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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.
| 7151. |
What you mean by this symbol \'*\' |
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Answer» When we write a answer A word that is not good for us it appears as *****you can see when you give answer and it contains some words like that Product |
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| 7152. |
how many mark need to pass the paper? |
| Answer» Maximun 27 marks means 33% | |
| 7153. |
If 2x ²+x+k is a zero of polynomial .find k |
| Answer» Since 3 is zero of f(x) =\xa02x2\xa0+ x + k, we havef(3) = 0{tex}\\Rightarrow {/tex}\xa02(3)2\xa0+ 3 + k = 0\xa0{tex}\\Rightarrow {/tex}\xa018 + 3 + k = 0{tex}\\Rightarrow {/tex}\xa021 + k = 0{tex}\\Rightarrow {/tex}\xa0k = -21 | |
| 7154. |
Lemma |
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Answer» Lemma is a proving statment which can be used to prove another statment. a = bq+r |
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| 7155. |
What is hcf and lcm of two prime number A and B. |
| Answer» HCF is 1 and LCM is AB means product of two prime numbers | |
| 7156. |
don\'t no |
| Answer» | |
| 7157. |
Draw a line segment of length 7.6c.m and divides in it the ratio 5:8 measure the two parts |
| Answer» Given: A line segment of length 7.6 cm.Required: To divide it in the ratio 5 : 8 and to measure the two parts.Steps of construction :\tFrom any ray AX, making an acute angle with AB.\tLocate 13 (= 5 + 8) points A1,\xa0A2, A3,..... and A13\xa0on AX such that\tAA1\xa0= A1A2\xa0= A2A3\xa0= A3A4\xa0=\xa0A4A5\xa0= A5A6\xa0= A6A7\xa0= A7A8\xa0= A8A9\xa0= A9A10\xa0= A10A11\xa0= A11A12\xa0= A12A13\tJoin BA13\tThrough the point A5, draw a line parallel to A13B intersecting AB at the point C.\tThen, AC : CB = 5 : 8\tOn measurement, AC = 3.1 cm, CB = 4.5 cm.Justification :{tex}\\because{/tex}\xa0A5C || A13B [ By Construction]{tex}\\because{/tex}\xa0{tex}\\frac { A A _ { 5 } } { A _ { 5 } A _ { 13 } } = \\frac { A C } { C B }{/tex}\xa0[By the Basic proportionality theorem]But,\xa0{tex}\\frac { A A _ { 5 } } { A _ { 5 } A _ { 13 } } = \\frac { 5 } { 8 }{/tex}\xa0[ By Construction[Therefore,\xa0{tex}\\frac { A C } { C B } = \\frac { 5 } { 8 }{/tex}This shows that C divides AB in the ratio 5 : 8. | |
| 7158. |
CotA-1÷2-(sec)2A |
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| 7159. |
Sin× cosec |
| Answer» 1 | |
| 7160. |
cotA + tanB=cotA tanB |
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| 7161. |
Prove the identity √tan2A%1+tan2A=sinA |
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| 7162. |
The distance of the point P(2,3) from the x-axis isa) 2b) 3c) 1d) 5 |
| Answer» 3 | |
| 7163. |
tanA/1-cotA+cotA/1-tanA=1+tanA+cotA=1+secAcosecA |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA [ since (1/tanA) =cotA ].= (1 + tanA + cotA){tex}\\therefore \\quad \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA ...........(1)Now, 1 + tanA + cotA = 1 +\xa0{tex}\\frac { \\sin A } { \\cos A } + \\frac { \\cos A } { \\sin A }{/tex}\xa0= 1 +\xa0{tex}\\frac { \\sin ^ { 2 } A + \\cos ^ { 2 } A } { \\sin A \\cos A }{/tex} = 1 +\xa0{tex}\\frac { 1 } { \\sin A \\cos A }{/tex}\xa0[{tex}\\because{/tex}Sin2A + Cos2 A = 1 ]\xa0= 1 + cosecAsecASo, 1 + tanA + cotA = 1+ cosecAsecA.......(2)From (1) and (2), we obtain{tex}\\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}\xa0= 1 + tanA + cotA = 1 + cosecAsecA | |
| 7164. |
(TanA +CosecB)² -(cotB- secA)² =2tanA.CotB(CosecA+SecB) |
| Answer» We have,LHS = (tanA + cosec B)2 - (cotB - sec A)2{tex}\\Rightarrow{/tex}\xa0LHS = (tan2A + cosec2B + 2tanA cosecB ) -\xa0(cot2B + sec2A\xa0- 2cotB secA){tex}\\Rightarrow{/tex}\xa0LHS = (tan2A - sec2A)+(cosec2B - cot2B)+2tanA cosecB +2cotB secABut, Sec2A - tan2A =1 & cosec2A - cot2\xa0A = 1{tex}\\therefore{/tex}\xa0LHS = -1 + 1 + 2 tanA cosecB + 2cotB secA{tex}\\Rightarrow{/tex}\xa0LHS = 2 (tanA cosecB + cotB secA){tex}\\Rightarrow{/tex}\xa0LHS = 2 tanA cotB{tex}\\left( \\frac { cosec\\: B } { \\cot B } + \\frac { \\sec A } { \\tan A } \\right){/tex}\xa0[Dividing and multiplying by tanA cotB]{tex}\\Rightarrow{/tex}\xa0LHS = 2tan A cotB{tex}\\left\\{ \\frac { \\frac { 1 } { \\sin B } } { \\frac { \\cos B } { \\sin B } } + \\frac { \\frac { 1 } { \\cos A } } { \\frac { \\sin A } { \\cos A } } \\right\\}{/tex}\xa0[Since, CosecA.SinA =1 , SecA.cosA =1, (sinA/cosA)= tanA & (cosA/SinA) =cotA ]{tex}\\Rightarrow{/tex}\xa0LHS = 2 tanA cotB{tex}\\left( \\frac { 1 } { \\cos B } + \\frac { 1 } { \\sin A } \\right){/tex}\xa0= 2tanA cotB ( secB + cosecA ) = RHS. Hence, proved. | |
| 7165. |
If 24×-20=150-× then×=? |
| Answer» 24x-20=150-x24x+x=150+2025x=170x=170/25x=34/5 | |
| 7166. |
Find the 11th term from the last term (towards the first term ) of the AP:10,7,4...,-62 |
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Answer» an=a + (n-1)d-62=10+(n-1)(-3)-62=10-3n+33n=75n=25term 15 from the first term = term 11 from last term\xa01,2,3,4,5,6,7,8,9,10,11,12,13,14, 15, 16,17,18,19,20,21,22,23,24,25a15=a+14x-3a15=10-42 = - 32 Use this formula for your ans ;l-(n-1)d in which l is the last term i.e -62 |
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| 7167. |
Diagonals of a square? Formula.... |
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| 7168. |
Which site provides latest version of sample papers for class 10? |
| Answer» You can check sample papers here : https://mycbseguide.com/cbse-sample-papers.html | |
| 7169. |
Find the quadratic polynomial whose zeroes are 7+√3 and 7-√3 |
| Answer» Take alpha=7+√3 and beta=7-√3Then put the value in the formula k[(x)2 -(alpha+beta)x + (alpha×beta) | |
| 7170. |
How to make project on linear equations in two variable |
| Answer» I suggest that the best idea of your project is that you use basic things | |
| 7171. |
Show that 2 root 3 + 3 root 2 I irrational. |
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| 7172. |
prove that √2 +√7 is not rational number |
| Answer» Let the √2+√7 is a rational no then we can write it in the form of √2+√7=a/b.........(1) √2=a/b-√7. Now square on both side | |
| 7173. |
What is identity of (a+b)³ |
| Answer» (a+b)³=a³+b³+3ab(a+b) | |
| 7174. |
Mode foemula |
| Answer» l+{fm-f1÷2fm-f1-f2}*h | |
| 7175. |
Describe about mathematician ramanujan |
| Answer» Srinivas ramanujan FRS was an Indian mathematician who lived during the British rule in India Born :22 Dec 1887,ErodeDied:26 April 1920,kumbakonam Award :fellow of the royal society Education :trinity college, Cambridge University of Cambridge Govt arts college, kumbakonam Town higher secondary school University of madras Pachaiyappa\'s college | |
| 7176. |
For what value of p, are 2p -1,7and 3p three consecutive terms an A.P |
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Answer» Given that :T1 = (2p-1)T2=7T3 =3pThe common difference d for two consecutive terms in AP is same soT3-T2=T2-T1 =d3p-7=7-(2p-1)3p-7=7-2p+15p=15p=3 Please explain how p=3??? P = 3 |
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| 7177. |
For what value of p, are 2p -1,7and 3p three consecutive terms of an A.P |
| Answer» P = 3 | |
| 7178. |
(sec theeta + cos theeta). (sec theeta- cos theeta) =tan square theeta + sin square theeta. |
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| 7179. |
RMO ke liye book suggest karo plz.. |
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| 7180. |
If a and b are acuit angles and sinb=cosa then value of a+b=how much |
| Answer» Sinb=cos(90-b)ThereforeCos(90-b)=cosa90-b=aa+b=90 | |
| 7181. |
-4+-1+2....+x=437 |
| Answer» (-4) + (-1) + 2 + 5 + ---- + x = 437.Now,-1 - (-4) = -1 + 4 = 32 - (-1) = 2 +\xa01 = 35 - 2 = 3Thus, this forms an A.P. with a = -4, d = 3,l = xLet their be n terms in this A.P.Then,Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] {/tex}{tex}\\Rightarrow 437 = \\frac { n } { 2 } [ 2 \\times ( - 4 ) + ( n - 1 ) \\times 3 ]{/tex}{tex}\\Rightarrow{/tex}\xa0874 = n[-8 + 3n - 3]{tex}\\Rightarrow{/tex}874 = n[3n - 11]{tex}\\Rightarrow{/tex}874 = 3n2\xa0- 11n{tex}\\Rightarrow{/tex}3n2\xa0- 11n - 874 = 0{tex}\\Rightarrow{/tex}3n2\xa0- 57n + 46n - 874 = 0{tex}\\Rightarrow{/tex}3n(n - 19) + 46(n - 19) = 0{tex}\\Rightarrow{/tex}3n + 46 = 0 or n = 19{tex}\\Rightarrow n = - \\frac { 46 } { 3 }{/tex}\xa0or n\xa0= 19Numbers of terms cannot be negative or fraction.{tex}\\Rightarrow{/tex}\xa0n = 19Now, Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ a + l ]{/tex}{tex}\\Rightarrow 437 = \\frac { 19 } { 2 } [ - 4 + x ]{/tex}{tex}\\Rightarrow - 4 + x = \\frac { 437 \\times 2 } { 19 }{/tex}{tex}\\Rightarrow - 4 + x = 46{/tex}{tex}\\Rightarrow x = 50{/tex} | |
| 7182. |
Is trigonometry necessary for students |
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Answer» Give me reason Every thing is important for student |
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| 7183. |
What is a maths |
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Answer» Mathematics is an old, broad, and deep discipline (field of study). People working to improve math education need to understand "What is Mathematics?" No |
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| 7184. |
Use of identity -sin^4+cos^4 |
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| 7185. |
Set of easy questions for weak students |
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| 7186. |
1+1=? |
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Answer» 2 2 2 No. Of oo in your body |
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| 7187. |
Show that n^3-n is divisible by 6 for any (+) n |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) | |
| 7188. |
Explain why 7×6×5×4×3×2×1+5 is a composite number |
| Answer» Since it has more than 2 factors therefore it is a composite number | |
| 7189. |
Find the value of k,. If the points A(2,3), B(4,k), and C(9,-3) |
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| 7190. |
Bp thm. / py thm |
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| 7191. |
Bp thm/py thm |
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| 7192. |
If x=2 and x=3 are roots of equation (3x)2 - 2kx +2km= 0, find the value of k, m |
| Answer» It is given that x = 2 and x = 3 are roots of the equation 3x2 - 2kx + 2m = 0.{tex}\\therefore{/tex} 3 {tex}\\times{/tex}\xa022 - 2k {tex}\\times{/tex}\xa02 + 2m = 0 and 3 {tex}\\times{/tex}\xa032 - 2k {tex}\\times{/tex}\xa03 + 2m = 0{tex}\\Rightarrow{/tex}\xa012 - 4k\xa0+ 2m = 0 and 27 - 6k\xa0+ 2m = 0{tex}\\Rightarrow{/tex}\xa012 = 4k\xa0- 2m...(i) and 27 = 6k - 2m...(ii)Solving i and ii equation, we get k =\xa0{tex}\\frac { 15 } { 2 }{/tex}\xa0and m = 9 | |
| 7193. |
calculate the mean of the first prime number |
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| 7194. |
152811ie519e18616291671 |
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| 7195. |
2a+b=5Find a |
| Answer» a=5-b/2 | |
| 7196. |
Anyone plzz explain introduction to trigonometry |
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| 7197. |
8.4 |
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| 7198. |
If one root of the quadratic equation 2x2+kx_6=0, find the value of k |
| Answer» Since x = 2 is a root of the equation 2x2 + kx - 6 = 0.{tex}\\therefore{/tex}\xa0{tex}2(2)^2+k(2)-6=0{/tex}\xa0{tex}8+2k-6=0{/tex}{tex}2k+2=0{/tex}{tex}2(k+1)=0{/tex}\xa0{tex}\\Rightarrow k+1=0{/tex}\xa0{tex}k=-1{/tex}Putting k\xa0= -1 in the equation 2x2 + kx - 6 = 0, we get{tex}\\Rightarrow 2x^2+(-1)x-6=0{/tex}{tex}\\Rightarrow 2x^2-x-6=0{/tex}{tex}\\Rightarrow{/tex}\xa02x2 - 4x + 3x - 6 = 0{tex}\\Rightarrow{/tex}\xa02x(x - 2) + 3(x - 2) = 0{tex}\\Rightarrow{/tex}\xa0(x - 2) (2x + 3) = 0{tex}\\Rightarrow{/tex} x - 2 = 0 and\xa02x + 3 = 0\xa0{tex}\\therefore x=2 \\ and \\ \\frac{-3}{2}{/tex}Hence, the other root is\xa0{tex}\\frac{-3}{2}{/tex} | |
| 7199. |
Are two triangle with equal corresponding side always similar |
| Answer» Two triangles having all the corresponding sides equal are congruent.Two congurent triangles are always similar. | |
| 7200. |
Solve for x if. 1/2a+b+2x=1/2a+1/b+1/2x |
| Answer» {tex}\\frac{1}{2a + b + 2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0+\xa0{tex}\\frac{1}{2x}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2a + b + 2x}{/tex}\xa0-\xa0{tex}\\frac{1}{2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}\\frac { 2 x - 2 a - b - 2 x } { ( 2 a + b + 2 x ) ( 2 x ) }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2a \\times b}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { - ( 2 a + b ) } { ( 2 a + b + 2 x ) 2 x }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2ab}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { - 1 } { 4 a x + 2 b x + 4 x ^ { 2 } }{/tex}\xa0=\xa0{tex}\\frac{1}{2ab}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax = -2ab{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax + 2ab = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2x(2x + b) + 2a(2x + b) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0(2x + b)(2x + 2a) = 0{tex}\\Rightarrow{/tex}\xa0x = -{tex}\\frac{b}{2}{/tex} or x = -a | |