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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
5tanA=4 |
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Answer» tanA=4÷5 tanA=4/5 |
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| 2. |
(sec theta+tan theta)=1/sec theta-tan theta ) |
| Answer» L.H.S.\xa0{tex}\\sec \\theta + \\tan \\theta {/tex}Rationalizing with\xa0{tex}\\sec \\theta - \\tan \\theta {/tex}{tex}\\sec \\theta + \\tan \\theta \\times {{\\sec \\theta - \\tan \\theta } \\over {\\sec \\theta - \\tan \\theta }}{/tex}=\xa0{tex}{{{{\\sec }^2}\\theta - {{\\tan }^2}\\theta } \\over {\\sec \\theta - \\tan \\theta }}{/tex}=\xa0{tex}{1 \\over {\\sec \\theta - \\tan \\theta }}{/tex} [Since\xa0{tex}{\\sec ^2}\\theta - {\\tan ^2}\\theta = 1{/tex}] | |
| 3. |
Find the value of K if the equation has equal roots 1.x2-2x(1+3k)+7(3+2k) |
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Answer» If the quadratic equation has two equal roots then D=0.⇒ b2-4ac=0Here, a=1, b=-2(1+3k), c=7(3+2k)⇒(-2(1+3k))2-4×1×7(3+2k) =0⇒4(1+2×1×3k+9k2)-28(3+2k) =0⇒4+24k+36 k2 -84-56k=0⇒36 k2 -32k-80=0⇒9 k2 -8k-20=0⇒9 k2 -18k+10k-20=0⇒ 9k(k-2)+10(k-2)=0⇒(9k+10) (k-2)=0⇒(9k+10)=0 (k-2)=0⇒ k=-10/9 k=2. Given:\xa0{tex}{x^2} - 2x\\left( {1 + 3k} \\right) + 7\\left( {3 + 2k} \\right) = 0{/tex}Here,\xa0{tex}a = 1,b =- 2\\left( {1 + 3k} \\right),c = 7\\left( {3 + 2k} \\right){/tex}Since, for equal roots, the condition is\xa0{tex}{b^2} - 4ac = 0{/tex}Therefore,\xa0{tex}{\\left\\{ {-2\\left( {1 + 3k} \\right)} \\right\\}^2} - 4 \\times 1 \\times 7\\left( {3 + 2k} \\right) = 0{/tex}=> {tex}4\\left( {1 + 9{k^2} + 6k} \\right) - 84 - 56k = 0{/tex}=> {tex}4 + 36{k^2} + 24k - 84 - 56k = 0{/tex}=> {tex}36{k^2} - 32k - 80 = 0{/tex}=> {tex}9{k^2} - 8k - 20 = 0{/tex}=> {tex}9{k^2} - 18k + 10k - 20 = 0{/tex}=> {tex}9k\\left( {k - 2} \\right) + 10\\left( {k - 2} \\right) = 0{/tex}=> {tex}\\left( {k - 2} \\right)\\left( {9k + 10} \\right) = 0{/tex}=> {tex}k-2=0{/tex}\xa0or\xa0{tex}9k+10=0{/tex}=> {tex}k=2{/tex}\xa0or\xa0{tex}k = {{ - 10} \\over 9}{/tex} |
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| 4. |
2018-19 changed maths syllabus |
| Answer» You can check syllabus here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 5. |
What is real number |
| Answer» A real number is any positive or negative number. This includes all integers and all rational and irrational numbers. For example, a program may limit all real numbers to a fixed number of decimal places. | |
| 6. |
Chapter 123458 |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 7. |
Heigh of cylinder 10cm and radius 10cm find area of cone of same volume |
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| 8. |
Euclid algorithms |
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| 9. |
If cot square a -5cot a=-1, find the value of cot square a + tan square a |
| Answer» Cot square a - 5cot a= - 1Cot a (cot a - 5) = - 1Cot a - 5 = - 1/cot aCot a - 5 = - tan aCot a + tan a = 5 Squaring both sides (cot a + tan a) squre = 5 squareCot square a + tan square a + 2 cot a tan a = 25Cot square a + tan square a + 2*1 = 25Cot square a + tan square a + 2 = 25Cot square a + tan square a =23 | |
| 10. |
Why sin30 = 1/2 |
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Answer» Why there is reason behind every thing in this world!! We need not prove sin 30 is 1/2But just know the values for each angle |
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| 11. |
4 x square - 4 x minus 3 |
| Answer» 4x²-4x-34x²-x(6-2)-34x²-6x+2x-32x(2x-3)+1(2x-3)(2x-3)(2x+1)X=3/2X=-1/2 | |
| 12. |
Find the area or shaded region |
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| 13. |
Evaluate2tan2 45•+cos2 30•-sin2 60• |
| Answer» {tex}2{\\tan ^2}{45^ \\circ } + {\\cos ^2}{30^ \\circ } - {\\sin ^2}{60^ \\circ }{/tex}=\xa0{tex}2 \\times {\\left( 1 \\right)^2} + {\\left( {{{\\sqrt 3 } \\over 2}} \\right)^2} - {\\left( {{{\\sqrt 3 } \\over 2}} \\right)^2}{/tex}=\xa0{tex}2 + {3 \\over 4} - {3 \\over 4}{/tex}= 2 | |
| 14. |
Prove root 2 is irrational |
| Answer» Root 2 is irrational Hence, proved | |
| 15. |
2x2..dbjd |
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| 16. |
PS is the bisector of angle QPR of |
| Answer» Given: In figure. PS is the bisector of\xa0{tex}\\angle QPR{/tex} of\xa0{tex}\\triangle PQR{/tex}To prove :{tex}\\frac { Q S } { S R } = \\frac { P Q } { P R }{/tex}Construction: Draw RT || SP to meet QP produced in T.Proof:\xa0{tex}\\because {/tex}\xa0RT||SP and transversal PR intersects them{tex}\\therefore \\angle 1 = \\angle 2{/tex}\xa0(1) ..... Alt. Int.\xa0{tex}\\angle s{/tex}{tex}\\because {/tex}\xa0RT||SP and transversal QT intersects them{tex}\\therefore \\angle 3 = \\angle 4{/tex} (2), ..... corres.\xa0{tex}\\angle s{/tex}But\xa0{tex}\\angle 1 = \\angle 3{/tex}\xa0...... Given{tex}\\therefore \\angle 2 = \\angle 4{/tex}\xa0......From (1) and (2){tex}\\therefore P T = P R{/tex}\xa0(3) ......{tex}\\because {/tex}\xa0Sides opposite to equal angles of a triangle are equal\xa0Now in\xa0{tex}\\Delta \\mathrm { QRT }{/tex}PS = RT .......By construction{tex}\\therefore \\frac { Q S } { S R } = \\frac { P Q } { P T }{/tex}\xa0....... By basic proportionally theorem\xa0{tex}\\Rightarrow \\frac { Q S } { S R } = \\frac { P Q } { P R }{/tex}\xa0......From (3) | |
| 17. |
Ex. 5.3 Q. 3 ka 8 part |
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| 18. |
If common difference of an AP is 3 then,a20 - a25 = ? |
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Answer» a = 1st termd = 3then a20\xa0= a + (20 - 1) d = a + 19x3= a + 57 ......... i a\xa025 = a + 24 d = a + 24 x 3 = a + 72 ............................ ii as per question i - ii a20\xa0\xa0- a25 = a + 57 - a - 72 = - 15 Let first term of an A.P. is a and common difference is d.Given common difference =3 d=3Now a20-a25=a+(20-1)d -(a+(25-1)d) [by using an=a+(n-1)d] =a+19d -(a + 24d) =a+19d - a -24d =-5d =-5×3 (since d=3 given) = -15 |
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| 19. |
X/A + y/B = A+B And X/A2 +Y/B2 =2 is solve by cros multiply method ? |
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| 20. |
What is the rule for pythagorean triplet ? |
| Answer» (2M, M^2-1, M^2+1) where ^ represent power and the value of 2M is given in questions , Equate the value with 2M to find M and then proceed the question... | |
| 21. |
Cos - sin= |
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| 22. |
Plz Plz can anyone provide me 2016-17 (last year) papers of Mathematics. Plz plz I request. |
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| 23. |
2x-2x-3=9 |
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| 24. |
What is Ecilid lamma |
| Answer» Euclid\'s Lemma = if the prime p divides the product ab of two integers a and b , then p must divide at least one of the integers or bothe.g.suppose p = 19 , a = 133 , b = 143 and ab = 133 x 143 = 19019and since this is divisible by 19 , and Lemma implies if one or both of 133 or 143 are divisible by 19 ,in fact 133 = 19 x 7\xa0 | |
| 25. |
To prove Pythagoras theorem |
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| 26. |
2+2÷2×2-=2 |
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Answer» 2+2÷2×2=2+{tex}2\\over 2{/tex}×2=2+1×2=2+2=4 [USE BODMASS] 2 + (2x 1/2) x 2 - 2= 2 + 1 x2 - 2= 2 + 2 - 2= 2\xa0 |
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| 27. |
prove that 5=3 |
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| 28. |
In adjoining figure? Find tan p - cot r |
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| 29. |
Find p which (p+1)x square - 6(p+1)x + 3(p+q) = 0 has equal roots . Also find roots. |
| Answer» The correct equation is {tex}\\left( {p + 1} \\right){x^2} - 6\\left( {p + 1} \\right)x + 3\\left( {p + 1} \\right) = 0{/tex}Here a = p + 1, b = -6(p + 1), c = 3(p + 1)For equal root,\xa0{tex}{b^2} - 4ac = 0{/tex}=> {tex}{\\left[ { - 6\\left( {p + 1} \\right)} \\right]^2} - 4 \\times \\left( {p + 1} \\right) \\times 3\\left( {p + 1} \\right) = 0{/tex}=> {tex}36\\left( {{p^2} + 1 + 2p} \\right) - 12\\left( {{p^2} + 1 + 2p} \\right) = 0{/tex}=> {tex}3\\left( {{p^2} + 1 + 2p} \\right) - \\left( {{p^2} + 1 + 2p} \\right) = 0{/tex}=> {tex}3{p^2} + 3 + 6p - {p^2} - 1 - 2p = 0{/tex}=> {tex}{p^2} + 4p + 2 = 0{/tex}=> {tex}p = {{ - 4 \\pm \\sqrt {16 - 8} } \\over 2}{/tex}=> {tex}p = {{ - 4 \\pm 2\\sqrt 2 } \\over 2}{/tex}=> {tex}p = - 2 \\pm \\sqrt 2 {/tex} | |
| 30. |
ABC is s triangle right angled at C.If angle A=30°\'AB=12cm,determine BC and AC |
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Answer» sin 30 = BC/ABBC = AB X Sin\xa030 = 12 X 1/2 = 6 cmcos 30 = AC / AB AC = AB X COS 30 = 12 X {tex} \\sqrt{3}/2{/tex} = 6\xa0{tex} \\sqrt{3}{/tex} cm sin30°= BC/12BC=1/2×12 =6cm.6/AC=1/√3AC=6√3 cm |
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| 31. |
Find the remainder when 19 to the power 92 is divided by 92 |
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| 32. |
Check wether (5,-2)(6,4)(7,-2)are the vertices of an isosceles triangle |
| Answer» let ABC be the triangle whose vertices are A (6 , 4) , B ( 5 , - 2) , C (7 , - 2)AB = {tex} \\sqrt{}{/tex}(6 - 5)2\xa0+ (4 + 2)2 = {tex} \\sqrt{}{/tex}12\xa0+ 62\xa0= {tex} \\sqrt{}{/tex}1 + 36 = {tex} \\sqrt{}{/tex}37AC = {tex} \\sqrt{}{/tex}(6 - 7)2\xa0+ (4 + 2)2\xa0= {tex} \\sqrt{}{/tex}\xa0(-1)2\xa0+ 62\xa0= {tex} \\sqrt{}{/tex}1 + 36 = {tex} \\sqrt{}{/tex}37\xa0as AB = ACTherefore the vertices A B C are\xa0of an isosceles triangle | |
| 33. |
What is discrimination |
| Answer» • Discrimination refers to practices, acts or activities resulting in the unjustified exclusion of the members of a particular groups from access to goods, services, jobs, resources, etc. • Discrimination is behavioural component of prejudice. Both are not the same. • Discrimination refers to the behaviours, that makes a distinction between the rich and poor. It can be seen in the matters of social interaction, education, gender and employment. • This refers to actual behaviour towards another group or individual. | |
| 34. |
The nth term of AP is 6n+2 find its common difference |
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Answer» 6n + 26(n - 1) + 6 + 2= 8 + (n -1)6 ...................... ia + (n - 1)d general form.... iicomparing i and iia = 8 , d = 6therefore common difference = 6 D=6 |
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| 35. |
2x+4y=4,find value of 1/x and 1/y |
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| 36. |
Send the syllabus |
| Answer» It is available on myCBSEguide app | |
| 37. |
If sin theta =3/5,find the value of (tan theta +sin theta)to the power2. |
| Answer» 729/400 | |
| 38. |
[1-tanA/1+tanA]^2=TAN^2A |
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| 39. |
In a ∆ABC, angle C=3angle B=2(angle A+ angle B).find the three Angles. |
| Answer» Each angle is of 60 | |
| 40. |
Show that the number 18 can never end with digit 0 for any natural number N |
| Answer» We have to find out that whether 18n can end with the digit 0 or 5 for any natural number nIf any number ends with the digit 0 then its\xa0factors must be in the form : 2m{tex} \\times {/tex}\xa05nIf\xa0any number ends with the digit 5 then its factors must be in form : 5m {tex} \\times {/tex}\xa03n or 5m{tex} \\times {/tex}\xa07n\xa0So for any number to end\xa0with the digit 0 or 5 its factors must be in form 5m {tex} \\times {/tex}\xa03n or 5m{tex} \\times {/tex}\xa07n\xa0\xa0or\xa02m{tex} \\times {/tex}\xa05n\xa0(where m and n are positive integers)Now, 18 = 2 × 9 = 2 × 3 × 3 = 2\xa0× 32So, 18n\xa0= (2\xa0× 32)n = 2n\xa0× (32)n= 2n\xa0× 32n18n is not in the form of 5m × 3n or 5m × 7n\xa0\xa0or\xa02m × 5n\xa0So, 18n cannot end with 0 or 5 for any natural number n. | |
| 41. |
Sample question paper of new pattern |
| Answer» Check Sample Papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 42. |
Sin A *Cos A=? |
| Answer» 1 | |
| 43. |
SinA= |
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Answer» Perpendicular /Hypotenuse P/H P/B |
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| 44. |
The mth term of an AP in and nth term of AP m. Find the erg term of the AP |
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| 45. |
Define trigonometry |
| Answer» Trigno means triangle and metry means measurement therefore trignometry means measurement of triangles | |
| 46. |
I have so much confusion in proving trigonometry identities. What should I do? |
| Answer» | |
| 47. |
1/x-1-1/x+5=6/7 |
| Answer» {tex}{1\\over x—1}—{1\\over x+5}={6\\over7}{/tex}{tex}{x+5—x+1\\over (x—1)(x+5)}={6\\over7}{/tex}{tex}{6\\over (x—1)(x+5)}={6\\over7}{/tex}{tex}(x—1)(x+5)=7{/tex}x2+5x—1x—5=7x2+4x—12=0x2+6x—2x—12=0x(x+6)—2(x+6)=0(x—2)(x+6)=0Either, x—2=0 or x+6=0 x=2 or —6 | |
| 48. |
If the sum of 3 number in AP be 24 and their product is 440 find their no |
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Answer» Let No. Be (a-x) (a) (a+x)a-x+a+a+x=24a=8 (a-x) (a) (a+x)=440Put value of aThen,xsquare = 9X=3No.\'s are 8-3=5,8,8+3=11 answer Let the three numbers be a, a+d, a+2da+a+d+a+2d=243a+3d=24a+d=8d=8-a--------(i)a(a+d)(a+2d)=440a(8)(a+d+d)=4408a(8+d)=4408a(8+8-a)=4408a(16-a)=440128a-8a2=4408a2-128a+440=0a2-16a+55=0(÷by 8)a2-11a-5a+55=0a(a-11)-5(a-5)=0(a-11)(a-5)=0a-11=0 or, a-5=0a=11 or a=5If a =11, d=8-11=-3If a=5, d=8-5=3So the numbers are 11,8,5or, 5,8,11 |
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| 49. |
Find the circumcentre of the triangle whose vertices at the are (-2,3),(-1,0),(7,-6). |
| Answer» The co - ordinates are(-2,3)(-1,0) (7,-6)so, the x- coordinate of the circumcenter is given by\xa0{tex}x1+x2+x3\\over3{/tex}={tex}-2-1+7\\over3{/tex}={tex}4\\over3{/tex}The y- coordinate of the circumcenter is given by\xa0{tex}y1+y2+y3\\over3{/tex}={tex}3+0-6\\over3{/tex}={tex}-3\\over3{/tex}=-1So the coordinate of the circumcenter is. ({tex}4\\over3{/tex},-1) | |
| 50. |
Probablity |
| Answer» Probability={tex}Possible ( outcomes)\\over total number of (outcomes){/tex} | |