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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.
| 4551. |
1/(x-1)(x-2)+1/(x-2)(x-3)=2/3.find for x |
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| 4552. |
From which chapter more questions will come in board examination |
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| 4553. |
If f(x) = x3 + x2 - ax + b is divisible by x2 -x write the value of a and b |
| Answer» Since f(x) =\xa0x3\xa0+ x2\xa0- ax + b is divisible by (x2\xa0- x), we havex2\xa0- x = 0{tex}\\Rightarrow{/tex}\xa0x(x - 1) = 0{tex}\\Rightarrow{/tex}\xa0x = 0 or x = 1Hence,f(0) = 0{tex}\\Rightarrow{/tex} x3\xa0+ x2\xa0- ax + b = 0{tex}\\Rightarrow{/tex}\xa003\xa0+ 02\xa0- a(0) + b = 0{tex}\\Rightarrow{/tex}\xa0b = 0Also,f(1) = 0{tex}\\Rightarrow{/tex} x3\xa0+ x2\xa0- ax + b = 0{tex}\\Rightarrow{/tex}\xa013\xa0+ 12\xa0- a(1) + 0 = 0{tex}\\Rightarrow{/tex}\xa01 + 1 - a = 0{tex}\\Rightarrow{/tex} 2 - a = 0{tex}\\Rightarrow{/tex}\xa0a = 2Hence , the value of a and b in given polynomial are a = 2 and b = 0. | |
| 4554. |
Find the value of x if 12th term of the AP is 81 and AP is x-7,x-2,x+3.........find S12 also |
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| 4555. |
2√6x² - 5x - 3√6 |
| Answer» So easy solve by .middle split.......2√6x2 - (9-4)x -3√6 .....solve further.... | |
| 4556. |
Ab is the diameter of the larger semicircle ab=21 |
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| 4557. |
Why only maths |
| Answer» Maths is a good subject. | |
| 4558. |
2√6x²-5x-3√6 |
| Answer» 3√6/4 | |
| 4559. |
What should be the timings of the examinations held in March of class10th?? |
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Answer» In my opinion it should be 10am to 1 am TIME OF COMMENCEMENT -10:30 A.M. |
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| 4560. |
Prove that 1/sec x - tan x - 1/cos x =1/cos x - 1/sec x+ tan x |
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Answer» To Prove: [1/(secx – tanx)] – [1/cosx] = [1/cosx] - [1/(secx + tanx)]Proof: LHS = [1/(secx – tanx)] – [1/cosx]But [1/cosx] = sec x = [1 ×(secx + tanx)/ (secx - tanx)(secx + tanx)] – [sec x] = [(sec x + tan x) / (sec2x – tan2x)] – [sec x]But (sec2x – tan2x) = 1LHS = sec x + tan x – [sec x] = tan x ………. (1)RHS = [1/cosx] - [1/(secx + tanx)] = [sec x] - [1 ×(secx - tanx)/ (secx + tanx)(secx - tanx)] = [sec x] - [(sec x - tan x) / (sec2x – tan2x)]But (sec2x – tan2x) = 1RHS = [sec x] - [(sec x - tan x)] = tan x …………… (2)As LHS = RHS , Hence Proved So esy bt likhu kaise |
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| 4561. |
Where can I find answers of lab manual. |
| Answer» Guve to me also if u will find ...thanks in advance ? | |
| 4562. |
if the two zeros of the quadratic polynomial 7x^2-15x-k are reciprocal of each other then find k |
| Answer» type answer | |
| 4563. |
Which gas is released when concentrated and dilute hno3 react to zinc???? |
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Answer» I think u r wrong tanu...because I am sure that gas is released. Yeah h2 isn\'t released... HNO^3 IS NITRIC ACID AND it\'s a strong oxidising agent as soon as the hydrogen gas is produced during the reaction it oxidises it into water. That\'s why NO GAS is released. Hydrogen.. Come on none have the guys??? |
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| 4564. |
2+2=5? |
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Answer» 2 +2=5 galti se hota Hai Ok got it ap fir sa busy ho gaye What By mistake Hiii bff ji No it is 4 |
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| 4565. |
Maths paper is on which date? |
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Answer» Ok ???????? 28 march 2018 Yes.. What 28 March ???Are u sure Thanks 28 march 2 8 March |
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| 4566. |
Find the probability of getting 53 Sundays in a non leap year. |
| Answer» 1/7 | |
| 4567. |
I want newly launched sampel papers of Cbse board |
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| 4568. |
CosA=3/5 find the value of 4+4tansquareA |
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Answer» 100/9 cosa= base/hypotenusBase ko 3 man lo and hypo. Ko 5 .P.G.T. lagao Perpendicular = 4 Then apply tan= p/hThen find the value of 4+4sin square theeta Please tell me kal mera paper hai 100/9 How 100/9 |
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| 4569. |
2√6x² - 5x - 3√6 find the both zeros |
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| 4570. |
Sum of all natural number |
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Answer» n(n+1)/2 An as there are infinite natural numbers. -1/12 |
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| 4571. |
What is a square +b square |
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Answer» asq.+bsq.=csq. Asq+bsq)+2ab (a+b)²-2ab |
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| 4572. |
How we find that any no. is a composite number??? |
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Answer» If it has at least 1 factor other than 1 and it self If any number divided by 1 ,itself and any other factor then it is a composite number. For ex. 9 divisible by 1,3,9 It means 4,6,8,9,10e.t.c are composite numbers because these numbers are not prime numbers No. that are not prime number are composite no. |
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| 4573. |
Prove that sin^4 alpha / sin^2 beta + cos^4 / cos^2 beta=1 |
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| 4574. |
Friends what is the minimum percentage we should get to take medical stream |
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Answer» I am not sure above 60℅ to 70% It must be more than 85% I think 8.5 cgpa |
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| 4575. |
Find the area of triangle A(-3,4),B(3,0) and C(5,0) |
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Answer» By using the formula of a triangle which is used in ch-7 of ncert book.Moreover, it\'s answere is 4 unitsq. By using the ar of triangle formula |
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| 4576. |
The product of my age two years ago with my age 3 year hence is 50 .find my present age |
| Answer» Your present age will be 7yrs | |
| 4577. |
Find a relation between x and y if the points (x,y),(7,0) are collinear |
| Answer» Google pa sarch karlo | |
| 4578. |
Is Using black and blue gel pen in board exam compulsory? |
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Answer» Gel pen ka koi protocol nahin hai Use good quality ball pen Teachers suggested to only use simple blue pens Mixed reactions! What should i do ? Yes.. No only blue simple pen allowed Be use ball pens in board exam... Noo I think no cuz gel pen can make your condition worse |
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| 4579. |
In an equilateral triangle ABC, is drawn perpendicular vto BC meeting BC in D.prove that AD^=3BD^ |
| Answer» What is your answer | |
| 4580. |
CosA=2/5,then find the value of 4+4tanb*tanb |
| Answer» 25 | |
| 4581. |
If n=a-1 /a +1 and m=a+1 /a-1, m² +n²-3mn is equal to : |
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| 4582. |
Find the sum of 15th term multipel of 8 |
| Answer» 960 | |
| 4583. |
Solve for quadratic equation by factorisation x-a +x-b = a+b. x-b. X-b. b. a |
| Answer» We have,{tex}\\frac{{x - a}}{{x - b}} + \\frac{{x - b}}{{x - a}} = \\frac{a}{b} + \\frac{b}{a}{/tex}{tex}\\Rightarrow \\frac{{(x - a)(x - a) + (x - b)(x - b)}}{{(x - b)(x - a)}} = \\frac{{{a^2} + {b^2}}}{{ab}}{/tex}{tex} \\Rightarrow \\frac{{{x^2} + {a^2} - 2ax + {x^2} + {b^2} - 2bx}}{{{x^2} - bx - ax + ab}}{/tex}{tex} = \\frac{{{a^2} + {b^2}}}{{ab}}{/tex}{tex}\\Rightarrow \\frac{{2{x^2} - 2ax - 2bx + {a^2} + {b^2}}}{{{x^2} - bx - ax + ab}} = \\frac{{{a^2} + {b^2}}}{{ab}}{/tex}{tex}\\Rightarrow{/tex}{tex} (2x^2 - 2ax - 2bx + a^2 + b^2)ab = (a^2 + b^2)(x^2 - bx - ax + ab){/tex}{tex}\\Rightarrow{/tex}{tex}2abx^2 - 2a^2bx - 2ab^2x + a^3b + ab^3 = a^2x^2 - a^2bx -a^3x + a^3b + b^2x^2 - b^3x - ab^2x + ab^3{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}2abx^2 - a^2x^2 - a^2bx - ab^2x + a^3x + b^3x - b^2x^2 = 0{/tex}{tex}\\Rightarrow{/tex}{tex}(2ab - a^2 - b^2)x^2 + (-a^2b - ab^2 + a^3 + b^3)x = 0{/tex}{tex}\\Rightarrow{/tex} x[(2ab - a2 - b2)x + (a3 + b3 - a2b - ab2)] = 0{tex}\\Rightarrow{/tex} x = 0 or (2ab - a2 - b2)x + a3 + b3 -a2b - ab2 = 0Now,(2ab - a2 - b2)x + a3 + b3 - a2b - ab2 = 0{tex}\\Rightarrow{/tex} (2ab - a2 - b2)x = a2b + ab2 - a3 - b3{tex}\\Rightarrow{/tex} -(a2 - b2 - 2ab)x = a2b - b3 + ab2 - a3{tex}\\Rightarrow{/tex} -(a - b)2x = b(a2 - b2) + a(b2 - a2){tex}\\Rightarrow{/tex} (a - b)2x = -b(a2 - b2) - a(b2 - a2){tex} \\Rightarrow x = \\frac{{ - b({a^2} - {b^2}) + a({a^2} - {b^2})}}{{{{(a - b)}^2}}}{/tex}{tex}\\Rightarrow x = \\frac{{\\left( {{a^2} - {b^2}} \\right)(a - b)}}{{{{(a - b)}^2}}}{/tex}{tex} \\Rightarrow x = \\frac{{{a^2} - {b^2}}}{{a - b}} = \\frac{{(a - b)(a + b)}}{{(a - b)}}{/tex}{tex}\\Rightarrow{/tex} x = a + b{tex}\\therefore{/tex}\xa0x = 0 or x = a + b | |
| 4584. |
(a-b)x+(a+b)y=2a×a-2b×b(a+b)(x+y)=4abFind the value of x by cross multiplication method |
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| 4585. |
If Alpha and beta are the roots of the 2x2+5x_4=0 find the value of Alpha 3+ beta 3 |
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| 4586. |
Pythagoras theorem proof |
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Answer» Se* see U can se it in mathematics ncert Not possible here ....but easiest ? |
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| 4587. |
What is diameter of radius ?? Without Google be honest |
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Answer» 2r twice Don\'t know Twice of it ( 2r ) 2r 2r 2r Give ans |
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| 4588. |
2chairs and3 tablescost RS5650,whereasv3chairs and 2tables cost rupee 7,500 |
| Answer» Chair= 2240Table cloth= 390 | |
| 4589. |
find next term.. 1 1 2 6 .... (any genius present?) |
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Answer» So simple... ??? Firstly we multiply one from one then answer came one, then we multiply one from two then answer came two, then we multiply two from three we get six then lastly we multiply six from four and we get final answer 24 Okk I will tell How it is 24 The answer is 24 59 ? Aur solution bhi Yees jaldi se if 24 how.. tell the basic idea behind 24 hi hoga My final ans 24... can I say the answer? 1 answer h ya bas 1 hi Banda sahi h???? Who??? Vrna 25 ydi alag way se dekhe toh question is complete and correct Must be 24 ...?? Multication .. 24 I think question is not complete one is right 1 1 2 7?? Is it 24 or 25 ??.. ab jyada nhi matha lgana chta .. kl english ka hai... no.. wrong 41??? Rahul 31?? Ap nhi mental ability hai.. This is not in A.P. 31?? Are question pura do no 16? |
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| 4590. |
If p(E)= 0.05 what is probability of not E |
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Answer» 1-0.05 19/20 .95 |
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| 4591. |
I have get my roll no. So how can I find center of examination |
| Answer» Skul will tell u.... Ur center | |
| 4592. |
If 2sin²A- cos²A=2 , find the value of A |
| Answer» Value of A is 90degree | |
| 4593. |
1+cot square A/1+cosec squareA = cosecA |
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Answer» 1/sin square A +1 ? |
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| 4594. |
a+b=0, prove that a=-b |
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Answer» Kya mazak h....??? What ?????? Already proved yrr... Already proved !! |
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| 4595. |
Find area of shadded region in AC. =24cm bc=10cm. |
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| 4596. |
Write the following in decimal form and say what kind of decimal expansion each has: 1/11 |
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Answer» 0.09090909090909 Terminating expansion |
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| 4597. |
If 15cot P=8 ,then find sin P and sec P |
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Answer» Check ncert examples from this chapter Take cot P = 8/15 and cot P = b/p so b = 8 and p = 15 then by Pythagoras theorem bsquare + psquare = hsquare hence b and p is given therefore find h.... Then sin P = p/h and sec P = h/b put the value of p, h and h,b respectively and find the solution.. ??? Very easy |
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| 4598. |
√5+√5=? |
| Answer» 2√5 | |
| 4599. |
Solve for x: (4x-3/2x+1)-10(2x+1/4x-3)-3 |
| Answer» Take 4x-3/2x+1 =p ....so 2x+1/4x-3 = 1/p.....hence the eqns are..p-10/p-3 = p2-10-3p....solve it then put the value of p in 4x-3/2x+1 | |
| 4600. |
Cos A =7/25 .Find tan A +cot A |
| Answer» cos A\xa0{tex}=\\frac{7}{25}{/tex}\xa0We know that sin2 A + cos2 A = 1{tex}\\Rightarrow{/tex}\xa0sin2\xa0A = 1 - cos2\xa0A{tex}\\Rightarrow{/tex}\xa0sin2\xa0A = 1\xa0{tex}-\\frac{49}{625}=\\frac{576}{625}{/tex}\xa0sin A\xa0{tex}=\\sqrt{\\frac{576}{625}}=\\frac{24}{25}{/tex}Now tan A\xa0{tex}=\\frac{\\sin A}{\\cos A}=\\frac{24 / 25}{7 / 25}{/tex}{tex}=\\frac{24}{7}{/tex}and cot A\xa0{tex}=\\frac{1}{\\tan A}=\\frac{7}{24}{/tex}{tex}\\therefore{/tex}\xa0tan A + cot A\xa0{tex}=\\frac{24}{7}+\\frac{7}{24}{/tex}{tex}=\\frac{625}{168}{/tex} | |