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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.
| 21951. |
If one root of 5x2+13x+k=0 is the reciprocal of the other root then find the value of k |
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Answer» Answer Answer plz |
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| 21952. |
If (x-2)(x+3)is hcf of p(x)=(2x-3x+2)(4+7x+a) and g(x)=(2x+4x+3)(6-7x+b) find a and b |
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| 21953. |
(1+cotA+tanA)(sinA-cosA)= sinA×tanA-cotA×cosA |
| Answer» L.H.S\xa0= (1 + cotA + tanA) (sinA - cosA)= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA{tex}= \\sin A - \\cos A + \\frac{{\\cos A}}{{\\sin A}} \\times \\sin A - \\cot A\\cos A + \\sin A\\;\\tan A - \\frac{{\\sin A}}{{\\cos A}} \\times \\cos A{/tex}= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA= sinA tanA - cotA cosA........(1)Now taking ;{tex}\\quad \\frac{{\\sec A}}{{\\cos e{c^2}A}} - \\frac{{\\cos ecA}}{{{{\\sec }^2}A}}{/tex}{tex} = \\frac{{\\frac{1}{{\\cos A}}}}{{\\frac{1}{{{{\\sin }^2}A}}}} - \\frac{{\\frac{1}{{\\sin A}}}}{{\\frac{1}{{{{\\cos }^2}A}}}}{/tex}{tex} = \\frac{{{{\\sin }^2}A}}{{\\cos A}} - \\frac{{{{\\cos }^2}A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\frac{{\\sin A}}{{\\cos A}} - \\cos A \\times \\frac{{\\cos A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\tan A - \\cos A \\times \\cot A{/tex}.......(2)From (1) & (2),(1 + cotA + tanA) (sinA - cosA) =\xa0{tex}\\frac { \\sec A } { cosec ^ { 2 } A } - \\frac { cosec A } { \\sec ^ { 2 } A }{/tex}\xa0= sinA.tanA - cosA.cotA\xa0Hence, Proved. | |
| 21954. |
(-2) (-2) (-2)? |
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Answer» -8) -8 (-8) |
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| 21955. |
Nth of ap 3\'6;5799 |
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| 21956. |
How many irrational numbers lie between √2 and √3 ? Write any two of them |
| Answer» Sq rt of 2 = 1.41.....Sq rt of 3 = 1.73.......U can take any 2 nos between these 2 values eg 1.423...., 1.65473........ | |
| 21957. |
Open this identity\' a cube - b cube\' (a3-b3) |
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Answer» Most Welcome Thnx once again (a+b)(a^2-ab-b^2) If we have a^3+b^3 (a-b)(a^2+ab+b^2) |
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| 21958. |
5000 red tiles and 2000 white tile find the probability of blue tiles |
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Answer» 0, this is impossible event 0, zero...... 0 0 Is this the complete question? |
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| 21959. |
CosecA-cotA=q then show thatq2-1\\q2+1+cosA =0 |
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| 21960. |
500 red toffeies and 200 blue toffeies find the probability of not red toffies |
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Answer» Total toffees = red toffees +blue toffees = 500 + 200 =700Not red toffees = blue toffees = 200P(E) =not red toffees /total toffees = 200/700 = 2/7 P(getting not a red toffies)=200/700=2/7 Probability of not red toffies=200/700=2/7 |
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| 21961. |
Sin 130 value |
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| 21962. |
6x+5y=0 |
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| 21963. |
20 red candies and 5 blue candies find the probability of red candies |
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Answer» P(getting red candies)=20/25=4/5 20/25 20/25 4/5 4/5 Probability of red candies=20/25 =4/5 4/5 |
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| 21964. |
Sin 2A = 2 sin A, when A= |
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Answer» A=0° A=0° A=0 I just guessed...... 90 degree.... A=0° |
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| 21965. |
Ch-4 question is ex-4.3 question no-3 |
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Answer» 2》1/x+4-1/x-7=11/30 x-7-x-4/x+4 (x-7)=11/30 -11/x^2-3x-28=11/30 -11 (30)/11 (x^2-3x-28) -30/x^2-3x-28 x^2-3x+2 x^2-x-2x+2 x (x-1)-2 (x-1) (x-2)(x-1)x=2x=1 1》x-1/x=3 x^2-1=3x x^2-3x-1D=b^2-4ac = (-3)^2-4 (1)(-1) =9+4 =13x=-b+root D/2a x=-b+rootD/2a =3+root13/2 =3-root13/2 If it is of. Ncert then ncert solutions is given in this app...... |
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| 21966. |
If sin4x/a+cos4x/b = 1/a+b. Prove that sin8x/a³+cos8x/b³ = (1/a+b)³ |
| Answer» Given : Sin⁴x/a + cos⁴x/b = 1/a+bTo prove :sin⁸x/a³ + cos⁸x/b³ =?Solution:Sin⁴x/a + cos⁴x/b = 1/a+bAs we know that Sin²x +Cos²x=1∴Cos²x=1-sin²xCos⁴x=(Cos²x)²=(1-sin²x)²=1+sin⁴x-2sin²x⇒Sin⁴x/a +[1+sin⁴x-2sin²x]/b =1/a+b⇒Sin⁴x/a+1/b +\xa0sin⁴x/b -2sin²x/b=(1/a+b)⇒Sin⁴x/a+\xa0sin⁴x/b -2sin²x/b=(1/a+b)-1/b⇒Sin⁴x(1/a +1/b)-2sin²xa/ab=[b-a-b]/b(a+b)⇒Sin⁴x(1/a +1/b)-2asin²x/ab=-a/b(a+b)⇒Sin⁴x(a+b)/ab-2asin²x/ab=-a/b(a+b)⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x= - a²⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x + a² =0This equation is similar to a²-2ab+b²=0\xa0⇒(a-b)²[(a+b) Sin²x-a]²=0sin²x=a/a+bCos²x=1-sin²x= 1- a/(a+b)= b/a+b∴sin²x=a/a+b and Cos²x=\xa0b/a+bNow :sin⁸x=a⁴/(a+b)⁴ and Cos⁸x= b⁴/(a+b)⁴sin⁸x/a³= a/(a+b)⁴ and Cos⁸x/b³=b/(a+b)⁴Now\xa0sin⁸x/a³ + Cos⁸x/b³ = a/(a+b)⁴ + b/(a+b)⁴=a+b/(a+b)⁴⇒sin⁸x/a³ + Cos⁸x/b³=1/(a+b)³ | |
| 21967. |
1.73 which angle value of tan |
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Answer» Yesss, its Tan 60°..... \'Tan 60\' as tan 60 stands fr √3 and value of it is 1.73 Complete ur question! |
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| 21968. |
Divide 24 in 3 parts such that they are in AP and their product is 440 ?? |
| Answer» Solution:-Let the required terms of the given AP be a-d, a and a+d *Where the first term is a-d*The common difference = d*Given : The sum of the three parts = 24*∴ (a-d)+(a)+(a+d) = 243a = 24 #a = 8#Given : The product of these three terms = 440∴ (a-d) (a) (a+d) = 440#(8-d) (8) (8+d) = 440#- 8d² + 512 = 440#- 8d² = 440 - 512#- 8d² = - 72#d² = 72/8#d² = 9#d = √9#d= 3#So the three required terms of AP is 8 - 3 = 5 ; 8 and 8 + 3 = 11Three terms are 5, 8, 11 | |
| 21969. |
In an ap if the common different -4,and the seventh term is 4, then find the first term |
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Answer» Givend=-4 and T7=4a+ 6d=4a+6(-4)=4a=4+24a=28 Given that a7=4a+6d=4 a+6(-4)=4 (d=-4)a-24=4a=24+4a=28 A=12 28 _12 A=28 |
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| 21970. |
If tanA+cosA=p,than Find value of cosecA. |
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Answer» Question me tan A +cosA=p h yrr And again plz I have already answered it earlier |
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| 21971. |
CosA/2-sinA=1+sinA/cosA |
| Answer» / means what ? Then the answer comes | |
| 21972. |
Find all zeroes of the polynomial if two zeros are (2+root3) and (2-root 3) |
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Answer» Bt, polynomial hai kahan...????? Where is the polynomial |
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| 21973. |
Solve b2x/a - a2y/b = ab(a + b) b2x - a2y = 2a2b2 |
| Answer» hbj | |
| 21974. |
How can be 2+2 will equal to 5. |
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Answer» Ram I appreciate you Mr Yuvraj ,This is useless thing 2+2=5 is impossible,behave like a good student and do not waste your time.May be somebody prove this by using 0\\0=1,which is again wrong and useless. |
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| 21975. |
If the areas of two similiar triangles are equql, prove that they are congruent |
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Answer» Q.If the areas of two similar triangles are equal, prove that they are congruent. Solution: |
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| 21976. |
Sample papper 12 answer |
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| 21977. |
If cos A = 2/5, find the value of 4 + 4 tan2 A. |
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Answer» CosA =2/5=b/hP= √h²-b²=√5²-2²=√25-4=√21Now, tan²A=(√21)²/2²=21/4According to question,4+4×21/4=4+21=25So the right answer is 25....Hope it helps you a lot.... 25 25 |
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| 21978. |
Tan 20 ta40 tan 8o is equal to tan 60 prove |
| Answer» Given, tan 20 * tan 40 * tan 80= tan 40 * tan 80 * tan 20= [{sin 40 * sin 80}/{cos 40 * cos 80}] *(sin 20/cos 20)= [{2 * sin 40 * sin 80}/{2 * cos 40 * cos 80}] *(sin 20/cos 20)= [{cos 40 - cos 120}/{cos 120 + cos 40}] *(sin 20/cos 20)= [{cos 40 - cos (90 + 30)}/{cos (90 + 30) + cos 40}] *(sin 20/cos 20)= [{cos 40 + sin30}/{-sin30 + cos 40}] *(sin 20/cos 20)= [{(2 * cos 40 + 1)/2}/{(-1 + cos 40)/2}] *(sin 20/cos 20)= [{2 * cos 40 + 1}/{-1 + cos 40}] *(sin 20/cos 20)= [{2 * cos 40 * sin 20 + sin 20}/{-cos 20 + cos 40 * cos 20}]= (sin 60 - sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20)= sin 60/cos 60= tan 60... | |
| 21979. |
Tan 20 tan 6o tan 80 is equal tan 60 |
| Answer» LHS = tan 20° tan 40° tan 80°\xa0{tex}=\\frac{\\sin 20^{\\circ} \\sin 40^{\\circ} \\sin 80^{\\circ}}{\\cos 20^{\\circ} \\cos 40^{\\circ} \\cos 80^{\\circ}}{/tex}{tex}=\\frac{\\left(2 \\sin 20^{\\circ} \\sin 40^{\\circ}\\right) \\sin 80^{\\circ}}{\\left(2 \\cos 20^{\\circ} \\cos 40^{\\circ}\\right) \\cos 80^{\\circ}}{/tex}\xa0{tex}=\\frac{\\left(\\cos 20^{\\circ}-\\cos 60^{\\circ}\\right) \\sin 80^{\\circ}}{\\left(\\cos 60^{\\circ}+\\cos 20^{\\circ}\\right) \\cos 80^{\\circ}}{/tex}\xa0[{tex}\\because{/tex}\xa02 sin a sin b = cos (a-b) - cos (a+b), 2 cos a cos b = cos (a+b) + cos (a-b)]{tex}=\\frac{\\sin 80^{\\circ} \\cos 20^{\\circ}-(1 / 2) \\sin 80^{\\circ}}{(1 / 2) \\cos 80^{\\circ}+\\cos 80^{\\circ} \\cos 20^{\\circ}}{/tex}\xa0[\xa0{tex}\\because \\cos\\;60^o=\\frac12{/tex}]{tex}=\\frac{2 \\sin 80^{\\circ} \\cos 20^{\\circ}-\\sin 80^{\\circ}}{\\cos 80^{\\circ}+2 \\cos 80^{\\circ} \\cos 20^{\\circ}}{/tex}{tex}=\\frac{\\sin 100^{\\circ}+\\sin 60^{\\circ}-\\sin 80^{\\circ}}{\\cos 80^{\\circ}+\\cos 100^{\\circ}+\\cos 60^{\\circ}}{/tex}[{tex}\\because{/tex}\xa02 sin a cos b = sin (a+b) + sin (a-b)]{tex}=\\frac{\\sin \\left(180^{\\circ}-80^{\\circ}\\right)+\\sin 60^{\\circ}-\\sin 80^{\\circ}}{\\cos 80^{\\circ}+\\cos \\left(180^{\\circ}-80^{\\circ}\\right)+\\cos 60^{\\circ}}{/tex}{tex}=\\frac{\\sin 80^{\\circ}+\\sin 60^{\\circ}-\\sin 80^{\\circ}}{\\cos 80^{\\circ}-\\cos 80^{\\circ}+\\cos 60^{\\circ}}{/tex}\xa0{tex}=\\frac{\\sin 60^{\\circ}}{\\cos 60^{\\circ}}{/tex}\xa0= tan 60° = RHS | |
| 21980. |
Find the sum of all two digit no. Which when dividd by 4 yield 3 as remainder |
| Answer» 1254 | |
| 21981. |
(k-12)x+2(k-12)x+2=0 |
| Answer» Anewer | |
| 21982. |
x/0.125 = x/0.75 + 20 |
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| 21983. |
Formula pf Sn |
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Answer» n/2(2a+(n-1)d) Another formula for Sn is n/2(a+an)............. Sn= n/2{2a+(n-1)d} |
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| 21984. |
Plz tell me how i do ch-9 of maths osme diagram ke galti Ho jati hay or fir Sara que galt |
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Answer» Statement ko point to point and word to word padho aur according to given points steps to step figure banate raho and figure bilkul correct banagi Thanks Learn from CBSE guide may it help u i had learn all difficult chapters from this site this is really helpful ?? |
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| 21985. |
What is the maximum value of 1/cosecQ |
| Answer» 1.because 1/cosec=sin.In trigonometry ratios table the maximum value for sin is 1 | |
| 21986. |
CosA -sinA+ 1÷ cosA +sinA -1=cosecA+ cotA |
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Answer» By which identity First divide each term by SinA and then solve it by trignometric identity Plz solve this problem fastly |
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| 21987. |
In an AP the sum of first n term is √3n/2+13n/2 find the 25th term |
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| 21988. |
(p-3)x+3y=ppx+py=12 |
| Answer» Consider equations (p - 3)x + 3y = pand px + py = 12For infinitely many solutions,{tex}\\frac{p - 3}{p}{/tex}\xa0=\xa0{tex}\\frac{3}{p}{/tex}\xa0=\xa0{tex}\\frac{p}{12}{/tex}...........(i)Consider,\xa0{tex}\\frac{3}{p}{/tex}\xa0=\xa0{tex}\\frac{p}{12}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0p2 = 36 {tex}\\Rightarrow{/tex}\xa0p =\xa0{tex}\\pm{/tex}6For p = 6, from (i)\xa0{tex}\\frac{3}{6}{/tex}\xa0=\xa0{tex}\\frac{3}{6}{/tex}\xa0=\xa0{tex}\\frac{6}{12}{/tex}, trueFor p = -6, from (i)\xa0{tex}\\frac{-9}{-6}{/tex}\xa0=\xa0{tex}\\frac{3}{-6}{/tex}\xa0=\xa0{tex}\\frac{-6}{12}{/tex}, false.Hence, for p = 6, pair of linear equations has infinitely many solutions. | |
| 21989. |
Sec theta + tan theta is equals to p find the value of cosec theta |
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| 21990. |
Pls any one tell me how to make a board exam easy on maths what preparation need |
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Answer» Yes it is true that maths need practise and concept understanding...And one more thing be confident every time...think positive and hope that i can do everything. Maths need practice and concept understanding. |
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| 21991. |
How to make trigonometric ratios easily .. |
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Answer» \tLearn sin and cos formula properly.\tThen Value of tan is value of sin divide by value of cos.\tCot is inverse of tan.\tSec is inverse of cos.\tcosec is inverse of sin. And the most important thing is practice.....without practice u cant do anything? Its soo easy u just have to convert everything into sin and cos and then simplify...or take the LCM and solve it....u just hve to think nd apply mind and thats all.....hope it helps yuu? |
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| 21992. |
If the sum of nth term of two A.P. is in ratio 4x+27:7x+14.Then find the ratio of their 9th term |
| Answer» 24:19 | |
| 21993. |
How to solve ratio\'s question in ap chapter |
| Answer» Putting= sign between them | |
| 21994. |
I need re paper 2018 |
| Answer» Which subject...???? | |
| 21995. |
I need question papers since 2013 |
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Answer» Get it from gooooogle. You can download fromhttps://www.topperlearning.com/learn/previous-year-question-papers/cbse/class-10/b101c2e8 U can get it from Google |
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| 21996. |
Cos 45 /sec 30 + cosec 30 please explain me how to do this???? |
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Answer» {tex}\\begin{array}{l}\\frac{\\cos45}{sec30}+\\cos ec30\\\\=\\frac{\\displaystyle\\frac1{\\sqrt2}}{\\displaystyle\\frac2{\\sqrt3}}+2=\\frac1{\\sqrt2}\\times\\frac{\\sqrt3}2+2\\\\=\\frac{\\sqrt3}{2\\sqrt2}+2\\\\=\\frac{\\sqrt3{\\displaystyle+}{\\displaystyle4}{\\displaystyle\\sqrt2}}{2\\sqrt2}=\\frac{1.73+4\\times1.41}{2\\times1.41}=\\frac{1.73+5.64}{2.82}\\\\=\\frac{7.37}{2.82}=2.613\\\\\\end{array}{/tex} I had puteen thia value but i m not able to solve please explain it Hope it will help you Just put the values and then solve, cos 45= 1/root2 , sec 30= 2/root3, and cosec 30=2. |
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| 21997. |
How to find the missing frequency |
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| 21998. |
How can we proof that 0=100 |
| Answer» I think ?in the question there is 0/0 = 100 not only 0 = 100 | |
| 21999. |
7sin+3cos=4,then prove that sec÷cosec=2÷2/√3 |
| Answer» Question is wrong. Go and check it once again because answer cannot be derived from your given question | |
| 22000. |
Can I get all india 2017 board paper |
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Answer» you are fool Yup it is given in the net ..u can do google |
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