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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.
| 21551. |
If seca + tana =p, then find the value of coseca. |
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Answer» cosec a = p^2+1 / p^2-1 Cosec a =P-cosa. Cosec a/cosa. Cosec a |
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| 21552. |
What is the probability of each letters in the word mathematics? |
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Answer» The probability of each letters in the word mathematics is 1/11 For M=2/11 ,A=2/11, T=2/11and for H,E, I,C,S there is probability of=1/11 1/11 Sorry 1/26 |
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| 21553. |
For what value of k will k+9,2k - 1 and 2k+7 are the consecutive term of an AP ? |
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Answer» If they are in A.P. Then,. 2b=a+c (where a=first term,b=second term and c=third term). Or,. 2(2k-1)=k+9+2k+7. Or,. 4k-2= 3k+16. Or,. K=18. Hence, For K=18 the following terms are in A.P. The value of k will be 18. K = 18 |
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| 21554. |
Sutra |
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| 21555. |
Can anyone solve this 2a+1/3a=6 then 3a+1/2a=? |
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Answer» Yes, the answer is 19/2 Wrong answer . Correct answer is 9. Answer is19 / 2 |
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| 21556. |
Use euclid\'s division lemma to prove that n3-n is divisible by 6 |
| Answer» I think its given in NCERT | |
| 21557. |
बहुपद ज्ञात करने का सूत्र क्या है |
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| 21558. |
Cot 45-cot45 |
| Answer» 0 | |
| 21559. |
1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)=1/6 .SOLVE FOR x. |
| Answer» We have,{tex}\\frac{1}{{(x - 1)(x - 2)}} + \\frac{1}{{(x - 2)(x - 3)}}{/tex}{tex}+ \\frac{1}{{(x - 3)(x - 4)}} = \\frac{1}{6}{/tex}{tex}\\Rightarrow{/tex}{tex} (x - 3)(x - 4) + (x - 1)(x - 4) + (x - 1)(x - 2) = {/tex}{tex}\\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}[{tex}\\because{/tex} Multiplying both sides by (x -1)(x - 2)(x - 3)(x - 4)]{tex}\\Rightarrow{/tex}\xa0{tex}x^2 - 4x - 3x + 12 + x^2 - 4x - x + 4 + x^2 - 2x - x + 2 ={/tex}{tex}\\frac{1}{6}{/tex}{tex}[(x - 1)(x - 2)(x - 3)(x - 4)]{/tex}{tex}\\Rightarrow{/tex}{tex}3x^2 - 15x + 18 ={/tex}{tex}\\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}3(x^2 - 5x + 6) =\u200b\u200b\u200b\u200b\u200b\u200b\u200b{/tex}{tex}\\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}18[x^2 - 3x - 2x + 6] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}18[x(x - 3) - 2(x - 3)] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}18(x - 3)(x - 2) = (x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex} 18 = (x - 1)(x - 4){tex}\\Rightarrow{/tex} 18 = x2 - 4x - 1x + 4{tex}\\Rightarrow{/tex} x2 - 5x + 4 - 18 = 0{tex}\\Rightarrow{/tex} x2 - 5x - 14 = 0In order to factorize x2 - 5x - 14, we have to find two numbers \'a\' and \'b\' such that.a + b = - 5 and ab = -14Clearly, -7 + 2 = -5 and (-7)(2) = -14{tex}\\therefore{/tex}\xa0a = -7 and b = 2Now,x2 - 5x - 14 = 0{tex}\\Rightarrow{/tex} x2 - 7x + 2x - 14 = 0{tex}\\Rightarrow{/tex} x(x - 7) + 2(x - 7) = 0{tex}\\Rightarrow{/tex} (x - 7)(x + 2) = 0{tex}\\Rightarrow{/tex} x - 7 = 0 or x + 2 = 0{tex}\\Rightarrow{/tex} x = 7 or x = -2 | |
| 21560. |
Solve 2x+3y=11 and 2×-4y=-24 and hence find the value of \'m\' for which y=m×+3 |
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Answer» First solve for algebraic method, from it u have the value of x & y and then put the value of x & y in given eqn....... Y=mx+3 .... ??? . x=-2 ,y=5 and m=-1 |
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| 21561. |
If m and n are integers prove that root m + root n are irrational |
| Answer» First assume √m + √n are rational numberP/q are co-primes √m+√n =p/q√n=p/q -√m√n is irrational therefore √m +√n are irrarional | |
| 21562. |
2a+1/3a=6 then 3a+1/2a=? |
| Answer» No one is there to answer my question. | |
| 21563. |
Do anyone solves RDSharma here?If yes , i want a help pls.. |
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Answer» Thank u guys!! Problem is of page 13.33 exercise 13.3 ques 9 of level 2 I ll solve Yes I can. |
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| 21564. |
Show that any positive odd integer is of the form 6q +1 or 6q+3or6q+5where q is integer some integer |
| Answer» Let a any +ve integer and b=6. Bet. them integers will be 0,1,2,3,4,5 so a=6q,6q+1,6q+2,...............6q+5. But we need only odd integers so 6q+1,6q+3,6q+5 are positive odd integer | |
| 21565. |
Shraddha Nawale your answer for the question 3a+1/2a=? is wrong . Correct answer is 9. |
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| 21566. |
The points A(1,-2) B(2,3) C(k,-2) D(-4,-3) are the vertices of a parallelogram. Find value of k |
| Answer» A(1, -2), B(2, 3), C(a, 2) and D(-4, -3)are the vertices of a parallelogram.{tex}\\Rightarrow{/tex}AB = CD and AD = BCConsider AD = BC{tex}\\Rightarrow{/tex}AD2 = BC2{tex}\\Rightarrow{/tex}(-4 -1)2 + (-3 + 2)2 = (a - 2)2 + (2 - 3)2{tex}\\Rightarrow{/tex}(-5)2 = (a - 2)2{tex}\\Rightarrow{/tex}\xa0a - 2 = -5{tex}\\Rightarrow{/tex}\xa0a = -3Area of\xa0{tex}\\Delta A B C = \\frac { 1 } { 2 } | 1 ( 3 - 2 ) + 2 ( 2 + 2 ) - 3 ( - 2 - 3 ) |{/tex}{tex}= \\frac { 1 } { 2 } | 1 + 8 + 15 |{/tex}{tex}= \\frac { 1 } { 2 } \\times 24{/tex}= 12 sq. units.Diagonal of a parallelogram divides it into two equal triangles.Area of parallelogram ABCD = 2\xa0{tex}\\times{/tex}12=24 sq. units{tex}\\Rightarrow {/tex}\xa0AB\xa0{tex}\\times{/tex}Height = 24{tex}\\Rightarrow {/tex}\xa0AB{tex}\\times{/tex}Height = 24{tex}\\Rightarrow \\left[ \\sqrt { ( 2 - 1 ) ^ { 2 } + ( 3 + 2 ) ^ { 2 } } \\right] \\times \\text { Height } = 24{/tex}{tex}\\Rightarrow [ \\sqrt { 1 + 25 } ] \\times \\text { Height } = 24{/tex}{tex}\\Rightarrow \\text { Height } = \\frac { 24 } { \\sqrt { 26 } } = \\frac { 12 \\times \\sqrt { 2 } \\times \\sqrt { 2 } } { \\sqrt { 13 } \\times \\sqrt { 2 } }{/tex}{tex}= \\frac { 12 \\sqrt { 2 } } { \\sqrt { 13 } }{/tex} | |
| 21567. |
A gulab jamun |
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Answer» Hb gungun What is the question??? ??? What\'s it???? |
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| 21568. |
Show that the product of two consecutive poditive integers id divisible by 2 |
| Answer» Let n-1 and n be consecutive positive integers,Let P be their productThen {tex}\\style{font-family:Arial}{\\style{font-size:12px}{\\mathrm n(\\mathrm n-1)=\\mathrm n^2-1\\;.................(1)}}{/tex}We know that any positive integers is of the form 2q or 2q + 1, where q is a positive integerCase I: When n = 2q, thenP=n2\xa0- n = (2q)2\xa0- 2q = 4q2\xa0- 2q = 2q(2q - 1)-----(2)Case II: When n = 2q + 1, thenP=n2\xa0- n = (2q + 1)2\xa0- (2q + 1)= 4q2\xa0+ 4q + 1 - 2q - 1= 4q2\xa0+ 2qP = 2q(2q + 1)..........(3)From (2) and (3) we conclude thatThe product of n-1 and n is divisible by 2 | |
| 21569. |
Tan*a+d-c=cotc |
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| 21570. |
2a +1 / 3a = 6 then 3a + 1 / 2a = ? |
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Answer» Shraddha Nawale well tried but your answer is wrong. The correct answer is 9. I think iys 19/2 |
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| 21571. |
If the roots of the quadratic equation ax^2 + bx + c = 0 are equal roots thwn show that b^2 = 4ac |
| Answer» Here given that roots are equal so D or discriminent will 0. hence b^2-4ac=o and b^2=4ac | |
| 21572. |
How to score good marks in mathematics |
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Answer» By more practice of questions and do many workbooks By more and more practicing of questions. |
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| 21573. |
All the formula of chap: pair of linear equation in two variables.. . |
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Answer» It is very easy.see your ncert book?? Simply tell that you want formulas of all the chapters. |
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| 21574. |
All the Formula of chapter : polynomial .......In nicely way . . Plzzz give |
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Answer» It is in your ncert book Which polynomia quadratic or cubic Given in book |
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| 21575. |
Chapter 9 solution |
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Answer» Are given in this app All |
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| 21576. |
Yeah Pythagoras Theorem Chapter 6 ke |
| Answer» Pythagoras and converce of pythagoras all in ncert book | |
| 21577. |
All the Formula of chapter : Real number . In nicely way want plzzzzzzz...... A humble request...... |
| Answer» There are not much formulas in this chaptr | |
| 21578. |
Define circumcentre and orthocentre |
| Answer» Circumcentre is the point where all the perpendicular bisectors of a triangle meets and orthocentre is the point of intersection of all the altitudes of a triangle. | |
| 21579. |
P(x)=4x3-12x2+14x-3by g(x)=2x-1 |
| Answer» 2x - 1= 0x = 1/2So, by putting the value of x = 1/24x3 - 12x2 + 14x - 3= 4 (1/2 )3 - 12 (1/2 )2 + 14 (1/2 ) - 3= 4 (1/8) - 12 (1/4) + 14/2 - 3= 4/8 - 12/4 + 14/2 - 3= 1/2 - 3 + 7 - 3= 1/2 + 1= 1+2 / 2= 3/2 - AnsSo the remainder is 3/2 . | |
| 21580. |
Why sun is not twinkle while it became a star? |
| Answer» becoz sun is much nearer in comparision with other star | |
| 21581. |
The volume of a hemisphere is 2425_1/2 cm3 .. Find its curved surface area |
| Answer» As, volume of hemisphere = 2425{tex}\\frac { 1 } { 2 } c m ^ { 3 }{/tex}{tex}\\Rightarrow \\frac { 2 } { 3 } \\pi r ^ { 3 } = 2425 \\frac { 1 } { 2 }{/tex}{tex}\\Rightarrow \\frac { 2 } { 3 } \\times \\frac { 22 } { 7 } r ^ { 3 } = \\frac { 4851 } { 2 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 4851 \\times 3 \\times 7 } { 2 \\times 2 \\times 22 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 441 \\times 3 \\times 7 } { 2 \\times 2 \\times 2 }{/tex}{tex}\\Rightarrow r ^ { 3 } = \\frac { 21 ^ { 3 } } { 2 ^ { 3 } }{/tex}{tex}\\Rightarrow r = \\frac { 21 } { 2 } c m{/tex}So, the curved surface area of the hemisphere ={tex}2 \\pi r ^ { 2 }{/tex}{tex}= 2 \\times \\frac { 22 } { 7 } \\times \\frac { 21 } { 2 } \\times \\frac { 21 } { 2 } = 693 \\mathrm { cm } ^ { 2 }{/tex} | |
| 21582. |
Find the maximum sum of an ap is 2,4,6,8. |
| Answer» If it is upto infinity then sum is -1/6 | |
| 21583. |
If secA+tanA=p then prove that P square - 1/p square +1=sinA |
| Answer» Shift tan to rhs then sq. both the sides then find the value of tan in terms of only p . Put that value in a rt angle? and then prove it. | |
| 21584. |
How to find the root of irrational number easily? |
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Answer» Yaa no shortcuts you have to follow all steps No shortcuts available..... |
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| 21585. |
√3 value |
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Answer» 1.732 1.73 |
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| 21586. |
Is full circle guide beneficial for maths?? |
| Answer» Ncert is enough...if u want 99.9 percent in maths then u can prefer.. | |
| 21587. |
Cubic formula |
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| 21588. |
Formula of cubic polynomial |
| Answer» Let α, β and ɣ be the roots of the cubic polynomialThe cubic polynomial with roots α, β and ɣ is x3 – (α + β + ɣ)x2 + (αβ+ βɣ + ɣα)x – (αβɣ) = 0 | |
| 21589. |
Find perimeter of triangle whose coordinat are (0,4) ,(0,0),(3,0). |
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Answer» First find length of each sides of ∆ Let A( 4 , 0) B(0, 0) and C (0 , 3) use distance formula, AB =√(4²+0) =4 BC= √(0+3²) = 3CA =√(4²+3²) =5 now ,perimeter of ∆ = 3 + 4 + 5= 12 unit........, hope it will help you.....?? 12 unit |
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| 21590. |
Can you prove collinearity of three points by Section formula?explaine with an example? |
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Answer» Yes mam,You are correct but by taking m:n u can prove mam We need to prove the points (3,-2),(5,2) and(8,8) are collinear. A=(3,-2) B=(5,2) C=(8,8) Let The points B divides AC in the ratio of k+1 Then the coordinates will be, (8k+2/k+1,8k+5/k+1) Coordinates of B are (4,6) Comparing we get, 8k+2/k+1=4 and 8k+5/k+1=6 8k+2=4k+4 and 8k+5=6k+6 4k=2 and 2k=1 k=2/4 and k=1/2 k=1/2 and k=1/2 Value of k is same in both x and y directions.Therefore Points A,B,C are collinears.hope it helps u.... |
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| 21591. |
How to derive divisibility rules of 7,13,17,19,23,27,31,37,41,43......etc? |
| Answer» This can derive by the concept of Osculator. | |
| 21592. |
What is the draw back of Remainder and Factors theorem?How to over come it? |
| Answer» If the divisor is not linear polynomial we are unable to apply R.T and F.T ,so we can convert given polynomial into linear factors | |
| 21593. |
What is the draw back of Basic propornality theorem ?How it is Overcome? |
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Answer» BPT does not explain lengths of parallel sides in a triangle.To over come this we have concept of Criteria for similarity b/w Triangles Kya, BPT theorem ka drawback bhi hai.....? |
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| 21594. |
Maths sare chapter kitne marks ke aate hai |
| Answer» Ch- 1 (6 marks) 3 question totalCh- 2,3,4,5 (20marks) 8 question totalCh-7 (6 marks) 3 queston total Ch- 8,9 (12 marks) 4 question total Ch- 6,10,11 (15 marks) 5 question totalCh 12,13 (10 marks) 3 question totalCh- 14, 15 (11 marks) 4 question total | |
| 21595. |
25x^4-20x^2+4=0 |
| Answer» 25x^4 - 20x^2 + 4=025x^4 - 10x^2 - 10x^2 +4 =05x^2(5x^2 - 2) - 2(5x^2 - 2) = 0(5x^2 -2 ) , (5x^2 - 2 ) = 0x^2 = 2/5 , x^2 = 2/5x = √2/5 ,. x= √2/5 | |
| 21596. |
Z^4-13z^2+36=0 |
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Answer» (z -3)(z + 3) (z - 2)(z + 2) = 0z = 3,-3,2 or -2 z4 - 9z2 - 4z2 + 36 = 0z2 (z2 - 9) - 4(z2 - 9) = 0(z2 - 9) (z2 - 4) = 0(z - 9) (z + 9) (z - 4)(z + 4) = 0z = 9,-9, 4 and -4 |
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| 21597. |
Z^4-10z^2+9=0 |
| Answer» z4 - 10z2 + 9 = 0z4 - 9z2 - 1z2 + 9 = 0z2 (z2 - 9) -1(Z2 - 9) = 0(z2 - 9) (z2 - 1) = 0(z -3) (z+3) (z - 1) (z + 1) = 0z = -3, 3, 1, -1 | |
| 21598. |
If x=3 in one root of the quadrate equation xsquare_2kx_6=0 then find the value of k |
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Answer» Hey... I think this can be ur answer!!Polynomial - x2 - 2kx - 6 =0P (3)=(3)2 - 2(3)k =6=9 - 6k =6=-6k =-3=k =1/2Hope it helps you dear ☺️☺️ The value of k will be 1 |
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| 21599. |
If three coins are tossed together then find the probability of getting two head |
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Answer» 5/8 Possible outcomes= 8 (H,H,H) (H,T,T) (T,T,T) (T,T,H) (T,H,T) (H,T,H) (H,H,T) (T,H,H)Probability of getting two heads = 4 (H,H,H) (H,H,T) (H,T,H) (T,H,H)Therefore, probability of getting two heads = 4÷8 = 1÷2 1/2 |
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| 21600. |
CosA+SinA=√2CosAthen prove thatCosA-SinA=√2SinA |
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Answer» From given eqn 1SinA = √2 cosA - cos A SinA = cosA(√2 - 1)SinA /√2-1 = cosARationalising SinA × √2 + 1----------------------- = cosA√2 - 1 × √2 +1√2 sinA + sinA. = CosA------------------------- 2 - 1 √ 2 sinA =. CosA - SinA Hence proved? By squaring both sides: (cosA+sinA)2=(√2cosA)2By adding (cosA +sinA)2 +( cosA -sinA)2----------->cos2A+sin2A+2cosAsinA+cos2A+sin2A-2cosAsinA---------->cos2A+sin2A+cos2A+sin2A------>1+1=2. Now,(cosA-sinA)2=2-(cosA+sinA)2-------->so, (cosA-sinA)2=2-2cos2A------>(cosA-sinA)2=2(1-cos2A)------>(cosA-sinA)2=2sin2A-------->(cosA-sinA)=√2sinA.... Hence,proved. (Here 2 is the power.) |
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