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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.
| 11701. |
Prove that( -)×(-)=(+) |
| Answer» | |
| 11702. |
Trignometry |
| Answer» Trigonometry\xa0is a branch of mathematics that studies relationships involving lengths and angles of triangles | |
| 11703. |
If the point P (x,y) is equidistant from the point a (5,1) b (1,5 )prove that x_ y |
| Answer» PA = PB (Given){tex}\\therefore {/tex}\xa0PA2 = PB2{tex} \\Rightarrow {/tex}\xa0(5 - x)2 + (1 - y)2 = (1 - x)2 + (5 - y)2{tex} \\Rightarrow {/tex}\xa025 + x2 - 10x + 1 + y2 - 2y = 1 + x2 - 2x + 25 + y2 - 10y{tex} \\Rightarrow {/tex}\xa0-8x = -10y + 2y{tex} \\Rightarrow {/tex}\xa0-8x = -8y{tex} \\Rightarrow {/tex}\xa0x = y\xa0 | |
| 11704. |
x-1/x-2+x-3/x-4=10/3 |
| Answer» The given equation is{tex}\\frac { x - 1 } { x - 2 } + \\frac { x - 3 } { x - 4 } = 3 \\frac { 1 } { 3 } ( x \\neq 2,4 ){/tex}{tex}\\Rightarrow \\frac { ( x - 1 ) ( x - 4 ) + ( x - 3 ) ( x - 2 ) } { ( x - 2 ) ( x - 4 ) } = \\frac { 10 } { 3 }{/tex}{tex}\\Rightarrow \\frac { x ^ { 2 } - 4 x - x + 4 + x ^ { 2 } - 2 x - 3 x + 6 } { x ^ { 2 } - 4 x - 2 x + 8 } = \\frac { 10 } { 3 }{/tex}{tex}\\Rightarrow \\frac { 2 x ^ { 2 } - 10 x + 10 } { x ^ { 2 } - 6 x + 8 } = \\frac { 10 } { 3 }{/tex}{tex}\\Rightarrow 3 \\left( 2 x ^ { 2 } - 10 x + 10 \\right) = 10 \\left( x ^ { 2 } - 6 x + 8 \\right){/tex}{tex}\\Rightarrow 6 x ^ { 2 } - 30 x + 30 = 10 x ^ { 2 } - 60 x + 80{/tex}{tex}\\Rightarrow 4 x ^ { 2 } - 30 x + 50 = 0{/tex}{tex}\\Rightarrow ( 2 x ) ^ { 2 } - 2 ( 2 x ) \\left( \\frac { 15 } { 2 } \\right) + \\left( \\frac { 15 } { 2 } \\right) ^ { 2 } - \\left( \\frac { 15 } { 2 } \\right) ^ { 2 } + 50 = 0{/tex}{tex}\\Rightarrow \\left( 2 x - \\frac { 15 } { 2 } \\right) ^ { 2 } - \\frac { 225 } { 4 } + 50 = 0{/tex}{tex}\\Rightarrow \\left( 2 x - \\frac { 15 } { 2 } \\right) ^ { 2 } - \\frac { 25 } { 4 } = 0{/tex}{tex}\\Rightarrow 2 x - \\frac { 15 } { 2 } = \\pm \\frac { 5 } { 2 } \\Rightarrow 2 x = \\frac { 15 } { 2 } \\pm \\frac { 5 } { 2 }{/tex}{tex}\\Rightarrow 2 x = \\frac { 15 } { 2 } + \\frac { 5 } { 2 } , \\frac { 15 } { 2 } - \\frac { 5 } { 2 }{/tex}{tex}\\Rightarrow 2 x = 10,5 \\Rightarrow x = 5 , \\frac { 5 } { 2 }{/tex}Hence, the solutions of the given equation and 5 and {tex}\\frac { 5 } { 2 }{/tex}. | |
| 11705. |
Solve by factorization 4x2+(a4-b4)+4a2x=1 |
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| 11706. |
How we can write minus one\'s square is one |
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Answer» Hey sister I know that as well. But comparing the herons and mokutik formula how\'s that possible But as per the!(÷€÷6*3-2=0).so what can u say now Because (-1) *(-1)=1,as we all know the property -*-=+. |
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| 11707. |
The sum of two zeros of the polynomial f ( x )=2x² + (p+3)x +5 is zero then the value of P is |
| Answer» – 3 | |
| 11708. |
Find the roots of equation 5xsquare-6x-2 by completing the square method |
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Answer» =(5x2-6x-2)×5=0=25x2-30x-10=0=[(5x)2]-2×[5x×3]+[(3)2]-(3)2-2=0=(5x-3)2-9-2=0=(5x-3)2=11=5x-3=+_√11=1st root=5x-3=+√11 =5x=3+√11 =x=3+√11/5=2nd root=5x-3=_√11 =5x=3-√11 =x=3-√11/5 Where 5x2 is 5x square and in step 3rd it\'s in the form of a2+b2 - 2ab Nahi aata |
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| 11709. |
Sin2a + cos2a = |
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Answer» 1 1 |
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| 11710. |
1÷x-1÷x-2=3, x not equal to 0, 2 find the roots |
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| 11711. |
What is area theorum |
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| 11712. |
sec50×sin40+cos40×cosec50=2 prove it |
| Answer» By LHS=sec50×sin40+cos40×cossec50 =sin90-40×sin90-40+sec90-50×sec90-40=1+1=2 | |
| 11713. |
√512 |
| Answer» √2*2*2*2*2*2*2*2*2=16√2 | |
| 11714. |
Sin2¥+ cos2¥=??? |
| Answer» 1? | |
| 11715. |
My day shedule |
| Answer» | |
| 11716. |
Find value of 2:7 |
| Answer» 0.285 | |
| 11717. |
When the math\'s test paper will ve updated ? |
| Answer» | |
| 11718. |
(2√2)2 |
| Answer» 8 | |
| 11719. |
2+2%3 |
| Answer» 2.67 | |
| 11720. |
If one root of the polynomial 5x square +13x +K is reciprocal of the other find the value of k |
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| 11721. |
Prove that √2+√3 is irrational |
| Answer» | |
| 11722. |
Prove that cube root 6 is irrational |
| Answer» explain whyand are composite numbers1) 7×13×11+13 = 13 ( 7X11 + 1) = 13 ( 77+1) = 13X78 = 13x78x1 as it has more than two factors it is a composite number.2)\xa0(7×6×5×4×2×3×2×1)+3 = 3(\xa07×6×5×4×2×1 +1) = 3x1681x1 as it has more than two factors it is a composite number. | |
| 11723. |
What is least positive integer divisible by 20 and 24 |
| Answer» 480 | |
| 11724. |
explain why 7×13×11+13 and (7×6×5×4×2×3×2×1)+3 are composite numbers |
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Answer» 9.png (482×591)@ https://yho.com/bwqfm1 13(7×1×11)+1So as it is divisible by 13 which is a composite no. So the whole no. Will also be a composite no.U can do 2nd one urself Because they have factors rather than 1 |
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| 11725. |
In triangle ABC angle C=3\'"angle B=2 (angle A+angle B),find the 3 angles? |
| Answer» NCERT question | |
| 11726. |
Cot theta+cosec theta-1/cot theta-cosec theta+1 |
| Answer» (cotA+cosecA-1)/(cotA-cosecA+1)=(cotA+cosecA-1)²/(cotA+cosecA-1)(cotA-cosecA+1)=(Cot²A+cosec²A+1+2cotAcosecA-2cotA-2cosecA)/(cot²A-cosec²A-1+2cosecA)={2cosec²A-2cosecA+2cotA(cosecA-1)}/(2cosecA-2)={2cosecA(cosecA-1)+2cotA(cosecA-1)}/2(cosecA-1)=2(cosecA-1)(cosecA+cotA)/2(cosecA-1)=cosecA+cotA | |
| 11727. |
Find all the odd number between 30000 and 80000 |
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Answer» Urself* There are 2500odd no. b/w 3000&8000Find them all itself:p |
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| 11728. |
How to divide point numerical |
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| 11729. |
How to find missing no. Of traingle in ntse question chapter 4, 8 question |
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| 11730. |
What is general form ki of linear in two variables |
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Answer» the general form of an equation of a straight line is ax + by +c = 0a or b can be zero , but not both at the same time ax +by+c=0 |
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| 11731. |
Give the proof of quadratic formula |
| Answer» {tex}ax^2+bx+c=0{/tex}or,\xa0{tex}x^2+\\frac{{b}}{a}x+\\frac{{c}}{a}=0{/tex}or,\xa0{tex}x^2+\\frac{{b}}{a}x=−\\frac{{c}}{a}{/tex}or,\xa0{tex}x^2+\\frac{{b}}{a}x+\\frac{{b^2}}{4a^2}=\\frac{{b^2}}{4a^2}−\\frac{{c}}{a}{/tex}or,\xa0{tex}(x+\\frac{{b}}{2a})^2=\\frac{{b^2−4ac}}{4a^2}{/tex}or,\xa0{tex}x+\\frac{{b}}{2a}=\\frac{\\pm\\sqrt{b^2−4ac}}{2a}{/tex}or,\xa0{tex}x=-\\frac{{b}}{2a}\\pm\\frac{\\sqrt{b^2−4ac}}{2a}{/tex}or,\xa0{tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex}\xa0 | |
| 11732. |
Find two consecutive multiple of 3 whose product is 483 |
| Answer» Let the consecutive positive odd integers be x and (x + 2).According to question,{tex}x(x+2)=483{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2+2x-483=0{/tex}Factorise the equation{tex}\\Rightarrow{/tex}\xa0x2\xa0+ 23x - 21x - 483 = 0{tex}\\Rightarrow{/tex}\xa0x(x + 23) - 21(x + 23) = 0{tex}\\Rightarrow{/tex}\xa0x + 23 = 0 or x - 21 = 0{tex}\\Rightarrow{/tex}\xa0x = -23 or x = 21As, x is a positive integer, {tex}{/tex}{tex}\\Rightarrow{/tex}\xa0x = 21{tex}\\Rightarrow{/tex}\xa0x + 2 = 21 + 2 = 23Therefore, the consecutive positive odd integer are 21 and 23. | |
| 11733. |
Which book should be followed for better results ij board xams |
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Answer» Golden guide, xam idea S. Chand, all in one, pradeep ncert textbook Ncert book |
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| 11734. |
Cos 90 |
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Answer» Value of Cos 90=0 Cos 90=0 |
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| 11735. |
Can we write the distance formula like this √(x1-x2) + (y1-y2) ?in solving the problems |
| Answer» Yes you can solving the problems by this distance formula | |
| 11736. |
X + 3 upon X + 2 minus one upon X upon X is equals to 17 upon x |
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| 11737. |
If the sum of first 7 terms aof an A.P. is 10 and sum of next 7 terms is 17 THEN find the A.P. |
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Answer» Let first term of AP = aCommon difference of AP = dSum of first 7 terms = 10So,\xa0{tex}10={7\\over 2}[2a+6d]\\\\=> 20 = 14a+42d \\ .......(1){/tex}Sum of next 7 terms = sum of 14 terms - sum of first 7 terms{tex}17 = {14\\over 2}[2a+13d]-10\\\\=> 27 = 14a+91d\\ \\ .....(2){/tex}Subtract (1) from (2), we get=> 7 = 49d=> d =\xa0{tex}1\\over 7{/tex}Put value of d in (1), we get=>\xa0{tex}20 = 14a + 6\\\\=> 14a = 14\\\\=> a = 1{/tex}AP :\xa0{tex}1, {8\\over 7},{15\\over 7},{22\\over 7},.....{/tex}\xa0\xa0 sum of first 7 terms = 10sum of next 7 terms = 17sum of first 14 terms = (10 + 17) = 27let first term = a and common difference = dthen S7\xa0= n/2(2a + (n = 1)d 10 = 7/2(2a + (7 - 1)d ............. 10 = 7a + 21d ........ ialso S14\xa0= 14/2 (2a + (14 - 1)d ...........27 = 14a + 91d ...... ii from i and ii we get 7 = 49d ,........ d = 1/710\xa0= 7a + 3 ................................... a = 1Thus the AP is defined by a , a+d , a+2d, ................. 1 , 8/7 , 9/7, ...............\xa0 |
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| 11738. |
6sin23+sec79+3tan48-----------------------------------------Cosec11+3cot42+6cos67 evaluate |
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| 11739. |
LCM of 255 and 625 by euclid leman devision |
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| 11740. |
m/n x2 + n/m = 1- 2x, solve by factorization method? |
| Answer» According to the question,{tex}\\frac{m}{n}{x^2} + \\frac{n}{m} = 1 - 2x{/tex}{tex}\\Rightarrow \\frac{m}{n}{x^2} + 2x + \\frac{n}{m} - 1 = 0{/tex}{tex} \\Rightarrow {x^2} + \\frac{{2nx}}{m} + \\frac{{{n^2}}}{{{m^2}}} - \\frac{n}{m} = 0{/tex} [multiplying both sides by \'n\' and dividing both sides by \'m\']{tex}\\Rightarrow {x^2} + \\frac{{2nx}}{m} + \\frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}To factorize {tex}{x^2} + \\frac{{2nx}}{m} + \\frac{{{n^2} - mn}}{{{m^2}}}{/tex}, we have to find two numbers {tex}\'a\'\\ and\\ \'b\'{/tex} such that.{tex}a + b = \\frac{{2n}}{m}{/tex} and {tex}a b = \\frac { n ^ { 2 } - m n } { m ^ { 2 } }{/tex}Clearly, {tex}\\frac{{n + \\sqrt {mn} }}{m} + \\frac{{n - \\sqrt {mn} }}{m} = \\frac{{2n}}{m}{/tex} and {tex}\\frac{{\\left( {n + \\sqrt {mn} } \\right)}}{m}\\times\\frac{{\\left( {n - \\sqrt {mn} } \\right)}}{m} = \\frac{{{n^2} - mn}}{{{m^2}}}{/tex} ({tex}\\therefore a = \\frac{{n + \\sqrt {mn} }}{m}{/tex} and {tex}b = \\frac{{n - \\sqrt {mn} }}{m}{/tex}){tex}\\Rightarrow {x^2} + \\frac{{2nx}}{m} + \\frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}{tex} \\Rightarrow {x^2} + \\frac{{\\left( {n + \\sqrt {mn} } \\right)}}{m}x + \\frac{{\\left( {n - \\sqrt {mn} } \\right)}}{m}x{/tex}{tex} + \\frac{{{n^2} - mn}}{{{m^2}}} = 0{/tex}{tex} \\Rightarrow x\\left[ {x + \\frac{{n + \\sqrt {mn} }}{m}} \\right] + \\frac{{n - \\sqrt {mn} }}{m}{/tex}{tex}\\left[ {x + \\frac{{n + \\sqrt {mn} }}{m}} \\right] = 0{/tex}{tex} \\Rightarrow \\left( {x + \\frac{{n - \\sqrt {mn} }}{m}} \\right)\\left( {x + \\frac{{n + \\sqrt {mn} }}{m}} \\right) = 0{/tex}{tex} \\Rightarrow x + \\frac{{n - \\sqrt {mn} }}{m} = 0{/tex} or {tex}x + \\frac{{n + \\sqrt {mn} }}{m} = 0{/tex}{tex} \\Rightarrow x = \\frac{{- n - \\sqrt {mn} }}{m}{/tex} or {tex}x = \\frac{{ - n + \\sqrt {mn} }}{m}{/tex} | |
| 11741. |
solve for x and y: bx + ay =a+b. |
| Answer» ax + by = a - b multiply by abx - ay = a + b multiply by b{tex}\\Rightarrow{/tex} x = 1{tex}\\therefore{/tex} a + by = a - bby = -by = -1{tex}\\therefore{/tex} x = 1y = -1 | |
| 11742. |
If angle b and angle q are acute angles such that sinB=sinQ,then prove that angle b =angle 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. | |
| 11743. |
ABCD is a quadrilateral in which {A + D} = 90°. Then prove that 2 2. 2. 2 AC. + BD. = AD + BC |
| Answer» Given: In quadrilateral ABCD,{tex}\\angle A + \\angle D = 90 ^ { \\circ }{/tex}AC and BD are joinedTo prove: AC2 + BD2 = AD2 + BC2Construction: Produce AB and DC to meet at P.Proof: In\xa0{tex}\\triangle{/tex}APD,{tex}\\angle A + \\angle D = 90 ^ { \\circ }{/tex}(given){tex}\\therefore \\angle P = 90 ^ { \\circ }{/tex}\xa0{tex}\\left( \\because \\angle A + \\angle P + \\angle D = 180 ^ { \\circ } \\right){/tex}Now in right\xa0{tex}\\triangle A C P , \\angle A P D = 90 ^ { \\circ }{/tex}AC2 = PA2 + PC2 ....(i)(Pythagoras Theorem)and in\xa0{tex}\\triangle{/tex}BPDBD2 = PB2 + PD2Adding (i) and (ii)AC2 + BD2 = PA2 + PC2 + PB2 + PD2= (PA2 + PD2) + (PC2 + PB2)= AD2 + BD2({tex}\\therefore{/tex}\xa0In right\xa0{tex}\\triangle{/tex}APD, PA2 + PD2 = AD2 and similarly PC2 + PB2 = BD2) | |
| 11744. |
sin |
| Answer» | |
| 11745. |
Sec theta (1-sin theta)(sec theta+tan theta)=1 |
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| 11746. |
Solve 2sec²A-sec⁴A-2cosec²A+cosec⁴A=cot⁴A-tan⁴A |
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| 11747. |
cosA-cos3A/sin3A+sinA=tanA |
| Answer» ok | |
| 11748. |
If sin square theeta +3coss quare theeta=4 show that tan theeta =1÷√3 |
| Answer» | |
| 11749. |
1 term syllabus |
| Answer» Different schools go for different syllabus for first term after the end of CCE pattern. | |
| 11750. |
Sum of integers from1 to100 which are not divisible by 3or5 is1)24892)47353)23174)2632 |
| Answer» Sum of the integers 1 to 100 =100(2+99×1)÷2=5050Sum of Integers which are divisible by 3 is=33(6+32×3)÷2=33×51=1683\xa0Sum of Integers which are divisible by 5 is=20(10+19×5)÷2=10×105=1050Sum of Integers which are divisible by 15 is=6(30+5×15)÷2=3×105=315Therefore, the number is,5050-1683-1050+315=4000-1368=2632So the answer is (4) 2632. | |