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InterviewSolution
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.
| 34901. |
What is Shridharacharya\'s rule?? |
| Answer» X=–b± under root b²–4ac/2a. | |
| 34902. |
Is question from optional exercise of ncert book has been asked by board exam |
| Answer» You must practise NCERT completely. They may be asked. | |
| 34903. |
if the pth term of an AP is q and its qth term is p then show that (p+q)th term is zero |
| Answer» Let a be the first term and d be the common differencepth\xa0term = a +(p - 1)d = q(given)-----(1)qth\xa0term = a +(q - 1) d = p(given)-----(2)subtracting (2) from (1)(p-q)d=q-p(p-q)d=-(p-q){tex}\\therefore{/tex}\xa0d = -1putting d=-1 in (i)a - (p - 1) = q{tex}\\therefore{/tex}\xa0a=p+q-1{tex}\\therefore{/tex}\xa0(p + q)th term = a + (p + q - 1)d= (p + q - 1) - (p + q - 1) = 0 | |
| 34904. |
If sec A +tan A =x then find SecA |
| Answer» secA=x-tanA | |
| 34905. |
Cosec a-cot a = q then proved that q^2-1/q^2+1+cos a =0 |
| Answer» | |
| 34906. |
If any study in v.m vishvas schooladampur |
| Answer» So | |
| 34907. |
What is Formula of frustum of cone |
| Answer» V=pie.h/3(R^2 +Rr+r^2) | |
| 34908. |
2 cubes each of volume 64cm3 are joined end to end .find the surface area of resulting cuboid. |
| Answer» a^3=64a=4L =8cmB=4H=4Surface area =2(lb+bh+lh)=2(32+16+32)=2(80)=160cm2 | |
| 34909. |
28#&3* |
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| 34910. |
Question no. 18 |
| Answer» | |
| 34911. |
Difference between tangent and secent |
| Answer» When the line touches the circle at two points: It is called a secant.When the line touches the circle at exactly one point: It is called a tangent. The tangent segments to a circle from the same external\xa0point are equal. | |
| 34912. |
Is Theorem is asked in board exam |
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Answer» Yaaa Yes theorems are very important. Specially the theorems in Triangles |
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| 34913. |
If alpha and beta are the zeroes of quadratic polynomial f(x)=5y^2-7y+1 |
| Answer» Given that, α and β are the zeroes of the quadratic polynomial.p(y) = 5y2 - 7y + 1Here, sum of the zeroes = α + β =\xa0{tex}\\frac{{ - ( -7 ) }}{ 5} = \\frac {{ 7 }}{ 5 }{/tex}and product of the roots = αβ =\xa0{tex}\\frac{{ 1 }}{ 5 }{/tex}So,{tex}\\frac{1}{\\alpha } + \\frac{1}{\\beta } = \\frac{{\\alpha + \\beta }}{{\\alpha \\beta }} = \\frac{{ 7 }}{ 5 }\\times\\frac{{ 5 }}{ 1 } = \\frac{7}{1} = 7 {/tex}Hence, required value of\xa0{tex}\\frac{{ 1 }}{ \\alpha } + \\frac{{ 1 }}{ \\beta }{/tex}\xa0is 7 | |
| 34914. |
How to find d in statistics |
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Answer» Second term - first termA2-A1 or A3-A2 It is given in the text book read it from there |
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| 34915. |
In triangle ABC angle A =60. Prove that BC sq=AB sq+ACsq - AB.AC |
| Answer» Given: In {tex}\\triangle{/tex}ABC,To prove: BC2\xa0= AB2\xa0+ AC2\xa0- AB.ACConstruction: Draw CE\xa0{tex}\\perp{/tex} ABProof: In {tex}\\triangle{/tex}BECBC2\xa0= CE2\xa0+ BE2\xa0..(i)In {tex}\\triangle{/tex}ACE, AC2\xa0= CE2\xa0+ AE2CE2\xa0= AC2\xa0- AE2\xa0..(ii)from (i) and (ii)BC2\xa0= AC2\xa0- AE2\xa0+ (AB - AE)2\xa0({tex}\\because{/tex} BE = AB - AE)BC2\xa0= AC2\xa0+ AB2\xa0- 2AB.AE - 2AB{tex}\\frac{1}{2}{/tex}ACSince side opposite to 30° angel is half hypoteneuse.In {tex}\\triangle{/tex}ACE, AE = {tex}\\frac{1}{2}{/tex}ACBC2 = AB2 + AC2 - AB.AC | |
| 34916. |
If common different of an AP is –6,find a16-a12 |
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Answer» a16= a+15d and a12=a+11d ; a16-a15= a+15d-a+11d = 26 d = 26 (6)=15 6 a+15d-(a + 11d)=4d d=-6=>4d=-24 -24 |
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| 34917. |
Find the positive root of 3x²+6=9 |
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Answer» 1 3x2 + 6x = 93x2 + 6x - 9 = 03x2 + 9x - 3x - 9 = 03x(x + 3) - 3 (x +3) = 0(x + 3) ( 3x - 3)= 0x + 3 = 0 or 3x - 3 = 0x = -3 or 3x = 3x = -3 or x = 3/3 = 1 3x^2 + 6 =93x^2 = 9-63x^2 =3x^2 =3/3x^2 =1x= root 1So positive root is +1 and negative root is -1 |
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| 34918. |
What is syllabs |
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Answer» Thank the education which we learn for succeding in our life |
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| 34919. |
Which one is a best sample paper for 10th board maths |
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Answer» No one Oswal\'s paper are very good. Just do all the questions of ncert and ncert examplear U like isuceed or sahitya bhawan for best preparation |
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| 34920. |
Which one is best book |
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Answer» Ocean publication RD sharma or RS Aggarwal and NCERT exemplar is best because it is recommended by cbse Examguru Xam idea and U like In maths.. Rd sharma |
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| 34921. |
Send me solution of question 5of exercise 13.1 immediately |
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Answer» Why we plus csa ofhemisphere inthis question Diameter of hemisphere = Edge of cube =LRadius of hemisphere =L/2Total surface area of solid = Surface area of cubical part + CSA of hemispherical part − Area of base of hemispherical part= 6 * L*L\xa0+ 2*.314* L/2*L/2- 3.14* L/2*L/2=6 L2\xa0+ 3.14 *L2/4 Surface area of solid=Total surface area of cube + Curved surface area of hemisphere --Area of circle Use this formula for fInding answer. |
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| 34922. |
State and proof area theorem |
| Answer» Theorems on the area of similar trianglesTheorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.To prove this theorem, consider two similar triangles ΔABC and ΔPQR;\xa0As, Area of triangle =\xa01/2\xa0× Base × HeightTo find the area of ΔABC and ΔPQR, draw the altitudes AD and PE from the vertex A and P of ΔABC and ΔPQR as shown in the figure given below:Now, area of ΔABC =\xa01/2\xa0× BC × ADarea of ΔPQR =\xa01/2\xa0× QR × PEThe ratio of the areas of both the triangles can now be given as:area\xa0of\xa0ΔABC / area\xa0of\xa0ΔPQR\xa0= (1/2×BC×AD) / (1/2×QR×PE)⇒\xa0area\xa0of\xa0ΔABC / area\xa0of\xa0ΔPQR\xa0=\xa0BC\xa0×\xa0AD / QR\xa0×\xa0PE\xa0……………. (1)Now in ∆ABD and ∆PQE, it can be seen that:∠ABC = ∠PQR (Since ΔABC ~ ΔPQR)∠ADB = ∠PEQ (Since both the angles are 90°)From AA criterion of similarity ∆ADB ~ ∆PEQ⇒\xa0AD/PE\xa0=\xa0AB/PQ …………….(2)Since it is known that ΔABC~ ΔPQR,AB/PQ\xa0=\xa0BC/QR\xa0=\xa0AC/PR\xa0…………….(3)Substituting this value in equation (1), we getarea\xa0of\xa0ΔABC/area\xa0of\xa0ΔPQR\xa0=\xa0AB/PQ\xa0×\xa0AD/PEUsing equation (2), we can writearea\xa0of\xa0ΔABC / area\xa0of\xa0ΔPQR\xa0=\xa0ABPQ\xa0×\xa0ADPE⇒area\xa0of\xa0ΔABCarea\xa0of\xa0ΔPQR\xa0=(AB/PQ)2Also from equation (3),area\xa0of\xa0ΔABC / area\xa0of\xa0ΔPQR\xa0=\xa0(AB/PQ)2\xa0=(BC/QR)2\xa0=\xa0(CA/RP)2This proves the theorem\xa0Note I could not post triangles pl draw yourself | |
| 34923. |
tan A + sec A -1÷tan A-secA+1=1+sinA÷cosA |
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Answer» Proof:LHS=(tanA +secA) - (sec2A - tan2A) / tanA + 1 - secA as sec2A - tan2A=1=(tanA+ secA) (1 - (secA - tanA))/tanA + 1 - secA=(tanA +secA) (1 - secA + tanA) /\xa0(1 - secA+tanA )=sec A + tan A=1/cosA + sinA/cosA\xa0=(1 + sinA) / cosA =RHS\xa0\xa0 It is in this app itself. It is in the exercise 8.3 of exemplar problems of chapter intro to trigonometry Prove this |
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| 34924. |
Find the area of a sector sf a circle with radius 6 cm if angle of the sector is 60 |
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Answer» The formula is theeta/360xπrsquare60/360x22/7x6x6= 132/7 cm square 18.84 cm2 Radius = 6cmAngle=60°Area of sector= 22 * 6* 6 * 60 7 360 =154/7 cm2 |
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| 34925. |
How to draw trignometry table |
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| 34926. |
A mettalic cylinder has radius 3cm...?... |
| Answer» | |
| 34927. |
The surface area is 616 cm square find radius |
| Answer» Shape? | |
| 34928. |
Tan theta + cot theta is equal to 7 and prove find that 10 square theta + cos square theta |
| Answer» Please correct your Question. | |
| 34929. |
Find the value of m if the points are (5,1) , (-2,-3) , (8, 2m) |
| Answer» | |
| 34930. |
Proof:- sin theta - cos theta + 1/sin theta + cos theta - 1 = 1/sec theta - tan theta |
| Answer» We have to prove that,{tex}\\frac{{\\sin \\theta - \\cos \\theta + 1}}{{\\sin \\theta + \\cos \\theta - 1}} = \\frac{1}{{\\sec \\theta - \\tan \\theta }}{/tex} using identity\xa0{tex}sec^2\\theta=1+tan^2\\theta{/tex}LHS = {tex}\\frac{{\\sin \\theta - \\cos \\theta + 1}}{{\\sin \\theta + \\cos \\theta - 1}} {/tex}{tex} = \\frac{{\\tan \\theta - 1 + \\sec \\theta }}{{\\tan \\theta + 1 - \\sec \\theta }}{/tex} [ dividing the numerator and denominator by {tex}\\cos{\\theta}{/tex}.]{tex} = \\frac{{(\\tan \\theta + \\sec \\theta)-1 }}{{(\\tan \\theta - \\sec \\theta )+1}}{/tex}{tex}=\\frac{\\{{(\\tan\\theta+\\sec\\theta)-1\\}}(tan\\theta-\\sec\\theta)}{\\{{(\\tan\\theta-\\sec\\theta)+1\\}}(\\tan\\theta-\\sec\\theta)}{/tex} [ Multiplying and dividing by {tex}(\\tan{\\theta}-\\sec{\\theta}){/tex}]{tex}=\\frac{{(\\tan^2\\theta-\\sec^2\\theta)-}(tan\\theta-\\sec\\theta)}{\\{{(\\tan\\theta-\\sec\\theta)+1\\}}(\\tan\\theta-\\sec\\theta)}{/tex} [{tex}\\because (a-b)(a+b)=a^2-b^2{/tex}]{tex} = \\frac{{-1-\\tan \\theta + \\sec \\theta }}{{(\\tan \\theta - \\sec \\theta+1)(\\tan{\\theta}-\\sec{\\theta}) }}{/tex}[{tex}\\because \\tan^2\\theta-\\sec^2\\theta=-1{/tex}]{tex}=\\frac{-(\\tan\\theta-\\sec\\theta+1)}{(\\tan\\theta-\\sec\\theta+1)(\\tan\\theta-\\sec\\theta)}{/tex}{tex}=\\frac{-1}{\\tan{\\theta}-\\sec{\\theta}}{/tex}{tex} = \\frac{1}{{\\sec \\theta - \\tan \\theta }}{/tex}=RHSHence Proved. | |
| 34931. |
Total surface area of frustum |
| Answer» Pie l (r1+r2)+pie r1 square+ pie r2 square | |
| 34932. |
Use Euclid divisions lemma to show That the cube of any positive integer is of form 9m, 9m+1,or 9m+8 |
| Answer» Let a positivi integer x when divided by 3 gives remainder rThen by using euclids division lemmaX=3q+r. And. r=0,1or2Case 1:-. r=0 X^3= 3×3×3×q×q×q+0×0×0 =27q^3 =9×3q^3 =9m. ( Say 3q^3=m)Similarly in case 2 &3 we find 9m+1 and 9m+8 respectively. | |
| 34933. |
2 add 3 is equal to |
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Answer» 5 Obviously, it is equal to 5. 5 Bkwas questionAnd answer is 5 |
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| 34934. |
Ex 5.3 4 |
| Answer» | |
| 34935. |
What is samroop organs |
| Answer» Medical Definition of Organ. Organ: A relatively independent part of the body that carries out one or more special functions. Examples of organs include the eyes, ears, heart, lungs, and liver. | |
| 34936. |
100 +678913 |
| Answer» 100 +678913= 679013 | |
| 34937. |
Solve for x 1/x+1 + 2/x+2 = 4/x+4 Where x≠-1, -2,-4 |
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Answer» X=2+8^1/2 or X=2-8^1/2 Kya masala marr ke Masala marr ke |
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| 34938. |
I want lacture on circle theorem |
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Answer» Mtlab Area related to circles muje nahi samajra Lacture kya?? |
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| 34939. |
All formula of trignometry |
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Answer» [Exercise:-8.3]....sin (90°-A)=cos A,,,,,cos(90°-A)=sinA,,,,,Tan(90°-A)=cot,,,,Cot(90°-A)=tanA,,,SecA(90°-A)=cosecA,,,,Cosec(90°-A)=secA....... [Exercise:-8.1].....sin= p\\h,,,cos=B\\H,,,tan=P\\B......nd [Exercise:-8.2]....u can read formula on NCERT book ...... |
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| 34940. |
How to prepare for science exam perfectly |
| Answer» Read the full ncert | |
| 34941. |
Exercise 14.3 for class 10 |
| Answer» | |
| 34942. |
Find the area of shaded degine in fig when ABCD IS sq of shaded 10cm |
| Answer» | |
| 34943. |
Cos(90-x)sec x tan(90- x)/sin^2(90- x) sec+90-x) |
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Answer» 1/cos^4 Sorry, can u type the question again clearly. Coz i am not sure where does the bracket open but closes at the end |
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| 34944. |
Plz can you give me important questions of chapter 4 NCERT Maths |
| Answer» √2,√8,√18,√32.... Find the common difference d and write three more terms | |
| 34945. |
If sec theta + tan theta = p find the value of cosec theta |
| Answer» Given,{tex}sec\\ \\theta+ tan\\ \\theta = p{/tex} ...(i)Also, we know that,\xa0{tex}sec^2\xa0\\theta - tan^2 \\theta = 1{/tex}{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0- tan\xa0{tex}\\theta{/tex}) (sec\xa0{tex}\\theta{/tex}\xa0+ tan\xa0{tex}\\theta{/tex}) = 1 [{tex}\\because a^2-b^2=(a+b)(a-b){/tex}]{tex}\\Rightarrow{/tex}\xa0(sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex})p = 1 [using equation (i)]{tex}\\Rightarrow{/tex}\xa0sec\xa0{tex}\\theta{/tex}\xa0-\xa0tan\xa0{tex}\\theta{/tex}\xa0{tex}=\\frac{1}{p}{/tex}\xa0...(ii)(i)+(ii), we get,{tex}sec\\theta + tan\\theta+ sec\\theta - tan\\theta = p+ \\frac{1}{p}{/tex}{tex}\\Rightarrow 2sec\\theta = \\frac{p^2+1}{p}{/tex}{tex}\\Rightarrow sec\\theta = \\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow \\frac{1}{cos\\theta} =\\frac{p^2+1}{2p}{/tex}{tex}\\Rightarrow cos\\theta =\\frac{2p}{p^2+1}{/tex}------(iii)Now, we know that,{tex}sin\\theta = \\sqrt( 1- cos^2\\theta) {/tex}put the value of\xa0{tex}cos\\theta{/tex}\xa0from eq. (iii), we get,{tex}sin\\theta = \\sqrt(1-(\\frac{2p}{p^2+1})^2){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(1-\\frac{4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{(p^2+1)^2-4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{p^4+1+2p^2-4p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{p^4+1-2p^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\sqrt(\\frac{(p^2-1)^2}{(p^2+1)^2}){/tex}{tex}\\Rightarrow sin\\theta = \\frac{p^2-1}{p^2+1}{/tex}{tex}cosec\\theta = \\frac{p^2+1}{p^2-1} [\\because cosec\\theta =\\frac{1}{sin\\theta}]{/tex}hence, {tex}cosec\\\xa0\\theta{/tex}\xa0{tex}=\\frac{1+p^{2}}{1-p^{2}}{/tex} | |
| 34946. |
If two vertices of an equilateral triangle are(3,0) and (6,0), then find the third vertex |
| Answer» Let ABC be the equilateral triangle such that,A = (3,0), B=(6,0) and C=(x,y)Distance between:{tex}\\sqrt {( x_{2}-x_{1})^2+(y_{2} -y_{1})^{2} }{/tex}we know that,AB=BC=ACBy distance formula we get,AB=BC=AC=3unitsAC=BC{tex}\\sqrt{(3-x)^2+y^2}=\\sqrt{(6-x)^2+y^2}{/tex}{tex}9+x^2-6 x+y^2=36+x^2-12 x+y^2{/tex}{tex}6 x=27{/tex}{tex}x=27 / 6=9 / 2{/tex}BC = 3 units{tex}\\sqrt{(6-\\frac{27}{6})^2+y^2}=3{/tex}{tex}(\\frac{(36-27)}{6})^2+y^2=9{/tex}{tex}(\\frac{9}{6})^2+y^2=9{/tex}{tex}(\\frac{3}{2})^2+y^2=9{/tex}{tex}\\frac{9}{4}+y^2=9{/tex}{tex}9+4 y^2=36{/tex}{tex}4 y^2=27{/tex}{tex}y^2=\\frac{27}{4}{/tex}{tex}y=\\sqrt{(\\frac{27}{4})}{/tex}{tex}y=3 \\sqrt{\\frac{3}{2}}{/tex}{tex}(x, y)=(9 / 2,3 \\sqrt{\\frac{3}{2}}){/tex}Hence third vertex of equilateral triangle = C =\xa0{tex}(9 / 2,3 \\sqrt{\\frac{3}{2}}){/tex} | |
| 34947. |
tan 60 cos 60 + tan 60 sec 30 |
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| 34948. |
What is rel number and what is poly nomial and triangle and statistics |
| Answer» | |
| 34949. |
Prove that :sinA-2sin3A/2cos3A-cosA=tanA |
| Answer» sinA (1-2sin2A)/cosA ( 2cos2A -1)SinA (1-2 (1-cos2A)/cosA (2cos2A -1)=sinA (2cos2A -1)/cosA ( 2cos2A -1)=sinA/ cosA =tanA | |
| 34950. |
(3+4+1)×2 |
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Answer» 16 16 (3 + 4 + 1) × 2= (8)\xa0× 2= 16 |
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