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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.
| 33301. |
Express the hcf of 234 and111 as 234x+111y ,where x and y are integers |
| Answer» x = -9 and y = 19 | |
| 33302. |
Define zeros of a polynomial |
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| 33303. |
What is the meanig of this sign |
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| 33304. |
If sum of square of zeroes of quadratic polynomial f(x) x2-8x+k is 40 then find the value of k |
| Answer» Let {tex}\\alpha,\\beta{/tex}\xa0be the zeros of the polynomial {tex}f(x)=x^2-8x+k{/tex}.Sum of zeroes =\xa0{tex} \\alpha + \\beta = - \\left( \\frac { - 8 } { 1 } \\right) = 8{/tex}\xa0and, Product of zeroes =\xa0{tex} \\alpha \\beta = \\frac { k } { 1 } = k{/tex}Now,\xa0{tex} \\alpha ^ { 2 } + \\beta ^ { 2 } = 40{/tex}{tex} \\Rightarrow \\alpha ^ { 2 } + \\beta ^ { 2 }+2 \\alpha\\beta-2 \\alpha\\beta= 40{/tex}{tex} \\Rightarrow \\quad ( \\alpha + \\beta ) ^ { 2 } - 2 \\alpha \\beta = 40{/tex}{tex} \\Rightarrow \\quad 8 ^ { 2 } - 2 k = 40{/tex}{tex} \\Rightarrow \\quad 2 k = 64 - 40 {/tex}{tex}\\Rightarrow 2 k = 24 {/tex}{tex}\\Rightarrow k = 12{/tex} | |
| 33305. |
X2_2x_8=0 |
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| 33306. |
If h is the hcf of 609 and 957.find x and y, satisfy h=609x+957y? |
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| 33307. |
55x + 67y =31167 x + 55y = 299 |
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Answer» Ayush Mishra i don\'t think that your answer is correct plz check by substitution method55x +67y =311 ----------eq. 167x +55y =299 ----------eq. 2from eq. 1 we obtainx = 311 - 67y / 55 ----------eq. 3substituting the value of x in eq. 267(311 - 67y / 55) + 55y = 299(20837 - 4489y /55) + 55y =29920837 - 4489y + 3025y = 299 * 5520837 - 1464y = 16445-1464y = 16445 - 20837-1464y = -4392y = -4392 / -1464y = 3 putting the value of y in eq. 3x = 311 - 67(3) / 55x = 311 - 201 / 55x = 110/55x = 2x = 2 and y = 3 x = 4 and y = 3 |
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| 33308. |
If one zero of tje polynomial (a^2+9)x^2+13x+6a is reciprocal of the other find value of a. |
| Answer» Let {tex} \\alpha{/tex}\xa0and\xa0{tex} \\frac { 1 } { \\alpha }{/tex} be the zeros of\xa0(a2\xa0+ 9)x2\xa0+ 13x\xa0+ 6a.Then, we have{tex} \\alpha \\times \\frac { 1 } { \\alpha } = \\frac { 6 a } { a ^ { 2 } + 9 }{/tex}⇒\xa01 =\xa0{tex} \\frac { 6 a } { a ^ { 2 } + 9 }{/tex}⇒\xa0a2\xa0+ 9 = 6a⇒ a2 - 6a + 9 = 0⇒\xa0a2\xa0- 3a - 3a + 9 = 0⇒\xa0a(a - 3) - 3(a - 3) = 0⇒\xa0(a - 3) (a - 3) = 0⇒\xa0(a - 3)2\xa0= 0⇒\xa0a - 3 = 0⇒ a = 3So, the value of a in given polynomial is 3. | |
| 33309. |
If d is the hcf of 56 and 72 find x and y satisfying d egual to 56x+72y |
| Answer» x=3 y=1 | |
| 33310. |
What r the zeroes of poly. f(x)=x^2-2x+3? |
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Answer» x^2-2x+3=x^2-3x-x+3=x(x-3)-1(x-3)=(x-3)(x-1)Zeroes of polynomial are +3 and +1 Hii |
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| 33311. |
9,13,22,? |
| Answer» 35 | |
| 33312. |
Root2 +root2 |
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Answer» 2√2 2root2 2root2 2root2 |
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| 33313. |
splitting middle term of 2x2+7x-15 |
| Answer» 2x^2 + 7x - 15 = 2x^2 - 3x + 10x - 15 = x (2x - 3) + 5 (2x -3) = (x+5) (2x-3) | |
| 33314. |
Exercise 12.3. 9que |
| Answer» Given therw | |
| 33315. |
am/an = |
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Answer» ques do solve me m/n |
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| 33316. |
Factors of Xsquare+3root2x+4 are |
| Answer» kon se chapter ka question h ye maths m | |
| 33317. |
Hsisgsjz |
| Answer» ABCD | |
| 33318. |
2x+5y_12=0, 3x+7y-17=0 by cross mulipication method |
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Answer» Y=2 x=1 and y=2 |
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| 33319. |
(a+ b)x+(a-b)y=a^2+b^2 (a-b)x+(a+b)y=a^2 + b^2 |
| Answer» The given system of linear equation is(a – b)x + (a + b)y = a2 – 2ab – b2 ....(1)(a + b)(x + y) = a2 + b2 ....(2)Equation (2) can be written as(a + b)x + (a + b)y = a2 + b2 ....(3)Subtracting equation (3) from equation (1), we get– 2bx = – 2ab – 2b2{tex}\\Rightarrow{/tex} – 2bx = – 2b(a + b){tex}\\Rightarrow{/tex} x = a + bSubstituting this value of x in equation (1), we get(a – b)(a + b)(a + b)y = a2 – 2ab – b2{tex}\\Rightarrow{/tex} a2 – b2 + (a + b)y = a2 – 2ab – b2{tex}\\Rightarrow{/tex} (a + b)y = – 2ab{tex}\\Rightarrow \\;y = - \\frac{{2ab}}{{a + b}}{/tex}Verification: Substituting x = a + b, {tex}y = - \\frac{{2ab}}{{a + b}}{/tex},We find that both the equations (1) and (2) are satisfied as shown below:(a – b)x + (a + b)y = (a – b)(a + b) + (a + b) {tex}\\left\\{ { - \\frac{{2ab}}{{a + b}}} \\right\\}{/tex}= a2 – 2ab – b2(a + b)(x + y) = (a + b){tex}\\left\\{ {(a + b) + (- \\frac{{2ab}}{{a + b}})} \\right\\}{/tex}= (a + b)2 – 2ab= a2 + b2hence, the solution we have got is correct. | |
| 33320. |
Some one can give phone number plz for quering my dought of lesson polynomials |
| Answer» 9645235012 | |
| 33321. |
If the distance ofp[x,y]from A(6,2)and B(2,6) an equal prove that y=x |
| Answer» P(x, y), A(6,2), B(- 2, 6)PA = PB or, PA2 = PB2(x-6)2 + (y -2)2 = (x + 2)2 + (y - 6)2or, x2 - 12x + 36 + y2- 4y + 4 = x2+ 4x + 4 + y2- 12y + 36or, -12x - 4y = 4 x - 12yor,12y -4y = 4x + 12xor,8y = 16xor,y = 2xHence Proved. | |
| 33322. |
Show that (root3+root5)^2 is an irrational number |
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| 33323. |
How to prove cos45deegre=1/√2 |
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| 33324. |
How can i score ? in maths |
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Answer» By doing daily 1 hour practice Regular practice of 20-25 most hard questions from pearson and rd sharma By daily practise and proper practice Do more practice in maths |
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| 33325. |
2x2 +ax-a2=0 |
| Answer» -a ora/2 | |
| 33326. |
The sum and product of zeroes of polynomial 4x square-27x+3ksquare are eqUal find theValue of k |
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Answer» Sum of zeros =-(-27)÷4=27÷4.Product of zeroes = 3k ÷4.A.t.q=27÷4=3k÷4, =27÷4×4=3k =27=3k, =27÷3 =k, =9=k. Please send me full qnswet K=9 |
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| 33327. |
There are two classeA and B containg student |
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| 33328. |
It is important to do choching |
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Answer» That is coaching by mistek I have typed that According to me coaching is not important.... But self study is must I think she means to say coaching No I don\'t think. If we do regular practice of all subjects it is not necessary Tuition Choching means?? |
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| 33329. |
Prove that $13+)66/%3/%68^55=358 is equall to |
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| 33330. |
Find roots |
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| 33331. |
2018 question paper is their or not?? |
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| 33332. |
Tn=4n-5 find sum of first 25 terms of ap |
| Answer» T1=4*1-5=-1T25=4*25-5=95S25=n/2(t1+t25)S25=25/2{(-1)+95}S25=1175Ans=1175 | |
| 33333. |
Find the zero of the polynomial x square - √2x-12 |
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| 33334. |
For any positive integer n, then show that n2-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 | |
| 33335. |
Prove that sin^4x+cos^4x=1-2sin^2xcos^2x |
| Answer» To prove :\xa0sin4x\xa0+ cos4x = 1 - 2sin2x cos2xWe have,LHS = sin4x + cos4x= (sin2x)2 + (cos2x)2 + 2sin2x cos2x - 2 sin2x cos2x [By adding and subtracting 2sin2xcos2 x]= (sin2x + cos2x)2 - 2sin2x cos2x= 1 - 2sin2xcos2x\xa0= RHS | |
| 33336. |
Prove that cos0- sin0 + 1- ---------------------------= coscc0 + cot0 Cos0 + sin0 - 1 |
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| 33337. |
Please give 10 class timetable for 10cgpa |
| Answer» No need to study just enjoy your life with ur wishes ? | |
| 33338. |
Explaination all method to find the LCM and HCF |
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Answer» Equlid division lemma Prime factorisation |
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| 33339. |
write whether 2 root 45 +3 root 20 by 2 root 5 on simplification gives a rational or irrational no. |
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Answer» (2√45 + 3√20 )/2√5 = (2*3√5 + 3*2√5)/2√5= 12√5/2√5= 6 , so it gives a rational number Irrational |
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| 33340. |
(cosecΦ-sinΦ)=a³&(secΦ-cosΦ)=b³,prove That a²b²(a²+b²)=1 |
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| 33341. |
Find theta if sec(theta -36)= cosectheta |
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Answer» Q. Is wrong questions is wrong |
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| 33342. |
459 add 456 |
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Answer» 915. Write question complete |
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| 33343. |
Prove that 2+2÷√3 is an irrational number if √3 is irrational |
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| 33344. |
What is surface are? |
| Answer» The surface area of a solid object is a measure of the total area that the surface of the object occupies. | |
| 33345. |
135and225euclids division |
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| 33346. |
2x2+7/2x+3/4 and 7y²-11/3y-2/3 Answers this question any one please |
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| 33347. |
prove that 2b=a+c, if (b-c) x^2 + (c-a)x + (a-b) =0 has two real and equal roots |
| Answer» (b-c)x^2+(c-a)x+(a-b) =0 has real roots so D = 0B^2 - 4ac =0= (c-a)^2-4*(b-c)*(a-b)=0= C^2 +a^2-2ac -4* (ab-ac- b^2+bc) = 0 = C^2+a^2-2ac -4ab+4ac +4 b^2 -4 bc = 0 = C^2+a^2+2ac = 4ab - 4b^2 + 4 bc= (a+c)^ 2 = 4b( a - b + c)√[ a+ c] ^ 2 = √ 4b^2= a+ c = 2 b | |
| 33348. |
How to get answers of practice papers . |
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| 33349. |
Hddhdjftiymutgegwo |
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Answer» ? Shjfgdcdhjjjfsdhfdvhgsfhvcddhhcxvh |
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| 33350. |
sec4A - sec2A = tan4A + tan2A |
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