InterviewSolution
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InterviewSolution
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.
| 41501. |
Is it necessary to write the steps for construction in exam? |
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Answer» No,it is not necessary to write the steps for construction in exam Yes ...... but in case if you have less time then you can also left it because it only carries the weightage of 1 mark.... Yup there r marks on it ... |
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| 41502. |
Proove that SinQ-cosQ+1/sinQ+cosQ-1=1/secQ-tanQ , using the identity sec square Q=1+tan squareQ |
| Answer» pls answer | |
| 41503. |
Sin2A=2sin A is this true when A= A 0° B 30° C 45° D 60° |
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Answer» Step by step O how Sir please mention answer properly 0° |
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| 41504. |
Kya optional exercise solve karna zaruri hai |
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Answer» Choice Examination point of view se dekho to jaruri nhi hai Agar knowledge jayada chahiye to jaruri hai |
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| 41505. |
Tan 2A = Cot ( A - 18) where 2A is an acute angle find the value of A |
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Answer» Cot ( 90 - 2A) = cot A -1890-2A = A- 18 3 A =108 A = 36 Tan 2A= tan (90-(A-18)).....2A=90-A+18.....3A=108...A=108/3=36....ans Tan(90-2A)=Cot (A-18).......90-2A=A-18........90+18=A+2A........108=3A........A=108/3.......A=36 36 |
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| 41506. |
Converse of basic proportionality theorumPlzz tell Tomorrow is exam? |
| Answer» If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. | |
| 41507. |
2 sin |
| Answer» 2/ cosec | |
| 41508. |
Sin18°\\cos72° |
| Answer» 1 | |
| 41509. |
In chapter 15 probability After completing soln E.g.50/100So it is nesesary To write in 1/2 form |
| Answer» Yes, it is as the examiner checks for the simlest fraction. | |
| 41510. |
Tanc/sinc=? |
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Answer» Tanc/sinc = sinc/cos c/sin c /1 = 1/cos c = sec c Sec C Sec c Sec c |
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| 41511. |
How to find the derivative of cotx using first principle |
| Answer» | |
| 41512. |
How to find that this number is rational or irrational |
| Answer» If a number has more than 2 factor it is rational | |
| 41513. |
If A,B and C are interior angle of triangle than prove that Sin (B+C/2) =Cot (A/2) |
| Answer» A B C are interiors angle then A+B+C= 180PUT the value of B+C in sin( B+C÷ 2)Sin( 180-A/2)=sin(90-A/2)=cos(A/2)Your question is wrong it is not cot it is cos | |
| 41514. |
Math me pass hone bas ke liye koi most important question bata do |
| Answer» Chapter 14 ,13 ,11,6 isse section d me question aate hain | |
| 41515. |
Solve for x and y1/2(2x+3y)+12/7(3x-2y)=1/27/(2x+3y)+4/(3x-2y)=2 |
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Answer» Or isme 2 equal ke sign h to us hisaab se to first value or 2 ko bhi lekar answer nikal sakte h Sis i have also same doubt 1/27/(2x+3y) he kya bolo ya 1/2 +7/(2x+3y) he bolo |
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| 41516. |
cos45°/sec30°+cosec30°=? |
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Answer» cos45°=1/√2...sec30°=2/√3.....cosec30°=2..[(1/√2)/(2/√3)]+2.......(1/√2)×(√3/2)+2....(√3/2√2)+2......(√3+4√2)/2√2 write the step (√3+4√2)/2√2 |
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| 41517. |
2x+4x-3 |
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Answer» If u have to solve any eqn u have to equate it to zero Why u have equate it with zero by your self . The value of x is 1/2.... X=1/2 |
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| 41518. |
If sinA=cosA,find the value of A |
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Answer» =sinA=cos(90-A) sinA=sin90-A A=90-A 2A=90 A=45 1 A= 45 |
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| 41519. |
2 ×tansquare45° + cossquare30° -sin square 60° |
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Answer» What is answer tan 45 = 1 cos30= √ 3/2 sin 60 = √ 3/22 tan245 + cos230 - sin2 60= 2(1) + (√ 3/2)2 - (√ 3/2)2= 2 + 3/4 - 3/4= 2 |
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| 41520. |
sin ●(1 + tan● ) + cos● (1+cot●)=sec●+cosec● |
| Answer» | |
| 41521. |
Find value of k 3x2 -k3x+4=0 |
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Answer» 3x2-k3x+4=0. 3x(2-k+4)=0. 3x(6-k)=0. 6-k=0. K= -6 K= -6 |
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| 41522. |
If sec theta +tan theta+1=0 |
| Answer» Complete the question | |
| 41523. |
Find the median of given data:-5,11,8,7,9,12,19 SOLVE BY DOING ALL THE STEPS |
| Answer» Frequency not required in the given questionWe have to make it according to the odd and even formula | |
| 41524. |
If mth term of an A.p. is 1/n and nth term is 1/m , show the sum of its first (mn) term is 1/2(mn+1) |
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Answer» By solving am=1/n we get equation i.e.,a+dm-d=1/n. Similarly solving an=1/m we get equation i.e.,a+dn-d=1/m. By eliminating these two equations we d=1/mn. Substitute this value of d in equation 1 . We get a=1/mn. Then find Amn,we Amn=1. By this find Smn i.e., Smn=mn/2{a+Amn} we get 1/2{mn+1]. Hence proved.... I/n=a+(m-1)d......1/m=a+(n-1)d.........subtract......1/n-1/m=a-a+(m-1-n+1)d..........(m-n)/mn=(m-n)d.......d=1/mn......put the value of d in eq (1) you get a=1/mn.......now put the formulae of sn.......mn/2[(2/mn+(mn-1)1/mn]........mn/2(2+mn-1)1/mn....then we get sn=1/2(mn+1) Question galat h aapka |
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| 41525. |
If a point A(4,3)B(x,5) are on a circle with center O(2,3) Find x |
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Answer» A and B are on circleso OA=OB=rOA^2=OB^2(4-2)^2 +(3-3)^2=(x-2)^2+(5-3)^2or 4+0=(x-2)^2+4(x-2)^2=0x-2=0x=2 Given centre of circle is O(2, 3)\xa0Points on the circle are A(4, 3) and B(x, 5)\xa0OA =OB (Radii) à (1)\xa0Recall the distance between two points formula\xa0\xa0Squaring on both the sides, we get\xa0\xa0\xa0x – 2 = 0\xa0∴ x = 2 What is A and B radius,diametre or cord |
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| 41526. |
(p-3)x+3y=p, px+py=12 what value of p |
| Answer» P=6 | |
| 41527. |
Tan(3x-15)=1 value of x |
| Answer» Tan (3x-15°) =1Tan(3x-15°) =tan 45°So, on comparing3x-15°=45°3x=60°X=20° | |
| 41528. |
Given that LCM (150,100)=300,find HCF(150,100) |
| Answer» lcm * hcf = product of two numbersas per question\xa0 300 * hcf = 150 * 100\xa0 hcf = 15000/300 hcf = 50 | |
| 41529. |
X²-2x-8= 0 |
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Answer» x2 - 2x - 8 = 0x2 - 4x + 2x - 8 = 0x(x -4) + 2 ( x-4) = 0(x + 2) (x -4) = 0x + 2 = 0 or x - 4 = 0x = -2 or x = 4 Your answer =2 |
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| 41530. |
In wahr ratio does the point (-4,6) divide tye line segment joining the points A(-6,10) and B(3,-8)? |
| Answer» 7:2 | |
| 41531. |
Find the value of a so that the points {3,a}lies on the line represented by 2x-3y=5 |
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Answer» x = 3 and y = a2x - 3y = 52(3) - 3(a) = 56 - 3a = 53a = 6 - 53a = 1a = 1/3 1/3=a |
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| 41532. |
What is the sum of all integers |
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Answer» Sum of all integers is 0 0 Is the sum of n positive integers formula, n(n+1)2n(n+1)2, applicable to sets of consecutive integers as well?For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x 20(21)220(21)2a similar approach has been used for even numbers. 1+2+3+4.........infinity =-1/12So the sum of all the integer is -1/12 n(n+1)/2 |
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| 41533. |
SinΦ(1+tanΦ)+cosΦ(1+cotΦ)= secΦ+cosecΦ |
| Answer» | |
| 41534. |
If sinA- cosA=1 then find the value of 1/ sinA+cosA |
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Answer» But is que ka answer 2√7/7 hi hai SinA - CosA=1So, SinA=1+CosAUpon Squaring on both sides,Sin square A = (1+CosA) squareSince Sin square A = (1 - CosA)(1+CosA),(1 - CosA)(1+CosA)=(1+CosA)(1+CosA)So, 1 - CosA = 1+CosA,2CosA=0So, CosA=0.So A=90 degreesSubstituting A=90 degrees in 1/sinA+cosA,We get 1/sinA+cosA = 1/1+0=1/1=1Therefore the answer is 1. But iska ans 2√7/7 hai Answer is 1+cosA |
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| 41535. |
sin =cos |
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Answer» Sin45=cos45 Complete your question first... |
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| 41536. |
If sin (A+B)=1 and sin(A-B)=1÷2,0 |
| Answer» B=60A=30 | |
| 41537. |
If pth term of an A.P.is q and the qth term is p prove that its nth term is (p+q-n) |
| Answer» Let a be the first term and d be the common difference of the given A.P. Then,pth term = q {tex}\\Rightarrow{/tex}a + (p-1) d = q ...(i)qth term = p {tex}\\Rightarrow{/tex}a + (q-1) d = p ...(ii) Subtracting equation (ii) from equation (i), we get(p - q) d = (q - p) {tex}\\Rightarrow{/tex}\xa0d = -1Putting d = - 1 in equation (i), we geta + ( p-1) × (-1) = q⇒ a = (p + q - 1)\xa0nth term = a + (n -1 )d= (p + q - 1)+ (n -1) × (-1)= p + q - 1 -n + 1= (p + q - n) | |
| 41538. |
Simplify : 1/√2+√3-√5 + 1/√2-√3-√5 |
| Answer» | |
| 41539. |
How i can read for board exam?? |
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Answer» Work hard. Understand each topic perfectly. Practice more on previous year question papers What are u say |
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| 41540. |
What is main important question in chapter 4 quadratic equation |
| Answer» | |
| 41541. |
parso mera maths ka pre board h ? how should i practice |
| Answer» | |
| 41542. |
How to practice for pre board exams |
| Answer» Maths | |
| 41543. |
Tan 47 degre4 |
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Answer» -0.12Checked in calculator Clearly write what is the question |
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| 41544. |
Common questions from triangle |
| Answer» | |
| 41545. |
(cos20 cos20 + cos70 cos70) + cot25/tan65 + cot5 cot10 cot60 cot80 cot85 |
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Answer» Yes yes sorry answer is 2+1√3 Answer is 2+1/√3 Answer is 3 Sin² (90-70°)+cos² 70°+cot25/cot(90-25) + cot5cot85 cot10cot80 cot60Sin² 70+cos² 70 + cot25/cot25 + cot5tan5 tan80cot80 1√31+1+1.1.1/√32+1/√32√3+1/√3 Is this cos square 20 |
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| 41546. |
Find the value of k so that points A (-2,3) B (3,-1)C (5,K ) are collineaf |
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Answer» -13/5 -13/5 |
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| 41547. |
Triangles chapter me se sabse zyada questions ncert se aate h kya?? |
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Answer» I also don\'t know because my teacher hasn\'t taught that Last ex ke sare ques aur sari theorem Mostly......base is ncert |
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| 41548. |
How can i practice triangles ch in maths. Its really difficult??plzzz tell mee |
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Answer» Do practice more and more and ask the question to your teacher or friends.Also you can study in group so that your daubt will clear Do revision as much you can do Gungun first understand 3 important theorums unke base par hi question aate h You can try it firstly in ncert book When you are able to solve that you can use any refrence book |
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| 41549. |
How to solve height and distance |
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Answer» Easy We can solve it by trigonometry The only thing is correct figure.Once u drawn the correct fig then u could solve it So simple .....understand the question nicely then draw?? How to make image Which concept |
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| 41550. |
1+5=5 ,prove |
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Answer» Is 1+5=5 or 6? What do you mean? Go to class 1 and learn number |
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