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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.
| 11301. |
Find the quadratic polynomial ,the sum of whose roots is √2 and their product is 1/3 |
| Answer» Let the roots of quadratic equation be {tex}\\alpha{/tex}\xa0and {tex}\\beta{/tex}.Then Sum of roots = {tex}\\alpha{/tex}\xa0+ {tex}\\beta{/tex}\xa0= {tex}\\sqrt 2 {/tex}=> {tex}{-b \\over a} = {{\\sqrt 2 } \\over 1}{/tex}=> {tex}{-b \\over a} = {{\\sqrt 2 } \\over 1} \\times {3 \\over 3}{/tex} => {tex}{-b \\over a} = {{3\\sqrt 2 } \\over 3}{/tex} ..............(i)And, Product of roots = {tex}\\alpha{/tex}.{tex}\\beta{/tex}\xa0= {tex}{1 \\over 3}{/tex}=> {tex}{c \\over a} = {1 \\over 3}{/tex} ................(ii)On comparing both the equations, we have{tex}a = 3,b = - 3\\sqrt 2 ,c = 1{/tex}Putting these values in the general quadratic equaiton {tex}a{x^2} + bx + c = 0{/tex}, we get{tex}3{x^2} - 3\\sqrt 2 x + 1 = 0{/tex} | |
| 11302. |
Use Euclid\'s division algorithm to find the HCF 27727 and\xa053124 |
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Answer» we take 53124 as the dividend and 27727 as the divisorby Euclid\'s Div Lemma53124 = (27727 x 1) + 25397now we take 27727 as dividend and 25397 as the divisor27727 = (25397 x 1 ) + 2330now we take 25397 as dividend and 2330 as the divisor25397 = (2330 x 10) + 2097now we take 2330 as the dividend and 2097 as the divisor2330 = (2097 x 1) + 233now we take 2097 as the dividend and 233\xa0as the divisor2097 = (233 x 9\xa0) + 0Hence the HCF of 27727 and 53124 is 233\xa0\xa0\xa0 53124 =\xa027727 x 1\xa0+ 25397=> 27727 = 25397 x 1 + 2330=> 25397 = 2330 x 10 + 2097=> 2330 = 2097 x 1 + 233=> 2097 = 233 x 9 + 0Since on finding remainder 0, we have the divisor 233.Therefore, H.C.F.(53124, 27727) = 233 |
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| 11303. |
Find the median of the following data :X-6,5,8,10,7,12,15F-4,2,6,5,8,3,9 |
| Answer» | |
| 11304. |
The following are the marks of 9 students in a class find the median 34,32,48,38,24,30,27,21,35 |
| Answer» To find median write the number in either increasing or decreasing order21,24,27,30,32,34,35,38,48Number of terms = 9 [odd number]So median =\xa0{tex}({Number\\ of\\ terms + 1\\over 2})^{th} term{/tex}=\xa0{tex}({9+1\\over 2})^{th} term = 5^{th} term {/tex}= 32 | |
| 11305. |
Find the largest no. Which divides 546 and 764 , leaving remainder 6 and 8 respectively |
| Answer» Ans. Subtract the remainder from each number, we get546 - 6 = 540\xa0764 - 8 = 756Using Euclid Lemma Division, We ll find HCF of 540 and 756{tex}=> 756 = 540\xa0\\times 1 + 216 {/tex}{tex}=> 540 = 216\xa0\\times 2 + 108 {/tex}{tex}=> 216 = 108\xa0\\times 2 + 0 {/tex}So 108 is the required Number. | |
| 11306. |
If a and b are the zeros of p(x) x2-2x+3 . Find the polynomial whose zeros are a-1/a+1 and b-1/b+1 |
| Answer» Ans. Given: {tex}p\\left( x \\right) = {x^2} - 2x + 3{/tex}\xa0, a and b are zeroes of given polynomial.\xa0We KnowSum of zeroes = {tex}-b\\over a{/tex}=> a + b = 2 andproduct of zeroes = {tex}c\\over a{/tex}=> ab = 3For required polynomial,Sum of zeroes = {tex}{{a - 1} \\over {a + 1}} + {{b - 1} \\over {b + 1}}{/tex}= {tex}{{\\left( {a - 1} \\right)\\left( {b + 1} \\right) + \\left( {b - 1} \\right)\\left( {a + 1} \\right)} \\over {\\left( {a + 1} \\right)\\left( {b + 1} \\right)}}{/tex}= {tex}{{ab + a - b - 1 + ab + b - a - 1} \\over {ab + a + b + 1}}{/tex}\xa0= {tex}{{2ab - 2} \\over {ab + a + b + 1}}{/tex}\xa0= {tex}{{2 \\times 3 - 2} \\over {3 + 2 + 1}}{/tex}\xa0= {tex}{4 \\over 6}{/tex}{tex}\\Rightarrow{/tex}{tex}{{ - b} \\over a} = {4 \\over 6}{/tex}Product of zeroes = {tex}\\left( {{{a - 1} \\over {a + 1}}} \\right)\\left( {{{b - 1} \\over {b + 1}}} \\right){/tex}\xa0= {tex}{{ab - a - b + 1} \\over {ab + a + b + 1}}{/tex}= {tex}{{ab - \\left( {a + b} \\right) + 1} \\over {ab + \\left( {a + b} \\right) + 1}}{/tex}\xa0= {tex}{{3 - 2 + 1} \\over {3 + 2 + 1}}{/tex}{tex}\\Rightarrow{/tex}{tex}{c \\over a} = {2 \\over 6}{/tex}On comparing, we get, a = 6, b = -4\xa0and c = 2Putting these values in the general polynomial {tex}a{x^2} + bx + c{/tex},the required polynomias is\xa0{tex}6x^2-4x+2{/tex} | |
| 11307. |
Find all the zeros of the polynomialx^4-3x^3-5x^2+22x-14 if two of the zeros are√7 and -√7. |
| Answer» Since {tex}\\sqrt 7{/tex}\xa0and {tex}-\\sqrt 7{/tex}\xa0are the zeroes of the given polynomial.This means (x−{tex}\\sqrt 7{/tex}) and (x+{tex}\\sqrt 7{/tex}) are the factors of the given polynomial.So dividing p(x) = x4−3x3−5x2+21x−14 by (x−{tex}\\sqrt 7 {/tex})(x+{tex}\\sqrt 7{/tex}) i.e. x2−7using long division method we get;\xa0So x2−3x+2 is the factor of the given polynomial. To find other zeroes equate it to 0 we get;x2−3x+2 =0=>x2−2x-x+2 =0=> x(x-2)-1(x-2)=0=> (x-2)(x-1)=0=> x= 1,2 | |
| 11308. |
From a quadtric polynomial whose zeroes are 3-√3÷5 and 3+√3÷5. |
| Answer» Ans. Given : Zeroes of Polynomial are\xa0{tex}{3-\\sqrt 3\\over 5 }\\ and \\ {\\ 3+\\sqrt 3 \\over 5}{/tex}Sum of zeroes=\xa0{tex}{b\\over a}{/tex} =\xa0{tex}{3-\\sqrt 3\\over 5 } + {\\ 3+\\sqrt 3 \\over 5} = {6\\over 5}{/tex}=> {tex}{-b\\over a } = {30\\over 25}{/tex}\xa0Product of Zeroes = {tex}{c\\over a}{/tex}=\xa0{tex}{3-\\sqrt 3\\over 5 }\\times {\\ 3+\\sqrt 3 \\over 5} = {6\\over 25}{/tex}On comparing, we get = b = -30a = 25 and c = 6So now Required quadratic polynomial = 25x2 - 30x + 6\xa0\xa0 | |
| 11309. |
if the zeroes of x2-kx+6 are in the ratio 3:2, find k. |
| Answer» Let roots are {tex}\\alpha \\ and \\ \\beta {/tex}{tex}{\\alpha \\over \\beta } = {3\\over 2}{/tex}=>\xa0{tex}\\alpha ={ 3\\over 2}\\beta{/tex}.....(1)ThenProduct of roots =\xa0{tex}c\\over a{/tex}=>\xa0{tex}\\alpha \\beta = 6{/tex}=>\xa0{tex}{3\\over 2}\\beta \\beta = 6{/tex}=>\xa0{tex}\\beta^2 = 4{/tex}=>\xa0{tex}\\beta = \\pm 2{/tex}Putting in (1) we get{tex}\\alpha = \\pm 3{/tex}Now sum of roots =\xa0{tex}-{b\\over a}{/tex}=> 3 + 2 = k or -3 + (-2) = k\xa0=> k = 5 or k = -5\xa0 | |
| 11310. |
In an A.P. the fifth term and tenth term are in the ratio of 1:2 if T12=36 find A.P.\xa0 |
| Answer» Let first term of AP = aCommon difference = dThenT12= a + (12-1)d=> 36 = a + 11d .....(1)Also{tex}{T_5\\over T_{10}} = {1\\over 2}{/tex}=>\xa0{tex}{a+4d\\over a+9d}={1\\over 2}{/tex}=> 2a+8d = a+9d=> a = d ...(2)Put a = d in (1) we get=> 36 = d + 11d=> 36 = 12d=> d = 3From (2) a = 3AP : 3,6,9,12,15 | |
| 11311. |
The distance between the points of co-ordinates (2,3) and (4,1) is |
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Answer» We know\xa0Distance formula =\xa0{tex}\\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}{/tex}So,\xa0={tex}\\sqrt {(4-2)^2 + (1-3)^2}{/tex}=\xa0{tex}\\sqrt {4+4} = 2\\sqrt 2{/tex} Distance between two coordinates = {tex}\\sqrt {{{\\left( {{x_2} - {x_1}} \\right)}^2} + {{\\left( {{y_2} - {y_1}} \\right)}^2}} {/tex}Given: {tex}\\left( {{x_1},{x_2}} \\right) = \\left( {2,3} \\right){/tex}\xa0and {tex}\\left( {{y_1},{y_2}} \\right) = \\left( {4,1} \\right){/tex}Then, Distance = {tex}\\sqrt {{{\\left( {4 - 2} \\right)}^2} + {{\\left( {1 - 3} \\right)}^2}} {/tex}\xa0= {tex}\\sqrt {{{\\left( 2 \\right)}^2} + {{\\left( { - 2} \\right)}^2}} {/tex} = {tex}\\sqrt {4 + 4} {/tex}\xa0= {tex}2\\sqrt 2 {/tex}\xa0units |
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| 11312. |
If the difference between the roots of the equation x2+px+8=0, then what is the value of p? |
| Answer» Ans, Let\xa0{tex}\\alpha \\ and \\ \\beta {/tex}\xa0be the roots of equation.Then\xa0{tex}\\alpha \\ - \\beta = 2 \\ \\ ........... (1) [Given ]{/tex}We know,\xa0Sum of roots =\xa0{tex}-{b\\over a}{/tex}{tex}\\alpha \\ + \\beta = -p \\ .............. (2) {/tex}Adding (1) and (2), we get{tex}2\\alpha = 2- p {/tex}=>\xa0{tex}\\alpha = {2-p\\over 2}{/tex}Subtracting (1) from (2), we get\xa0{tex}2\\beta = - p -2 {/tex}=>\xa0{tex}\\beta = {-p-2\\over 2}{/tex}Also,Product of roots =\xa0{tex}c\\over a{/tex}=>\xa0{tex}\\alpha \\times \\beta = 8{/tex}=>\xa0{tex}{2-p\\over 2} \\times {-2-p\\over 2} = 8 {/tex}=>\xa0{tex} -4 -2p+2p+p^2 = 32{/tex}=> p2\xa0= 36=> p = -6, 6\xa0 | |
| 11313. |
Check whether the digit 7 power n ends with the digit zero\xa0 |
| Answer» Every number that end with zero has 2 and 5 as its prime factor.As 7n don\'t have both 2 and 5 as prime factor. It ll not end with zero\xa0 | |
| 11314. |
L.c.m of two numbers is 2175 h.c.f is 144one number is 725 find the other number.\xa0 |
| Answer» Let other numebr be x.We know that,Product of two numbers = L.C.M. x H.C.F.725 x {tex}x{/tex}\xa0= 2175 x 144=> {tex}x = {{2175 \\times 144} \\over {725}}{/tex}=> {tex}x = 432{/tex}Therefore, the other number is 432.\xa0 | |
| 11315. |
If tan theta + sin theta=m, tan theta - sin theta = n,show that m2-n2=4{(mn)1\\2} |
| Answer» L.H.S. {tex}{m^2} - {n^2}{/tex}\xa0= {tex}{\\left( {\\tan \\theta + \\sin \\theta } \\right)^2} - {\\left( {\\tan \\theta - \\sin \\theta } \\right)^2}{/tex}= {tex}\\left( {\\tan \\theta + \\sin \\theta + \\tan \\theta - \\sin \\theta } \\right)\\left( {\\tan \\theta + \\sin \\theta - \\tan \\theta + \\sin \\theta } \\right){/tex} [Since {tex}{a^2} - {b^2} = \\left( {a + b} \\right)\\left( {a - b} \\right){/tex}]= {tex}\\left( {2\\tan \\theta } \\right)\\left( {2\\sin \\theta } \\right){/tex}= {tex}4 \\times {{\\sin \\theta } \\over {\\cos \\theta }} \\times \\sin \\theta {/tex}= {tex}{{4{{\\sin }^2}\\theta } \\over {\\cos \\theta }}{/tex}R.H.S. {tex}4\\sqrt {mn} {/tex}\xa0= {tex}4\\sqrt {\\left( {\\tan \\theta + \\sin \\theta } \\right)\\left( {\\tan \\theta - \\sin \\theta } \\right)} {/tex}= {tex}4\\sqrt {{{\\tan }^2}\\theta - {{\\sin }^2}\\theta } {/tex} [Since {tex}{a^2} - {b^2} = \\left( {a + b} \\right)\\left( {a - b} \\right){/tex}]= {tex}4\\sqrt {{{{{\\sin }^2}\\theta } \\over {{{\\cos }^2}\\theta }} - {{\\sin }^2}\\theta } {/tex}= {tex}4\\sqrt {{{{{\\sin }^2}\\theta - {{\\sin }^2}\\theta .{{\\cos }^2}\\theta } \\over {{{\\cos }^2}\\theta }}} {/tex}= {tex}4\\sqrt {{{{{\\sin }^2}\\theta \\left( {1 - {{\\cos }^2}\\theta } \\right)} \\over {{{\\cos }^2}\\theta }}} {/tex}= {tex}4\\sqrt {{{{{\\sin }^2}\\theta .{{\\sin }^2}\\theta } \\over {{{\\cos }^2}\\theta }}} {/tex}= {tex}{{4{{\\sin }^2}\\theta } \\over {\\cos \\theta }}{/tex}L.H.S. = R.H.S. | |
| 11316. |
What is procedure of using elimination method? |
| Answer» Ans. The elimination method of solving systems of equations is also called the addition method. To solve a system of equations by elimination we transform the system such that one variable "cancels out".Here are the steps to follow:Step 1 : Try to eliminate a variable as you add the left sides and the right sides of the two equationsStep 2 : Set the sum resulting from adding the left sides equal to the sum resulting from adding the right sidesStep 3 : Solve for the variable that was not cancelled or eliminatedStep 4 : Use the answer found in step 3 to solve for the other variable by substituting this value in one of the two equations | |
| 11317. |
tan²A+cot²A+2=sec²A cosec²A |
| Answer» tan2a + cot2a + 2= sec2a - 1 + cosec2a - 1 + 2= sec2a - 2+2 + cosec2a=sec2a + cosec2a | |
| 11318. |
Write the quadratic polynomial having -1/4,1 as it\'s zeroes\xa0 |
| Answer» Ans. Let equation is ax2\xa0+ bx + c = 0, roots are\xa0{tex}-{1\\over 4},1{/tex}.We know that,Sum of roots =\xa0{tex}-{b\\over a}{/tex}=>\xa0{tex}-{1\\over 4} + 1 = -{b\\over a}{/tex}{tex}=> {3\\over 4} = -{b\\over a}{/tex}=> b = -3 and a = 4AlsoProduct of roots =\xa0{tex}c\\over a{/tex}=>\xa0{tex}-{1\\over 4}\\times 1 = {c\\over a}{/tex}{tex}=> -{1\\over 4}= {c\\over a}{/tex}c = -1 and a = 4Put Values of a,b,c in equation we get,4x2\xa0- 3x -1 = 0 Requried equation | |
| 11319. |
If one zero of the quadratic polynomial x2+3x+k is 7,then find the value of k |
| Answer» Let p(x) = x2 + 3x + kIf one zero of the p(x) is 7, thenp(7) = 0=> (7)2 + 3 x 7 + k = 0=> 49 + 21 + k = 0=> 70 + k = 0=> k = -70 | |
| 11320. |
If smallest factor of a is 3.smallest factor of b is 7.find smallest factor of a+b |
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Answer» As a has 3 as smallest\xa0factor and b has 7 as smallest factor it means both numbers are odd. Sum of two odd numbers is always a even number so the smallest factor of (a+b) ll be 2. smallest factor of any nnumber is 1. therefore 3 and 7 can never be the smallest factor of required number |
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| 11321. |
What is the value of the following\xa01. 002. 0/0\xa0 |
| Answer» 00\xa0= 1{tex}{0\\over 0} = undefined {/tex} | |
| 11322. |
Does arihant publishes ncert exemplar. |
| Answer» Yes it publishes | |
| 11323. |
If alpha and beta are the zeroes of the polynomial 3x2-6x-7x.find alpha2\xa0and beta2 |
| Answer» 3x^2-6x-7x=03x^2-13x=0x(3x-13)=0therfore zeroe are 0 and 13/3alpha^2= o^2=0beta^2=(13/3) = 169/9........... | |
| 11324. |
find the common difference and next 3 terms of AP 3,-2,-7,-12 |
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Answer» Common difference = {tex}-2-3=-5{/tex}Therefore, next three terms are:{tex}-12+(-5)=-12-5=-17{/tex}{tex}-17+(-5)=-17-5=-22{/tex}{tex}-22+(-5)=-22-5=-27{/tex} its common difference is 5next 3 terms will be -17,-22,-27,-32 and so on\xa0 |
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| 11325. |
What will be value of (6561)0.25\xa0?Plz solve it also. |
| Answer» {tex}(6561)^{0.25}{/tex}{tex}= (6561)^{25\\over100}{/tex}{tex}= (6561)^{1\\over 4}{/tex}{tex}=[(81)^{2}]^{1\\over 4}{/tex}{tex}=(81)^{1\\over 2}{/tex}= 9 | |
| 11326. |
Factorise\xa0X2+2√2x-6\xa0 |
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Answer» Given expression=x2+2√2x-6 =x2+3√2x-√2x-6 =x(x+3√2)-√2(x+3√2) =(x+3√2)(x-√2) {tex}{x^2} + 2\\sqrt 2 x - 6{/tex}= {tex}{x^2} + 3\\sqrt 2 x - \\sqrt 2 x - 6{/tex}= {tex}x\\left( {x + 3\\sqrt 2 } \\right) - \\sqrt 2 \\left( {x + 3\\sqrt 2 } \\right){/tex}= {tex}\\left( {x + 3\\sqrt 2 } \\right)\\left( {x - \\sqrt 2 } \\right){/tex} √2,-3√2. |
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| 11327. |
\xa0Write the zeroes of the polynomial 4x2\xa0–4 x +1. |
| Answer» 4x2 - 4x + 1= (2x)2 - 2 . 2x . 1 + (1)2= (2x - 1)2 [Using identity a2 - 2ab + b2]= (2x - 1)(2x - 1)Taking, 2x - 1 = 0 and 2x - 1 = 0x = 1/2 and x = 1/2Therefore, the zeroes of given polynomial are 1/2 and 1/2 | |
| 11328. |
Please tell me how can i answer betterly in maths exam |
| Answer» Start with questions having high marks.Eg: 4 marks . then go for next .Highlight ur answer with pencil.don\'t take too much time for one sum go for next sum. | |
| 11329. |
How to manage time in maths exam?\xa0 |
| Answer» There is always a problem of time in maths exam paper. I will tell you the correct method of doing the maths paper.Always start the paper from last section, and then second last and so on. Because at the starting of paper your mind is fresh, you can handle the long questions easily which are given in last section.But if you start from first section, then sometimes 1 marks quetion take your 4 to 5 mins. And in last 1 hour you confuse what to do last section of long questions which 5 marks each. And you will leave those quetion because of short time. This effects your marks too much. But if you start from last section and in the end of the paper even your complete A section leave, then it matters only 5 marks(1 marks for each question very short answer type question). This not effects your marks too much. | |
| 11330. |
X2 _\xa02(a2+ b2)x + (a2_\xa0b2)2\xa0= 0Find the roots?\xa0 |
| Answer» {tex}{x^2} - 2\\left( {{a^2} + {b^2}} \\right) + {\\left( {{a^2} - {b^2}} \\right)^2} = 0{/tex}Using {tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex}{tex}x = {{2\\left( {{a^2} + {b^2}} \\right) \\pm \\sqrt {{2^2}.{{\\left( {{a^2} + {b^2}} \\right)}^2} - 4 \\times 1 \\times {{\\left( {{a^2} - {b^2}} \\right)}^2}} } \\over {2 \\times 1}}{/tex}{tex}x = {{2\\left( {{a^2} + {b^2}} \\right) \\pm \\sqrt {{{\\left\\{ {4\\left( {{a^2} + {b^2}} \\right)} \\right\\}}^2} - 4{{\\left( {{a^2} - {b^2}} \\right)}^2}} } \\over 2}{/tex}{tex}x = {{2\\left[ {\\left( {{a^2} + {b^2}} \\right) \\pm 2\\sqrt {{{\\left( {{a^2} + {b^2}} \\right)}^2} - {{\\left( {{a^2} - {b^2}} \\right)}^2}} } \\right]} \\over 2}{/tex}{tex}x = {a^2} + {b^2} \\pm \\sqrt {\\left( {{a^2} + {b^2} + {a^2} - {b^2}} \\right)\\left( {{a^2} + {b^2} - {a^2} + {b^2}} \\right)} {/tex}{tex}x = {a^2} + {b^2} \\pm \\sqrt {\\left( {2{a^2}} \\right)\\left( {2{b^2}} \\right)} {/tex}{tex}x = {a^2} + {b^2} \\pm 2{a}{b}{/tex}Taking positive sign Taking negative sign{tex}x = {a^2} + {b^2} + 2{a}{b}{/tex}{tex}x = {a^2} + {b^2} - 2{a}{b}{/tex}{tex}x = {\\left( {a + b} \\right)^2}{/tex}{tex}x = {\\left( {a - b} \\right)^2}{/tex} | |
| 11331. |
What r all the important concepts that we should look in for maths board exam class 10 ? |
| Answer» | |
| 11332. |
Find the value of k for which the equation x2+4x+k=0 has coincident roots |
| Answer» Ans. x2\xa0+ 4x + k = 0,\xa0On Comparing with ax2 + bx + c = 0,We get a = 1, b = 4 and c = k\xa0For Coincident roots D = 0=> b2 - 4ac = 0=> 42\xa0- 4 (1)(k) = 0=> 16 - 4k =0=> 4k = 16\xa0=> k = 4\xa0 | |
| 11333. |
If ratio of the roots of equation px2+qx+q=0 is a:b, prove that √a÷b + √b÷a + √q÷p=0. |
| Answer» Let the roots of the given equation be {tex}\\alpha {/tex}\xa0and {tex}\\beta{/tex}.Then, {tex}\\alpha {/tex}\xa0+ {tex}\\beta{/tex}\xa0= {tex}{{ - q} \\over p}{/tex} and {tex}\\alpha .\\beta = {q \\over p}{/tex}Given: {tex}{\\alpha \\over \\beta } = {a \\over b}{/tex}Now, {tex}\\sqrt {{a \\over b}} + \\sqrt {{b \\over a}} + \\sqrt {{q \\over p}} = 0{/tex}=> {tex}\\sqrt {{\\alpha \\over \\beta }} + \\sqrt {{\\beta \\over \\alpha }} + \\sqrt {\\alpha \\beta } = 0{/tex} [Since, {tex}{\\alpha \\over \\beta } = {a \\over b}{/tex}\xa0and {tex}\\alpha .\\beta = {q \\over p}{/tex}]=> {tex}{{\\sqrt \\alpha } \\over {\\sqrt \\beta }} + {{\\sqrt \\beta } \\over {\\sqrt \\alpha }} + \\sqrt {\\alpha \\beta } = 0{/tex}=> {tex}{{\\alpha + \\beta + \\alpha \\beta } \\over {\\sqrt {\\alpha \\beta } }} = 0{/tex}=> {tex}\\alpha + \\beta + \\alpha \\beta = 0{/tex}=> {tex}{{ - q} \\over p} + {q \\over p} = 0{/tex} [Since, {tex}\\alpha + \\beta = {{ - q} \\over p}{/tex}\xa0and {tex}\\alpha \\beta = {q \\over p}{/tex}]=> {tex}-q+q=0{/tex}=> 0 = 0Hence proved. | |
| 11334. |
is_/3,_/6_/9_/12 form an a.p. ??sir I need answer immediately\xa0 |
| Answer» Common difference,{tex}d_1 = \\sqrt { 6 } - \\sqrt { 3 }{/tex}{tex}= \\sqrt { 3 } ( \\sqrt { 2 } - 1 ){/tex}{tex}d_2= \\sqrt { 9 } - \\sqrt { 6 }{/tex}{tex}= \\sqrt { 3\\times3 } - \\sqrt{2\\times 3}{/tex}{tex}= 3 - \\sqrt { 6 }{/tex}{tex}d_3 = \\sqrt { 12 } - \\sqrt { 9 } {/tex}{tex}= \\sqrt { 4\\times3 } - \\sqrt { 9 } {/tex}{tex}= 2 \\sqrt { 3 } - 3{/tex}As common difference does not equal.Hence, The given series is not in A.P. | |
| 11335. |
Find four consecutive terms in AP whose sun is 20 and the sum of whose square is 120 ? |
| Answer» Ans. Let first term of A.P. is a-2dCommon difference is d,Then four terms are, a-2d, a-d, a and a+dAccording To Ques,=> a - 2d + a - d + a + a + d = 20=> 4a - 2d = 20=> 2a - d = 10=> d = 2a - 10 ……………(1)Also,=> (a-2d)2 + (a-d)2 + a2 +(a+d)2 = 120put value of d from (1)=> (a - 4a+20)2\xa0+(a-2a+10)2\xa0+\xa0a2 +(a+2a-10)2= 120=> (20-3a)2 + (10-a) + a2 + (3a-10)2\xa0= 120=> 400 + 9a2 - 120a + 100+ a2- 20a + a2+ 9a2\xa0+ 100 - 60a = 120=> 20a2 - 200a + 600 = 120divide by 20=> a2 - 10a + 30= 6=> a2 - 10a + 24\xa0= 0=> a2\xa0- 6a - 4a + 24 =0=> a(a-6) -4(a-6) = 0=> (a-6)(a-4) = 0=> a = 6, 4if a = 6 then d = 2and if a = 4 then d = -2First A.P = 2, 4, 6, 8,....second A.P. = 8, 6, 4, 2,...\xa0\xa0 | |
| 11336. |
In a family of 3children the probability of having at least one boy is : |
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Answer» Ans. Three children can be as follow:BBB, BBG, BGB, BGG,\xa0GBB, GBG,\xa0GGB, GGG.No of total cases = 8No of favorable cases = 7Probability = 7/8 \t7/8 |
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| 11337. |
How many terms of AP65 60 55 be taken to give the sun zero\xa0 |
| Answer» Ans. First term = 65Common Difference = -5Let number of terms required = nSn\xa0= 0=>\xa0{tex}{n\\over 2}[{2\\times 65 +(n-1)\\times (-5)}] =0{/tex}=> n[130 -5n +5] = 0=> n (135-5n)=0=> 135-5n = 0=> 135 = 5n=> n = 27number of terms = 27 | |
| 11338. |
Theorem of sa 2.which will come in board exam .\xa0 |
| Answer» We have only 2 theorems so anything can be asked for the safer side learn all the theorems..... | |
| 11339. |
If in the clock the given time is 2:30 then what\'s the angle formed at the given time 2:30.\xa0 |
| Answer» Angle formed by clock hands =360/12*4 =30*4 =120 degrees | |
| 11340. |
Factorise:11x^2-122x+11=0 |
| Answer» Its easy bro:11x2_122x+11=011x2 _121x - x+11=011x(x-11)-1(x-11)=0Rest u r smart...Best of luck | |
| 11341. |
Find the sum of all 3 digit numbers which leave remainder 3 when divided by 5. |
| Answer» Three digit numbers which leave the remainder 3 when divided by 5 are 103, 108, 113,......., 998.103, 108, 113,......, 998 is an A.PFirst term of the A.P,\xa0a\xa0= 103Common different of the A.P,\xa0d\xa0= 5Let 998 be the\xa0nth\xa0term of the A.P.an\xa0=\xa0a\xa0+ (n\xa0– 1)\xa0d∴ 103 + (n\xa0– 1) × 5 = 998⇒ 5 (n\xa0– 1) = 998 – 103 = 895 ⇒ (n\xa0– 1) = 179⇒ n\xa0= 180Sum of all three digit numbers which leaves remainder 3 when divided by 5was this answer useful??????\xa0 | |
| 11342. |
\xa0Find the sum of (1− 1) + (1− 2) + (1− 3) ……. upto n terms. n n n |
| Answer» We can write the given series as:(1+1+1+.....+n) - (1+2+3+........+n)= 1 X n - Sn= n -\xa0{tex}{ n (n+1) \\over 2}{/tex}=\xa0{tex}{n-n^2\\over 2}{/tex} | |
| 11343. |
Prove that am + n + am - n =2am |
| Answer» Let the first term and common difference of the A. P. be\xa0a\xa0and\xa0d\xa0respectively.Then (m\xa0+\xa0n)th\xa0term (am\xa0+\xa0n\xa0) =\xa0a\xa0+ (m\xa0+\xa0n\xa0– 1)\xa0dand (m\xa0–\xa0n)th\xa0term (am –\xa0n\xa0) =\xa0a\xa0+ (m\xa0–\xa0n\xa0– 1)\xa0d\xa0am +\xa0n\xa0+\xa0am –\xa0n\xa0=\xa0a\xa0+ (m\xa0+\xa0n\xa0– 1)\xa0d\xa0+\xa0a\xa0+ (m\xa0–\xa0n\xa0– 1)\xa0d= 2a\xa0+ (m\xa0+\xa0n\xa0– 1\xa0+\xa0m\xa0–\xa0n\xa0– 1)\xa0d= 2a\xa0+ (2m\xa0– 2)\xa0d\xa0= 2a\xa0+ 2 (m\xa0– 1)\xa0d= 2[a\xa0+ (m\xa0– 1)\xa0d] = 2am\xa0= 2 (m\xa0th\xa0term) | |
| 11344. |
if the quadratic equation x2+4x+k=0, has rreal and distinct roots, find the value of k |
| Answer» Ans. For real and distinct roots,D> 0=> b2-4ac > 0=> 16-16k > 0=> 1 - k > 0=> k < 1\xa0 | |
| 11345. |
If the sum of first n terms of an ap is n2 find the 5th\xa0term of the ap\xa0 |
| Answer» | |
| 11346. |
If the sum of first n terms of an ap is n2\xa0find the 5th\xa0term of the ap\xa0 |
| Answer» {tex}sum\\,of\\,first\\,n\\,terms\\,{S_n} = {n^2}{/tex}{tex}nth\\,term\\,{a_n} = {S_n} - {S_{n - 1}}{/tex}{tex} = {n^2} - {\\left( {n - 1} \\right)^2}{/tex}{tex} = {n^2} - \\left( {{n^2} - 2n + 1} \\right){/tex}{tex} = 2n - 1{/tex}{tex}5th\\,term\\,{a_5} = 2 \\times 5 - 1 = 9{/tex}\xa0 | |
| 11347. |
If one root of the quadratic Equation 2xsq+kx-6=0 is 2 then find the value of k+1. |
| Answer» As we know that root of an equation is one of the values which satisfies the equation, we have -2(2)2\xa0+ 2k - 6 =08+2k-6=02k+2=02 (k+1)=0k+1=0 Ans.\xa0 | |
| 11348. |
die is thrown twice. The probability that two will not come up either time is\xa0 |
|
Answer» Naha ke aa pehle\xa0 No. of favourable outcome (two will not come) = 25Total outcome = 36Required probability = 25/36 |
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| 11349. |
A circle with centre O has tangent PQ and PB. PQ is 9cm .Angle PAB is 60*. Find length of AB.\xa0 |
| Answer» The angle is PQB\xa0 | |
| 11350. |
Which term of the sequence 20,77/4,37/2,67/4 is the 1st negative term\xa0\xa0\xa0 |
| Answer» For the first negative term,an < 0{tex}20 + (n - 1) \\cdot \\left( {\\frac{{ - 3}}{4}} \\right) < 0{/tex}\xa0{tex}\\left[ {\\because d = \\frac{{77}}{4} - 20 = \\frac{{ - 3}}{4}} \\right]{/tex}{tex} \\Rightarrow \\frac{{20}}{1} - \\frac{3}{4}n + \\frac{3}{4} < 0{/tex}{tex} \\Rightarrow \\frac{{80 - 3n + + 3}}{4} < 0{/tex}{tex} \\Rightarrow 83 - 3n < 0{/tex}{tex} \\Rightarrow - 3n < - 83{/tex}{tex} \\Rightarrow n > \\frac{{83}}{3}{/tex}28th term is first negative term. | |