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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.
| 7101. |
What is the maximum value of 1÷cosec theta? |
| Answer» Answer is 1 | |
| 7102. |
Afrab5 |
| Answer» | |
| 7103. |
CosecA+cotA=P,then find cosA |
| Answer» CosecA+cotA=P1/cosA+cosA/sinA=PSinA+cos^2A/cosAsinA=PSinA+cos^2A=PcosAsinACosA=sinA+cos^2A/PsinsA | |
| 7104. |
find the least positive integer which on adding 1is exactlY Divisible by 126 and600 |
| Answer» | |
| 7105. |
What is the formula of primter of triangle |
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Answer» Sum of all its three sides 2(L+B) |
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| 7106. |
Show that there is no positive integer n for which √(n-1)+√(n+1) is Rational |
| Answer» Let us assume that there is a positive integer n for {tex}\\sqrt{n-1}+\\sqrt{n+1}{/tex}which is rational and equal to {tex}\\frac pq{/tex}, where p and q are positive integers and (q\xa0{tex}\\neq{/tex}\xa00).{tex}\\sqrt { n - 1 } + \\sqrt { n + 1 } = \\frac { p } { q }{/tex}......(i)or,\xa0{tex}\\frac { q } { p } = \\frac { 1 } { \\sqrt { n - 1 } + \\sqrt { n + 1 } }{/tex}on multiplication of numerator and denominator by\xa0{tex}\\sqrt{n-1}-\\sqrt{n+1}{/tex}\xa0we get{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( \\sqrt { n - 1 } + \\sqrt { n + 1 } ) ( \\sqrt { n - 1 } - \\sqrt { n + 1 } ) }{/tex}{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( n - 1 ) - ( n + 1 ) } = \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { - 2 }{/tex}or,\xa0{tex}\\sqrt { n + 1 } - \\sqrt { n - 1 } = \\frac { 2 q } { p }{/tex} ........(ii)On adding (i) and (ii), we get{tex}2 \\sqrt { n + 1 } = \\frac { p } { q } + \\frac { 2 q } { p } = \\frac { p ^ { 2 } + 2 q ^ { 2 } } { p q }{/tex}{tex}\\sqrt{n+1}\\;=\\frac{p^2+2q^2}{2pq}{/tex}...............(iii)From (i) and (ii),{tex}\\style{font-family:Arial}{\\sqrt{n-1}\\;=\\frac{p^2-2q^2}{2pq}}{/tex}........(iv)In RHS of (iii) and (iv)\xa0{tex}\\frac{p^2+2q^2}{2pq}\\;and\\;\\frac{\\displaystyle p^2-2q^2}{\\displaystyle2pq}\\;are\\;rational\\;number\\;because\\;p\\;and\\;q\\;are\\;positive\\;integers{/tex}But it is possible only when (n + 1) and (n - 1) both are perfect squares.Now n+1-(n-1)=n+1-n+1=2Hence they differ by 2 and two perfect squares never differ by 2.So both (n + 1) and (n -1 ) cannot be perfect squares. Hence there is no positive integer n for which\xa0{tex}\\style{font-family:Arial}{\\sqrt{n-1\\;}+\\sqrt{n+1}}{/tex} is rational | |
| 7107. |
Easiest way to prove pythagores therom |
| Answer» | |
| 7108. |
If the zeroes of the polynomial f(x)=x^3-3x^2+x+1 are (a-b),a and(a+b),find a and b. |
| Answer» f(x) = x3 - 3x2 + x + 1It is given that a - b, a and a + b are the zeroes of f(x).Now, Sum of the zeroes ={tex} - \\frac { \\text { Coeff. of } x ^ { 2 } } { \\text { Coeff. of } x ^ { 3 } }{/tex}⇒ a - b + a + a +b =\xa0{tex}- \\frac { - 3 } { 1 } {/tex}⇒ 3a = 3\xa0⇒ a = 1and, Product of zeros ={tex} - \\frac { \\text { Constant term } } { \\text { Coeff. of } x ^ { 3 } }{/tex}⇒ (a -b )(a) (a + b) ={tex} - \\frac { 1 } { 1 }{/tex}⇒ a(a2\xa0- b2) = -1⇒ 1 - b2\xa0= -1 (∵ a = 1)⇒b2\xa0= 2\xa0⇒\xa0b =\xa0{tex} \\pm \\sqrt { 2 }{/tex}Hence the value of a = 1 and b = {tex} \\pm \\sqrt { 2 }{/tex} | |
| 7109. |
1234567+98765 |
| Answer» 1333332 | |
| 7110. |
How to open this square root √7,225 please ans |
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Answer» Please ans The square root is wr3ydyeyeyysyyeysyywyysyywyysywyzywysywhshsuhwhdyhwusuyewyYshwJhshwhzshshzgysjUdushhxsuwuwjduwusuxyfnckepqpjfuwjzhfucudwnzifusywjdjgheakdjharshitsjwufhwjdjejajxjwjdh2yhsysudueisu2hrhsjzjzjdubeunsjsjsjsjsjsjsjsjsjsjdubeyxhdjejzkwodnbxhafrsbirnzjdjcjsjkwjcj3j2ioqod skshcusiwifufusuduejzkwodnbskshcusiwifufysu Iwjkwjfjjsu2ufyd rtrthsbx2ufydI changed for bday the day of the Crown be abit. For hd 2nd, do I need for your help hd the best way to be just as hfu2ughxj to the jxj3ucjkeu, hd ud to the jf jxjskakvjkskgjzkwjhckfkihjjajcjjsjwu the hsjwkbbbvbzjjshdhiajsershit sjdjejJehHjdibdeme |
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| 7111. |
(sec-tan) ² = cosec-1 . Cosec+1 prove that |
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| 7112. |
Find mean median and mode of the data as you take it |
| Answer» | |
| 7113. |
Sec x = 2x ; tan x = 1/2x ; x*x - 1/ x*x |
| Answer» | |
| 7114. |
If cosecA-sinA=m and secA-cosA=n then prove that m×m×n×n{(m×m)+(n×n)+3}=1 |
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| 7115. |
X square + 4 x + 4 |
| Answer» =x2+4x+4=x2+2x+2x+4=x(x+2)+2(x+2)=(x+2)2x=-2 | |
| 7116. |
Find the value of \'a\' and \'b\' so that xpower4 +xpower3+8xpower2+ax+b is divisible by xpower2+1. |
| Answer» If {tex}x^4+x^3+8x^2+ax +b{/tex}\xa0is exactly divisible by x2 + 1, the remainder after division should be zero.Now let us perform long divisionWe get, remainder = x (a - 1) + (b - 7)\xa0x (a - 1) + (b - 7 ) = 0{tex}\\Rightarrow{/tex}\xa0x (a - 1) + (b - 7) = 0x + 0{tex}\\Rightarrow{/tex}\xa0a - 1 = 0 and b - 7 = 0\xa0[On equating the coefficients of like powers of x]{tex}\\Rightarrow{/tex}a = 1 and b = 7 | |
| 7117. |
Write form of mid point of class interval |
| Answer» | |
| 7118. |
What is the square foot of 3 |
| Answer» 1.73 | |
| 7119. |
What is square root if 8 |
| Answer» 2root2 | |
| 7120. |
If tanA=0.75 find the value of sinA+cosA. |
| Answer» {tex}% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGG0b% Gaaiyyaiaac6gacaaMc8Uaamyqaiabg2da9iaaicdacaGGUaGaaG4n% aiaaiwdaaeaaciGG0bGaaiyyaiaac6gacaaMc8Uaamyqaiabg2da9m% aalaaabaGaaG4maaqaaiaaisdaaaaabaGaciiDaiaacggacaGGUbGa% amyqaiabg2da9maalaaabaGaamiCaiaadwgacaWGYbGaamiCaiaadw% gacaWGUbGaamizaiaadMgacaWGJbGaamyDaiaadYgacaWGHbGaamOC% aaqaaiaadkgacaWGHbGaam4CaiaadwgaaaaabaGaamiAaiabg2da9m% aakaaabaGaamiCamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadkga% daahaaWcbeqaaiaaikdaaaaabeaaaOqaaiaadIgacqGH9aqpdaGcaa% qaaiaaiodadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI0aWaaWba% aSqabeaacaaIYaaaaaqabaaakeaacaWGObGaeyypa0JaaGynaaaaaa!6AF2!\\eqalign{ & \\tan \\,A = 0.75 \\cr & \\tan \\,A = {3 \\over 4} \\cr & \\tan A = {{perpendicular} \\over {base}} \\cr & h = \\sqrt {{p^2} + {b^2}} \\cr & h = \\sqrt {{3^2} + {4^2}} \\cr & h = 5 \\cr} {/tex}{tex}% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGGZb% GaaiyAaiaac6gacaWGbbGaeyypa0ZaaSaaaeaacaWGWbaabaGaamiA% aaaacqGH9aqpdaWcaaqaaiaaiodaaeaacaaI1aaaaaqaaiGacogaca% GGVbGaai4CaiaaykW7caWGbbGaeyypa0ZaaSaaaeaacaWGIbaabaGa% amiAaaaacqGH9aqpdaWcaaqaaiaaisdaaeaacaaI1aaaaaqaaiGaco% hacaGGPbGaaiOBaiaaykW7caWGbbGaey4kaSIaci4yaiaac+gacaGG% ZbGaaGPaVlaadgeacqGH9aqpdaWcaaqaaiaaiodaaeaacaaI1aaaai% abgUcaRmaalaaabaGaaGinaaqaaiaaiwdaaaaabaGaci4CaiaacMga% caGGUbGaaGPaVlaadgeacqGHRaWkciGGJbGaai4BaiaacohacaaMc8% Uaamyqaiabg2da9maalaaabaGaaG4naaqaaiaaiwdaaaaaaaa!67BD!\\eqalign{ & \\sin A = {p \\over h} = {3 \\over 5} \\cr & \\cos \\,A = {b \\over h} = {4 \\over 5} \\cr & \\sin \\,A + \\cos \\,A = {3 \\over 5} + {4 \\over 5} \\cr & \\sin \\,A + \\cos \\,A = {7 \\over 5} \\cr} {/tex} | |
| 7121. |
g(x)=a(x^2+1)-x(a^2+1) |
| Answer» | |
| 7122. |
If coseco+cotfita =x find the value of cosecfita-cotfita |
| Answer» | |
| 7123. |
what is the weightage of class10board exam 2017-18 ? |
| Answer» It is of 80 marks | |
| 7124. |
Find the point on the x axis which is equidistant from (-2,5) and(-2,9)? |
| Answer» point on x axis =x,0also this point is equidistant from both given points (given)a.(-2,5) c.(-2,9) b .(x,0)ab=bcso, by putting distance formula for ab and bc this problem will solve | |
| 7125. |
how we can find trigonometric ratio of 30° |
| Answer» Inncert | |
| 7126. |
Cos square theta+tan square theta-1 ÷sin square theta =tan square theta prove it |
| Answer» | |
| 7127. |
How we put the linear equations value in cross multipication method? |
| Answer» | |
| 7128. |
Ex. 8.2solution video |
| Answer» | |
| 7129. |
slove 2xsquare+4x_8=0 by completing the square method . |
| Answer» | |
| 7130. |
find the zeroes of root3xsquare +10x +7root3. |
| Answer» -7 , +3 | |
| 7131. |
Olympiad syllabus |
| Answer» Yes, it is easy | |
| 7132. |
prove that 15+17root3 be an irrational Number. |
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Answer» Suppose\xa0{tex}\\sqrt { 3 } = \\frac { a } { b }{/tex}, where a and b are co-prime integers, {tex}b\\ne 0{/tex}Squaring both sides,\xa0{tex}\\Rightarrow 3 ={/tex}\xa0{tex}\\frac { a ^ { 2 } } { b ^ { 2 } }{/tex}Multiplying with {tex}b{/tex}\xa0on both sides,\xa0{tex}\\Rightarrow 3b ={/tex}\xa0{tex}\\frac { a ^ { 2 } } { b }{/tex}LHS = {tex}3\\times b{/tex}{tex}= Integer{/tex}RHS =\xa0{tex}\\frac { a ^ { 2 } } { b } = \\frac { \\text { Integer } } { \\text { Integer } }{/tex}{tex}= Rational\\ Number{/tex}{tex}\\Rightarrow \\mathrm { LHS } \\neq \\mathrm { RHS }{/tex}{tex}\\therefore{/tex}\xa0Our supposition is wrong.{tex}\\Rightarrow \\sqrt { 3 }{/tex}\xa0is irrational.Suppose\xa0{tex}15 + 17{/tex}{tex}\\sqrt 3{/tex}\xa0is a rational number.{tex}\\therefore 15 + 17 \\sqrt { 3 } = \\frac { a } { b }{/tex}, where {tex}a\\ and\\ b{/tex} are co-prime, {tex}b\\ne0{/tex}{tex}\\Rightarrow \\quad 17 \\sqrt { 3 } = \\frac { a } { b } - 15{/tex}{tex}\\sqrt { 3 } = \\frac { a - 15 b } { 17 b }{/tex}{tex}\\frac { a - 15 b } { 17 b }{/tex}\xa0is rational number,{tex}\\sqrt 3{/tex}\xa0is irrational.{tex}\\therefore \\quad \\sqrt { 3 } \\neq \\frac { a - 15 b } { 17 b }{/tex}{tex}\\therefore{/tex} Our supposition is wrong.{tex}\\Rightarrow \\quad 15 + 17 \\sqrt { 3 }{/tex}\xa0is irrational. thnx |
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| 7133. |
show that any positive odd integer is of the form 4q+1 or 4q+3,where a is a positive integer. |
| Answer» By Euclid\'s division algorithm,a = bq + r = 4q + rTake b = 4.Since, 0 {tex}\\leqslant{/tex}\xa0r < 4, r = 0,1, 2, 3{tex} a=4q,4q+1,4q+2 ,4q+3{/tex}Clearly, a =4q=2(2q) and\xa04q+2=2×(2q+1)So 4q and 4q+2 are evenTherefore 4q + 1, 4q + 3 are odd, as they are proceeding numbers of even numbers 4q and 4q+2.{tex}\\therefore{/tex}\xa0Any positive odd integer is of form 4q+1 or 4q+3 .Where q is a positive integer. | |
| 7134. |
The p Term of an a.p. is q and q term of a.p. is p . Find its p+q tetm |
| Answer» ap=a+(p-1)d=q ................. iaq=a+(q-1)d=p...................iisubtracting ii\xa0from i ( ii - i)(q-1 -p+1)d=p - q(q-p)d=p - qd=p - q / -(p - q) d = - 1\xa0adding i and ii (i + ii)2a+(p-1+q-1) (-1)=p+q2a+(-p-q+2)=p+q2a-p-q+2=p+q2a=2(p+q-1)a = p+q-1tp+q\xa0=a+(p+q-1)d ......... iii putting the values of a and d in iii\xa0tp-q = p+q-1 + (p+q-1)(-1) = p+q-1-p-q+1 = 0p+q term =0\xa0 | |
| 7135. |
show that _/3 is irrational number. |
| Answer» We have to prove √3 is irrational Let us assume the opposite, i.e., √3 is rational Hence, √3 can be written in the form ?/? where a and b (b≠ 0) are co-prime (no common factor other than 1) Hence, √3 = ?/? √3 b = a Squaring both sides (√3b)2 = a2 3b2 = a2 ?^2/3 = b2 Hence, 3 divides a2 So, 3 shall divide a also Hence, we can say ?/3 = c where c is some integer So, a = 3c Now we know that 3b2 = a2 Putting a = 3c 3b2 = (3c)2 3b2 = 9c2 b2 = 1/3 × 9c2 b2 = 3c2 ?^2/3 = c2 Hence 3 divides b2 So, 3 divides b also By (1) and (2) 3 divides both a & b Hence 3 is a factor of a and b So, a & b have a factor 3 Therefore, a & b are not co-prime. Hence, our assumption is wrong ∴ By contradiction, √3 is irrational | |
| 7136. |
4s+4s=8 |
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Answer» If it is problem then s=1 No, it will be 8s |
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| 7137. |
Sin2degree+sin4degree+......+sin180degree=? |
| Answer» 1 | |
| 7138. |
Alpha2+beta2= |
| Answer» | |
| 7139. |
Find the least number that is divisible by all the numbers between 1 and 10. |
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Answer» LEAST NO DIVISIBLE BY 1,2,3,4,5,6,7,8,9,10FIND THE LCM OF THESE NOS1=12=2X13=3X14=2X25=5X16=2X37=7X18=2X2X29=3X310=5X2LCM = 5X7X8X9=2530 1 is the number |
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| 7140. |
if 4sinA + 3cosA=4 Then find 3sinA - 4cosA=? |
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| 7141. |
State and prove table\'s theorem |
| Answer» | |
| 7142. |
Find the HCF of 117 and 65 |
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Answer» 117=13 x 3\xa0x 365=13 x 513 is highest common factor (HCF) 13 13 |
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| 7143. |
If m,n and t are in ap then prove(M+2n-t) (2n+t-m) (t+m-n) = 4mnt |
| Answer» | |
| 7144. |
Tan theeta + cot theeta = 2 it is given . To find tan square theeta + cot square theeta.? |
| Answer» Tan+Cot =2(Tan+Cot)^2=2^2Tan ^2+ Cot^2+ 2tan×cot =4Tan^2+Cot^2+2×tan×1/tan=4Tan^2+Cot^2=4-2=2Hence,Tan^2+Cot^2=2 | |
| 7145. |
Find hcf of 793 and 305 by Euclid method |
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Answer» continuing from abovenow check ur answer 793/61=13 , 305/61=5 (\xa0u may contact me on 9818299692 if u have any doubt) step 1\xa0take bigger no (793) as dividend and smaller no (305) as divisor793=2x305+183step 2take 305 as dividend and 183 as divisor305=1x183+122step 3take 183 as dividend and 122 as divisor183=1x122+61step 4take 122 as dividend and 61 as divisor122=2x61+0 (we get zero as remainder to stop as hcf is 61)hence HCF of 793 and 305 is 61 |
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| 7146. |
Find all the trigonometric ratio of 30degree geometrically |
| Answer» | |
| 7147. |
1265*65321 |
| Answer» 82631065 | |
| 7148. |
in the following APs, find the missing the terms in the boxes _,13,_3 |
| Answer» 18 &8 | |
| 7149. |
What is optical lobe? |
| Answer» The occipital lobe is one of the four major lobes of the cerebral cortex in the brain of mammals. The occipital lobe is the visual processing center of the mammalian brain containing most of the anatomical region of the visual cortex. | |
| 7150. |
What is appolinuos thm |
| Answer» | |