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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.
| 10701. |
Find the value of Cos 70/sin20+cosec20/sec70-2tan30tan60 |
| Answer» 0 | |
| 10702. |
4sinA+4cosA |
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| 10703. |
Is there any easy method for solving completing square method? |
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| 10704. |
Dgh |
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| 10705. |
1/sinA+1/1-sinA = 2sec²A |
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| 10706. |
What is the value of :√3/sin 10°-1/cos 10°. |
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| 10707. |
If a+b+c=1 a^2+b^2+c^2=2 a^3+b^3+c^3=3 Then find the value of a^4+b^4+c^4. |
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| 10708. |
Exercise 5.3 question 16 |
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| 10709. |
Solve the following quadratic equations by factorization.: (1) ( X - 4) (X+2)= 0 |
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Answer» X-4=0. X+2=0X=4. X=-2 Let, (x-4)(x+2)(1)=0X²+2x-4x-8=0X²-2x-8=0X²-(4-2)x-8=0X²-4x+2x-8=0X(x-4)+2(x-4)=0(X+2)(x-4)=0X+2=0. X-4=0X=-2. X=4 |
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| 10710. |
How to use trigonometry identities? I really confuse? |
| Answer» When the quetion has specific 0,30,45,60,90. | |
| 10711. |
What is the value of cos 30 degree |
| Answer» root 3 by 2 | |
| 10712. |
What is the value of sin 30 |
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Answer» 1/2 root 1/2 1/2 |
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| 10713. |
Thals theoram |
| Answer» The ratio of any two corresponding sides in two equangular triangles is always same | |
| 10714. |
What type of question paper pattern will come? NCERT or CBSE made questions |
| Answer» Helen character | |
| 10715. |
Reasons to contrust Ex11.1 question 2 |
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| 10716. |
Easy form to find x valu |
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| 10717. |
Which easy methods to find x value |
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| 10718. |
(sin⁴ theta-cos4 theta) cosec2 theta=2 |
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| 10719. |
Show that any positive odd integer is of the form 6q+1,or 6q+3, or 6q+5, where q is some integer |
| Answer» Let a be any positive integer and b = 6∴ by Euclid’s division lemmaa = bq + r, 0≤ r and q be any integer q ≥ 0∴ a = 6q + r,where, r = 0, 1, 2, 3, 4, 5If a is even then then remainder by division of 6 is 0,2 or 4Hence r=0,2,or 4or A is of form 6q,6q+2,6q+4As, a = 6q = 2(3q), ora = 6q + 2 = 2(3q + 1), ora = 6q + 4 = 2(3q + 2).If these 3 cases a is an even integer.but if the remainder is 1,3 or 5 then r=1,3 or 5or A is of form 6q+1,,6q+3 or,6q+5Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,\xa0Case 2: a = 6q + 3 = 6q + 2 + 1,= 2(3q + 1) + 1 = 2n + 1,\xa0Case 3: a = 6q + 5 = 6q + 4 + 1= 2(3q + 2) + 1 = 2n + 1This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. | |
| 10720. |
If m th term of an AP is 1/n and n th term is 1/m. Show that the sum of the mn terms is 1/2 (mn+1) |
| Answer» Given {tex}m^{th}term=\\frac{1}{n}{/tex}\xa0i.e. {tex}{/tex}\xa0{tex}a_m=\\frac{1}{n}{/tex}{tex}\\Rightarrow (a+(m-1)d)=\\frac{1}{n}{/tex} ......(1)and{tex}n^{th}term=\\frac{1}{m}{/tex}i.e. {tex}a_n=\\frac{1}{m}{/tex}{tex}\\Rightarrow a+(n-1)d=\\frac{1}{m}{/tex}.....(2)subtracting Eq(1) from\xa0Eq(2) , we get\xa0{tex}\\Rightarrow[ a+(n-1)d] -[(a+(m-1)d)] =\\frac{1}{m} - \\frac{1}{n}{/tex}{tex}\\Rightarrow a+(n-1)d -a+(m-1)d =\\frac{n - m}{mn} {/tex}{tex}\\Rightarrow nd- d -md+d =\\frac{n - m}{mn} {/tex}{tex}\\Rightarrow (n -m)d =\\frac{n - m}{mn} {/tex}{tex}d=\\frac{1}{mn}{/tex}\xa0and {tex}a=\\frac{1}{mn}{/tex}Now, sum of mn terms,{tex}{S_{mn}} = \\frac{{mn}}{2}[2a + (mn - 1)d]{/tex}{tex}\\Rightarrow[{S_{mn}} = \\frac{{mn}}{2}[\\frac{2}{{mn}} + (mn - 1)\\frac{1}{{mn}}]{/tex}={tex}\\frac{1}{2}(mn+1){/tex}\xa0Hence Proved. | |
| 10721. |
Find the probability of getting 53 Sundays in (a) leap year , (b) a non leap year |
| Answer» (a)2/53 (b)1/53 | |
| 10722. |
Sin theta |
| Answer» Sin$=p/h | |
| 10723. |
10_1000 |
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| 10724. |
An |
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| 10725. |
Ram has 2 pens and 3eraser how many total thing has Ram? |
| Answer» 5 | |
| 10726. |
45 |
| Answer» | |
| 10727. |
Q. 3 If the sum of zeroes of quadratic polynomial 3x square - Kx +6 is 3, then fine the value of k |
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Answer» Solve it in proper manner.. 9 Can you please solve it properly 8.48 |
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| 10728. |
Sec A +cosec A -tan A=??? |
| Answer» Answer is sin A Cos A | |
| 10729. |
What is 4q+1 and 4q |
| Answer» Let a = 4q + r, when r = 0, 1, 2 and 3{tex}\\therefore{/tex}Numbers are 4q, 4q + 1, 4q + 2 and 4q + 3{tex}( a ) ^ { 2 } = ( 4 q ) ^ { 2 } = 16 q ^ { 2 } = 4 ( 4 q ) ^ { 2 } = 4 m{/tex}{tex}( a ) ^ { 2 } = ( 4 q + 1 ) ^ { 2 } = 16 q ^ { 2 } + 8 q + 1 = 4 \\left( 4 q ^ { 2 } + 2 q \\right) + 1 = 4 m + 1{/tex}{tex}( a ) ^ { 2 } = ( 4 q + 2 ) ^ { 2 } = 16 q ^ { 2 } + 16 q + 4 = 4 \\left( 4 q ^ { 2 } + 4 q + 1 \\right) = 4 m{/tex}{tex}( a ) ^ { 2 } = ( 4 q + 3 ) ^ { 2 } = 16 q ^ { 2 } + 24 q + 9 = 4 \\left( 4 q ^ { 2 } + 6 q + 2 \\right) + 1 = 4 m + 1{/tex}{tex}\\therefore {/tex}\xa0the square of any +ve integer is of the form 4q or 4q + 1\xa0 | |
| 10730. |
Show that the points (1,1),(4,4),(4,8),(1,5) form a parallelogram |
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| 10731. |
Please send chapter wise marking system of new case syllabus 2018 |
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| 10732. |
Find roots of equation |
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| 10733. |
In board which type of question ask?hard or easy.. |
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Answer» They are easy but they will change the questions and answer will be the same. Don\'t worry, it will be easy if you study |
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| 10734. |
880 is in root than what comes out |
| Answer» 4√55 | |
| 10735. |
Find the 5th term the end of the A.P 17, 14, 11....., -40 |
| Answer» Here, a= 17 and d=(14-17)=-3 Since, a+(n-1)d17+(5-1)×-3=17+(4)×-3=17-12=5Therefore,5th term is 5 | |
| 10736. |
Sec a +Tan a _1/Tan a _Sec a +1=Cos a /1+Sin a |
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| 10737. |
What are the composite number |
| Answer» Composite Number:A whole number that can be divided evenly by numbers other than 1 or itself. Example: 9 can be divided evenly by 3 (as well as 1 and 9), so 9 is a composite number. But 7 cannot be divided evenly (except by 1 and 7), so is NOT a composite number (it is a prime number). | |
| 10738. |
solve 3x + y=3 and 9x-3y=9. |
| Answer» 3x - y = 39x - 3y = 9The given pair of linear equations is\xa03x - y = 3..............(1)9x - 3y = 9.............(2)From equation(1),\xa0y = 3x - 3...................(3)9x - 3(3x - 3) = 9{tex}\\Rightarrow{/tex}\xa09x - 9x + 9 = 9{tex}\\Rightarrow{/tex}\xa09 = 9which is true. Therefore, equation (1) and (2) have infinitely many solutions. | |
| 10739. |
Find the root of the following quardtic equations by the method of completing square 2x²-7x+3=0 |
| Answer» We have 2x2 - 7x + 3 = 0{tex}\\implies2( x^2 - {7 \\over 2}x + {3\\over 2}) = 0{/tex}{tex}\\implies\u200b\u200b x^2 - {7 \\over 2}x + {49 \\over 16} = {-3 \\over 2} +{ 49 \\over 16}{/tex} (Adding 49/16 to both sides){tex}\\implies x^2 -2 \\times x \\times {7 \\over 4} + ({7 \\over 4})^2 = {-24 +49 \\over 16}{/tex}{tex}\\implies (x-{7\\over4})^2 = {25 \\over 16}{/tex}{tex}\\implies x-{7\\over 4}= \\pm \\sqrt({25 \\over 16}){/tex}{tex}\\implies x={7\\over 4} \\pm {5 \\over 4}{/tex}{tex}\\implies x={7\\over 4} + {5 \\over 4}\\, and \\,x={7\\over 4} - {5 \\over 4}{/tex}{tex}\\implies x=3\\, and \\,{1\\over 2}{/tex}{tex}\\therefore{/tex}the roots of the given equation are {tex}3{/tex} and {tex}1\\over 2{/tex}. | |
| 10740. |
What is Bpt teorm |
| Answer» It state that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points the other 2 sides are divided in the same ratio | |
| 10741. |
Using Euclid division algorithm find hcf of135 and 225 |
| Answer» On applying the division lemma to 225 and 135\xa0We get225=135×1+90135=90×1+4590=45×2+0Hence HCF(225,135)=45 | |
| 10742. |
If sinA+cosA=x then prove that sin6A+cos6A= 1+(x2-1)2/4 |
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| 10743. |
Solve tsquare +underoot15 with quadatic formula |
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| 10744. |
find the value of p when the distance between the A(3,p) B(4,1)is underroot 10 |
| Answer» We have P(3, a) and Q(4,1)Here,\xa0x1 = 3, y1 = ax2 = 4, y2 = 1PQ =\xa0{tex}\\sqrt { 10 }{/tex}{tex}P Q = \\sqrt { \\left( x _ { 2 } - x _ { 1 } \\right) ^ { 2 } + \\left( y _ { 2 } - y _ { 1 } \\right) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\sqrt { 10 } = \\sqrt { ( 4 - 3 ) ^ { 2 } + ( 1 - a ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\sqrt { 10 } = \\sqrt { ( 1 ) ^ { 2 } + ( 1 - a ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\sqrt { 10 } = \\sqrt { 1 + 1 + a ^ { 2 } - 2 a }{/tex}\xa0{tex}\\Rightarrow \\quad \\sqrt { 10 } = \\sqrt { 2 + a ^ { 2 } - 2 a }{/tex}Squaring both sides{tex}\\Rightarrow ( \\sqrt { 10 } ) ^ { 2 } = \\left( \\sqrt { 2 + a ^ { 2 } - 2 a } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa010 = 2 + a2 - 2a{tex}\\Rightarrow{/tex}\xa0{tex}a^2 - 2a + 2 - 10 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}a^2 - 2a - 8 =0{/tex}Splitting the middle term.{tex}\\Rightarrow{/tex}\xa0a2 - 4a + 2a - 8 = 0{tex}\\Rightarrow{/tex}\xa0{tex}a(a - 4) + 2 (a - 4)=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(a - 4)(a + 2) = 0{/tex}{tex}\\Rightarrow{/tex}{tex}\xa0a = 4, a = -2{/tex} | |
| 10745. |
Cosec31 -sec59ka man gyat kare |
| Answer» Cosec(90-59)-sec59=sec59-sec59=0 | |
| 10746. |
Sin A -Cos B ज्ञात कीजिए A+B और कोण A और B न्यून कोण है |
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| 10747. |
I want half yearly sample paper with solution. |
| Answer» samples | |
| 10748. |
How can we find area of isosceles triangle. If two sides are 20cmand12cm.Area is 60cmsquare? |
| Answer» Yes | |
| 10749. |
Wht is the syllabus of 10th class for SA 1?? |
| Answer» ******* mam ne na likh wayo | |
| 10750. |
Good morning friends next mera mathematics ka exam hai ? |
| Answer» | |