Saved Bookmarks
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. |
Full form of a.p |
|
Answer» Thanks for sample paper Arithmetic progression, not arithmetic propagation? Arithmetic propagation.. Arithmatic propogation Arithmetic progression |
|
| 2. |
(sinA + cosecA)2 + (cosA + secA)2 =7+tan2A + cot2A |
| Answer» LHS=>(sin2A+cosec2A+2×sinA×1/sinA)+(cos2A+sec2A+2×cosA×1/cosA)(sin2A+cosec2A+2+cos2A+sec2A+2)=(sin2A+cos2A)+(1+tan2A)+(1+cot2A)+2+21+1+1+4+tan2A+cot2A7+tan2A+cot2ALHS=RHS Hence proved | |
| 3. |
Find the roots of the equation ax^2+a=a^2+a. |
| Answer» | |
| 4. |
Three digit number which are divisible by3 are |
| Answer» 102, 105, 108...999 | |
| 5. |
What is frustum ..... |
|
Answer» Hii .. Hi Top When a to of cone is cut horizontally it makes frustum When the top cut from the cone this shape is known as frustom. Ex- steel ki balti Cone * Frustum is a part of come |
|
| 6. |
In an A.P if the 5th and 12th terms are 30 and 65 respectively . What is the sum of first 20 terms |
| Answer» 1150 | |
| 7. |
3sin A+4cos B=6 and 6sinB+4cosA=1 then show that sinA = |
|
Answer» sanket Just go aside |
|
| 8. |
Which of the following cannot be the probability of an event ?A. 1.1B. 0.1C. 0.9D. 5% |
|
Answer» 1.1 A |
|
| 9. |
The Value of 9sec{square}A - 9Tan{square}A is ______. |
|
Answer» Sorry.. 9(sec^2 - tan^2 ) = 9×1 = 9. 9(1)=9 1 |
|
| 10. |
Ratianal number |
| Answer» | |
| 11. |
(69)³ + (69)³ +(69)³ |
| Answer» 985,527 | |
| 12. |
(1-cos A)(1+cos A)(1+cos²A)=1 |
|
Answer» And the solution isTaking LHS :Using identity (a-b) (a+b) =(a² - b²) ( 1-Cos²A)( 1 + Cot²A)As we know:[1 - Cos² A = Sin²A 1 + Cot²A = Cosec²A] identity Now, on putting the values, we have( Sin²A ) ( Cosec²A)[Sin²A = 1/ Cosec²A] identity =1=RHSPROVED I think it should be ( 1 - CosA )( 1 + CosA)( 1 + Cot²A)=1 |
|
| 13. |
Which of the following is not a zero of the polynomial P(x)= x3 - 7x + 6 (A) 1 (B) 2 (C) -2 (D) -3 |
| Answer» (C) -2 is not a zero of P(x) | |
| 14. |
Median formula |
| Answer» L+{n/2-cf/f}×h | |
| 15. |
If angle A and angle B are acute angle such that cos A = cos B , then show that Angle A = Angle B . |
| Answer» Cos will get eliminated from both the lhs and rhs | |
| 16. |
How can we solve the question The mirror image of (3,9) is.? |
|
Answer» Ulta 9,3 ulta banana Where do we suppose the mirror to be.? It needs construction |
|
| 17. |
Simplify. 2+2+4 |
|
Answer» 8 8 |
|
| 18. |
0 is a integer or not |
|
Answer» Yes 0 is an integer because it does not have a fractional part. It is constant |
|
| 19. |
How less than ogive made |
| Answer» | |
| 20. |
(-13+8)-(-6) simplify |
| Answer» Answer is 1⃣-13+8+6=-13+14=1 | |
| 21. |
How to prove converse of Pythagoras th. |
|
Answer» Hamne apko galat answer diya hai......???don\'t mind......kal apke room pe aa ke sahi batayenge. Gupta g boliya....kaha hai abhi motihari ki Ramgarhwa......name mat lijiya ga mera....aap agar pahchan gay ho to call kijiya Abhishek bhai shantiniketan ke.... ka hal ba???? Draw a perpendicular on hypotenous of a right angle triangle and similar both new triangle to the main triangle.......hope this help you Friend u will find it in the Google in much easier wayy |
|
| 22. |
If (cos theta+sin theta)=2,show that cos theta -sin theta=√2sin theta |
| Answer» | |
| 23. |
prove that :- root( 1- cos theta )÷root (1-cos theta)=cosec theta +cot theta |
|
Answer» sorry, typing mein galti ho gayi thi first wale mein (1+cos theta) RHS=[1/sin -cos/sin]. ||. [1-cos/sin]². ||. (1-cos)(1-cos)/1²-cos². ||. (1-cos)(1-cos)/(1+ cos)(1-cos). ||. 1-cos/1+cos. ||. To phir question ko vasi likh kr bejo m solve kr dugi Yes kuhu is right May b ek me minus ki jgh plus aaye ga Riddhi may b aapka question glt h |
|
| 24. |
If (sin theta + cos theta)=p and (sec theta +cosec theta) =q,show that q(p²-1)=2p |
| Answer» | |
| 25. |
(1+cot theta-cosec theta)(1+tan theta +sec theta)=2 |
| Answer» Multiply kar do yrr.... | |
| 26. |
If tan theta +sin theta =m and tan theta--sin theta =n, show that m²-n²=4√mn. |
| Answer» LHS = (Tan thetha+sin theta)^2 - (Tan thetha+sin theta)^2 RHS=4√mn 4√(tan thetha+sin theta)(tan theta-sin theta) 4√(tan^2theta-sin^2theta) 4.√(sin^2theta/cos^2theta-sin^2theta) 4.√sin^2theta-sin^2theta cos^2theta/cos theta = 4.sin theta/cos theta.1√1-cos^2 theta= 4tan.√sin^2 theta = 4 tan theta sin theta. | |
| 27. |
2020 cbse sample paper class 10 |
| Answer» | |
| 28. |
How( 2+2=5) ?? |
| Answer» NOT POSSIBLE | |
| 29. |
What is the formula of a ki power 4 plus b ki power 4 upon a square into b square |
| Answer» | |
| 30. |
Please send basic paper |
| Answer» Basic paper..?? Sample paper aap website se download kr lo.. | |
| 31. |
Find two consecutive positive integers sum of whose square is 365 |
|
Answer» Let the two consecutive integers be x and x + 1.Then according to the question, we have:x2 + (x + 1)2 = 365⇒ x2 + x2 + 2x + 1 - 365 = 0⇒ 2x2 + 2x - 364 = 0⇒ 2(x2 + x - 182) = 0⇒x2 + x - 182 = 0⇒x2 + 14x - 13x - 182 = 0 ⇒ (x + 14) (x - 13) = 0⇒ x = -14 or x = 13Since the integers are positive, so neglecting x = -14 , we have x = 12.Hence 13, 14 are two consecutive positive integers. Thanks?? 14 and 13. |
|
| 32. |
2x-y=5Kx-8y=7 Find the unique solution |
| Answer» For unique solution 2/k should not be equal to 1/8 So except 16 we can put any value | |
| 33. |
Find the 20th term from lest term of the ap 3,8,13...253 |
|
Answer» 158 Here d= 5 but if we take añ as a and a as añ Then d=-5An=a+(n-1)d =253+19(-5) =253+(-95) =253-95 =158 T²0= 253+(20-1)×(-5)=253-95=158 158 |
|
| 34. |
If m=cosA-sinA and n=cosA+sinA show that m^2 +n^2/m^2-n^2=-1/2secA ×cosecA= (cotA+tanA)/2 |
| Answer» | |
| 35. |
Is coincident lines consistent or depended consistent |
| Answer» Consistent and dependent consistent both are same | |
| 36. |
What is PGT theorem |
| Answer» Hypoteneuse square equal to base square and perpendicular square | |
| 37. |
Find k,if -1 is a zero of equation 3kxsquare- 2x+3=0 |
|
Answer» Hii -5/3 |
|
| 38. |
Extra maths ques karne k lia kon sai book karne hai |
|
Answer» Only and only rs agarwal best book of maths but after completing NCERT BOOK Manjit singh Rs aggarwal and Ncert RD Sharma and Xam Idea...but after completing whole NCERT book? R D sharma |
|
| 39. |
Products of two zeros of -2x+kx+6 |
|
Answer» -3 is the ans. because alpha × beta = c/a= 6/-2=-3 By ar 2 |
|
| 40. |
what is 13+690 |
|
Answer» 703 703 700+3? 703 703 |
|
| 41. |
Sec theta+tan theta=x,prove that sin theta=x2-1/x2+1 |
|
Answer» Sec theta +tan theta =x [given]. [1].|| Sec²theta -- tan² theta = 1. ||. (Sec-tan)(sec +tan)=1.||. (Sec-tan)=1/x [using 1]. Adding both eq. ||. Sec=x²+1/x. || So, cos theta =x/x²+1. ||. Sin theta =√1-cos²theta ||. Sin = √1-x/x²+1. ||™ Sin =x²-1/x²+1. Answer |
|
| 42. |
How to find the coordinate of X on collinear points |
| Answer» | |
| 43. |
How to prove the theorem of chapter 6 triangle |
| Answer» Just go through ncert | |
| 44. |
If B(x,y)is point on line segment joining points A(a,b)and C(b,a)then prove that x+y=a+b |
| Answer» | |
| 45. |
Every positive odd integer is of the form what if q is some integer |
| Answer» | |
| 46. |
Tanthethacot(90-thetha)-secthethacosec(90-thetha)+3root3tan13tan30tan77 |
| Answer» | |
| 47. |
Find the area of a quadrant of a circle whose circumference is 22 cm. |
|
Answer» 77/8 22=2πr22=2*22\\7*rr=7/2 77/8 cm square. Circumference=2πr2*22/7*r=22Therefore r=7/2Area of quadrant=1/4*π*r*rTherefore answer 77/8 |
|
| 48. |
Find the area of rohombus ABCD,if it\'s vertices taken in order are (2,-1),B(3,4),C(-2,3)D, (-3,-2) |
| Answer» | |
| 49. |
How to score 90+ in mathematics |
| Answer» Solve ncert.....and practice sample papers. | |
| 50. |
If x+y=a+b and ax-by=(a² -- b²) then value of x and y is??Please reply fast |
|
Answer» Thnq The solution of the equations is x=a and y=b.Step-by-step explanation:Given :\xa0Equations\xa0\xa0and\xa0To find :\xa0Solve the equations ?Solution :Let\xa0\xa0.......(1)and\xa0\xa0......(2)Substitute x from (1) in equation (2),Substitute y in equation (1),Therefore, The solution of the equations is x=a and y=b. |
|