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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.
| 11501. |
mx\u200b\u200b\u200b\u200b2/x+x/m=1-2x |
| Answer» {tex}{mx^2\\over x } + {x\\over m} = 1-2x{/tex}{tex}=> {m^2x^2 + x^2\\over xm} = 1-2x{/tex}{tex}=> m^2x^2 + x^2 = xm-2x^2m{/tex}{tex}=> m^2x^2 + x^2 + 2x^2m = xm{/tex}{tex}=> x^2(m^2 + 1 + 2m )= xm{/tex}{tex}=> x(m^2 + 1 + 2m )= m{/tex}{tex}=> x= {m\\over (m^2 + 1 + 2m )}{/tex}{tex}=> x= {m\\over (m + 1)^2}{/tex} | |
| 11502. |
the sum of two naturals number is8and sum of their reciprocal is 8/15. find the numbers. |
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Answer» Let first no. be x and second no. 8-xAccording to question1/x + 1/8-x = 8/158/8x - x2\xa0= 8/158x - x2\xa0\u200b\u200b\u200b\u200b\u200b\u200b= 15x2\xa0- 8x + 15 = 0x2\xa0- 5x - 3x - 15 = 0x = 3, 5Therefore first no. = 3 ; second no = 5 Let One number = xThen 2nd number = 8-xAccording to question{tex}{1\\over x} + {1\\over 8-x}= {8\\over 15}{/tex}{tex}=> {8-x+x\\over x(8-x)} = {8\\over 15}{/tex}{tex}=> {8\\over x(8-x)} = {8\\over 15} {/tex}{tex}=> {1\\over 8x-x^2} = {1\\over 15} {/tex}{tex}=> 8x-x^2 = {15} {/tex}{tex}=>x^2 -8x+ {15} =0{/tex}{tex}=>x^2 -5x-3x+ {15} =0{/tex}=> x(x-5)-3(x-5)=0=> (x-5)(x-3)=0=> x = 5 or 3If first number is 3 then 2nd number is 5.If first number is 5 then 2nd number is 3. |
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| 11503. |
make a daily life situation of pair of linear equation and show it garphically. |
| Answer» | |
| 11504. |
If one zero of the polynomial x2\xa0+ a is (-3) then find its other zero. |
| Answer» If -3 is a zero of x2 + a, then(-3)2 + a = 0=> 9 + a = 0=> a = -9Then the given polynomial will be x2\xa0 - 9.=> (x + 3)(x - 3)=> x + 3 = 0 and x - 3 = 0=> x = -3 and x = 3Therefore the other zero of given polynomial is 3. | |
| 11505. |
tan A ÷ ( 1 + tan2 A )2 + cot A ÷ ( 1 + cot2\xa0A )2\xa0= sin A cos A |
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Answer» Take LHS{tex}{tan A\\over (1+tan^2A)^2}+{cotA\\over (1+cot^2A)^2}{/tex}{tex}= {tan A\\over sec^4A}+{cotA\\over cosec^4A}{/tex}{tex}={sin A\\over cosAsec^4A}+{cosA\\over sinAcosec^4A}{/tex}{tex}= {sin A\\over sec^3A}+{cosA\\over cosec^3A}{/tex}{tex}= {sin Acos^3A}+{cosAsin^3A}{/tex}{tex}= {sin AcosA}({cos^2A+sin^2A}){/tex}=> sinAcosA = RHSHence Verified\xa0 {tex}{tan A\\over (1+tan^2A)^2}+{cotA\\over (1+cot^2A)^2}= sinAcosA{/tex}Taking LHS{tex}{tan A\\over (1+tan^2A)^2}+{cotA\\over (1+cot^2A)^2}{/tex}=\xa0{tex}{tan A\\over (sec^2A)^2}+{cotA\\over (cosec^2A)^2}{/tex}=\xa0{tex}{tan A\\over sec^4A}+{cotA\\over cosec^4A}{/tex}=\xa0{tex}{sin A\\over cosAsec^4A}+{cosA\\over sinAcosec^4A}{/tex}{tex}= {sin A\\over sec^3A}+{cosA\\over cosec^3A}{/tex}\xa0{tex}= {sin Acos^3A}+{cosAsin^3A}{/tex}{tex}= {sin AcosA}({cos^2A+sin^2A}){/tex}= sinAcosA = RHSHence proved\xa0 |
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| 11506. |
Cot A + tan B ÷ cot B + tan A = cot A tan B |
| Answer» {tex}\\frac { cotA+tanB }{ cotB+tanA } =\\frac { \\frac { 1 }{ tanA } +tanB }{ \\frac { 1 }{ tanB } +tanA } =\\frac { \\frac { 1+tanB.tanA }{ tanA } }{ \\frac { 1+tanA.tanB }{ tanB } } \\quad \\quad \\quad \\quad \\left[ \\frac { 1 }{ tan\\theta } =cot\\theta \\right] \\quad \\\\ =\\frac { 1+tanB.tanA }{ tanA } .\\frac { tanB }{ 1+tanA.tanB } =\\frac { tanB }{ tanA } =tanB.\\frac { 1 }{ tanA } =cotA.tanB{/tex} | |
| 11507. |
Cot2 A ( sec A - 1 ) / 1 + sin A = sec2 A ( 1 - sin A ) / (1 + sec A ) |
| Answer» L.H.S. = {tex}{{{{\\cot }^2}{\\rm{A}}\\left( {\\sec {\\rm{A}} - 1} \\right)} \\over {1 + \\sin {\\rm{A}}}}{/tex}= {tex}{{{{\\cot }^2}{\\rm{A}}\\left( {\\sec {\\rm{A}} - 1} \\right)} \\over {1 + \\sin {\\rm{A}}}} \\times {{\\sec {\\rm{A}} + 1} \\over {\\sec {\\rm{A}} + 1}} \\times {{1 - \\sin {\\rm{A}}} \\over {1 - \\sin {\\rm{A}}}}{/tex}= {tex}{{{{\\cot }^2}{\\rm{A}}\\left( {{{\\sec }^2}{\\rm{A}} - 1} \\right)\\left( {1 - \\sin {\\rm{A}}} \\right)} \\over {\\left( {1 - {{\\sin }^2}{\\rm{A}}} \\right)\\left( {1 + \\sec {\\rm{A}}} \\right)}}{/tex}= {tex}{{{{\\cot }^2}{\\rm{A}}{\\rm{.}}{{\\tan }^2}{\\rm{A}}\\left( {1 - \\sin {\\rm{A}}} \\right)} \\over {{{\\cos }^2}A \\left( {1 + \\sec {\\rm{A}}} \\right)}}{/tex}= {tex}{{{{\\sec }^2}{\\rm{A}}\\left( {1 - \\sin {\\rm{A}}} \\right)} \\over {1 + \\sec {\\rm{A}}}}{/tex}= R.H.S. | |
| 11508. |
Lab manual work is not defined.\xa0 |
| Answer» | |
| 11509. |
How to download your papers from your app?\xa0 |
| Answer» What problem did you face? | |
| 11510. |
Sinthita - costhita + 1 ÷ sinthita + costhita - 1 = 1 ÷ secthita - tanthita |
| Answer» L.H.S. = {tex}{{\\sin \\theta - \\cos \\theta + 1} \\over {\\sin \\theta + \\cos \\theta - 1}}{/tex}Dividing all terms by {tex}\\cos \\theta {/tex}= {tex}{{\\tan \\theta - 1 + \\sec \\theta } \\over {\\tan \\theta + 1 - \\sec \\theta }}{/tex}= {tex}{{\\tan \\theta + \\sec \\theta - 1} \\over {\\tan \\theta - \\sec \\theta + 1}}{/tex}= {tex}{{\\tan \\theta + \\sec \\theta - \\left( {{{\\sec }^2}\\theta - {{\\tan }^2}\\theta } \\right)} \\over {\\tan \\theta - \\sec \\theta + 1}}{/tex}= {tex}{{\\tan \\theta + \\sec \\theta - \\left( {\\sec \\theta + \\tan \\theta } \\right)\\left( {\\sec \\theta - \\tan \\theta } \\right)} \\over {\\tan \\theta - \\sec \\theta + 1}}{/tex}= {tex}{{\\left( {\\tan \\theta + \\sec \\theta } \\right)\\left( {1 - \\sec \\theta + \\tan \\theta } \\right)} \\over {\\tan \\theta - \\sec \\theta + 1}}{/tex}= {tex}\\sec \\theta + \\tan \\theta {/tex}= {tex}\\sec \\theta + \\tan \\theta \\times {{\\sec \\theta - \\tan \\theta } \\over {\\sec \\theta - \\tan \\theta }}{/tex}= {tex}{{{{\\sec }^2}\\theta - {{\\tan }^2}\\theta } \\over {\\sec \\theta - \\tan \\theta }}{/tex}= {tex}{1 \\over {\\sec \\theta - \\tan \\theta }}{/tex}= R.H.S. | |
| 11511. |
1 + costhita + sinthita ÷ 1 + costhita - sinthita = 1 + sinthita ÷ costhita |
| Answer» L.H.S. = {tex}{{1 + \\sin \\theta + \\cos \\theta } \\over {1 + \\cos \\theta - \\sin \\theta }}{/tex}Dividing all terms by {tex}{\\cos \\theta }{/tex}= {tex}{{\\sec \\theta + \\tan \\theta + 1} \\over {\\sec \\theta + 1 - \\tan \\theta }}{/tex}= {tex}{{\\sec \\theta + \\tan \\theta + 1} \\over {\\sec \\theta - \\tan \\theta + 1}} \\times {{\\sec \\theta + \\tan \\theta } \\over {\\sec \\theta + \\tan \\theta }}{/tex}= {tex}{{\\left( {\\sec \\theta + \\tan \\theta + 1} \\right)\\left( {\\sec \\theta + \\tan \\theta } \\right)} \\over {\\left( {\\sec {\\theta ^2} - {{\\tan }^2}\\theta } \\right) + \\left( {\\sec \\theta + \\tan \\theta } \\right)}}{/tex}= {tex}{{\\left( {\\sec \\theta + \\tan \\theta + 1} \\right)\\left( {\\sec \\theta + \\tan \\theta } \\right)} \\over {\\left( {1 + \\sec \\theta + \\tan \\theta } \\right)}}{/tex}= {tex}\\sec \\theta + \\tan \\theta {/tex}= {tex}{1 \\over {\\cos \\theta }} + {{\\sin \\theta } \\over {\\cos \\theta }}{/tex}= {tex}{{1 + \\sin \\theta } \\over {\\cos \\theta }}{/tex}R.H.S. | |
| 11512. |
If the time at this moment is 9pm what be the time 23999999992 hours later |
| Answer» At every 24hr It will be 9pm.So, In this way After 24000000000 Hr. it will be 9 pm.Now we know 24000000000-8=23999999992So, deducting 8hrs from 9 pm we get 1pm.So, answer is 1pm. | |
| 11513. |
1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)=1/6 |
| Answer» [{tex}1\\over(x-1)(x-2){/tex}+{tex}1\\over(x-2)(x-3){/tex}]+{tex}1\\over(x-3)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}(x-3)+(x-1)\\over(x-1)(x-2)(x-3){/tex}+{tex}1\\over (x-3)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}2x-4\\over(x-1)(x-2)(x-3){/tex}+{tex}1\\over(x-3)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}2(x-2)\\over(x-1)(x-2)(x-3){/tex}+{tex}1\\over(x-3)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}2\\over(x-1)(x-3){/tex}+{tex}1\\over(x-3)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}2(x-4)+(x-1)\\over(x-1)(x-3)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}2x-8+x-1\\over(x-1)(x-3)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}3x-9\\over(x-1)(x-3)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}3(x-3)\\over(x-1)(x-3)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}3\\over(x-1)(x-4){/tex}={tex}1\\over6{/tex}=>{tex}(x-1)(x-4)=18{/tex}=>{tex}x\\ ^2-4x-x+4=18{/tex}=>{tex}x\\ ^2-5x-14=0{/tex}=>{tex}x\\ ^2-7x+2x-14=0{/tex}=>{tex}x(x-7)+2(x-7)=0{/tex}=>{tex}(x+2)(x-7)=0{/tex}=>{tex}either ,x+2=0 or, x-7=0{/tex}{tex}x=-2 or7{/tex} | |
| 11514. |
Prove that tanA +sinA/tanA-sinA = secA +1/secA-1=1+cosA/1- cosA\xa0 |
| Answer» {tex}{{\\tan {\\rm{A}} + \\sin {\\rm{A}}} \\over {\\tan {\\rm{A}} - \\sin {\\rm{A}}}}{/tex}= {tex}{{{{\\sin {\\rm{A}}} \\over {\\cos {\\rm{A}}}} + \\sin {\\rm{A}}} \\over {{{\\sin {\\rm{A}}} \\over {\\cos {\\rm{A}}}} - \\sin {\\rm{A}}}}{/tex}= {tex}{{\\sin {\\rm{A}} + \\cos {\\rm{A}}.\\sin {\\rm{A}}} \\over {\\sin {\\rm{A}} - \\cos {\\rm{A}}.\\sin {\\rm{A}}}}{/tex}= {tex}{{\\sin {\\rm{A}}\\left( {1 + \\cos {\\rm{A}}} \\right)} \\over {\\sin {\\rm{A}}\\left( {1 - \\cos {\\rm{A}}} \\right)}}{/tex}= {tex}{{1 + \\cos {\\rm{A}}} \\over {1 - \\cos {\\rm{A}}}}{/tex}Proved.Also, {tex}{{1 + \\cos {\\rm{A}}} \\over {1 - \\cos {\\rm{A}}}}{/tex}= {tex}{{1 + {1 \\over {\\sec {\\rm{A}}}}} \\over {1 - {1 \\over {\\sec {\\rm{A}}}}}}{/tex}= {tex}{{\\sec {\\rm{A}} + 1} \\over {\\sec {\\rm{A}} - 1}}{/tex}Proved. | |
| 11515. |
(SinA +1-cosA)/(cosA-1+sinA )\xa0=1+sinA/cosA\xa0 |
| Answer» divide numerator and denomintor of LHS by\xa0cos A{tex}LHS=\\frac{tan A+sec A-1}{1-sec A+tanA} but 1=sec^2A-tan^2ALHS=\\frac{tan A+sec A-(sec^2A-tan^2A)}{1-sec A+tanA}=\\frac{tan A+sec A-(secA-tanA)(secA+tanA)}{1-sec A+tanA} from the numerator take (sec A+tan A)=\\frac{(tan A+sec A)(1-(secA+tanA)}{1-sec A+tanA} cancel (1-(secA+tanA)=tanA+secA=\\frac{sinA}{cosA}+\\fract{1}{cosA}=\\fract{sinA+1}{cosA{/tex} | |
| 11516. |
Find the value of 7/cot2A - 7/cos2A |
| Answer» {tex}\\frac { 7 }{ cot2A } -\\frac { 7 }{ cos2A } =7\\left[ \\frac { 1 }{ cot2A } -\\frac { 1 }{ cos2A } \\right] =7\\left[ \\frac { sin2A }{ cos2A } -\\frac { 1 }{ cos2A } \\right] \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\left[ \\because cotA=\\frac { cosA }{ sinA } \\right] \\\\ \\frac { -7 }{ cos2A } \\left[ 1-sin2A \\right] =\\frac { -7 }{ cos2A } \\left[ \\sin ^{ 2 }{ A } +\\cos ^{ 2 }{ A } -2sinAcosA \\right] \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\left[ \\because \\quad \\sin ^{ 2 }{ A } +\\cos ^{ 2 }{ A=1, } sin2A=2sinAcosA \\right] \\\\ =\\frac { -7 }{ cos2A } { \\left( sinA-cosA \\right) }^{ 2 }=\\quad \\frac { -7 }{ \\cos ^{ 2 }{ A } -\\sin ^{ 2 }{ A } } { \\left( sinA-cosA \\right) }^{ 2 }\\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\quad \\left[ \\because \\quad cos2A=\\cos ^{ 2 }{ A } -\\sin ^{ 2 }{ A } \\right] \\\\ \\\\ =\\frac { -7{ \\left( sinA-cosA \\right) }^{ 2 } }{ \\left( cosA-sinA \\right) \\left( cosA+sinA \\right) } \\quad =\\quad \\frac { 7{ \\left( sinA-cosA \\right) }^{ 2 } }{ \\left( sinA-cosA \\right) \\left( cosA+sinA \\right) } =\\frac { 7{ \\left( sinA-cosA \\right) } }{ \\left( sinA+cosA \\right) } \\\\ {/tex} | |
| 11517. |
If the zeros of the polynomial f(x)=2xcube -15xsquare +37x -30 are in A.P.,find them. |
| Answer» Given\xa0{tex}f(x)=2{ x }^{ 3 }-15{ x }^{ 2 }+37x-30{/tex}Let the zeroes of the polynomial in A.P be a-d,a,a+dUsing the relationship between the zeroes and coefficients of a cubic polynomial ,we getSum of the roots = {tex}a-d+a+a+d=\\frac { -\\left( -15 \\right) }{ 2 } {/tex}\xa0{tex}\\Rightarrow 3a=\\frac { 15 }{ 2 } \\\\ \\Rightarrow a=\\frac { 5 }{ 2 } {/tex}Now the product of the roots=\xa0{tex}\\left( a-d \\right) a\\left( a+d \\right) =\\frac { -\\left( -30 \\right) }{ 2 } {/tex}{tex}\\Rightarrow a\\left( { a }^{ 2 }-{ d }^{ 2 } \\right) =\\frac { 30 }{ 2 } \\\\ \\Rightarrow \\frac { 5 }{ 2 } \\left( \\frac { 25 }{ 4 } -{ d }^{ 2 } \\right) =15\\\\ \\Rightarrow \\left( \\frac { 25 }{ 4 } -{ d }^{ 2 } \\right) =15\\times \\frac { 2 }{ 5 } =6\\\\ \\Rightarrow { d }^{ 2 }=\\frac { 25 }{ 4 } -6=\\frac { 25-24 }{ 4 } =\\frac { 1 }{ 4 } \\\\ \\Rightarrow { d }=\\pm \\frac { 1 }{ 2 } {/tex}Hence the roots are\xa0{tex}\\frac { 5 }{ 2 } -\\frac { 1 }{ 2 } ,\\frac { 5 }{ 2 } ,\\frac { 5 }{ 2 } +\\frac { 1 }{ 2 } =2,2.5,3{/tex} | |
| 11518. |
The 6th and 17th terms of an A.P. are 19 & 41,find the 40th term? |
| Answer» We have nth\xa0term of an A.P is {tex}{ T }_{ n }=a+\\left( n-1 \\right) d{/tex}Now\xa0{tex}{ T }_{ 6 }=19\\Rightarrow a+5d=19.............(i)\\\\ { T }_{ 17 }=41\\Rightarrow a+16d=41..........(ii){/tex}Subtracting equation(i) from(ii), we get{tex}11d=22\\Rightarrow d=\\frac { 22 }{ 11 } =2{/tex}Now from equation (i) we get\xa0{tex}a+10=19\\Rightarrow a=9{/tex}Hence we get\xa0{tex}{ T }_{ 40 }=a+39d=9+39\\left( 2 \\right) =9+78=87{/tex}So the 40th\xa0term is 87.\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0 | |
| 11519. |
Is 302 a term of the A.P. 3,8,13,...? |
| Answer» Let 302 be the nth term of the A. P.\xa0an=a+(n-1)d302=3+(n-1)5299/5=n-1As n is not a whole number, so 302 is not a term for the given A. P.\xa0 | |
| 11520. |
16.4x+2-16.2x+1+1=0 |
| Answer» {tex}16.4\\ ^ {x+2}-16.2\\ ^ {x+1}+1=0{/tex}{tex}16(4\\ ^ x\\times4\\ ^2)-16(2\\ ^ x\\times2\\ ^1)+1=0{/tex}{tex}16\\times16\\times4\\ ^x-16\\times2\\times2\\ ^x+1=0{/tex}{tex}256\\times4\\ ^ x-32\\times2\\ ^x+1=0{/tex}{tex}256\\times(2\\times2)\\ ^ x-32\\times2\\ ^ x+1=0{/tex}Let,{tex}2\\ ^x=p{/tex}{tex}256\\times p\\times p-32\\times p+1=0{/tex}{tex}256p\\ ^2-32p+1=0{/tex}{tex}256p\\ ^2-16p-16p+1=0{/tex}{tex}16p(16p-1)-1(16p-1)=0{/tex}{tex}(16p-1)(16p-1)=0{/tex}Either\xa0{tex}16p-1=0{/tex},or,{tex}16p-1=0{/tex}{tex}p={1\\over16},{1\\over16}{/tex}Now,\xa0{tex}p=2\\ ^ x{/tex}{tex}{1\\over 16}={ 2\\ ^ x}{/tex}, {tex}({1\\over2})\\ ^ 4=2\\ ^ x{/tex}{tex}(2)\\ ^ {-4}=2\\ ^ x{/tex}{tex}x= -4{/tex} | |
| 11521. |
Comment upon the nature of roots of the following quadratic equation. (x-2a)(x-2b)=4ab\xa0 |
| Answer» (x-2a)(x-2b)=4abx(x-2b)-2a(x-2b)=4abx2-2bx-2ax+4ab=4abx2-(2b+2a)x+4ab-4ab=0x2-(2a+2b)x+0=0So on comparing with quadratic eqaution Ax2+Bx+C=0We get A=1 B=-(2a+2b) C=0Discriminent D=B2-4AC =(-(2a+2b))2-4×1×0 =(2a+2b)2-0 =(2a+2b)2\xa0=(2(a+b))2=4(a+b)2Since square of any number cannot be negative, therefore D cannot be negative.So either D>0 or D=0Case1:- D=4(a+b)2>0 when a+b>0 , in this case roots are real and distinct.Case 2:- D=4(a+b)2=0 when\xa0a+b=0, in this case roots are real and repeated. | |
| 11522. |
If SECa =X+1/4x, prove that SECa+ TANa =2x+1/2x |
| Answer» Given:\xa0{tex}\\sec {\\rm{A}} = x + {1 \\over {4x}}{/tex}Then,\xa0{tex}\\tan {\\rm{A}} = \\sqrt {{{\\sec }^2}{\\rm{A}} - 1} {/tex}\xa0=> {tex}\\tan {\\rm{A}} = \\sqrt {{x^2} + {1 \\over 2} + {1 \\over {16{x^2}}} - 1} {/tex}=> {tex}\\tan {\\rm{A}} = \\sqrt {{{16{x^4} + 8{x^2} + 1 - 16{x^2}} \\over {16{x^2}}}} {/tex}=> {tex}\\tan {\\rm{A}} = \\sqrt {{{16{x^4} - 8{x^2} + 1} \\over {16{x^2}}}} {/tex}=> {tex}\\tan {\\rm{A}} = \\sqrt {{x^2} - {1 \\over 2} + {1 \\over {16{x^2}}}} {/tex}=> {tex}\\tan {\\rm{A}} = \\sqrt {{{\\left( {x - {1 \\over {4x}}} \\right)}^2}} {/tex}=> {tex}\\tan {\\rm{A}} = \\pm \\left( {x - {1 \\over {4x}}} \\right){/tex}Therefore,\xa0{tex}\\sec {\\rm{A}} + \\tan {\\rm{A}} = x + {1 \\over {4x}} + x - {1 \\over {4x}} = 2x{/tex}Or\xa0{tex}\\sec {\\rm{A}} + \\tan {\\rm{A}} = x + {1 \\over {4x}} - x + {1 \\over {4x}} = {1 \\over {2x}}{/tex} | |
| 11523. |
If one zero of the polynomial, (a⁴+4)x² + 13x + 4a, is reciprocal of the othe,Find \' a\' |
| Answer» Let one root =\xa0{tex}\\alpha {/tex}Then second root =\xa0{tex}1\\over \\alpha{/tex}We know\xa0Product of roots =\xa0{tex}c\\over a{/tex}{tex}=> \\alpha\\times {1\\over \\alpha}= {4a\\over a^2+4}{/tex}=>\xa0{tex}a^2+4=4a{/tex}{tex}=> a^2-4a+4= 0{/tex}{tex}=> (a-2)^2= 0{/tex}=> a-2= 0=> a = 2 | |
| 11524. |
if x and y are two irrational nubers then tell whether x-y is always irrational or not? |
| Answer» It may be rational or irrational both. | |
| 11525. |
what is the difference between CaO and CaOH. |
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Answer» CaO is Calcium Oxide, otherwise known as Quick Lime.CaOH is Calcium Hydroxide otherwise known as slaked lime which is obtained when calcium oxide is mixed or slaked with water. CaO is known as quick lime while Ca(OH)2 is slaked lime , aqueous form of slaked lime is known as lime water which is used to show the presence of CO2.\xa0 |
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| 11526. |
In chapter introduction to trignometry how we will know where we have to which formula\xa0 |
| Answer» understand the concpt apply it | |
| 11527. |
1/secA-1+1/secA+1= 2CosecA*cotA |
| Answer» | |
| 11528. |
2x2-3x+1=0my question is how to spilt middle term???? |
| Answer» | |
| 11529. |
Explain why 7×11×13+13 & 7×6×5×4×3×2×1+5?My doubt is how to find this\xa0\xa0 |
| Answer» Numbers are of two types - prime and composite.Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.It can be observed that7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)= 13 × (77 + 1)= 13 × 78= 13 ×13 × 6The given expression has 6 and 13 as its factors.Therefore, it is a composite number.7 × 6 × 5 × 4 × 3 × 2 × 1 + 5= 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)= 5 × (1008 + 1)= 5 ×10091009 cannot be factorized furtherTherefore, the given expression has 5 and 1009 as its factors.Hence, it is a composite number. | |
| 11530. |
Find first there terms of an A.P whose nth term is -5+2x |
| Answer» [I think nth term of an A.P. is -5+2n not -5+2x.]Given nth term of an A.P. an=-5+2nPut n=1 we get a1=-5+2×1=-5+2=-3Put n=2, a2=-5+2×2=-5+4=-1Put n=3, a3=-5+2×3=-5+6=1So the first three term are -3,-1,1 | |
| 11531. |
In CBSE exam of maths.is Rd SHARMA books prescribed by CBSE .for new session 2017_18\xa0 |
| Answer» No\xa0\xa0\xa0\xa0\xa0 | |
| 11532. |
If Sum of the first n terms of an AP is given by Sum of n terms=5n^2+3n,then find its nth term |
| Answer» Give Sn=5n2+3n ---------(1)Then Sn-1=5(n-1)2+3(n-1) [on replacing n by n-1 in eq.(1)] =5(n2-2n+1)+3n-3 =5n2-10n+5+3n-3 =5n2-7n+2So nth\xa0term of A.P. is given by an=Sn\xa0- Sn-1 an=5n2+3n-(5n2-7n+2) =5n2+3n-5n2+7n-2 =10n-2[CONCEPT USED:- IF THE SUM OF n terms OF AN A.P. IS GIVEN Sn\xa0THEN THE nth term IS GIVEN BY an=Sn\xa0- Sn-1\xa0\xa0] | |
| 11533. |
Is there is same paper for class 10 all our india? For all subject. |
| Answer» Check sample papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 11534. |
Find hcf of 56,96,324 by euclids algorithum |
| Answer» Euclid\'s formula is given by ,a=bq+r, where a= dividend, b= divisor,q=quotient, r= remainder324=96{tex}\\times{/tex}3+3696=36{tex}\\times{/tex}2+2436=24{tex}\\times{/tex}1+1224=12{tex}\\times{/tex}2+0HCF =12Again,96=56{tex}\\times{/tex}1+4056=40{tex}\\times{/tex}1+1640=16{tex}\\times{/tex}2+816=8{tex}\\times{/tex}2+0HCF=8So, HCF of 56,96,324= 12-8=4 | |
| 11535. |
What is sa1syllabus for mathematics\xa0 |
| Answer» SA 1 and SA 2 system is discountinued by CBSE. Now there will be only annual exam in class 10. The new syllabus is available in class 10 syllabus section in app and website. | |
| 11536. |
prove that √2\xa0is irrational\xa0 |
| Answer» Let us assume\xa0{tex}\\sqrt{2}{/tex}\xa0be a rational number and its simplest form be\xa0{tex}\\frac{a}{b}{/tex}, a and b are coprime positive integers and b\xa0{tex}\\ne{/tex}\xa00.So,\xa0{tex}\\sqrt{2}{/tex}\xa0=\xa0{tex}\\frac{a}{b}{/tex}{tex}\\Rightarrow{/tex}\xa0a2 = 2b2\xa0Thus, a2 is a multiple of 2{tex}\\Rightarrow{/tex} a is a multiple of 2Let a = 2m for some integer m{tex}\\therefore{/tex}\xa0b2 = 2m2Thus, b2 is a multiple of 2{tex}\\Rightarrow{/tex}\xa0b is a multiple of 2Hence 2 is a common factor of a and b.This contradicts the fact that a and b are coprimesHence\xa0{tex}\\sqrt{2}{/tex}\xa0is an irrational number. | |
| 11537. |
x=√5+√5+√5+.................. |
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| 11538. |
Efgd is a square and angle bac is 90 prove that bf×cg=gf square |
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| 11539. |
EFGD is a square and angle BAC is 90 prove that BF×CG=GF sq |
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| 11540. |
Show that the sequence defined by an=5n-7 is an A.P.,find its common difference |
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Answer» an = a1 + (n - 1) dan = - 2 + (n - 1) 5 ......... an = 5n - 7hence it is an APd = 5a = - 2 Common difference is 5 |
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| 11541. |
If the sum of Pth term of an Ap 3p2+4p, find its N th term? |
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| 11542. |
If the Pth term of an ap is 1/q and Qth term is 1/p ,show that the sum of pq term is (pq+1)/2? |
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Answer» Given ap={tex}1\\over q{/tex} aq={tex}1\\over p{/tex}\xa0let first term and common difference of given AP is a and d respectively.Then ap={tex}1\\over q{/tex}\xa0 a+(p-1)d={tex}1\\over q{/tex}\xa0--------(1)and aq=a+(q-1)d={tex}1\\over p{/tex} --------(2)On subtracting eq(2) from (1) we geta+(p-1)d -(a+(q-1)d) ={tex}1\\over q{/tex}\xa0{tex}-{1\\over p}{/tex}a+pd-d-a-qd+d={tex}p-q\\over pq{/tex}pd-qd={tex}p-q\\over pq{/tex}d(p-q)={tex}p-q\\over pq{/tex} d={tex}1\\over pq{/tex}--------(3)Substituting value of d in eq(1) we geta+(p-1)×{tex}1\\over pq{/tex}\xa0={tex}1\\over q{/tex}a+p×{tex}1\\over pq{/tex}\xa0-{tex}1\\over pq{/tex}\xa0={tex}1\\over q{/tex}\xa0a+{tex}1\\over q{/tex}\xa0-{tex}1\\over pq{/tex}=\xa0{tex}1\\over q{/tex} a={tex}{1\\over q}-{1\\over q}+{1\\over pq}{/tex} a={tex}1\\over pq{/tex} ------(4)Now Spq={tex}pq\\over 2{/tex}\xa0[2a+(pq-1)d] ={tex}pq\\over 2{/tex}\xa0[2×{tex}1\\over pq{/tex}\xa0+(pq-1)×{tex}1\\over pq{/tex}\xa0] =\xa0{tex}pq\\over 2{/tex}[{tex}{2+(pq-1)\\over pq}{/tex}] ={tex}{pq×(pq+1)\\over 2×pq}{/tex} ={tex}pq+1\\over 2{/tex} H.Proved P+q/2pq |
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| 11543. |
2x2-6x+3=0 find the root of the following quadratic equation by factorisation |
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| 11544. |
More than type cumulative distribution |
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| 11545. |
solve for x and y 5/x_1+1/Y_2 |
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| 11546. |
How we can make another equation from one given to us?? |
| Answer» By multiply or divide the whole equation by a any single number. | |
| 11547. |
What is the meaning of mathematics |
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| 11548. |
Sides ab and ac and median ad are properational to the sides of the |
| Answer» Given : In\xa0{tex}\\Delta A B C \\text { and } \\Delta P Q R{/tex} The AD and PM are their medians,such that\xa0{tex}\\frac { A B } { P Q } = \\frac { A D } { P M } = \\frac { A C } { P R }{/tex}To prove :\xa0{tex}\\Delta A B C \\sim \\Delta P Q R{/tex}Construction : Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join CE and RN.Proof : In\xa0{tex}\\Delta A B D \\text { and } \\Delta E D C{/tex}\xa0{tex}AD=DE{/tex}\xa0{tex}\\angle A D B = \\angle E D C{/tex}\xa0(vertically opposite angles){tex}BD=DC\\text{(as AD is a median)}{/tex}{tex}\\therefore \\quad \\Delta A B D \\equiv \\Delta E D C{/tex}\xa0(By SAS congruency)or,\xa0{tex}AB=CE{/tex} (By CPCT)Similarly,\xa0PQ = RN {tex}{/tex}{tex}\\frac { A B } { P Q } = \\frac { A D } { P M } = \\frac { A C } { P R }{/tex}\xa0(Given)or,\xa0{tex}\\frac { C E } { R N } = \\frac { 2 A D } { 2 P M } = \\frac { A C } { P R }{/tex}or\xa0{tex}\\frac{CE}{RN}=\\frac{AE}{PN}=\\frac{AC}{PR}{/tex}So\xa0{tex}∆ACE \\sim ∆PRN{/tex}\xa0{tex}\\angle 3=\\angle 4{/tex}Similarly\xa0{tex}\\angle 1=\\angle 2{/tex}\xa0{tex}\\angle 1+\\angle3=\\angle2+\\angle4{/tex}So\xa0{tex}\\angle A=\\angle P\\text{ and}{/tex}\xa0{tex}\\frac{AB}{PQ}=\\frac{AC}{PR}\\text{(given)} {/tex}Hence\xa0{tex}∆ABC\\sim ∆PQR{/tex} | |
| 11549. |
BL and CM are median of a triangle ABC right angled at A. Prove that 4(BL^2 + CM^2)=5BC^2. |
| Answer» BL and CM are medians of a {tex}\\triangle {/tex} ABC in which {tex}\\angle{/tex} A=900From {tex}\\triangle {/tex}ABC, BC2 = AB2 + AC2 ....(i)From right angled {tex}\\vartriangle {/tex} ABL,BL2 = AL2 + AB2i.e {tex}B{L^2} = {\\left( {\\frac{{AC}}{2}} \\right)^2} + A{B^2}{/tex}{tex}\\Rightarrow {/tex} 4BL2= AC2 + 4AB2 .....(ii)From right-angled {tex}\\triangle {/tex}CMA,CM2 = AC2 + AM2i.e {tex}C{M^2} = A{C^2} + {\\left( {\\frac{{AB}}{2}} \\right)^2}{/tex}[mid-point]{tex}\\Rightarrow {/tex}{tex}C{M^2} = A{C^2} + \\frac{{A{B^2}}}{4}{/tex}{tex}\\Rightarrow {/tex}{tex}4C{M^2} = 4A{C^2} + A{B^2}{/tex} .....(iii)Adding (ii) and (iii), we geti.e.4(BL2 + CM2) = 5(AC2 + AB2) = 5BC2 [From (i)] | |
| 11550. |
How the formula of mean,median and mode is formed? |
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