theta—tan\\ ^2\\theta=1{/tex}{tex}cosec\\ ^2\\theta—cot\\ ^2\\theta=1{/tex} | |
| 11254. |
sin90 |
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Answer» Sin90degee is 1 sin 90 = 1angle in degrees 0 30 45 60 90 sin {tex} \\sqrt{}{/tex}0/4 {tex} \\sqrt{}{/tex}1/4 {tex} \\sqrt{}{/tex}2/4 {tex} \\sqrt{}{/tex}3/4 {tex}\\sqrt{}{/tex}4/4 sin 0 1/2 1/{tex} \\sqrt{}{/tex}2 {tex} \\sqrt{}{/tex}3/2 1from the table above we get sin 90 = 1 sin 60 = {tex} \\sqrt{}{/tex}3/2 sin 45 = 1/{tex} \\sqrt{}{/tex}2 sin 30 = 1/2 sin 0 = 0 1 0 0 |
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| 11255. |
If AD and PM are medians of ∆s ABC and PQR respectively where ∆ABC~∆PQR prove that AB/PQ =AD/PM |
| Answer» Given: AD and PM are median of triangles ABC PQR respectively where {tex}\\triangle {/tex} ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}PQRTo prove: {tex}\\frac{{AB}}{{PQ}} = \\frac{{AD}}{{PM}}{/tex}Proof: {tex}\\triangle {/tex}ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}PQR ........Given{tex}\\therefore {/tex}{tex}\\frac{{AB}}{{PQ}} = \\frac{{BC}}{{QR}} = \\frac{{CA}}{{RP}}{/tex} .......(1).....[{tex}\\because {/tex} Corresponding sides of two similar triangles are proportional]and{tex}\\angle{/tex} A = {tex}\\angle{/tex} P, {tex}\\angle{/tex} B = {tex}\\angle{/tex} Q, {tex}\\angle{/tex} C ={tex}\\angle{/tex} R, ..........(2) [ {tex}\\because {/tex} corresponding sides of two similar triangles are proportional]But BC = 2BD and QR = 2QM.............. {tex}\\because {/tex} AD and PM are mediansSo, from(1), {tex}\\frac{{AB}}{{PQ}} = \\frac{{2BD}}{{2QM}}{/tex}{tex}\\Rightarrow {/tex}{tex}\\frac{{AB}}{{PQ}} = \\frac{{BD}}{{QM}}{/tex} ........(3)Also, {tex}\\angle{/tex} ABD = {tex}\\angle{/tex} PQM .........(4).......... From (2){tex}\\therefore {/tex}{tex}\\triangle {/tex} ABD {tex} \\sim {/tex}{tex}\\triangle {/tex}PQM .......SAS similarity criterion{tex}\\therefore {/tex}{tex}\\frac{{AB}}{{PQ}} = \\frac{{AD}}{{PM}}{/tex} ........{tex}\\because {/tex}[Corresponding sides of two similar triangles are proportional] | |
| 11256. |
Solve the following pairs of linear equation using elimination method:-3x+5y-4=0 and 9x=2y+7 |
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Answer» 3x + 5y - 4 = 0 .......... i9x - 2y - 7 = 0 ..... iimultiply ( i) by 3 and subtract ii from i\xa09x + 15y - 12 = 09x - 2y - 7 = 0+ 9x - 9x +15y - (-2y) - 12 - (-7) = 017y - 12 + 7 = 017y = 5y = 5/17subsituting the value of y in (i ) we get 3x + 5x5/17 - 4 = 0 3x + 25/17 - 4 = 03x - 43/17 = 03x = 43/17x = 43/51therefore x = 43/51 , y = 5/17 3x+5y—4=0-------(i)9x=2y+79x—2y=7------------(ii)multiplying equation (i) by 3 we get,9x+15y—12=09x+15y=12---------(iii)subtracting (iii) from(ii) we get,—2y—15y=7—12—17y=—5y={tex}5\\over17{/tex}, putting this value in (i) we get,3x+5y—4=03x+5({tex}5\\over17{/tex})—4=03x=4—{tex}25\\over17{/tex}3x={tex}68—25\\over17{/tex}3x={tex}43\\over17{/tex}x={tex}43\\over51{/tex} |
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| 11257. |
a,b,c are in AP proove that b+c,c+a,a+b are in AP |
| Answer» Since a, b, c are in AP, the common difference will be sameb—a=c—bb+b=c+a2b=c+a2b—c=a-------(i)Now to show b+c, c+a,a+b in A.P , the common difference should be same in each case1st common difference = (c+a)—(b+c) = c+a—b—c = a—b =2b—c—b(from (i)) = b—c2nd common difference = ( a+b)—(c+a) = a+b—c—a =b—cSince, both the common difference is same , so (b+c), (c+a),(a+b) are in AP | |
| 11258. |
Sum of arithmetic series with 15th term is 180 then 8 th term is |
| Answer» LetNo.\'s are a-7b a-6b .......a.......a+6b a+7bThen sum = 15a15a=180a=12and the middle term or 8th term =a Then,8th term = a = 12 ans | |
| 11259. |
If 2x,x+10,3x+2 are in A.P find the value of x. |
| Answer» Since 2x,x+10 and 3x+2 are in APTherefore the common difference must be same.So x+10-2x=3x+2-(x+10) -x+10=3x+2-x-10 -x+10=2x-8 -x+10-2x+8=0 -3x+18=0 -3x=-18 x={tex}-18\\over -3{/tex} x=6\xa0 | |
| 11260. |
If H.P T4=1/12andT10=1/42 find the common difference |
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| 11261. |
2÷10 |
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Answer» 2 x 1/10 = 1/5 = 0.2 0.2 5 |
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| 11262. |
Trick of multiple 9 |
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| 11263. |
In maths ex5.3 question 4 how |
| Answer» By applying the formula sn=2a+(n-1)d we get an quadratic equation .On solving it you get the answer. | |
| 11264. |
Can anyone provide me all the formulas of trigonometry??? Plzzz...... |
| Answer» Trigonometric FormulasReciprocal Properties:Quotient Properties:Odd/Even Identities1. sin (-x) = -sin x2. cos (-x) = cos x3. tan (-x) = -tan x4. csc (-x) = -csc x5. sec (-x) = sec x6. cot (-x) = -cot xCofunction Identity - radiansCofunction Identities - degreesPeriodicity Identities - radiansPeriodicity Identities - degreesSum/Difference Identitiessin (x + y) = sin(x) cos(y) + cos(x) sin(y),cos(x + y) = cos(x) cos(y) - sin(x) sin(y),sin(x - y) = sin(x) cos(y) - cos(x) sin(y),cos(x - y) = cos(x) cos(y) + sin(x) sin(y),Double Angle Identitiessin(2x) = 2 sin(x) cos(x),cos(2x) = cos2(x) - sin2(x),cos(2x) = 2 cos2(x) - 1,cos(2x) = 1 - 2 sin2(x),tan(2x) = [2 tan(x)]/[1-tan2(x)],Half Angle IdentitiesProduct identitiesSum to Product Identities | |
| 11265. |
(a+b)2 |
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Answer» a^2+b^2+2ab (a+b)(a-b) |
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| 11266. |
CosA-sinA+1/cosA+sinA-1=cosecA+cotA |
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Answer» Thank you so much. {tex}{{\\cos {\\rm{A}} - \\sin {\\rm{A}} + 1} \\over {\\cos {\\rm{A}} + \\sin {\\rm{A}} - 1}}{/tex}Dividing all terms by sin A, we get{tex}{{\\cot {\\rm{A}} - 1 + \\cos ec{\\rm{A}}} \\over {\\cot {\\rm{A}} + 1 - \\cos ec{\\rm{A}}}}{/tex}=\xa0{tex}{{\\cot {\\rm{A}} + \\cos ec{\\rm{A}} - 1} \\over {\\cot {\\rm{A}} - \\cos ec{\\rm{A}} + 1}}{/tex}=\xa0{tex}{{\\cot {\\rm{A}} + \\cos ec{\\rm{A}} - \\left( {\\cos e{c^2}{\\rm{A}} - {{\\cot }^2}{\\rm{A}}} \\right)} \\over {\\cot {\\rm{A}} - \\cos ec{\\rm{A}} + 1}}{/tex}=\xa0{tex}{{\\cot {\\rm{A}} + \\cos ec{\\rm{A}}\\left( {1 - \\cos ec{\\rm{A}} + \\cot {\\rm{A}}} \\right)} \\over {\\cot {\\rm{A}} - \\cos ec{\\rm{A}} + 1}}{/tex}=\xa0{tex}{\\cot {\\rm{A}} + \\cos ec{\\rm{A}}}{/tex}Hence proved cosA-sinA+1/cosA+sinA-1= 1/sinA+cosA/sinAcosA-sinA+1/cosA+sinA-1=sinA+cosAsinA/sin (square)AcosA-sinA+1/cosA+sinA-1=sina(1+cosA)/sin(square)AcosA-sinA+1/cosA+sinA-1=1+cosA/sinAsinAcosA-sin(square)A+sinA=cosA+sinA-1+cos(square)A+sinAcosA-cosA(by cross multiplication)sincosA-sin(square)A+sinA=cosA+sinA-1+1-sin(square)A+sinAcosA-cosA0=0Hence proved Yes |
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| 11267. |
2+2=5 |
| Answer» Not defined | |
| 11268. |
(a+b)x + (a-b)y = a*2+b*2 (a-b)x + (a+b)y = a*2+b*2*:this number squre with varible |
| Answer» (a+b)x+(a-b)y=a2+b2\xa0---------(1)(a-b)x+(a+b)y=a2+b2\xa0----------(2)On adding eq(1)and (2) we get(a+b)x+(a-b)y+(a-b)x+(a+b)y=2a2+2b2\xa0(a+b+a-b)x+(a-b+a+b)y=2(a2+b2)2ax+2ay=2(a2+b2)2(ax+ay)=2(a2+b2) ax+ay=a2+b2\xa0-------(3)Now subtracting eq(1) and(2)(a+b)x+(a-b)y -[(a-b)x+(a+b)y]=0(a+b-a+b)x+(a-b-a-b)y =0 2bx-2by=0 2bx=2by x=y ------(4)Subtitute value of x from eq(4) into eq(3) we getay+ay=a2+b2 2ay=a2+b2 y=\xa0{tex}{a^2+b^2\\over 2a}{/tex}From eq (4)\xa0x=y={tex}{a^2+b^2\\over 2a}{/tex}\xa0 | |
| 11269. |
f (x)=sec (1-x×x) |
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| 11270. |
x^2=√5 then x=??? |
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| 11271. |
(2)2 |
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Answer» How hard question is this?????????? 4 |
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| 11272. |
After which decimal place |
| Answer» Ali was a coachman and also avery good hunter. | |
| 11273. |
Prove that √7 is an irrational number |
| Answer» let us assume that √7 be rational.then it must in the form of p / q [q ≠ 0] [p and q are co-prime]√7 = p / q=> √7 x q = psquaring on both sides=> 7q2= p2\xa0\xa0------> (1)p2\xa0is divisible by 7p is divisible by 7p = 7c [c is a positive integer] [squaring on both sides ]p2\xa0= 49 c2\xa0--------- > (2)subsitute p2\xa0in equ (1) we get7q2\xa0= 49 c2q2\xa0= 7c2=> q is divisble by 7thus q and p have a common factor 7.there is a contradictionas our assumsion p & q are co prime but it has a common factor.so that √7 is an irrational. | |
| 11274. |
If tanA= tanB × n , sinA= sinB × m Then, prove that: cos²A = m -1 / no -1 |
| Answer» {tex}\\sin {\\rm{A}} = \\sin {\\rm{B}} \\times m{/tex} => {tex}m = {{\\sin {\\rm{A}}} \\over {\\sin {\\rm{B}}}}{/tex} => {tex}{1 \\over {\\sin {\\rm{B}}}} = {m \\over {\\sin {\\rm{A}}}}{/tex}=> {tex}\\cos ec{\\rm{B}} = {m \\over {\\sin {\\rm{A}}}}{/tex}{tex}\\tan {\\rm{A}} = \\tan {\\rm{B}} \\times n{/tex} => {tex}n = {{\\tan {\\rm{A}}} \\over {\\tan {\\rm{B}}}}{/tex} => {tex}{1 \\over {\\tan {\\rm{B}}}} = {n \\over {\\tan {\\rm{A}}}}{/tex}=> {tex}\\cot {\\rm{B}} = {n \\over {\\tan {\\rm{A}}}}{/tex}Using identity,\xa0{tex}\\cos e{c^2}{\\rm{B}} - {\\cot ^2}{\\rm{B}} = 1{/tex}=> {tex}{{{m^2}} \\over {{{\\sin }^2}{\\rm{A}}}} - {{{n^2}} \\over {{{\\tan }^2}{\\rm{A}}}} = 1{/tex}=> {tex}{{{m^2}} \\over {{{\\sin }^2}{\\rm{A}}}} - {{{n^2}.{{\\cos }^2}{\\rm{A}}} \\over {{{\\sin }^2}{\\rm{A}}}} = 1{/tex}=> {tex}{m^2} - {n^2}.{\\cos ^2}{\\rm{A}} = {\\sin ^2}{\\rm{A}}{/tex}=> {tex}{m^2} - {n^2}.{\\cos ^2}{\\rm{A}} = 1 - {\\cos ^2}{\\rm{A}}{/tex}=> {tex}{m^2} - 1 = - {\\cos ^2}{\\rm{A}} + {n^2}.{\\cos ^2}{\\rm{A}}{/tex}=> {tex}{\\cos ^2}{\\rm{A}} = {{{m^2} - 1} \\over {{n^2} - 1}}{/tex} | |
| 11275. |
How can we solve quadratic equation who have in root |
| Answer» And if the word problem is given in root try to take the variable in square.Like instead of taking something as (x) take it as(x^2). | |
| 11276. |
Trignometery |
| Answer» Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. | |
| 11277. |
What is square root of611150 |
| Answer» 781.760832992 | |
| 11278. |
2x plus 5 |
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| 11279. |
This year totally new concept of exam what is pattern of questions |
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| 11280. |
Find the roots 1/x - 1/×-2 =3 |
| Answer» {tex}{1\\over x}-{1\\over x-2}=3{/tex}\xa0{tex}{x-2-x\\over x(x-2)}=3{/tex} [by taking L.C.M.]-2=3×(x(x-2))-2=3(x2-2x)-2=3x2-6x3x2-6x+2=0On comparing with ax2+bx+c=0 we get\xa0a=3 b= -6 c=2By using quadratic formula\xa0{tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex}{tex}x = {-(-6) \\pm \\sqrt{(-6)^2-4×3×2} \\over 2×3}{/tex}{tex}x = {6 \\pm \\sqrt{36-24} \\over 6}{/tex}{tex}x = {6 \\pm \\sqrt{12} \\over 6}{/tex}{tex}x = {6 \\pm 2\\sqrt{3} \\over 6}{/tex}{tex}x = {2(3\\pm \\sqrt{3}) \\over 6}{/tex}{tex}x = {3\\pm \\sqrt{3} \\over 3}{/tex}So roots are\xa0{tex}{3 +\\sqrt{3} \\over 3}{/tex} and {tex}{3 - \\sqrt{3} \\over 3}{/tex} | |
| 11281. |
√x+y=7. x+√y=11 |
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Answer» first rearrange the equations as followsy - 4 = 3 - {tex}\\sqrt{x}{/tex} ....... i\xa0x - 9 = 2 - {tex} \\sqrt{y}{/tex}\xa0....... iimultiplying i and\xa0ii(y - 4)(x - 9) = (3 - {tex} \\sqrt{x}{/tex})(2 - {tex} \\sqrt{y}{/tex})we can factor LHS({tex} \\sqrt{y}{/tex}\xa0- 2)({tex} \\sqrt{y}{/tex}\xa0+ 2)({tex}\\sqrt{x}{/tex}\xa0+3)({tex} \\sqrt{x}{/tex}\xa0- 3) = (3 - {tex} \\sqrt{x}{/tex})(2 - {tex} \\sqrt{y}{/tex})({tex} \\sqrt{y}{/tex}\xa0- 2)({tex} \\sqrt{x}{/tex}\xa0- 3)({tex} \\sqrt{y}{/tex}\xa0- 2)({tex} \\sqrt{x}{/tex}\xa0- 3) - 1 = 0at least one of the factors must be zero{tex} \\sqrt{y}{/tex}\xa0- 2 = 0 ...... {tex} \\sqrt{y}{/tex}\xa0= 2 ............ y = 4substituting in i3\xa0- {tex} \\sqrt{x}{/tex}\xa0= 0 ........{tex} \\sqrt{x}{/tex}\xa0= 3 ........... x = 9therefore x = 9 and y = 4 first rearrange the equations as followsy - 4 = 3 - sqrt x ....... i\xa0x - 9 = 2 - sqrt y ....... iimultiplying i and\xa0ii(y - 4)(x - 9) = (3 - sqrt x)(2 - sqrt y)we can factor LHS(sqrt y - 2)(sqrt y + 2)(sqrt x +3)(sqrt x - 3) = 3 - sqrt x)(2 - sqrt y)(sqrt y - 2)(sqrt x - 3)(sqrt y - 2)(sqrt x - 3) - 1 = 0at least one of the factors must be zerosqrt y - 2 = 0 ...... sqrt y = 2 ............ y = 4substituting in i3\xa0- sqrt x = 0 ........ sqrt x = 3 ........... x = 9therefore x = 9 and y = 4 |
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| 11282. |
Root of x-1 |
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| 11283. |
2x2-3x+5=0 find the root of the following quadratic equation by factorisation |
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| 11284. |
(1-sinx)÷(1-cosx) |
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| 11285. |
Why we take sin theta =p/h |
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| 11286. |
HCF of two number is 23 and LCM is 1449 if one number is 161 then find the other number? |
| Answer» We know that,{tex}L C M \\times H C F = a \\times b{/tex}{tex}\\Rightarrow 1449 \\times 23 = 161 \\times b{/tex}{tex}\\Rightarrow b = \\frac { 1449 \\times 23 } { 161 } = 207{/tex} | |
| 11287. |
Find the zeroes of √3x2-8x+4√3 |
| Answer» {tex}\\sqrt 3x^2-8x+4\\sqrt 3=0{/tex}{tex}=> \\sqrt 3x^2-6x-2x+4\\sqrt 3=0{/tex}{tex}=> \\sqrt 3x(x-2\\sqrt 3)-2(x-2\\sqrt 3)=0{/tex}{tex}=> (\\sqrt 3x-2)(x-2\\sqrt 3)=0 {/tex}{tex}=> (\\sqrt 3x-2)= 0 ,(x-2\\sqrt 3)=0 {/tex}{tex}=> x ={2\\over \\sqrt 3},2\\sqrt 3{/tex} | |
| 11288. |
Prove that (5-2√3)2 is irrational number? |
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Answer» We will prove it in a same way as we prove that\xa0{tex} \\sqrt{3}{/tex}\xa0is an irrational number..Let us assume that\xa0{tex}({5-2\\sqrt{3}})^2{/tex}\xa0is a rational number.\xa0Then {tex}( {5-2\\sqrt{3}})^2={p\\over q}{/tex} where p and q are co-prime.{tex} =>{/tex}{tex}25-2×5×2{\\sqrt 3}+(2{\\sqrt 3})^2={p\\over q}{/tex} [by using\xa0{tex}(a-b)^2=a^2-2ab+b^2{/tex}{tex}25-20{\\sqrt 3} +4×3={p\\over q}{/tex}{tex}25-20{\\sqrt 3}+12={p\\over q}{/tex}{tex}37-20{\\sqrt 3}={p\\over q}{/tex}{tex}37+{p\\over q}=20{\\sqrt3 }{/tex}{tex}{37\\over 20} +{p\\over 20q}={\\sqrt 3}{/tex} Clearly L.H.S. is a sum of two rational number and therefore L.H.S is rational.So\xa0{tex}{\\sqrt 3}{/tex}\xa0is a rational number.But we know that\xa0{tex}{\\sqrt3}{/tex}\xa0is an irrational number.So our assumption is wrong.Hence\xa0{tex}({5-2\\sqrt{3}})^2{/tex}is an irrational number. prove (5-2 root3)2\xa0=\xa025-20root 3+12=p/q consider it as rational firstbut solving it we get that root 3 could be represented as rational but it is irrational number thus our assumption that (5-2 root 3)2was wrong and hence it is irrational |
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| 11289. |
can you please solve the important questions part\xa03 in introduction of trignometry |
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| 11290. |
2/x^2-5/x+2=0Bye using the method of compeleting the square. |
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Answer» {tex}{2\\over x^2}{/tex}-{tex}{5\\over x}{/tex}+2=0{tex}{2-5x+2x^2\\over x^2}{/tex}=02-5x+2x2=02x2-5x+2=0 ------(1)\xa0Now first of all we will make coefficient of x2\xa0to 1So divide eq.(1) by 2 we getx2-{tex}{5\\over 2}{/tex}x+1=0Now add and subtract the square of half of the coefficient of x we getx2-{tex}{5\\over2}{/tex}x+{tex}({5\\over 4})^2 -({5\\over 4})^2{/tex}+1=0(x-{tex}{5\\over 4})^2{/tex}-{tex}{25\\over 16}{/tex}+1=0 [ by using identity (a-b)2=a2-2ab+b2](x-{tex}{5\\over 4})^2{/tex}-{tex}{25+16\\over 16}{/tex}=0(x-{tex}{5\\over 4})^2{/tex}-{tex}{9\\over 16}{/tex}=0(x-{tex}{5\\over 4})^2{/tex}={tex}{9\\over 16}{/tex}(x-{tex}{5\\over 4}{/tex})={tex}{\\pm \\sqrt{9 \\over 16}}{/tex}(x-{tex}{5\\over 4}){/tex}={tex}{\\pm 3\\over 4}{/tex}So x-{tex}{5\\over 4}{/tex}={tex}{3\\over 4}{/tex}\xa0or x-{tex}{5\\over 4}{/tex}={tex}{-3\\over 4}{/tex}x={tex}{3\\over 4} +{5\\over 4}{/tex} or x={tex}{-3\\over 4}+{5\\over 4}{/tex}x={tex}{3+5\\over 4}{/tex}\xa0or x={tex}{-3+5\\over 4}{/tex}x={tex}{8\\over 4}{/tex}\xa0or x={tex}{2\\over 4}{/tex}x=2 or x=\xa0{tex}{1\\over 2}{/tex} {tex}2x^2-5x+2=0{/tex}{tex}=> (\\sqrt 2x)^2-2\\times \\sqrt 2 \\times {5\\over 2\\sqrt2}x+2 + ({5\\over 2\\sqrt 2})^2-({5\\over 2\\sqrt 2})^2=0{/tex}{tex}=> (\\sqrt 2x)^2-2\\times \\sqrt 2 \\times {5\\over 2\\sqrt2}x + ({5\\over 2\\sqrt 2})^2+2 -{25\\over 8}=0{/tex}{tex}=> (\\sqrt 2x)^2-2\\times \\sqrt 2 \\times {5\\over 2\\sqrt2}x + ({5\\over 2\\sqrt 2})^2-{9\\over 8}=0{/tex}{tex}=> (\\sqrt 2x-{5\\over 2\\sqrt 2})^2-({3\\over 2\\sqrt 2})^2=0{/tex}{tex}=> (\\sqrt 2x-{5\\over 2\\sqrt 2} +{3\\over 2\\sqrt 2})(\\sqrt 2x-{5\\over 2\\sqrt 2} -{3\\over 2\\sqrt 2}){/tex}{tex}=> (\\sqrt 2x-{1\\over \\sqrt 2} )(\\sqrt 2x-{\\sqrt 2}){/tex} |
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| 11291. |
If sinA +sin (2)^A=1 Prove that cos (2)^A+cos (4)^A=1 |
| Answer» {tex}SinA + Sin\\ ^ 2A = 1{/tex}{tex}SinA = 1-Sin\\ ^ 2A{/tex}{tex}SinA = Cos\\ ^ 2 A{/tex}( Using identity)Squaring both sides,{tex}(SinA)\\ ^ 2 = ( Cos\\ ^2 A )\\ ^ 2{/tex}{tex}Sin\\ ^ 2 A=Cos\\ ^ 4 A{/tex}{tex}1- Cos\\ ^ 2 A= Cos \\ ^ 4 A{/tex}( Using identity){tex}1 = Cos \\ ^ 2A + Cos \\ ^ 4 A{/tex}{tex}Cos \\ ^ 2A + Cos \\ ^ 4 A = 1{/tex} | |
| 11292. |
If the zeros of the polynomial f(x)=2xcube +15xsquare 37x-30 are in A.P.,Find them. |
| Answer» Let, {tex}α{/tex} = a - d, {tex}β{/tex} = a and {tex}\\gamma {/tex} = a + d be the zeroes of the polynomial.f(x) = 2x3 - 15x2 + 37x - 30{tex}\\alpha + \\beta + \\gamma = - \\left( {\\frac{{ - 15}}{2}} \\right) = \\frac{{15}}{2}{/tex} ....... (i){tex}\\alpha \\beta \\gamma = - \\left( {\\frac{{ - 30}}{2}} \\right) = 15{/tex} ......... (ii)From (i)a - d + a + a + d = {tex}\\frac{{15}}{2}{/tex}\xa0So, 3a = {tex}\\frac{{15}}{2}{/tex}a = {tex}\\frac{{5}}{2}{/tex}and From (ii)a(a - d)(a + d) = 15So, a(a2 - d2) = 15{tex}⇒ \\frac{{5}}{2}{/tex}{tex}\\left[\\left(\\frac52\\right)^2\\;-d^2\\right]{/tex} = 15{tex}⇒ \\frac{25}4\\;-d^2{/tex}= 6{tex}⇒ \\;d^2\\;=\\;\\frac{25}4-6\\;{/tex}{tex}⇒ {d^2} = \\frac{1}{4}{/tex}{tex}⇒ d = \\frac{1}{2}{/tex}Therefore, {tex}\\alpha = \\frac{5}{2} - \\frac{1}{2} = \\frac{4}{2} = 2{/tex}{tex}\\beta = \\frac{5}{2}{/tex}{tex}\\gamma = \\frac{5}{2} + \\frac{1}{2} = 3{/tex}. | |
| 11293. |
Cos(x)(tan(x)+1)(2tan(x)+1)= 2sec(x)+5sin(x) |
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| 11294. |
Prove that\xa0{tex}4\\sqrt{3} - 3\\sqrt{2}{/tex}\xa0is an irrational number. |
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| 11295. |
Prove that cot(2)^A(secA-1/1+sinA)+sec(2)^A (sinA-1/1+secA)=0 |
| Answer» L.H.S. = {tex}{{{{\\cot }^2}{\\rm{A}}\\left( {\\sec {\\rm{A}} - 1} \\right)} \\over {\\left( {1 + \\sin {\\rm{A}}} \\right)}} +{{{{\\sec }^2}{\\rm{A}}\\left( {\\sin {\\rm{A}} - 1} \\right)} \\over {\\left( {1 + \\sec {\\rm{A}}} \\right)}}{/tex}= {tex}{{{{\\cot }^2}{\\rm{A}}\\left( {\\sec {\\rm{A}} - 1} \\right)} \\over {\\left( {1 + \\sin {\\rm{A}}} \\right)}} \\times {{1 - \\sin {\\rm{A}}} \\over {1 - \\sin {\\rm{A}}}} \\times {{\\sec {\\rm{A}} + 1} \\over {\\sec {\\rm{A}} + 1}} + {{{{\\sec }^2}{\\rm{A}}\\left( {\\sin {\\rm{A}} - 1} \\right)} \\over {\\left( {1 + \\sec {\\rm{A}}} \\right)}}{/tex}= {tex}{{{{\\cot }^2}{\\rm{A}}\\left( {{{\\sec }^2}{\\rm{A}} - 1} \\right)\\left( {1 - \\sin {\\rm{A}}} \\right)} \\over {\\left( {1 - {{\\sin }^2}{\\rm{A}}} \\right)\\left( {1 + \\sec {\\rm{A}}} \\right)}} + {{{{\\sec }^2}{\\rm{A}}\\left( {\\sin {\\rm{A}} - 1} \\right)} \\over {\\left( {1 + \\sec {\\rm{A}}} \\right)}}{/tex}= {tex}{{{{\\cot }^2}{\\rm{A}}{\\rm{.ta}}{{\\rm{n}}^2}{\\rm{A}}\\left( {1 - \\sin {\\rm{A}}} \\right)} \\over {{{\\cos }^2}{\\rm{A}}\\left( {1 + \\sec {\\rm{A}}} \\right)}} + {{{{\\sec }^2}{\\rm{A}}\\left( {\\sin {\\rm{A}} - 1} \\right)} \\over {\\left( {1 + \\sec {\\rm{A}}} \\right)}}{/tex}= {tex}{{{{\\sec }^2}{\\rm{A}}\\left( {1 - \\sin {\\rm{A}}} \\right)} \\over {\\left( {1 + \\sec {\\rm{A}}} \\right)}} - {{{{\\sec }^2}{\\rm{A}}\\left( {1 - \\sin {\\rm{A}}} \\right)} \\over {\\left( {1 + \\sec {\\rm{A}}} \\right)}}{/tex}= 0= R.H.S. | |
| 11296. |
Factorise -2x2\xa0-5x + 7 |
| Answer» | |
| 11297. |
If the polynomial x^3+2x^2+ax+b has factors x+1 and x-1, find the value of a and b |
| Answer» p(x) =\xa0{tex}x^3+2x^2+ax+b{/tex}As x+1 is factor of p(x)Thenp(-1)=0=>\xa0{tex}(-1)^3+2(-1)^2+a(-1)+b=0{/tex}{tex}=>-1+2-a+b=0{/tex}=> b-a = -1 .......(1)Also x-1 is factor of p(x)p(1)=0=>\xa0{tex}(1)^3+2(1)^2+a(1)+b=0{/tex}=> 1+2+a+b=0=> b+a=-3 .......(2)Adding (1) and (2), we get=> 2b= -4=> b =-2From (2)=> 2+a=-3=> a = -5\xa0 | |
| 11298. |
Sir, please give me sample paper of latest patterns\xa0 |
| Answer» Check Sample Papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 11299. |
The value of quadratic polynomial f(x)=2x² -3x-2 at x=-2 is ...... |
| Answer» f(x) = 2x2 - 3x - 2=> f(-2) = 2(-2)2 - 3(-2) - 2=> f(-2) = 2 x 4\xa0+ 6 - 2=> f(-2) = 8 + 6 - 2=> f(-2) = 14 - 2=> f(-2) = 12 | |
| 11300. |
find all the common zeroes of the polynomial x3+8x2+15x and x3+5x2-9x-45 |
| Answer» {tex}x^3+8x^2+15x{/tex}To find zero equate it with zero{tex}=> x^3+8x^2+15x= 0{/tex}{tex}=>x[ x^2+8x+15]= 0{/tex}{tex}=>x[ x^2+5x+3x+15]= 0{/tex}{tex}=>x[ x(x+5)+3(x+5)]= 0{/tex}{tex}=>x(x+5)(x+3)= 0{/tex}=> x = 0, -5,-3Similarly,\xa0{tex}x^3+5x^2-9x-45= 0{/tex}{tex}=> x^2(x+5)-9(x+5)= 0{/tex}{tex} =>( x^2-9)(x+5)= 0{/tex}{tex} =>( x+3)(x-3)(x+5)= 0{/tex}x = -3,-5,3Common zeroes are -3 and -5 | |