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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.
| 34351. |
Splitting the middle term |
| Answer» Given, 2x2 - 7x - 15= 2x2 - 10x + 3x - 15{tex}= 2x (x - 5) + 3(x - 5)\\\\= (x - 5)(2x + 3){/tex} | |
| 34352. |
Two different dice are tossed together find the probability of getting a doublet |
| Answer» 50:50 | |
| 34353. |
Obtain the zeros of 3x+4 |
| Answer» X=-4/3 | |
| 34354. |
Verify that 1/2 1 - 2 are zeroes of cubic polynomial |
| Answer» {tex}2 \\mathrm { x } ^ { 3 } + \\mathrm { x } ^ { 2 } - 5 \\mathrm { x } + 2 ; \\frac { 1 } { 2 } , 1 , - 2{/tex}Comparing the given polynomial with\xa0{tex}a x ^ { 3 } + b x ^ { 2 } + c x + d{/tex}, we geta = 2, b = 1, c = 5, d = 2Let\xa0{tex}p ( x ) = 2 x ^ { 3 } + x ^ { 2 } - 5 x + 2{/tex}Then,{tex}P \\left( \\frac { 1 } { 2 } \\right) = 2 \\left( \\frac { 1 } { 2 } \\right) ^ { 3 } + \\left( \\frac { 1 } { 2 } \\right) ^ { 2 } - 5 \\left( \\frac { 1 } { 2 } \\right) + 2{/tex}{tex}= \\frac { 1 } { 4 } + \\frac { 1 } { 4 } - \\frac { 5 } { 2 } + 2 = 0{/tex}{tex}p ( 1 ) = 2 ( 1 ) ^ { 3 } + ( 1 ) ^ { 2 } - 5 ( 1 ) + 2{/tex}= 2 + 1 - 5 + 2 = 0{tex}p ( - 2 ) = 2 ( - 2 ) ^ { 3 } + ( - 2 ) ^ { 2 } - 5 ( - 2 ) + 2{/tex}= -16 + 4 + 10 + 2 = 0Therefore,\xa0{tex}\\frac{1}{2}{/tex}, 1 and -2 are the zeroes of{tex}2 x ^ { 3 } + x ^ { 2 } - 5 x + 2{/tex}So,\xa0{tex}\\alpha = \\frac { 1 } { 2 } , \\beta = 1 \\text { and } \\gamma = - 2{/tex}Therefore,{tex}\\alpha + \\beta + \\gamma = \\frac { 1 } { 2 } + 1 + ( - 2 ) = - \\frac { 1 } { 2 } = - \\frac { b } { a }{/tex}{tex}\\alpha \\beta + \\beta \\gamma + \\gamma \\alpha = \\left( \\frac { 1 } { 2 } \\right) \\times ( 1 ) + ( 1 ) \\times ( - 2 ) + ( - 2 ) \\times \\left( \\frac { 1 } { 2 } \\right){/tex}{tex}= \\frac { 1 } { 2 } - 2 - 1 = - \\frac { 5 } { 2 } = \\frac { c } { a }{/tex}{tex}\\alpha \\beta \\gamma = \\left( \\frac { 1 } { 2 } \\right) \\times ( 1 ) \\times ( - 2 ) = - 1 = \\frac { - 2 } { 2 } = \\frac { - d } { a }{/tex} | |
| 34355. |
2x+3y=8 and4×+6y=7 solve by elimination method |
| Answer» | |
| 34356. |
Is /2a polynomial |
| Answer» | |
| 34357. |
Find the largest number which divides 615and963 leaving remainder 6in each case |
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| 34358. |
If a2(square) +b2(square)=10 and a-b=2 then find the value of a and b |
| Answer» | |
| 34359. |
Is in the final examination of class 10 question from whole book is asked? |
| Answer» Yes because cce pattern was removed from session 2017-18 | |
| 34360. |
What definitions should be written in real no. |
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Answer» The numbers are divided into two part real number and non real number. In math we always use real number. Non real number are imaginary number so we can use it.so we say that all the whole no, integer, natural no, rational or irretional no,are real number. Real no is the combination or group of ratinal and irratinal no called real no I want to ask that what definitions in the chapter of real no. set of rational and irrational no. |
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| 34361. |
Is cbse class 10 sylabuss reduced by half |
| Answer» No, according to our school full book is coming | |
| 34362. |
Cos(90°-A)sin(90°-A)/tan(90-A) =sin×sinA |
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Answer» From which school you belong to??? Mujhe nahi pata |
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| 34363. |
What is round off |
| Answer» | |
| 34364. |
Prove underroot 2 is a irrational number |
| Answer» let underroot 2 is the rational number.underroot 2 =a/b (co prime number)2b square= a square2 divides a 2divides a squarewhere,a=2c2b square= 4csquaeb square= 2c square2 divides b 2 divides b squarehence proved underroot 2 is a irrational number. | |
| 34365. |
If zero of the polynomial xcube - 16xsquare + 11x + 364 are (a-b) , a (a+b) find a and b |
| Answer» | |
| 34366. |
Divide 2xsq +3x+1 by x+2 |
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Answer» Quotient --- 2x-1 and remainder --- 3. The quotient will be 2x-1 and the reminder will be 3 |
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| 34367. |
What is the difference between lemma and algorithm? |
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Answer» Lemma is a proven statement used for proving another statement while algorithm is a series of well defined steps which gives a procedure for solving a type of a problem. Mast punjabi |
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| 34368. |
Prove under root 7 is irrational |
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Answer» Let us assume that |
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| 34369. |
7.6 daigram |
| Answer» | |
| 34370. |
Simplify :23×122×207*52 85×103*182×814 |
| Answer» | |
| 34371. |
Sin 37 value |
| Answer» Sin 37= Cos 53 =3/5 | |
| 34372. |
Explain me Cartesian product |
| Answer» | |
| 34373. |
Arithmetic progression 1 and 2 marks questions and answers |
| Answer» | |
| 34374. |
Show that the square of an odd positive integer is of the form 8q+1, for some integer q. |
| Answer» This is the question of NCERT | |
| 34375. |
√2x plus √3y =0√3x plus √y = 0Solve by substitution method |
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| 34376. |
4x+3y=9 ploting points |
| Answer» | |
| 34377. |
My full name:- Samayma Venkata Rama Naga Sai Shankara Subramanya Lakshmi Krishna Vivek |
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Answer» Are you from south india?????Bcoz i think that this type of name are only in south india........sorry if you are hurted....????? Just asking for kind information....... Fake |
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| 34378. |
Which book should i study to get 100 out of 100 |
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Answer» NCERT exampler Ncert& do previous year question paper ... I recommend you oswaal. It concepts are always clear. You have to study urself.... You can study any book Evergreen Only NCERT? |
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| 34379. |
To verify the given sequence of A.P |
| Answer» a2-a1&a3-a2 .......If their difference is same then it is an Ap | |
| 34380. |
For what value of the system of linear eq. |
| Answer» | |
| 34381. |
Find the 67th term of an AP 0,5,10,15,1\\6 |
| Answer» | |
| 34382. |
If d=36a+24b find the value of a and b |
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Answer» Plzz understand clearly Let d be the HCF of 24 and 36.24 = 23{tex}\\times{/tex}336 = 22{tex}\\times{/tex}32HCF = 22{tex}\\times{/tex}3 = 12{tex}\\Rightarrow{/tex}{tex}d = 12{/tex}Now {tex}d = 24a + 36b{/tex}{tex}12 = 24a + 36b{/tex}When\xa0{tex} a = -1{/tex} and {tex}b = 1{/tex}, we get12 = 24{tex}\\times{/tex}(-1) + 36{tex}\\times{/tex}(1)= -24 + 3612 = 12So, a = -1 , b = 1 satisfy the equation d = 24a + 36b{tex}\\therefore{/tex} One possible value of a and b is -1 and 1. |
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| 34383. |
Mathematics book has total 15 chapter so you think more marks question which question |
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Answer» At max..most of the questions are frm the two chapters of trigonometry...not most bt I should say most of the weightage is frm the two chapters of trigono..that is approx 8- 9 marks ! Trigonometry Sorry chapter |
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| 34384. |
f (x + k) is a factor of 2x² + 2kx + 5x + 10, find k. |
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Answer» If (x+k) is a factor then x = -\xa0kNow , 2x²+2kx+5x+10 = 0 -----------------(1)Now, substituting -\xa0k in the place of x in equation (1)⇒2(-k)²+2k(-k)+5(-k)+10 = 0⇒2k²-2k²-5k+10 = 0⇒-5k=-10⇒5k=10⇒k=10/5⇒k=2 . The value of k is 2 Cannot answer on this |
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| 34385. |
For any positive integer ‘n’ prove that n³– n is divisible by 3. |
| Answer» n³-n=n(n²-1)Let the positive integer be N=3q,3q+1,3q+2Since 0 | |
| 34386. |
For any positive integer ‘n’ prove that n3– n is divisible by 6. |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)\xa0\xa0 | |
| 34387. |
What are the most important questions in cbse class 10 exam 1 |
| Answer» All are important bro | |
| 34388. |
The 4th term of an AP is zero .Prove that its 25th term is triple its 11th term |
| Answer» We have,a4\xa0= 0{tex}a + 3d = 0{/tex}{tex}3d = -a{/tex}or {tex}-3d = a{/tex}..........(i)Now,a25\xa0= {tex}a + 24d{/tex}= {tex}-3d + 24d{/tex} [Putting value of a from eq(i)]= {tex}21d{/tex}...........(ii)a11\xa0= {tex}a + 10d{/tex}= {tex}-3d + 10d{/tex}= {tex}7d{/tex}.........(iii)From eq(ii) and (iii), we geta25 = 21 da25 = 3(7d)a25\xa0= 3a11Hence Proved | |
| 34389. |
Ap 5 d6 find the last termof A.p is 6 |
| Answer» Question is not written properly | |
| 34390. |
Are you taught me how can l easily understand that how to make diagram in this chapter |
| Answer» | |
| 34391. |
find the greatest number of 6 digits exactly divisible by 18,24 and 36 |
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Answer» Solution: First find the LCM of all the given numbers. LCM ( 18, 24, 36 ) = 72Now the greatest 6-digit number is 99,99,99.Thus, 99,99,99\xa0/ 72\xa0=\xa0{tex}\\large13888 \\frac 78 \\sim13888{/tex}and 13888\xa0x 72\xa0= 99,99,36.which is our required answer. The answer is = 999936 |
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| 34392. |
Find the least 6 digit number which is exactly divisible by 12 ,16 and 36 |
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Answer» Sorry previous answer is for the 5-digit\xa0greatest number.The correct answer for the given question is 100080.Solution: First find the LCM of all the given numbers. LCM ( 12, 16, 36 ) = 144Now the smallest 6-digit number is 10,00,00.Thus, 10,00,00 / 144 =\xa0{tex}\\large 694\\frac49\\space \\sim 695{/tex}and 695 x 144 = 10,08,00.which is our required answer. The answer is 99936. |
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| 34393. |
Cos squaretheta +cos squarethetacotsquretheta=cot squaretheta |
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Answer» Here, I replaced the Theta "θ"\xa0with "x".Please take θ in your solution. To prove : cos2x + (cos2x X cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx) = cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx ?Solution : Taking the LHS => cos2x + (cos2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx X cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx)\xa0 (Taking\xa0cos2x common from both\xa0the terms) =>\xa0cos2x (1 +\xa0cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx)\xa0 (Since 1 +\xa0cot2\u200b\u200b\u200b\u200b\u200b\u200b\u200bx = cosec2x ) =>\xa0cos2x (cosec2x\xa0)\xa0 (Since cosec2x = 1/ sin2x ) =>\xa0 cos2x / sin2x => (cos x / sin x )2 (Since cosx / sinx = cotx ) => (cot\xa0x )2=> cot2xwhich is equals to RHSThus, LHS = RHS\u200b\u200b\u200b\u200b\u200b\u200b\u200bHence proved. |
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| 34394. |
prove that 1\\√3 is irrational |
| Answer» {tex}\\frac { 1 } { \\sqrt { 3 } } = \\frac { 1 } { \\sqrt { 3 } } \\times \\frac { \\sqrt { 3 } } { \\sqrt { 3 } } = \\frac { \\sqrt { 3 } } { 3 } {/tex} .............(i)If possible, let {tex}\\frac { 1 } { \\sqrt { 3 } }{/tex} be rationalThen, from (i) it follows that {tex}\\frac { \\sqrt { 3 } } { 3 } {/tex} is rational.Let {tex}\\frac { \\sqrt { 3 } } { 3 } = \\frac { a } { b }{/tex} where a and b are non-zero intergers having no common factor other than 1.Now, {tex}\\frac { \\sqrt { 3 } } { 3 } = \\frac { a } { b }{/tex}{tex} \\Rightarrow \\sqrt { 3 } = \\frac { 3 a } { b }{/tex} ...............(ii)But 3a and b are non-zero integers{tex}\\therefore \\frac { 3 \\mathrm { a } } { \\mathrm { b } }{/tex} is rational. So, {tex}\\sqrt { 3 }{/tex} is rational.This contradicts the fact that {tex}\\sqrt { 3 }{/tex} is irrationalThe Contradiction arises by assuming that {tex}\\frac { 1 } { \\sqrt { 3 } }{/tex} is rational.Hence {tex}\\frac { 1 } { \\sqrt { 3 } }{/tex} is irrational. | |
| 34395. |
Any one know only prime is rational no. |
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Answer» what is "s". No. This is wrong. Not only prime no.s but even composite no.s are rational no.s |
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| 34396. |
If b and q are acute angles such that sin b =sin q then prove that b=q |
| Answer» Consider two right triangles ABC and PQR in which\xa0{tex} \\angle B{/tex} \xa0and\xa0{tex}\\angle Q{/tex} are the right angles.We have,In\xa0{tex}\\triangle ABC{/tex}{tex}\\sin B=\\frac{AC}{AB}{/tex}\xa0and, In\xa0{tex}\\triangle PQR{/tex}\xa0{tex}\\sin Q=\\frac{PR}{PQ}{/tex}{tex} \\because \\quad \\sin B = \\sin Q{/tex}{tex} \\Rightarrow \\quad \\frac { A C } { A B } = \\frac { P R } { P Q }{/tex}{tex} \\Rightarrow \\quad \\frac { A C } { P R } = \\frac { A B } { P Q } = k{/tex}(say) ...... (i){tex} \\Rightarrow {/tex} AC = kPR and AB = kPQ .....(ii)Using Pythagoras theorem in triangles ABC and PQR, we obtain\xa0AB2 = AC2 + BC2 and PQ2 = PR2 + QR2{tex} \\Rightarrow \\quad B C = \\sqrt { A B ^ { 2 } - A C ^ { 2 } } \\text { and } Q R = \\sqrt { P Q ^ { 2 } - P R ^ { 2 } }{/tex}{tex} \\Rightarrow \\quad \\frac { B C } { Q R } = \\frac { \\sqrt { A B ^ { 2 } - A C ^ { 2 } } } { \\sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = \\frac { \\sqrt { k ^ { 2 } P Q ^ { 2 } - k ^ { 2 } P R ^ { 2 } } } { \\sqrt { P Q ^ { 2 } - P R ^ { 2 } } }{/tex}\xa0[ using (ii) ]{tex} \\Rightarrow \\quad \\frac { B C } { Q R } = \\frac { k \\sqrt { P Q ^ { 2 } - P R ^ { 2 } } } { \\sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = k{/tex}...(iii)From (i) and (iii), we get{tex} \\frac { A C } { P R } = \\frac { A B } { P Q } = \\frac { B C } { Q R }{/tex}{tex} \\Rightarrow \\quad \\Delta A C B - \\Delta P R Q{/tex}\xa0[By S.A.S similarity]{tex} \\therefore \\quad \\angle B = \\angle Q{/tex}\xa0Hence proved. | |
| 34397. |
Show px^2+(2q-p^2)x-2pq have p&-2q/p as zero |
| Answer» | |
| 34398. |
If x =2 and x=3 are the roots of the equation 3xsquare-2mx+2n=0,find the value of m and n |
| Answer» F(2)=3×4-4m+2n0=12-4m+2n-4m+2n=-12....... eq (1)f(3)=27-6m+2n=0-6m+2n=-27....... eq (2)eq(1)-(2)2m=15m=15/2put m=15/2 in eq (1)-4×15/2+2n=-12-30+2n=-122n=18n=9m=15/2: n=9 | |
| 34399. |
Chapter 1-In statement problems how we will come to know that when to find LCM and when to find HCF? |
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Answer» If we are asked minimum thn LCMnd if we are asked maximum than HCF....EXCEPT IN SOME CASES....IN WHICH U HAVE TO BE LOGICAL It is already given in question |
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| 34400. |
Find the sum of 0.6,1,7,2,8,....to 100 terms |
| Answer» Given AP: 0.6, 1.7, 2.8, ............\xa0a = 0.6d = 1.7 - 0.6 = 1.1n = 100Sn = n/2[ 2a + (n-1)d] = 100/2 [ 2(0.6) + (100 - 1)×(1.1)] = 50[ 1.2 + (99\xa0× 1.1)] = 50[ 1.2 + 108.9] =50[110.1] = 5505 | |