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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.
| 27301. |
3x-243 |
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| 27302. |
Find the root of 2x square+ x-4 by quadratic formula |
| Answer» We have, 2x2 + x + 4 = 0Dividing both sides by 2, we get{tex}x^2 +{1 \\over 2}x + 2 = 0 {/tex}{tex}\\implies (x)^2 + {1 \\over 2}x + {1 \\over 16} = -2 + {1 \\over 16}{/tex}{tex}\\implies (x)^2 + 2(x) ({1 \\over 4})+ ({1 \\over4})^2= {- 32 +1 \\over 16}{/tex}{tex}\\implies (x + {1 \\over4})^2 = -{31 \\over 16} <0{/tex}Which is not possible, as square cannot be negative.So, there is no real value of x which satisfy the given equation.Therefore, the given equation has no real roots. | |
| 27303. |
What are the possible values of remainder n,when a positive integer \'a`is divided by 3? |
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| 27304. |
Can two number have 18 as their HCF and 380 as their LCM. Give reason |
| Answer» I hope soo noo | |
| 27305. |
Find the zeros of the polynomial 9x2+5 |
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| 27306. |
If 2 sin²0 - cos²0=2,find the value of 0 |
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| 27307. |
Prove that 1/ root 2 is an irrational number by contradiction method |
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| 27308. |
Find a cubic polynomial with the sum of the product of its zeros are 2,-7,-14 respectively |
| Answer» Let the cubic polynomial be ax3 + bx2 +cx + dand its zeroes be\xa0{tex}\\alpha ,\\beta {/tex}\xa0and\xa0{tex}\\gamma{/tex}.Then,\xa0{tex}\\alpha + \\beta + \\gamma= 2 = \\frac { - b } { a }{/tex}{tex}\\alpha \\beta + \\beta \\gamma + \\gamma \\alpha = - 7 = \\frac { C } { a }{/tex}and\xa0{tex}\\alpha \\beta \\gamma = - 14 = \\frac { - d } { a }{/tex}If a = 1, then b = -2, c = -7, d = 14So, one cubic polynomial which fits the givenconditions is x3 - 2x2 - 7x + 14. | |
| 27309. |
Write wheather 2√45+2√20/2√5 on simplification gives a rational or an irrational number |
| Answer» 2√45+2√20/2√52√3×3×5 + 2√2×2×5/2√52×3√5 + 2×2√5/2√56√5 + 4√5/2√5= 10√5/2√55hence it is rational number | |
| 27310. |
What is full form of z |
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| 27311. |
(x-3y-7)^2+(3x-3y-15)^2=10 |
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| 27312. |
How to study real number |
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| 27313. |
Proove that 3+2√5 is an irrational numberBy two methods |
| Answer» There is only one method | |
| 27314. |
Find the number when divided by 47 give 23 as qoutient and 37 as remainder |
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Answer» 1118 47×23+37=1118 |
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| 27315. |
Is this system consistent or inconsistent 2x + Y - 6 = 0 and 4 x - 2y - 5 = 0 |
| Answer» Inconsistent | |
| 27316. |
π is a rational number or irrational number? |
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Answer» Rational numbers can be written as ab where a and b are integers. Also remember that rational numbers include terminating decimal numbers. Therefore −√8;3,3231089...;3+√2;π are all irrational. It\'s irrational no. |
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| 27317. |
6x - 3 - 7x |
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Answer» -x-3 -x-3 |
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| 27318. |
How much chapters of maths we should end in may |
| Answer» 10chapters | |
| 27319. |
x-1/x+2+x-3/x-4=10/3 |
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| 27320. |
Find the smallest number which leaves remainder 8 and 12 when divided by 28 and 32 respectively. |
| Answer» The LCM of 28 and 3228 = 2× 2 × 7=22×732 = 2 × 2 × 2 × 2 × 2=25LCM = 25 × 7 = 224Smallest no: which leaves remainder 8 and 12 when divided by 28 and 32= LCM of 8 & 12 = 224 - 20 = 204Therefore, 204 is smallest number which when divided by 28 and 32 leaves remainder 8 and 12 respectively. | |
| 27321. |
Find the hcf of 196 and 38220 |
| Answer» 38220>196 we always divide greater number with smaller one.Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as38220 = 196 × 195 + 0As there is no remainder so deviser 196 is our HCF | |
| 27322. |
Find the quadratic polynomial whose sum and product of zeros are \'underroot2+1and 1/underroot2+1 |
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| 27323. |
Find the values of a & b so that x^4+x^3+8x^2+ax+b,is exactly divisible by x^2+1 |
| Answer» If {tex}x^4+x^3+8x^2+ax +b{/tex}\xa0is exactly divisible by x2 + 1, the remainder after division should be zero.Now let us perform long divisionWe get, remainder = x (a - 1) + (b - 7)\xa0x (a - 1) + (b - 7 ) = 0{tex}\\Rightarrow{/tex}\xa0x (a - 1) + (b - 7) = 0x + 0{tex}\\Rightarrow{/tex}\xa0a - 1 = 0 and b - 7 = 0\xa0[On equating the coefficients of like powers of x]{tex}\\Rightarrow{/tex}a = 1 and b = 7 | |
| 27324. |
Prove that 7√5 are irrational |
| Answer» We can prove\xa0{tex}7 \\sqrt { 5 }{/tex} irrational\xa0by contradiction.Let us suppose that {tex}7 \\sqrt { 5 }{/tex} is rational.It means we have some co-prime integers a and b (b≠ 0)such that{tex}7 \\sqrt { 5 } = \\frac { a } { b }{/tex}{tex}\\Rightarrow \\sqrt { 5 } = \\frac { a } { 7 b }{/tex}\xa0.......(1)R.H.S of (1) is rational but we know that{tex}\\sqrt { 5 }{/tex}is irrational.It is not possible which means our supposition is wrong.Therefore, {tex}7 \\sqrt { 5 }{/tex} cannot be rational.Hence, it is irrational. | |
| 27325. |
Which book is the best for class 10 maths, other than ncert |
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Answer» Ty????? Every book has its own uniqueness....But for my opinion K.C sinha ,S.chand are also good. R.D sharma is best R .D. Sharma 110% R.D. Sharma? |
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| 27326. |
Find hcf of 81 and 237.exp |
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| 27327. |
Can anybody send me the solutions of exercise 3.6 from maths NCERT book. |
| Answer» You can check NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 27328. |
prove that tana=sina/cosa |
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| 27329. |
Which is best book for class 10 all subjects |
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| 27330. |
What is the maximum value of Sin thita? |
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Answer» realy 1 1 |
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| 27331. |
Find the greatest number of four digits which is exactily divisible by 15 ,24 and 36 |
| Answer» first find lcm of 15 24and 36 and then divide by that number to 9999(greatest four digit number) then get remainder and substract from 9999 Hence that number will divide 15 24 and 36. | |
| 27332. |
Prove that if x and y are both odd positive integer then x×x+y×y is even but not divisible by 4 |
| Answer» Since x and y are odd positive integers, Then it should be in the form of 2x+1 (where x is a positive integer)Let x = 2m + 1 and y = 2n + 1 ( where m and n positive integers)Now x2 + y2= (2m + 1)2 + (2n + 1)2= 4m2 + 4m + 1 + 4n2 + 4n + 1= 4(m2 + n2 + m + n) + 2 ....... (1)from (1) we getx2 + y2 = 2 {m2 + n2 + m + n) + 1} = 2t (t = 2(m2 + n2 + m + n) + 1 is a positive integerTherefore it is clear that {tex}x^2+y^2{/tex} is an even number but not divisible by 4. | |
| 27333. |
Prove that if x and |
| Answer» Age | |
| 27334. |
Can i get questions of chapter 3 from ncert Exampler |
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Answer» Hii Aditi Yes ofcourse Yess ? Just visit that section in this AAP |
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| 27335. |
2x+3y=7( a+b+1)x+(2a_b)y=28 |
| Answer» 7-3y/2 | |
| 27336. |
Find hcf of 963 and 657 and express it as linear combination |
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Answer» Plzz express it in linear combination Using Euclid’s Division Lemmaa = bq+r , o ≤ r < b963 = 657×1 + 306657 = 306×2 + 45\xa0306 = 45×6+3645 = 36×1+936 = 9×4+0\xa0∴ HCF (657, 963) = 9. |
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| 27337. |
Why a^2+b^2 not equal to zero is given in definition of pair of linear equation in two variable |
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| 27338. |
6ax + 6by = 3a + 2b and 6bx + 6ay = -2a + 3b. Find the value of x and y using elimination method. |
| Answer» 6(ax + by) = 3a + 2b6ax + 6by = 3a + 2b.........(i)6(bx - ay) = 3b - 2a6bx - 6ay = 3b - 2a.........(ii)Multiplying (i) by a and (2) by b,So, 6a2x + 6aby = 3a2 + 2ab .......... (iii)And 6b2x - 6aby = 3b2 - 2ab ......... (iv)Add (iii) and (iv), we get6a2x + 6aby + 6b2x - 6aby = 3a2 + 2ab + 3b2 - 2ab⇒ 6a2x + 6b2x = 3a2\xa0+ 3b2⇒6 (a2x + b2x) = 3(a2\xa0+ b2)⇒{tex}x = \\frac { 3 \\left( a ^ { 2 } + b ^ { 2 } \\right) } { 6 \\left( a ^ { 2 } + b ^ { 2 } \\right) } = \\frac { 3 } { 6 } = \\frac { 1 } { 2 }{/tex}Substituting\xa0{tex}x = \\frac 12{/tex}\xa0in (i),we get{tex}6 a \\times \\frac { 1 } { 2 } + 6 b y = 3 a + 2 b{/tex}⇒ 3a + 6by = 3a + 2b⇒ 6by =3a + 2b -3a⇒ 6by = 2b⇒ {tex}y = \\frac { 2 b } { 6 b } = \\frac { 1 } { 3 }{/tex}Hence, the solution is\xa0{tex}x = \\frac { 1 } { 2 } , y = \\frac { 1 } { 3 }{/tex} | |
| 27339. |
What is the value of 0/0 . |
| Answer» Not define | |
| 27340. |
Find the zeros of polynomial 6xsquare-3 |
| Answer» Let x be 0.P(x)=(6x)²-3 then,P(0)=(6*0)²-3 =0-3 =-3Similarly, in this process only you have to find zero of the polynomial by putting values in x. | |
| 27341. |
What is cos thita |
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| 27342. |
For any positive integer n, prove that n cube -n is divisible by 6 |
| Answer» According to euclid division lemma any positive integer is of the form 6m, 6m+1, 6m+2, 6m+3, 6m+4, 6m+5.If a=6m (a)³= (6m)³ =216m³ =6(36m) =6n ( n=36m)If a=6m+1 (a)³=(6m+1) =(6m)³ + 3*(6m)² * (1) +3 *(6m) * (1)² + (1)³ =216m³ + 108m² + 18m + 1 =6(36m³ + 18m² + 3m) + 1 =6n+1 (n=36m³ + 18m² + 3m) + 1 Similarly you have to find if you can\'t than tell me latter ok baby | |
| 27343. |
Find the HCF of 96 and 228 euclid lemma |
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| 27344. |
prove the root 3 is the irrational |
| Answer» Let us assume that √3 is a rational numberThat is, we can find integers a and b (≠ 0) such that √3 = (a/b)Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.√3b = a⇒\xa03b2=a2 (Squaring on both sides) → (1)Therefore, a2 is divisible by 3Hence \x91a\x92 is also divisible by 3.So, we can write a = 3c for some integer c.Equation (1) becomes,3b2 =(3c)2⇒\xa03b2 = 9c2∴ b2 = 3c2This means that b2 is divisible by 3, and so b is also divisible by 3.Therefore, a and b have at least 3 as a common factor.But this contradicts the fact that a and b are co-prime.This contradiction has arisen because of our incorrect assumption that √3 is rational.So, we conclude that √3 is irrational. | |
| 27345. |
Koi mahan shaksh hai jisne 11th ki maths start kr di mujhe ek doubt hai yrr? |
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Answer» Aapne lagta Hai, Set kee Definition Naheen padhi Hai achche se.......\'cause.......Sets\xa0are usually symbolized by uppercase, italicized, boldface letters such as A, B, S, or Z. Each object or number in a\xa0set\xa0is called a member or element of the\xa0set.Examples\xa0include the\xa0set\xa0of all computers in the world, the\xa0set\xa0of all apples on a tree, and the\xa0set\xa0of all irrational numbers between 0 and 1......... toh Aapne Dekh hee liya hoga Definition se, kee Set ek Particular Thing ka hota hai, Agar differ hai, Toh woh Anoother set Hoga........ First thing A set Doesn\'t contains differ Elements.....and Second thing is that...... Agar kar raha hoga, Then\xa0........It\'ll be another set not the subset of the same....... Set me jaise ek......bhi element differ krta hoga toh kya subset mana jayega? Subset wale topic me .... |
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| 27346. |
X2-4root3x+3=0 please help..... |
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Answer» Your most welcome Thank you very much.... 2✓3 +3 and 2✓3-3 |
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| 27347. |
Solve xy/x+y=1/5 , xy/x+y =1/7 |
| Answer» {tex}\\frac{xy}{x + y}{/tex}\xa0=\xa0{tex}\\frac{1}{5}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{x + y}{xy}{/tex}\xa0=\xa0{tex}\\frac{5}{1}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{1}{y}{/tex}+\xa0{tex}\\frac{1}{x}{/tex}\xa0{tex}= 5{/tex}....(i)and\xa0{tex}\\frac{xy}{x - y}{/tex}\xa0=\xa0{tex}\\frac{1}{7}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{x - y}{xy}{/tex}\xa0{tex}= 7{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{y}{/tex}\xa0-\xa0{tex}\\frac{1}{x}{/tex}\xa0= 7...(ii)Now solve equation (i) and (ii) by assuming {tex}\\frac{1}{y}{/tex}\xa0{tex}= a{/tex} and\xa0{tex}\\frac{1}{x}{/tex}\xa0{tex}= b{/tex}{tex}\\therefore{/tex} eq.(i) and (ii) becomes{tex}\\Rightarrow{/tex}{tex}b = -1{/tex}.....(iii)Putting the value of\xa0{tex} b = -1{/tex}\xa0from\xa0eq. (iii) in equation (i), we get{tex}a - 1 = 5{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}a = 6{/tex}Now,\xa0{tex}\\frac{1}{y}{/tex}\xa0{tex}= 6{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}y ={/tex}\xa0{tex}\\frac{1}{6}{/tex}and\xa0{tex}\\frac{1}{x}{/tex}\xa0= -1\xa0{tex}\\Rightarrow{/tex}\xa0{tex}x = -1{/tex} | |
| 27348. |
Relationship in zeroes |
| Answer» X+0 | |
| 27349. |
F(x)=x2-2x-8 |
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Answer» x2 - 2x - 8x2 - 4x + 2x - 8= x (x - 4) + 2 ( x - 4)= ( x + 2) (x - 4) f (X)= X2-2X-8 = X2+X (4-2)-8 = X2+4X-2X-8 = X (X+4)-2 (X+4) = (X-2) (X+4) (X-2)= 0 X= 2 (X+4)= 0 X=-4 |
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| 27350. |
If cos A = 1/2, find the value of 2secA/1+tanSqure A |
| Answer» Apprx 7 | |