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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.
| 6451. |
if 3tanAtanB=1,then prove that 2cos(A+B)=cos(A-B) |
| Answer» Given: {tex}3\\tan {\\rm{A}}\\tan {\\rm{B}} = 1{/tex}=> {tex}{{3\\sin {\\rm{A}}\\sin {\\rm{B}}} \\over {\\cos {\\rm{A}}\\cos {\\rm{B}}}} = 1{/tex}=> {tex}3\\sin {\\rm{A}}\\sin {\\rm{B}} = \\cos {\\rm{A}}\\cos {\\rm{B}}{/tex}=> {tex}3\\left[ {{1 \\over 2}\\left\\{ {\\cos \\left( {{\\rm{A}} - {\\rm{B}}} \\right) - \\cos \\left( {{\\rm{A}} + {\\rm{B}}} \\right)} \\right\\}} \\right] = {1 \\over 2}\\left[ {\\cos \\left( {{\\rm{A}} - {\\rm{B}}} \\right) + \\cos \\left( {{\\rm{A}} + {\\rm{B}}} \\right)} \\right]{/tex}=> {tex}3\\cos \\left( {{\\rm{A}} - {\\rm{B}}} \\right) - 3\\cos \\left( {{\\rm{A}} + {\\rm{B}}} \\right) = \\cos \\left( {{\\rm{A}} - {\\rm{B}}} \\right) + \\cos \\left( {{\\rm{A}} + {\\rm{B}}} \\right){/tex}=> {tex} - 4\\cos \\left( {{\\rm{A}} + {\\rm{B}}} \\right) = - 2\\cos \\left( {{\\rm{A}} - {\\rm{B}}} \\right){/tex}=> {tex}2\\cos \\left( {{\\rm{A}} + {\\rm{B}}} \\right) = \\cos \\left( {{\\rm{A}} - {\\rm{B}}} \\right){/tex}Hence proved. | |
| 6452. |
2×4=? |
| Answer» 8 | |
| 6453. |
What is the importance of studying statistics?Solve the equation. Sin2x + cosx = 0 |
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Answer» Knowledge in statistics provides us\xa0with the necessary tools and conceptual foundations in quantitative reasoning to extract information intelligently from this sea of data. {tex}\\sin 2x + \\cos x = 0{/tex}=> {tex}\\sin 2x = - \\cos x{/tex}=> {tex}2\\sin x.\\cos x = - \\cos x{/tex}=> {tex}\\sin x = - {1 \\over 2}{/tex}=> {tex}\\sin x = \\sin \\left( {\\pi + {\\pi \\over 6}} \\right){/tex}=> {tex}\\sin x = \\sin {{7\\pi } \\over 6}{/tex}=> {tex}x = {{7\\pi } \\over 6} = {210^ \\circ }{/tex} |
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| 6454. |
Find domain of1/(x-4)(x-1) |
| Answer» Ans.\xa0{tex}f(x) = {1\\over (x-4)(x-1)}{/tex}\xa0We need\xa0to find where the\xa0expression\xa0is undefined. These values are not part of the\xa0domain.For the above function, if it is defined then{tex}(x-4)(x-1) \\ne 0 {/tex}=>\xa0{tex}x-4\\ne 0 \\space and\\space x-1\\ne 0 {/tex}{tex}x\\ne 4 \\space and\\space x\\ne 1 {/tex}So Domain of f(x) = R - {1,4} | |
| 6455. |
Find domain of(x-4)(x-1)? |
| Answer» {tex}\\left( {x - 4} \\right)\\left( {x - 1} \\right) = {x^2} - 5x + 4{/tex}\xa0is a polynomial.we know that Polynomial is defined for every real x.hence Domain of\xa0{tex}\\left( {x - 4} \\right)\\left( {x - 1} \\right){/tex}=R | |
| 6456. |
evaluate\xa0{tex}evaluate sigma superscript 13 subscript n=1 i^n + i^n+1) where n belongs to N{/tex} |
| Answer» | |
| 6457. |
If\xa0AunionB=AintersectionB then proved A=B |
| Answer» | |
| 6458. |
F(x) = x÷1+x² |
| Answer» X÷1+x*2x^\xa0 | |
| 6459. |
How to find the square root of -7-24i ? |
| Answer» Ans.\xa0Let\xa0{tex}\\sqrt {-7-24i} = x+yi{/tex}Squaring Both Sides,=> -7-24i = x2 - y2\xa0+ 2xyion comparing, we get=> x2 - y2\xa0= -7 ……(1)and 2xy = -24 ………(2)We know(x2+y2)2\xa0= (x2-y2)2\xa0+ 4x2y2=> (x2+y2)2 = 49 + 576=> (x2+y2)2\xa0= 625=> x2+ y2 = 25 …………(3)solving (2) and (3), We get{tex}x = \\pm 3 \\space \\space \\space and \\space y =\\pm4{/tex}from (2) as product of xy is -ve, so it means x and y are of opposite sign.So, when x = 3, y = -4when x = -3, y = 4Therefore,\xa0{tex}\\sqrt {-7-24i} = \\pm (3-4i){/tex} | |
| 6460. |
please what is the meaning of general and pricipal solution in trignometry. |
| Answer» | |
| 6461. |
Find the term independent of x in the expansion (x/√3)+(√3/2x²)^10\xa0 |
| Answer» {tex}{\\left( {\\sqrt {{x \\over 3}} + {3 \\over {2{x^2}}}} \\right)^{10}}{/tex}{tex}{T_{r + 1}}{ = ^{10}}{C_r}{\\left( {\\sqrt {{x \\over 3}} } \\right)^{10 - r}}{\\left( {{3 \\over {2{x^2}}}} \\right)^r}{/tex}{tex}{ = ^{10}}{C_r}{\\left( x \\right)^{5 - {r \\over 2} - 2r}}{\\left( 3 \\right)^{ - 5 + {r \\over 2} + r}}{\\left( 2 \\right)^{ - r}}{/tex}{tex}{ = ^{10}}{C_r}{\\left( x \\right)^{{{10 - 5r} \\over 2}}}{\\left( 3 \\right)^{{{3r - 10} \\over 2}}}{\\left( 2 \\right)^{ - r}}{/tex}{tex}for\\,independent\\,of\\,x{/tex}{tex}{\\left( x \\right)^{{{10 - 5r} \\over 2}}} = {x^0}{/tex}{tex}{{10 - 5r} \\over 2} = 0{/tex}{tex}10 - 5r = 0{/tex}{tex}r = 2{/tex}{tex}independent\\,term\\,{T_3}{ = ^{10}}{C_2}{\\left( 3 \\right)^{ - 2}}{\\left( 2 \\right)^{ - 2}}{/tex}{tex} = {{10!} \\over {2!8!}} \\times {1 \\over {36}}{/tex}{tex} = {{10 \\times 9} \\over {2 \\times 36}}{/tex}{tex} = {5 \\over 4}{/tex} | |
| 6462. |
Find Mode-20-2424-2930-3435-3940-44 |
| Answer» | |
| 6463. |
find the derivative of sin2x |
| Answer» {tex}let\\,y = \\sin 2x{/tex}{tex}differentiating\\,with\\,respect\\,to\\,x{/tex}{tex}{{dy} \\over {dx}} = {d \\over {dx}}\\left( {\\sin 2x} \\right){/tex}{tex} = \\cos 2x{d \\over {dx}}\\left( {2x} \\right){/tex}{tex} = 2\\cos 2x{/tex} | |
| 6464. |
Find derivative of function f(x)=sinx/sinx-cosx with respect to x. |
| Answer» Ans.\xa0{tex}f(x) = {sinx\\over sinx-cosx}{/tex}Using quotient Rule, we get{tex}f\'(x) ={ {(sinx-cosx){d(sinx)\\over dx}- sinx{d(sinx-cosx)\\over dx}}\\over (sinx-cosx)^2}{/tex}{tex}=> { (sinx-cosx)cosx - sinx(cosx-(-sinx)) \\over (sinx-cosx)^2}{/tex}{tex}=> {sinx cosx - cos^2x -sinxcosx -sin^2x\\over (sinx-cosx)^2}{/tex}{tex}=> {- (cos^2x +sin^2x)\\over (sinx-cosx)^2}{/tex}{tex}=> {-1\\over (sinx-cosx)^2}{/tex} | |
| 6465. |
find limits of limx→infinity (root a+x - root 9/x)\xa0 |
| Answer» | |
| 6466. |
let f:r-r be a function given by f(x)=x2+2.find f-1(27). |
| Answer» {tex}let\\,{f^{ - 1}}\\left( {27} \\right) = x{/tex}{tex} \\Rightarrow f\\left( x \\right) = 27{/tex}{tex} \\Rightarrow {x^2} + 2 = 27{/tex}{tex} \\Rightarrow {x^2} = 25{/tex}{tex} \\Rightarrow x = \\pm 5{/tex}{tex}Hence{/tex}{tex}{f^{ - 1}}\\left( {27} \\right) = \\left\\{ { - 5,5} \\right\\}{/tex} | |
| 6467. |
Tan a=1÷3, tan b=1÷2 prove that sin2(a+b)=1 |
| Answer» sin 2(a + b) = 12 tan(a + b)/{1 - tan2(a + b)} = 12[(tan a + tan b)/(1 - tana.tanb)]/[1 + {(tan a + tan b)/(1 - tana.tanb)}2] = 12{(1/3 + 1/2)/1 - 1/3 x 1/2)]/[1 +{(1/3 + 1/2)/(1 - tan a .tan b)}2] = 12[(5/6)/(5/6)]/[1 + {(5/6)/(5/6)}2] = 12/[1 + 1] = 12/2 = 11 = 1Hence proved.\xa0\xa0 | |
| 6468. |
Tan32+tan13+tan32tan13=1 |
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Answer» tan 32o + tan 13o + tan 32o.tan13o = 1tan 32o + tan 13o = 1 - tan 32o.tan13o(tan 32o + tan 13o)/(1 - tan 32o.tan13o) = 1tan (32o + 13o) = 1 [Since tan (A + B) = (tan A + tan B)/(1 - tan A. tan B)]tan 45o = 11 = 1 [Since tan 45o = 1]Hence proved. Hello friends,If we can recall the formulatan( A+B)= ( tan A + tan B) /(1- tan A tan B)so let us takeA= 32 and B= 13therefore tan( 32+13) = (tan 32+ tan 13)/(1-tan32\xa0tan13)i.e tan 45 =\xa0(tan 32+ tan 13)/(1-tan32\xa0tan13) \xa0[since 32+13 =45]\xa01=\xa0(tan 32+ tan 13)/(1-tan32\xa0tan13) [tan 45 =1](1-tan32\xa0tan13) =\xa0(tan 32+ tan 13)1= tan32\xa0tan13\xa0+ tan 32+ tan 13PROOVEDThanksS Mukherjee7864927899\xa0 |
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| 6469. |
Sin 780°.Sin120°+Cos240°.Sin390°=1/2 |
| Answer» Ans.\xa0\\(Sin 780°.Sin120°+Cos240°.Sin390°={1\\over 2}\\)Taking LHS\\(Sin 780°.Sin120°+Cos240°.Sin390°\\)=>\xa0\\(Sin( 2\\times 360°+60°).Sin(180°-60°)+Cos(180°+60°).Sin(360°+30°)\\)\\(=> Sin( 4\\pi+60°).Sin(\\pi-60°)+Cos(\\pi+60°).Sin(2\\pi+30°)\\)=>\xa0\\(Sin60°.Sin60°- Cos60°.Sin30°\\)=>\xa0\\({\\sqrt 3\\over 2}.{\\sqrt 3\\over 2} - {1\\over 2}.{1\\over 2} = {3\\over 4} - {1\\over 4} \\)=>\xa0\\({2\\over 4 } = {1\\over 2 } = RHS \\)Hence Proved | |
| 6470. |
Represent in the form of a + ib.\xa0i3/2 |
| Answer» Ans.\xa0\\(let \\space (a+ib) = i^{3\\over 2}\\)Squaring Both Sides, We get\xa0\\(=> a^2 +b^2.i^2 +2abi = i^3\\)\\(=> a^2 -b^2 +2abi = -i \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space [i^2 = -1]\\)On Comparing Both sides, We get\xa0\\(a^2-b^2 = 0 \\space \\space \\space \\space \\space \\space \\space and \\space \\space \\space \\space \\space \\space 2ab =-1\\)=>\xa0\\(a^2 = b^2 \\space \\space \\space ...(1) \\space \\space \\space \\space \\space and \\space \\space a = {-1\\over 2b} \\space \\space \\space \\space \\space ... (2)\\)Put value of a in (1)=>\xa0\\(({-1\\over 2b})^2 = b^2 => {1\\over 4b^2} = b^2 => {1\\over 4} = b^4 \\)=>\xa0\\(b^2 = {1\\over 2} => b = {1\\over \\sqrt 2}\\)Put value of b in (2), we get\xa0\\(a = {-1\\over 2 \\times {1\\over \\sqrt 2}} => a = {-1\\over \\sqrt2}\\)So=> \\({-1\\over \\sqrt2}+{1\\over \\sqrt 2}i = i^{3\\over 2}\\)\xa0 | |
| 6471. |
Prove that cos4π/8 + cos43π/8 + cos45π/8 + cos4\xa07π/8 = 3/2 |
| Answer» Ans.\xa0\\(cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}+cos^4{4\\pi \\over 8}+cos^4{7\\pi \\over 8} ={3\\over 2}\\)Taking LHS,\\(cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}+cos^4{4\\pi \\over 8}+cos^4{7\\pi \\over 8}\\)=>\xa0\\(cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}+cos^4[{{\\pi} -{3\\pi \\over 8}}]+cos^4[{{\\pi - {\\pi \\over 8}}}]\\)=>\xa0\\(cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}+({-cos{3\\pi \\over 8}})^4+({-cos{\\pi \\over 8}})^4\\)=>\xa0\\(2[cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}]\\)=>\xa0\\(2[cos^4{\\pi \\over 8}+cos^4({\\pi \\over 2}-{\\pi \\over 8})]\\)=>\xa0\\(2[cos^4{\\pi \\over 8}+sin^4{\\pi \\over 8}]\\)\\(=> 2[(cos^2{\\pi \\over 8}+sin^2{\\pi \\over 8})^2 -2 cos^2{\\pi \\over 8}.sin^2{\\pi \\over 8}]\\)\\(=> 2[(1)^2 -{1\\over 2}(2 cos{\\pi \\over 8}.sin{\\pi \\over 8})^2]\\)\\(=> 2[1 -{1\\over 2}(sin {\\pi \\over 4})^2]\\)\\(=> 2[1 -{1\\over 2}\\times {1\\over 2}]\\)\\(=> 2[1 -{1\\over 4}] => 2 \\times {3\\over 4} = {3\\over 2} = RHS \\)Hence Proved | |
| 6472. |
Find the range of the function f(x)=1/1-x2 |
| Answer» Ans. To Find Range,\\(Let \\space y = {1 \\over 1-x^2}\\)\\(=> {1-x^2} = {1\\over y} \\)\\(=> x^2 ={1-{1\\over y}}\\)\\(=> x^2 = {y-1\\over y}\\)\\(=> x= {\\sqrt{y-1\\over y}}\\)\\(Now,\\space this \\space is \\space defined \\space for \\space {y-1\\over y} \\geq 0 \\space except \\space y \\neq 0\\)\\(y \\in (-\\infty , 0) \\cup [1, \\infty)\\)So Domain is\xa0\\((-\\infty , 0) \\cup [1, \\infty)\\)\xa0\xa0 | |
| 6473. |
Tan3x - tan2x - Tanx = 0 . Solve the equation |
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Answer» \xa0It is wrong because we can\'t write tan3x = tan2x + tanx\xa0\xa0The and of this question is\xa0Tan(2x+x) = tan2 tan 3x = tan 2x + tan xtan (2x + x) = tan 2x + tan x(tan 2x + tan x)/(1 - tan 2x.tanx x) = tan 2x + tan x1 - tan 2x.tanx = 1tan 2x. tan x = 0tan 2x = 0 and tan x = 0tan 2x = tan 0o and tan x = tan 0o2x =\xa00o and x = 0ox = 0o and x = 0o |
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| 6474. |
Sin3x + sin3(2π/3 + x) + sin3(4π/3 + x) = -3/4sin3x |
| Answer» Ans. we know\\(sin^3x = sinx(sin^2x)\\)\\(=>{ sin x (1-cos2x)\\over2}\\)\\(=> {1\\over 2} [{sin x -sin x cos2x}]\\)\\(=> {1\\over 2} [{ sin x - {1\\over 2} [sin (x+2x) +sin (x-2x)]}]\\)\\(=> {1\\over 2} [{ sin x - {1\\over 2} [sin 3x -sin x}]]\\)\\(=> {1\\over 2}\\times{1\\over 2} [{2 sin x - sin 3x +sin x}]\\)\\(=> {3sin x - sin 3x \\over4}\\) (1)Similarly,\\(=> sin^3({{2\\pi\\over 3} +x})= {3sin ({{2\\pi\\over 3 }+x })- sin (2\\pi +3x )\\over4}\\)\\(=> {3[sin ({{2\\pi\\over 3 })cos x +sin xcos( {2\\pi\\over 3} })]- sin 3x\\over4}\\)\\(=> {3[{\\sqrt3\\over 2}cos x -{1\\over 2}sin x)]- sin 3x\\over4}\\)\\(=> {3\\sqrt3cos x -3sin x- 2sin 3x\\over 8}\\) (2)Similarly,\xa0\\(=> sin^3({{4\\pi\\over 3} +x})= {3sin ({{4\\pi\\over 3 }+x })- sin (4\\pi +3x )\\over4}\\)\\(=> {3[sin ({{4\\pi\\over 3 })cos x +sin xcos( {4\\pi\\over 3} })]- sin 3x\\over4}\\)\\(=> {3[(-{\\sqrt3\\over 2 })cos x +sin x({-1\\over 2})]- sin 3x\\over4}\\)\\(=> {-3\\sqrt3cos x -3sin x-2sin 3x\\over 8}\\) (3)From (1),(2) and (3)\\(=> sin^3{x}+ sin^3({{2\\pi\\over 3} +x}) + sin^3({{4\\pi\\over 3} +x})\\)\\(=>{3sin x - sin3x \\over 4}+ {3\\sqrt3cos x -3sin x- 2sin 3x\\over 8} + {-3\\sqrt3cos x -3sin x- 2sin 3x\\over 8}\\)\\(=>{1\\over 8}[{6sin x - 2sin3x + 3\\sqrt3cos x -3sin x- 2sin 3x -3\\sqrt3cos x -3sin x- 2sin 3x}]\\)\\(=>{1\\over 8}[ {- 6sin 3x}] = {-6\\over 8} {sin 3x} \\)\\(=> {-3\\over 4}{sin 3x} = RHS\\)Hence Proved | |
| 6475. |
Lim\xa0x tends to 0\xa0(Tanx - sinx )/x3 |
| Answer» Ans.\xa0\\(\\lim_{x \\to 0} {tan x - sin x \\over x^3}\\)=>\xa0\\(\\lim_{x \\to 0} {{sin x\\over cos x} - sin x \\over x^3}\\)=>\xa0\\(\\lim_{x \\to 0} {sin x({1\\over cosx } - 1) \\over x^3}\\)=>\xa0\\(\\lim_{x \\to 0} {sin x(1 - cosx) \\over x^3 . \\space cos x }\\)=>\xa0\\(\\lim_{x \\to 0} {[{1\\over cos x}. {sin x \\over x} .{ 1 - cosx \\over x^2 }}]\\)=>\xa0\\(\\lim_{x \\to 0} {[{1\\over cos x}. {sin x \\over x} .{ 2sin^2{x\\over 2} \\over x^2 }}]\\)=>\xa0\\(2\\lim_{x \\to 0} {[{1\\over cos x}. {sin x \\over x} .{ sin{x\\over 2} \\over x}.{ sin{x\\over 2} \\over x}}]\\)=>\xa0\\(2\\lim_{x \\to 0} {[{1\\over cos x}. {sin x \\over x} .{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}.{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}}]\\)=>\xa0\\(2[\\lim_{x \\to 0} {{1\\over cos x}\\times \\lim_{x \\to 0}{sin x \\over x} \\times \\lim_{x \\to 0}{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}\\times \\lim_{x \\to 0}{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}}]\\)=>\xa0\\(2[\\lim_{x \\to 0} {{1\\over cos x}\\times \\lim_{x \\to 0}{sin x \\over x} \\times \\lim_{{x\\over 2} \\to 0}{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}\\times \\lim_{{x\\over2} \\to 0}{1\\over 2}{ sin{x\\over 2} \\over {x\\over 2}}}]\\) [as x tends to zero, x/2 also tends to zero]=>\xa0\\(2\\times 1 \\times 1 \\times {1\\over 2}\\times {1\\over 2} = {1\\over 2}\\) | |
| 6476. |
Derivative of cube root of tanx by first principle |
| Answer» Ans.\xa0\\(Derivative \\space of \\space \\sqrt [3] {tan x} \\space by \\space First \\space Principle \\space is \\space given \\space by : \\)\\(=> f\'(x) = lim_{h \\to 0} \\space {{ \\sqrt [3]{tan(x+h)} - \\sqrt [3] {tan x}} \\over h}\\)\\(=> lim_{h \\to 0} \\space {{ \\sqrt [3]{sin(x+h)\\over cos(x+h)} - \\sqrt [3] {sinx\\over cosx }} \\over h}\\)\\(=> lim_{h \\to 0} \\space {{ \\sqrt [3]{sin(x+h) cos \\space x} - \\sqrt [3] {sinx \\space cos(x+h) }} \\over h {\\sqrt [3]{cos(x+h) cosx} }}\\)\\(=> lim_{h \\to 0} {1\\over {\\sqrt [3]{cos(x+h) cosx} }} . \\space lim_{h \\to 0} \\space {{ \\sqrt [3]{sin(x+h) cos \\space x} - \\sqrt [3] {sinx \\space cos(x+h) }} \\over h}\\)\\(=> {1\\over {\\sqrt [3]{cos(x+0) cosx} }} . \\space lim_{h \\to 0} \\space {{ \\sqrt [3]{sin(x+h) cos \\space x} - \\sqrt [3] {sinx \\space cos(x+h) }} \\over h}\\)\\(=> {1\\over {\\sqrt [3]{cos^2x } }}.\\space lim_{h \\to 0} \\space {{ \\sqrt [3]{sin(x+h) cos \\space x} - \\sqrt [3] {sinx\\space cos(x+h)}}\\over h} \\times{ {[{(sin(x+h) cos \\space x})^{2\\over 3}}+{({sinx \\space cos(x+h)})^{2\\over3}}+\\sqrt [3]{sin(x+h)cosx. \\space cox(x+h)sinx}]\\over {[{(sin(x+h) cos \\space x})^{2\\over 3}}+{({sinx \\space cos(x+h)})^{2\\over3}}+ \\sqrt [3]{sin(x+h)cosx. \\space cox(x+h)sinx}]}\\)\\(\\)\\(=> {1\\over {\\sqrt [3]{cos^2x } }}.\\space lim_{h \\to 0}{ \\space {{sin(x+h) cos \\space x - sinx\\space cos(x+h)}} \\over {h [{(sin(x+h) cos \\space x})^{2\\over 3}+({sinx \\space cos(x+h)})^{2\\over3}+ \\sqrt [3]{sin(x+h)cosx. \\space cox(x+h)sinx}}]}\\)\\([a^3-b^3 = (a-b)(a^2+b^2+ab)]\\)\\(=> {1\\over {\\sqrt [3]{cos^2x } }}.\\space lim_{h \\to 0}{ \\space {{sin(x+h -x )}} \\over {h [{(sin(x+h) cos \\space x})^{2\\over 3}+({sinx \\space cos(x+h)})^{2\\over3}+ \\sqrt [3]{sin(x+h)cosx. \\space cox(x+h)sinx}}]}\\)\\([sin (a-b) = sina.cosb - cos a. sin b ]\\)\\(=> {1\\over {\\sqrt [3]{cos^2x } }}.\\space lim_{h \\to 0}{ \\space {{sinh }} \\over h} . \\space lim_{h \\to 0}{ { 1\\over [{(sin(x+h) cos \\space x})^{2\\over 3}+({sinx \\space cos(x+h)})^{2\\over3}+ \\sqrt [3]{sin(x+h)cosx. \\space cox(x+h)sinx}]}}\\)\\(=> {1\\over cos^{2\\over3}x} . 1. {1\\over[(sin(x+0) cos \\space x)^{2\\over 3}+(sinx \\space cos(x+0))^{2\\over3}+ \\sqrt [3]{sin(x+0)cosx. \\space cox(x+0)sinx}]}\\)\\(=> {1\\over cos^{2\\over3}x} . 1. {1\\over[(sinx cos \\space x)^{2\\over 3}+(sinx \\space cosx)^{2\\over3}+ \\sqrt [3]{sinx.cosx. \\space cosx .sinx}]}\\)\\(=> {1\\over cos^{2\\over3}x} . 1. {1\\over[(sinx cos \\space x)^{2\\over 3}+(sinx \\space cosx)^{2\\over3}+ ({sinxcosx})^{2\\over3}]}\\)\\(=> {1\\over cos^{2\\over3}x} . {1\\over3(sinx cos \\space x)^{2\\over 3}}\\)\\(=> {1\\over 3}{1\\over cos^{4\\over3}x. sin^{2\\over 3}x}\\)\\(=> {1\\over 3}{{1\\over cos^2x}\\over ({ cos^{4\\over3}x. sin^{2\\over 3}x\\over cos^2x})}\\)\\(=> {1\\over 3}{{sec^2x}\\over ({ sin^{2\\over 3}x\\over cos^{2\\over 3}x})}\\)\\(=> {1\\over 3}.{1\\over { tan^{2\\over 3}x}}.{sec^2x}\\) | |
| 6477. |
4tanx(1-tan2x)/1-6tan2x+tan4x |
| Answer» The question is : Prove that tan 4x=\xa04 tanx(1-tan2x)/1-6tan2x+tan4x\xa0L.H.S.tan 4x = tan 2(2x)[We know that tan 2x = 2 tan x / 1 - tan\xa02\xa0x]= 2 tan 2x / 1 - tan2\xa0(2x)[now putting tan 2x = 2 tan x / 1 - tan2x]= 2[2 tan x/1-tan2x] / 1 - [2 tan x / 1 - tan2x]\xa02=[4 tan x / 1 - tan\xa02\xa0x] / [1 - 4 tan\xa02\xa0x / (1 - tan2\xa0x)2]=[4 tan x / 1 - tan\xa02\xa0x] / [ (1- tan\xa02\xa0x)2\xa0- 4 tan\xa02\xa0x / (1 - tan2\xa0x)2]= 4 tan x (1 - tan\xa02\xa0x) / (1- tan\xa02\xa0x)2\xa0- 4 tan\xa02\xa0x= 4 tan x (1 - tan\xa02\xa0x) / 1 - 2 tan2\xa0x +tan\xa04\xa0x - 4tan2\xa0x= 4 tan x (1 - tan\xa02\xa0x) / 1 - 6 tan\xa02\xa0x + tan\xa04\xa0x | |
| 6478. |
if |z|=1,z not equal to 1, then find re(z-1/z+1)pls reply |
| Answer» | |
| 6479. |
sin5x-2sin3x+sinx/cos5x-cosx=tanx |
| Answer» sin5x + sinx = 2sin3x cos2x\xa0cos5x - cosx = -2sin3x sin2x ..... use factorisation formulae of trigonometry.\xa0So the fraction is 2sin3x(cos2x-1)/ - 2sin3xsin2x\xa0= (1- cos2x)/sin2x\xa0= 2(sin2x)/ 2sinxcosx\xa0=sinx/cosx\xa0=tanx. | |
| 6480. |
Find dy/dx of log. Log sin x½ |
| Answer» Ans. If question is: log (log sin x1/2)then according to\xa0Chain rule\\({dy\\over dx} = { 1\\over log( sin x^{1\\over 2})} \\times { 1 \\over sin (x)^{1\\over2}} \\times cos (x)^{1\\over 2}\\times {1\\over2\\sqrt x}\\) | |
| 6481. |
Find dy/dx yx\xa0= xy |
| Answer» Ans. yx\xa0= xyTake log both side, we get\xa0log yx\xa0= log xy=> x log y = y log xdifferentiate w.r.t x, we get=>\xa0\\(x {1\\over y} {dy \\over dx} + log y = y {1\\over x} + log x {dy \\over dx}\\)=>\xa0\\( {x\\over y} {dy \\over dx} - log x {dy \\over dx} = {y\\over x} - log y\\)=>\xa0\\(( {{x\\over y} - log x} ) {dy \\over dx} = {y\\over x} - log y\\)=>\xa0\\( {dy \\over dx} = { ( { {y\\over x} - log y } ) \\over ( {{x\\over y} - log x} )}\\)\xa0 | |
| 6482. |
Find Sum of : -\xa0 8 + 88 + 888 + 8888 +.............upto n number of terms. |
| Answer» Ans. Sn\xa0= 8( 1 + 11 + 111 + 1111 + ..... Upto n terms )=> 8/9 (9\xa0+ 99 + 999 + 9999 + ……… upto n terms)=> 8/9 [ (10-1) + (100-1) + (1000-1) + upto n terms]=> 8/9 [ (10 + 100 + 1000 + ...... n terms ) -(1+ 1 + 1 + ………… n terms)]=> 8/9 [ 10(10n\xa0-1)/ (10-1) - n ]=> 8/9 [ 10(10n-1)/9 - n]=> 80(10n\xa0-1) /81 - 8n/9 | |
| 6483. |
find derivative of sin x. |
| Answer» cos x is the derivative of sin x\xa0 | |
| 6484. |
find derivative of sin x by first principle.\xa0 |
| Answer» The derivative of sin(x) from first principles.Setting aside the limit for now, our first step is to evaluate the fraction with\xa0f(x) = sin\xa0x.On the right hand side we have a difference of 2 sines, so we apply the formula in (A2) above:Simplifying the right hand side gives:Now to put it all together and consider the limit:We make use of (3), fraction on a fraction, to bring that 2 out front down to the bottom:Now, the limit of a product is the product of the limits, so we can write this as:\xa0Now, the first limit is in the form of\xa0Limit of sin θ/θ\xa0that we met in (A1) above.We know it has value 1.For the right hand limit, we simply obtain cos\xa0x.So we can conclude that | |
| 6485. |
if the sum of n terms of an A.P. is 3n2\xa0+ 5n and its mth\xa0 term is 164, find the value of m |
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Answer» It is given that sum of n terms of an A.P. is 3n2\xa0+ 5n=> S1\xa0= 3(1)2\xa0+ 5(1) = 3 + 5 = 8 which will also be 1st term as "a1".Further, S2\xa0\xa0= 3 (2)2\xa0+ 5 (2) = 12 + 10 = 22And, we know, a2\xa0=\xa0S2\xa0-\xa0S1\xa0= 22 - 8 = 14....\xa0So, d =\xa0a2\xa0- a1\xa0= 14 - 8 = 6Now, it\'s given that am\xa0= 164 =>\xa0a1\xa0+ (m - 1)d = 1648 +\xa0(m - 1)×6 = 164 =>\xa0(m - 1)×6 = 156\xa0=>\xa0(m - 1) = 26\xa0=> m = 27 (Answer)\xa0 According to question,S1 = a = 3(1)2 + 5 x 1 = 8S2 =\xa03(2)2 + 5 x\xa02 = 22 Then, a2 = S2 - S1 = 22 - 8 = 14S3 =\xa03(3)2 + 5 x\xa03 =\xa042 Then, a3 = S3 - S2 =\xa042 -\xa022 = 22Therefore, d = 14 - 8 = 6Now, am = a + (m - 1)d = 164=> 164 = 8 + (m - 1)6=> 156 = (m - 1)6=> 26 = m - 1=> m = 27 |
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| 6486. |
The coefficient of 5th,6th,7th term are in the a.p in the expansion of (1+x)power n.\xa0 |
| Answer» In the expension of (1+X)n, T(r+1)\xa0=\xa0nCr\xa0XrCofficient of (r+1)th term = nCrSo Cofficient of 5th, 6th and 7th term ll be\xa0nC4\xa0nC5\xa0nC6 respectivily\xa0As it is given these terms are in APSo, 2 * (\xa0nC5 ) =\xa0nC4\xa0+\xa0nC6(2 * n! ) / { (n-5)! 5!} = (n! ) / { (n-4)! 4!} + (n! ) / { (n-6)! 6!}i think you can proceed from here .\xa0\xa0 | |
| 6487. |
If 23\xa0+43+63+............+(2n)3\xa0=kn2(n+1)2, then k=? |
| Answer» The sum of cubes of even numbers is equal to 2 n^2 (n+1)^2So 2^3+ 4^3+ 6^3 +........+2n^2 = 2n^2(n+1)^2\xa02 n^2 (n+1)^2 = k n^2 (n+1) ^2Cancelling the common terms, we get\xa0k=2 | |
| 6488. |
if a,b,c are in a.p b,c,d are in gp and 1/c,1/d,1/e are in a.p prove that a,c,e are in g.p.\xa0 |
| Answer» Given: b = (a + b)/2. ........(i)And. c2\xa0= bd ..........(ii)And. d = 2ce/(c + e). ..........(iii)Substituting values of b and d from eq.(i) and (iii), in eq.(ii),c2\xa0= {(a + c)/2} × {2ce/(c + e)}c2\xa0= (ace + c2e)/(c + e)c3\xa0= acec2 = ae | |
| 6489. |
if g is the geometric mean b/w a and b show that 1/g+a,+ 1/g+b=1/g. |
| Answer» L.H.S. = [1/(ab)1/2+a] + [1/(ab)1/2+b] = [1/a1/2(a1/2 + b1/2)] + [1/b1/2(a1/2 + b1/2)] = (a1/2 + b1/2)/(ab)1/2(a1/2 + b1/2) = 1/(ab)1/2 = 1/g = R.H.S. | |
| 6490. |
Cos(3π/4+x)-cos(3π/4-x)=-√2sinx |
| Answer» L.H.S. = Cos(3π/4+x)-cos(3π/4-x)= cos 3π/4 cos x - sin 3π/4 sin x - [cos 3π/4 cos x\xa0+ sin 3π/4 sin x]= cos 3π/4 cos x - sin 3π/4 sin x - cos 3π/4 cos x\xa0- sin 3π/4 sin x]= -2sin 3π/4 sin x= -2 x (1/√2) sin x= -√2sin x | |
| 6491. |
If ( x + iy ) ( 3 – 4i ) = (5 + 12i ) , then root of(x2\xa0+ y2) =??\xa0 |
| Answer» Given: ( x + iy ) ( 3 – 4i ) = (5 + 12i )Solution:( x + iy ) =\xa0(5 + 12i ) /\xa0( 3 – 4i )By Rationalization,\xa0( x + iy ) =\xa0(5 + 12i ) /\xa0( 3 – 4i ) ×\xa0( 3 +\xa04i )/\xa0( 3 +\xa04i )=\xa0(5 + 12i )\xa0( 3 +\xa04i )\xa0/\xa0( 3 +\xa04i )\xa0( 3 -\xa04i )= (15 + 36i + 20i +\xa048i2)\xa0/\xa03^2 - (4i)2 {( 3 +\xa04i )\xa0( 3 -\xa04i )\xa0=\xa032 - (4i)2 because (a+b)(a-b)=a2 - b2}= (15 - 48 + 56i)\xa0/\xa09 - 16i2 (48i^2 = -48 because i2 = -1)= ( -33 + 56i)\xa0/\xa09 +16= ( -33 + 56i)\xa0/\xa025So, ( x + iy ) = -33/25\xa0+ 56i/25\xa0And, ( x -\xa0iy ) = -33/25 -\xa056i/25\xa0Now,\xa0( x + iy )( x -\xa0iy ) = (-33/25\xa0+ 56i/25 )\xa0(-33/25 -\xa056i/25 )x2\xa0+ y2\xa0=\xa0(-33/25)2\xa0- (56i/25)2= 1089/625 - 3136 i2/\xa0625= 1089/625 +\xa03136 /\xa0625= 4225\xa0/\xa0625 = 6.76Now Root of\xa0x2\xa0+ y2\xa0means root of\xa06.76 = 2.6 (Answer) | |
| 6492. |
What is binomial theorem |
| Answer» The Binomial Theorem is a quick way (okay, it\'s a less slow way) of expanding (or multiplying out) a binomial expression that has been raised to some (generally inconveniently large) power. | |
| 6493. |
Write the general term of the following series13/1 + 13+23/1+3,......\xa0 |
| Answer» | |
| 6494. |
Write the middle term in the expension of (x2\xa0- yx)12\xa0.\xa0 |
| Answer» total number of terms in expansion = (power of expression +1)so in this expension total number of terms ll be 13 and middle term ll be 7th termT7\xa0=\xa012C6\xa0({x)2}6\xa0(yx)6\xa0 | |
| 6495. |
What is dx of sin x/2cos^2x. |
| Answer» dx = sin x / 2cos2x = 1/2[sin x - sec2x] [By Product rule]dx = 1/2[sin x.2sec x.sec x.tan x + sec2x.cos x]dx = 1/2[sin x.2sec2x.tan x + sec x]dx = (1/2) sec x[2sin2x.sec2x + 1] = (1/2) sec x[2tan2x + 1] = 1/(2cos x) [(2sin2x/cos2x) + 1]dx = 1/(2cos3 x) [2sin2x + 1 - sin2x\xa0] = 1/(2cos3 x) [sin2x + 1] | |
| 6496. |
Find the domain and range of the function f(x)=(x2+2x+1)/(x2-8x+12) |
| Answer» Ans. f(x) = (x2+2x+1)/(x2-8x+12)To find domain as the function is rational so its denominator must not be equal to zero(0).x2-8x+12\xa0= 0=> x2-6x-2x+12=0=> x(x-6)-2(x-6)=0=> (x-6)(x-2)=0=> x = 2,6For these two values this function \'ll become undefined. So domain = R - {2,6}To Find Range:Let f(x)=y=> (x2+2x+1)/(x2-8x+12) = y=> x2+2x+1 = yx2+-8xy+12y=> yx2-8xy +12y -x2-2x-1\xa0=0\xa0=> (y-1)x2\xa0-(8y+2)x\xa0+(12y-1) = 0Now D\xa0≥ 0=> b2\xa0-4ac\xa0≥\xa00=> (8y+2)2\xa0- 4(y-1)(12y-1)\xa0≥ 0\xa0= (8y+2)2\xa0≥\xa04(12y2\xa0-13y +1)=> 64y2\xa0+ 4 + 32y\xa0≥ 48y2\xa0-52y +4=> 16y2\xa0+ 84y\xa0≥ 0divide by 4=> 4y2\xa0+ 21y\xa0≥ 0=> y(4y+21)\xa0≥ 0=> y\xa0≥ 0 and\xa04y +21\xa0≥ 0y\xa0≥ 0 and y\xa0≥ -21/4So domain = R - (-21/4, 0) | |
| 6497. |
Prove 2tanx - cotx = -1 |
| Answer» 2tan x - cot x + 1 = 02tan x - 1/tan x + 1 = 02tan2x - 1 + tan x = 02tan2x + tan x - 1 = 02tan2x + 2tan x - tan x - 1 = 02tan x(tan x + 1) - 1 (tan x + 1) = 0(tan x + 1)(2tan x - 1) = 0tan x + 1 = 0 or 2tan x - 1 = 0tan x = -1 or tan x = 1/2x = 3π/4, 7π/4 or x = 3.6052, 0.4636 | |
| 6498. |
I also wants to know the explanation of wrong answer in practice test\xa0 |
| Answer» In that case,\xa0post the question along with your answer and we will give the correct explanation for your wrong answers.!\xa0 | |
| 6499. |
How 0×infinity =1 ???\xa0 |
| Answer» Ans.\xa0Infinity\xa0multiplied by\xa0zero\xa0is UNDEFINED. It is one of the 7 indeterminate forms in limits. | |
| 6500. |
how to find domain and range |
| Answer» Domain: If f : X →Y is a function, then the set X is called the Domain and Y is called the co-domain of the function f. Example: Find domain of the function fx\xa0=\xa01x-5 The given function is not defined at x = 5. Therefore, Domain of the given function is R - {5}, where R is set of all real numbers.Range: The set of second elements of the ordered pairs defining a function is called the Range of the function. Example: Find the range of function fx\xa0=\xa01x-5 Now, y=\xa01x-5⇒\xa0\xa0\xa0\xa0\xa0\xa01y=x-5⇒\xa0\xa0\xa0\xa0\xa0\xa0x\xa0=\xa05y+1y Clearly, x is not defined when y = 0. Therefore, Range of given funciton is R - {0}. | |