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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.
| 6401. |
Draw appropriate venn diagram of A\'-B |
| Answer» It will not included set A and B | |
| 6402. |
cos 2 - |
| Answer» | |
| 6403. |
Is 0 is a integer? |
| Answer» Yes | |
| 6404. |
3. -3 |
| Answer» -9 | |
| 6405. |
If the ratio in the angles of triangles are 1 : 2 :3 then find the ratio in the corresponding sides |
| Answer» {tex}Let\\; the \\quad angles\\ be\\quad x,2x\\; and\\; 3x\\quad then,x+2x+3x={ 180^{ o } }\\Rightarrow 6x=180\\Rightarrow x{ { =\\frac { 180 }{ 6 } } }={ 30^{ o\\\\ } }\\\\ Using\\; sine\\quad rule\\quad we\\quad have\\frac { a }{ sinA } =\\frac { b }{ sinB } =\\frac { c }{ sinC } =k(say)\\\\ Herea=k\\sin 30,b=k\\sin 60,c=k\\sin 90\\\\ \\Rightarrow a=\\frac { { { k } } }{ 2 } ,b={ { \\frac { \\sqrt { 3 } }{ 2 } } }.k,c={ 1 }.k\\\\ a{ { :b{ { : } }c } }=\\frac { { { k } } }{ 2 } { { :\\frac { k\\sqrt { 3 } }{ 2 } { { : } } } }{ 1 }.k\\\\ \\Rightarrow a{ { :b{ { : } }c } }=1{ { :\\sqrt { 3 } { { : } }2 } }{/tex} | |
| 6406. |
prove tahat--sin³x+cos ³x=(sin x+cos x)(1-sin x•cos x) |
| Answer» | |
| 6407. |
What is cordinal number |
| Answer» | |
| 6408. |
Prove thata cosA+b cosB+c cosC=2a sinB sinC |
| Answer» | |
| 6409. |
y=3sin2x Sketch the graph |
| Answer» | |
| 6410. |
7+24i |
| Answer» | |
| 6411. |
Solve abc equation,a+b+c=0,-a-3b+2c=o,-2a-2c=1 |
| Answer» | |
| 6412. |
Find the value of tan pie divided by 8 |
| Answer» √2-1 | |
| 6413. |
Sin(x+y)/sin(x-y)=tan x+tan y/tan x- tan y (prove that) |
| Answer» | |
| 6414. |
Proove cosec squarex+sin square x >2 |
| Answer» | |
| 6415. |
the roots of the equation x^2+(p+2)x+2p=0 are distinct integers when |
| Answer» x2+(p+2)x+2p=0On comparing with ax2+bx+c=0 we geta=1 b=p+2 c=2pNow since roots of the given quadratic equation are distinct integer.Therefore, Discriminent D=b2-4ac > 0 (p+2)2-4×1×2p > 0 p2+2×p×2+22\xa0-8p > 0 p2+4p+4-8p > 0 p2\xa0-4p +4 > 0 p2-2p-2p+4 > 0 p(p-2)-2(p-2) > 0 (p-2)(p-2) > 0 (p-2)2\xa0> 0 p-2 > 0 p > 2So roots of the given quadratic eq. is distinct integer if p>2. | |
| 6416. |
Proove cosec2x |
| Answer» | |
| 6417. |
(b+c)*cos(B+C)/2=? |
| Answer» | |
| 6418. |
What is integral and differential |
| Answer» | |
| 6419. |
(a+b)/c=cos( (A-B) /2)/sinC/2 |
| Answer» | |
| 6420. |
What is the meaning of sin x |
| Answer» | |
| 6421. |
What is the value of tan -pie by 18 |
| Answer» | |
| 6422. |
an+1+bn+1/an +bn |
| Answer» | |
| 6423. |
How to get the value of pie/18 degree.Prove according to class 11 method |
| Answer» -Pie by 18 means 10 degree.....so u have to use the identities of 3x .Because thrice of -pie by 18 is 30 degree | |
| 6424. |
Tan70=tan50+tan20 |
| Answer» | |
| 6425. |
X-3/2x+1. Find domain and range |
| Answer» | |
| 6426. |
Find the rank of the word mathematics ? |
| Answer» | |
| 6427. |
cosec4π/15+cosec8π/15+cosec16π/15+cosec32π/15 |
| Answer» | |
| 6428. |
|x| |
| Answer» Its mod x | |
| 6429. |
Factorise: m^4+n^4+18m^2 n^2=0 |
| Answer» | |
| 6430. |
Prove that cot x cot 2x - cot 2x cot 3x - cot 3x cot x = 1 |
| Answer» | |
| 6431. |
tan(A+B) =1,tan(A-B)=1/7 find tanA and tanB. |
| Answer» | |
| 6432. |
Write in form of a + ibUnder the root 6-i2 |
| Answer» | |
| 6433. |
Why factorial 1 value is 1 |
| Answer» | |
| 6434. |
Prove:-cos2x = 1-tan^2x/1+tan^2x |
| Answer» | |
| 6435. |
Examples of trigonometric function |
| Answer» | |
| 6436. |
How angle can be greater than 360 or in negative |
| Answer» | |
| 6437. |
1/√1-x |
| Answer» | |
| 6438. |
Cos4 |
| Answer» | |
| 6439. |
How to find value of cos 20 degree\xa0 |
| Answer» {tex}\\cos 3\\theta = 4{\\cos ^3}\\theta - 3\\cos \\theta \\\\take \\theta =20 \\\\\\frac{1}{2} = 4{x^3} - 3x \\\\8{x^3} - 6x-1=0{/tex}solve the equation | |
| 6440. |
Supplement exercise with solutions of mathematics for class XI |
| Answer» | |
| 6441. |
How do determine the domain and range of a function. Please also give some examples |
| Answer» | |
| 6442. |
Method to write Cartesian product of two sets. And please give me a example of the question.\xa0 |
| Answer» In\xa0set theory\xa0(and, usually, in other parts of\xa0mathematics), a\xa0Cartesian product\xa0is a\xa0mathematical operation\xa0that returns a\xa0set\xa0(or\xa0product set\xa0or simply\xa0product) from multiple sets. That is, for sets\xa0Aand\xa0B, the Cartesian product\xa0A\xa0×\xa0B\xa0is the set of all\xa0ordered pairs\xa0(a,\xa0b)\xa0where\xa0a\xa0∈\xa0A\xa0and\xa0b\xa0∈\xa0B. Products can be specified using\xa0set-builder notation, e.g.{\\displaystyle A\\times B=\\{\\,(a,b)\\mid a\\in A\\ {\\mbox{ and }}\\ b\\in B\\,\\}.}[1]A table can be created by taking the Cartesian product of a set of rows and a set of columns. If the Cartesian product\xa0rows\xa0×\xa0columns\xa0is taken, the cells of the table contain ordered pairs of the form\xa0(row value, column value).More generally, a Cartesian product of\xa0nsets, also known as an\xa0n-fold Cartesian product, can be represented by an array of\xa0n\xa0dimensions, where each element is an\xa0n-tuple. An ordered pair is a\xa02-tuple or couple.The Cartesian product is named after\xa0René Descartes,[2]\xa0whose formulation of\xa0analytic geometry\xa0gave rise to the concept, which is further generalized in terms of\xa0direct product.ExamplesA deck of cardsStandard 52-card deckAn illustrative example is the\xa0standard 52-card deck. The\xa0standard playing cardranks {A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, 2} form a 13-element set. The card suits\xa0{♠,\xa0♥,\xa0♦, ♣}\xa0form a four-element set. The Cartesian product of these sets returns a 52-element set consisting of 52\xa0ordered pairs, which correspond to all 52 possible playing cards.Ranks\xa0×\xa0Suits\xa0returns a set of the form {(A, ♠), (A,\xa0♥), (A,\xa0♦), (A, ♣), (K, ♠), ..., (3, ♣), (2, ♠), (2,\xa0♥), (2,\xa0♦), (2, ♣)}.Suits\xa0×\xa0Ranks\xa0returns a set of the form {(♠, A), (♠, K), (♠, Q), (♠, J), (♠, 10), ..., (♣, 6), (♣, 5), (♣, 4), (♣, 3), (♣, 2)}.Both sets are distinct, even disjoint. | |
| 6443. |
-78/IxI-28>=1wherex belong to R |
| Answer» | |
| 6444. |
Show that for z belongs to C, |z|=0 if and only if z=0 |
| Answer» | |
| 6445. |
In a triangle ABC prove that a square+b square+c square =2(bc cosA+ca cosB+ab cosC) |
| Answer» | |
| 6446. |
If 1,w,w² are the cube roots of unity then w²(1+w)³-(1+w²)w=? |
| Answer» If 1,w,w² are the cube roots of unity then we have\xa0{tex}{ \\omega }^{ 3 }=1,1+\\omega +{ \\omega }^{ 2 }=0{/tex}w²(1+w)³-(1+w²)w=\xa0{tex}{ \\omega }^{ 2 }{ \\left( -{ \\omega }^{ 2 } \\right) }^{ 3 }-\\left( -\\omega \\right) .\\omega ={ \\omega }^{ 2 }.{ -\\omega }^{ 6 }+{ \\omega }^{ 2 }=-{ \\omega }^{ 8 }+{ \\omega }^{ 2 }=-{ \\omega }^{ 3 }.{ \\omega }^{ 3 }.{ \\omega }^{ 2 }+{ \\omega }^{ 2 }=-{ \\omega }^{ 2 }+{ \\omega }^{ 2 }=0{/tex}{tex}\\\\ \\left[ \\because { \\left( { a }^{ m } \\right) }^{ n }={ a }^{ mn },{ a }^{ m }.{ a }^{ n }={ a }^{ m+n } \\right] {/tex} | |
| 6447. |
Find the general solution of the equation sin3x+cos2x=0\xa0 |
| Answer» | |
| 6448. |
Find the sum of n-terms of the series 1^2+ +3^2+5^2+........ |
| Answer» Let {tex}{{\\rm{S}}_n} = \\sum {{{\\left( {{\\rm{odd\\ number}}} \\right)}^3}} {/tex}=> {tex}{{\\rm{S}}_n} = \\sum\\limits_{k = 1}^n {\\left( {2k - 1} \\right)} {/tex} [Sn = Sum of n odd natural numbers]=> {tex}{{\\rm{S}}_n} = \\sum\\limits_{k = 1}^n {\\left( {8{k^3} - 12{k^2} + 6x - 1} \\right)} {/tex}=> {tex}{{\\rm{S}}_n} = 8\\left( {\\sum\\limits_{k = 1}^n {{k^3}} } \\right) - 12\\left( {\\sum\\limits_{k = 1}^n {{k^3}} } \\right) + 6\\left( {\\sum\\limits_{k = 1}^n k } \\right) - n{/tex} [1 + 1 + 1 + ...... to n terms = n]=> {tex}{{\\rm{S}}_n} = 8.{1 \\over 4}{n^2}{\\left( {n + 1} \\right)^2} - 12.{1 \\over 6}n\\left( {n + 1} \\right)\\left( {2n + 1} \\right) + 6.{1 \\over 2}n\\left( {n + 1} \\right) - n{/tex}=> {tex}{{\\rm{S}}_n} = 2{n^2}{\\left( {n + 1} \\right)^2} - 2n\\left( {n + 1} \\right)\\left( {2n + 1} \\right) + 3n\\left( {n + 1} \\right) - n{/tex}=> {tex}{{\\rm{S}}_n} = n\\left( {n + 1} \\right)\\left[ {2n\\left( {n + 1} \\right) - 2\\left( {2n + 1} \\right) + 3} \\right] - n{/tex}=> {tex}{{\\rm{S}}_n} = n\\left( {n + 1} \\right)\\left[ {2{n^2} - 2n + 1} \\right] - n{/tex}=> {tex}{{\\rm{S}}_n} = n\\left[ {\\left( {n + 1} \\right)\\left( {2{n^2} - 2n + 1} \\right) - 1} \\right]{/tex}=> {tex}{{\\rm{S}}_n} = n\\left[ {2{n^3} - 2{n^2} + n + 2{n^2} - 2n + 1 - 1} \\right]{/tex}=> {tex}{{\\rm{S}}_n} = n\\left[ {2{n^3} - n} \\right]{/tex}=> {tex}{{\\rm{S}}_n} = {n^2}\\left[ {2{n^2} - 1} \\right]{/tex} | |
| 6449. |
difference between relation and function\xa0 |
| Answer» In Relation, each input has only one output whereas a\xa0Function\xa0is a\xa0relation\xa0in which no input relates to more than one output.\xa0 | |
| 6450. |
Let f = {(1,1),(2,3),(0,-1),(-1,-3)} be linear function from Z into Z . Find f (x) |
| Answer» Let the linear function from Z into Z be\xa0{tex}f\\left( x \\right) =ax+b{/tex}Given\xa0{tex}f\\left( 1 \\right) =1,f\\left( 2 \\right) =3,f\\left( 0 \\right) =-1,f\\left( -1 \\right) =-3{/tex}Now\xa0{tex}f\\left( 1 \\right) =1\\Rightarrow a+b=1\\\\ f\\left( 0 \\right) =-1\\Rightarrow 0+b=-1\\Rightarrow b=-1{/tex}Substituting\xa0{tex}b=-1{/tex}\xa0in\xa0a+b=1 we get\xa0a-1 =1\xa0{tex}\\Rightarrow{/tex}a=1+1=2So we have a = 2 and\xa0{tex}b=-1{/tex}Hence the linear function from Z into Z is\xa0{tex}f\\left( x \\right) =2x-1{/tex} | |
