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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.
| 32801. |
Yaar maths is very hard |
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Answer» Maths is interesting but sometimes boring No this is hard but interesting How can it be a piece of cake Alia.Maths is very interesting , it dont let us free to be bored. Yoú will alwaya enjoy with maths instead of gêtting bored.Its my opinion☝. Sometimes maths becomes interesting And sometimes it becomes boaring Maths is very interesting It\'s just ur own opinion.!!!! Its a piece of cake. |
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| 32802. |
Sin18 |
| Answer» Sin 18 | |
| 32803. |
2+5(-54+98) |
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Answer» 222 222 |
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| 32804. |
Is there any trick for splitting the middle term.Please tell if any |
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Answer» Or if u have any problem then use shri dharacharya method Ya , firstly multiply the coefficient of first and third term and split the middle term as the factors of it ( if you add or subtract then it will be the middle term ) |
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| 32805. |
If in an AP , tp=q and tq=p, find tp+q. |
| Answer» q = a + (p - 1)d ..... (i)p = a + (q - 1)d ...... (ii)q - p = (p - 1 - q + 1)d{tex}\\frac{{q - p}}{{p - q}} = d{/tex}{tex} \\Rightarrow d = - 1{/tex}Put the value of d in eq (i)q = a + (p - 1) (-1){tex} \\Rightarrow {/tex}\xa0q = a - p + 1{tex} \\Rightarrow {/tex}\xa0a = q + p - 1ap+q = a +(p + q - 1)d= (q + p - 1) + (p + q - 1) (-1)= q + p - 1 - p - q + 1= 0 | |
| 32806. |
CosA-sinA+1/cosA+sinA-1 |
| Answer» 1/secA-tanA | |
| 32807. |
Write the number of real roots of the equation x^2 +3|x| +2=0 |
| Answer» 2 | |
| 32808. |
4xsquare -4ax +(asquare - bsquare)=0 |
| Answer» According to the question,\xa0{tex}4 x^{2}-4 a x+\\left(a^{2}-b^{2}\\right)=0{/tex}{tex}\\Rightarrow x^{2}-a x+\\left(\\frac{a^{2}-b^{2}}{4}\\right)=0{/tex}{tex}\\Rightarrow x^{2}-2\\left(\\frac{a}{2}\\right) x=-\\left(\\frac{a^{2}-b^{2}}{4}\\right){/tex}{tex}\\Rightarrow x^{2}-2\\left(\\frac{a}{2}\\right) x+\\left(\\frac{a}{2}\\right)^{2}{/tex}=\xa0{tex}-\\left(\\frac{a^{2}-b^{2}}{4}\\right)+\\left(\\frac{a}{2}\\right)^{2}{/tex}{tex}\\Rightarrow\\left(x-\\frac{a}{2}\\right)^{2}{/tex}\xa0=\xa0{tex}-\\frac{a^{2}}{4}+\\frac{b^{2}}{4}+\\frac{a^{2}}{4}{/tex}{tex}\\Rightarrow\\left(x-\\frac{a}{2}\\right)^{2}=\\frac{b^{2}}{4}{/tex}{tex}\\Rightarrow x-\\frac{a}{2}=\\pm \\frac{b}{2}{/tex}\xa0{tex}\\therefore{/tex}\xa0x =\xa0{tex}\\frac{a+b}{2}{/tex},\xa0{tex}\\frac{a-b}{2}{/tex} | |
| 32809. |
which term of the AP is 7,11,14 is 121 |
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Answer» sumit bro 11-7 is 4 not 3 !2! is not a term of this AP i think a =7d =a" - a\'= 11 - 7 = 3 an = 121an =a+(n-1)d121=7+(n-1)3121-7=(n-1)3114=(n-1)3114/3=(n-1)36=n-1n=36+1n=37 |
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| 32810. |
Solve a^2x^2-3abx+2b^2 by completing square |
| Answer» The given equation is:a2x2\xa0- 3abx + 2b2 = 0{tex}\\Rightarrow{/tex}\xa0a2x2\xa0- 3abx = -2b2Adding\xa0{tex}\\left( \\frac { 3 b } { 2 } \\right) ^ { 2 }{/tex} on both sides, we have,{tex}(ax)^2-2\\times(ax)\\times\\frac{3b}2+\\left(\\frac{3b}2\\right)^2=-2b^2+\\left(\\frac{3b}2\\right)^2{/tex}{tex}(ax-\\frac{3b}2)^2=-2b^2+(\\frac{3b}2)^2{/tex}{tex}(ax-\\frac{3b}2)^2=\\frac{b^2}4{/tex}\xa0{tex}\\Rightarrow \\quad \\left( a x - \\frac { 3 b } { 2 } \\right) = \\pm \\frac { b } { 2 }{/tex}\xa0(By taking square root on both sides)So either,\xa0{tex} \\left( a x - \\frac { 3 b } { 2 } \\right) = \\frac { b } { 2 }{/tex}\xa0or\xa0{tex}\\left( a x - \\frac { 3 b } { 2 } \\right) = \\frac { - b } { 2 }{/tex}{tex}\\Rightarrow \\quad a x = \\left( \\frac { b } { 2 } + \\frac { 3 b } { 2 } \\right) = \\frac { 4 b } { 2 }{/tex}= 2b,thus,\xa0x = 2b/aor ax =\xa0{tex}\\left( \\frac { - b } { 2 } + \\frac { 3 b } { 2 } \\right) = \\frac { 2 b } { 2 }{/tex}\xa0= bthus, x = b/aSo,\xa0x =\xa0{tex}\\frac { 2 b } { a }{/tex}\xa0or x =\xa0{tex}\\frac { b } { a }{/tex}.Hence,\xa0the roots of the given equation are\xa0{tex}\\frac { 2 b } { a }{/tex}\xa0and\xa0{tex}\\frac { b } { a }{/tex}. | |
| 32811. |
What is the formula of distance in coordinate geometry |
| Answer» √ (X2-X1)^2+(Y2-Y1)^2 | |
| 32812. |
The pole of the straight line 9x+y-28=0 w.r.t the circle x^2+y^2=16 will be |
| Answer» | |
| 32813. |
If m and n are the zeros of the polynomial 3x^2+11x-4,find the value of m/n+n/m. |
| Answer» According to the question,\xa0m and n are the zeroes of the polynomial 3x2 + 11x - 4,\xa0Let p(x) = 3x2 + 11x - 4 = 0= 3x2 + 12x - x - 4= 3x(x + 4) - 1(x + 4)= (3x - 1) (x + 4)So, zeroes are m =\xa0{tex}\\frac 13{/tex}\xa0and n =-4Now,\xa0{tex}\\frac { m } { n } + \\frac { n } { m } = \\frac { \\left( \\frac { 1 } { 3 } \\right) } { - 4 } + \\frac { - 4 } { \\left( \\frac { 1 } { 3 } \\right) }{/tex}{tex}=\\frac { 1 } { - 12 } - 12{/tex}{tex}= \\frac { - 145 } { 12 }{/tex} | |
| 32814. |
222 |
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Answer» Sayad yeh puzzle suljhana padega ? Write properly please What??? |
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| 32815. |
√2=what |
| Answer» Non repeating and terminating decimal | |
| 32816. |
We have to RD for board exam or NCERT book for board exam |
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Answer» Only NCERT don\'t waste your time on another books bro Ncert and rs agrawal best for study in board exam best book Ncert book are best for study. |
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| 32817. |
3x+2x=54x+5x=9 |
| Answer» | |
| 32818. |
In maths board paper we have to solve audio only NCERT |
| Answer» | |
| 32819. |
Square root sum how we can find greater or less |
| Answer» | |
| 32820. |
Prove that 10165-287√99 is irrational. |
| Answer» | |
| 32821. |
Where i can search board last three years paper? |
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Answer» U can also take notes from here it is very useful website for board students Notes is completely based on ncert book and Very short notes is here. Www.tiwariacademy.com |
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| 32822. |
For what value of k, -4 is a zero of the polynomial x to the power 2 -x-(2k+2) |
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Answer» Easy ans for this question 2k=18k=9 x^2-x-(2k+2)Substitute (-4) instead of x because (-4) is one of the zeros of the polynomial .Then,(-4)^2-(-4)-(2k+2)=16+4-2k-2=18-2k |
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| 32823. |
justify 2 tan 30degree divide 1+ tan square 30 degree |
| Answer» sin 60° | |
| 32824. |
Cot thitha = 7/9 |
| Answer» | |
| 32825. |
Prove how Herons formula comes? |
| Answer» | |
| 32826. |
What is prime no. |
| Answer» Prime number are those no. which is divisible by 1& it self. | |
| 32827. |
This month unit test paper format |
| Answer» No idea | |
| 32828. |
(a2-b2)sin $ + 2ab cos $ = a2+b2Then show that tan $= a2+b2/2ab |
| Answer» | |
| 32829. |
The 6th term of an ap is -10 and its 10th term is -26. Determine the 15th term of the AP |
| Answer» -46 is 15th term | |
| 32830. |
If mth tetm of AP is 1 by n and th term of an AP is 1 bym.show that nmth term is 1 |
| Answer» Let a be the first term and d be the common difference of the given AP. Now, we know that in general mth and nth terms of the given A.P can be written asTm = a + (m-l)d and Tn =\xa0a + (n-1)d respectively.Now, Tm = {tex}\\frac{1}{n}{/tex}\xa0and Tn = {tex}\\frac{1}{m}{/tex}\xa0(given).{tex}\\therefore{/tex}\xa0a + (m-1)d = {tex}\\frac{1}{n}{/tex}\xa0......................(i)and a + (n-1)d ={tex}\\frac{1}{m}{/tex}................ ... (ii)On subtracting (ii) from (i), we get(m-n)d = ({tex}\\frac{1}{n}{/tex}-{tex}\\frac{1}{m}{/tex}) = ({tex}\\frac{{m - n}}{{mn}}{/tex}) {tex}\\Rightarrow{/tex}\xa0d =\xa0{tex}\\frac{1}{{mn}}{/tex}Putting d ={tex}\\frac{1}{{mn}}{/tex}\xa0in (i), we geta +\xa0{tex}\\frac{{(m - 1)}}{{mn}}{/tex}\xa0{tex} \\Rightarrow {/tex}\xa0a = {{tex}\\frac{1}{n}{/tex}-{tex}\\frac{{(m - 1)}}{{mn}}{/tex}} =\xa0{tex}\\frac{1}{{mn}}{/tex}Thus, a={tex}\\frac{1}{{mn}}{/tex}\xa0and d={tex}\\frac{1}{{mn}}{/tex}{tex}\\therefore{/tex}Now, in general (mn)th term can be written as Tmn\xa0= a +(mn-1)d= {{tex}\\frac{1}{{mn}}{/tex}+{tex}\\frac{{(mn - 1)}}{{mn}}{/tex}} [{tex}\\because {/tex}a={tex}\\frac{1}{{mn}}{/tex}]= 1.Hence, the (mn)th term of the given AP is 1. | |
| 32831. |
find a cubic polynomial whose zeros are 2,-3and 4 |
| Answer» Suppose {tex}\\text{α,β and γ}{/tex} are the zeros of the said polynomial p(x)Then, we have\xa0{tex}\\alpha{/tex}\xa0= 2,\xa0{tex}\\beta{/tex}\xa0= -3 and\xa0{tex}\\gamma{/tex}\xa0= 4Now,{tex}\\text{α+β +γ=2-3+4=3}{/tex}......(1){tex}\\text{αβ +βγ+γα=2(-3)+(-3)(4)+(4)(2)=-6-12+8=-10......(.2)}{/tex}{tex}\\text{αβγ=2(-3)(4)=-24 .......(3)}{/tex}Now, a cubic polynomial whose zeros are\xa0{tex}\\alpha, \\space \\beta \\space and \\space \\gamma{/tex}\xa0is given byp(x) = x3\xa0-\xa0{tex}\\text{(α+β+γ)x}^2{/tex}\xa0+\xa0{tex}\\text{(αβ+βγ+γα)x}{/tex}\xa0-\xa0{tex}\\alpha\\beta\\gamma{/tex}Now putting the values from (1),(2) and (3) we getp(x) = x3\xa0-(3)x2\xa0+ (-10)x - (-24)= x3\xa0-3x2 -10x +24 | |
| 32832. |
Find the roots of the equation x²+x -P(p+1)=0 where p is a constant |
| Answer» Thanks Shivani ? | |
| 32833. |
If x=asec teta+btan teta and y,=atan teta +bsec teta,prove that x2-y2 |
| Answer» | |
| 32834. |
on which day in august class 10 compartment result wiill be declared |
| Answer» Don\'t know brhtr!!!? | |
| 32835. |
Who all like mathematics |
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Answer» I love mathematics but i hate biology But I hate s.st I dont like mathematic Ya I like maths ???? |
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| 32836. |
If n is an odd positive integer show that n^2-1 is divisible by 8 |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 32837. |
Why is trignometry so difficult?????? |
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Answer» No its not difficult there r many chapters yaar which is more difficult than trigonometry That\'s your wish that is difficult or easy Hey...its not so difficult .Just try to understand the consepct It depends on you how you try to penetrate it |
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| 32838. |
Why the number of zeroes is 3 |
| Answer» The maximum number of zeroes that a polynomial of degree 3 can have is three | |
| 32839. |
How to find length of hypo when only one side and one angle is given |
| Answer» Find the third angle as two angles are given. Then use sin theata or cos theata and find hypotenyse | |
| 32840. |
RD Sharma Chapter 2 Exercise2A |
| Answer» | |
| 32841. |
Write all the formula used in mathmatics of class 10 |
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Answer» Stastics Which chapter |
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| 32842. |
I want to sow maths solution in hindi |
| Answer» What ?????????? | |
| 32843. |
What is the HCF of smallest prime number and smallest composite number in how to solve |
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Answer» 2 The smallest pri.e number is 2.And smallest composite number is 4.So the HCF of 2 and 4 is 2. |
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| 32844. |
Trignometry ratios |
| Answer» Its in maths book | |
| 32845. |
if one root of 2x^2+kx+1=0 is -1/2 then find the value of k. |
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Answer» K=3. 2tsbaisnw |
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| 32846. |
Who need my th12 ? meet me in coc. Take my th9 code y8cvo9qjr |
| Answer» | |
| 32847. |
Prove root 5 is an irrational number |
| Answer» Rd sharma has same q | |
| 32848. |
Maurya ji give ans of my ques |
| Answer» | |
| 32849. |
Prove that 0=2 |
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Answer» Then.....0=10+10/10+100=20/200=1 0=100-100/100-1000=10^2-10^2/10^2-10^20=(10+10)(10-10)/(10+10)(10-10) |
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| 32850. |
2a - 2d =152a+ 9d =25 |
| Answer» -10/11 | |