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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.
| 6501. |
Find domain and range of f(x)=-x |
| Answer» Domain=RRange=R | |
| 6502. |
2x-3/3x-7>0\xa0 |
| Answer» 2x-33x-7>0Case-1:\xa02x-3>0\xa0and\xa03x-7>0⇒x>32\xa0and\xa0x>73⇒x>73Case-22x-3<0\xa0and\xa03x-7<0⇒x<32\xa0and\xa0x<73⇒x<32-∞,32∪73,∞\xa0is\xa0the\xa0solution\xa0set | |
| 6503. |
Linear inequality miscellaneous problem 20th\xa0\xa0 |
| Answer» | |
| 6504. |
If tan=1/3 and tan b=1/2 . prove that sin2(a+b)=1 |
| Answer» {tex}Given\\,that{/tex}{tex}\\tan A = {1 \\over 3}\\,and\\,\\tan B = {1 \\over 2}{/tex}{tex}Now\\,\\,\\tan \\left( {A + B} \\right) = {{\\tan A + \\tan B} \\over {1 - tan A \\cdot \\tan B}}{/tex}{tex} = {{\\left( {{1 \\over 3} + {1 \\over 2}} \\right)} \\over {\\left( {1 - {1 \\over 3} \\cdot {1 \\over 2}} \\right)}}{/tex}{tex} = {{\\left( {{5 \\over 6}} \\right)} \\over {\\left( {{5 \\over 6}} \\right)}}{/tex}{tex}\\tan \\left( {A + B} \\right) = 1{/tex}{tex}A + B = {45^ \\circ }{/tex}{tex}Now\\,Sin2\\left( {A + B} \\right) = \\sin \\left( {2 \\times {{45}^ \\circ }} \\right){/tex}{tex} = \\sin {90^ \\circ }{/tex}{tex}\\sin 2\\left( {A + B} \\right) = 1{/tex} | |
| 6505. |
If A.M and G.M of two positive number a and b are 10 and 8 respectively find the number |
| Answer» {tex}let\\,two\\,numbers\\,be\\,a\\,and\\,b{/tex}{tex}A.M. = {{a + b} \\over 2} = 10{/tex}{tex}a + b = 20 \\ldots \\ldots \\left( 1 \\right){/tex}{tex}G.M. = \\sqrt {ab} = 8{/tex}{tex}ab = 64 \\ldots \\ldots \\left( 2 \\right){/tex}{tex}Now\\,{\\left( {a - b} \\right)^2} = {\\left( {a + b} \\right)^2} - 4ab{/tex}{tex} = {\\left( {20} \\right)^2} - 4 \\times 64{/tex}{tex} = 400 - 256{/tex}{tex}{\\left( {a - b} \\right)^2} = 144{/tex}{tex}a - b = \\pm 12 \\ldots \\ldots \\left( 3 \\right){/tex}{tex}solving\\,\\left( 1 \\right)\\,and\\,\\left( 3 \\right){/tex}{tex}a = 4,b = 16\\,or\\,a = 16,b = 4{/tex}{tex}hence\\,two\\,numbers\\,are\\,4\\,and\\,16\\,or\\,16\\,and\\,4.{/tex} | |
| 6506. |
Differentiate Sin2\xa0x with respect to x from 1st principle method\xa0 |
| Answer» Ans. By First Principle \\(f\'(x) = lim_{h\\to 0} \\space{ f(x+h) - f(x) \\over h}\\)\\(=> lim_{h\\to 0} \\space{ sin^2(x+h) - sin^2x \\over h}\\)\\(=> lim_{h\\to 0} \\space{ [sin(x+h) + sinx]\\times [sin(x+h) - sinx]\\over h}\\)\\([Using \\space (a^2 - b^2) = (a+b)(a-b)]\\)\\(=> lim_{h\\to 0} \\space{2 sin({x+h+x\\over 2}) cos({x+h-x\\over 2})\\times 2cos({x+h+x\\over 2}) sin({x+h-x\\over 2})\\over h}\\)\\([Using \\space sin a + sin b = 2 sin({a+b\\over 2})cos ({a-b\\over 2})\\)\xa0and\xa0\\( sin a - sin b = 2 cos({a+b\\over 2})sin ({a-b\\over 2})]\\)\\(=> lim_{h\\to 0} \\space4{ sin({2x+h\\over 2}) cos{h\\over 2} \\space cos({2x+h\\over 2}) sin{h\\over 2}\\over h}\\)\\(=> \\space4 sin({2x+0\\over 2}) cos{0\\over 2} \\space cos({2x+0\\over 2}) lim_{h\\to 0} {sin{h\\over 2}\\over {2h\\over 2}}\\)\\(=> 4 sin x. cos 0. cos x. lim_{{h\\over 2}\\to 0} {1\\over 2}{sin{h\\over 2}\\over {h\\over 2}}\\)\\([as \\space h \\to 0, then, {h\\over 2}\\to 0 ]\\)\\(=> 4 sin x. cos x. {1\\over 2}\\)\\([Using \\space Identity, lim_{x \\to 0 } \\space {sinx \\over x } = 1]\\)\\(=> 2 sin x. cos x\\) | |
| 6507. |
If tan a=1÷3,tan b=1÷2 then prove thatSin2(a+b)=1 |
| Answer» is this a correct question | |
| 6508. |
Write the domain of function of f(x)=[X^2-2x+3]÷[x^2- x-20] |
| Answer» f(x) =\xa0\\((x^2 - 2x +3)/((x+4)(x-5))\\)So Since denominator should not be zero therefore x should not be -4 and 5Hence domain x belongs to R - {-4,5} | |
| 6509. |
Find the term independent of x in the expansion of (x÷√3+√3÷2x^2) |
| Answer» | |
| 6510. |
If X and Y are acute angels such that sinX=1/√5 and sinY =1/√10 prove that (x+y)=π/4. |
| Answer» Ans. Given : {tex}sin X = {1\\over \\sqrt 5}{/tex},\xa0{tex}sin Y = {1\\over \\sqrt {10}}{/tex}X =\xa0{tex}sin^{-1}({1\\over \\sqrt 5}){/tex}Y =\xa0{tex}sin^{-1}({1\\over \\sqrt {10}}){/tex}=> X+Y =\xa0{tex}sin^{-1}({1\\over \\sqrt 5}) + sin^{-1}({1\\over \\sqrt {10}}){/tex}{tex}[using \\ \\ sin^{-1} a + sin^{-1} b = sin^{-1}(a\\sqrt{1-b^2}+b\\sqrt{1-a^2})]{/tex}{tex}=> X+Y = sin^{-1}\\left ( {1\\over \\sqrt 5} \\sqrt {1- {1\\over 10}} + {1\\over \\sqrt {10}} \\sqrt {1- {1\\over 5}}\\right ){/tex}\xa0{tex}=> X+Y = sin^{-1}\\left ( {3\\over \\sqrt {50}} + {2\\over \\sqrt {50}} \\right ){/tex}{tex}=> X+Y = sin^{-1}\\left ( {5\\over 5\\sqrt 2} \\right ){/tex}{tex}=> X+Y = sin^{-1}\\left ( {1\\over \\sqrt 2} \\right ){/tex}{tex}=> X+Y = sin^{-1}\\left ( sin {\\pi \\over 4} \\right ){/tex}{tex}=> X+Y = {\\pi \\over 4} {/tex}Hence Proved | |
| 6511. |
If cosecA+secA=cosecB+secB, then show that tanAtanB=cot(A+B)/2\xa0 |
| Answer» {tex}\\cos ec{\\rm{A}} + \\sec {\\rm{A}} = \\cos ec{\\rm{B}} + \\sec {\\rm{B}}{/tex}=> {tex}\\cos ec{\\rm{A}} - \\cos ec{\\rm{B}} = \\sec {\\rm{A}} - \\sec {\\rm{B}}{/tex}=> {tex}{1 \\over {\\sin {\\rm{A}}}} - {1 \\over {\\sin {\\rm{B}}}} = {1 \\over {\\cos {\\rm{A}}}} - {1 \\over {\\cos {\\rm{B}}}}{/tex}=> {tex}{{\\sin {\\rm{B}} - \\sin {\\rm{A}}} \\over {\\sin {\\rm{A}}{\\rm{.}}\\sin {\\rm{B}}}} = {{\\cos {\\rm{B}} - \\cos {\\rm{A}}} \\over {\\cos {\\rm{A}}{\\rm{.cos B}}}}{/tex}=> {tex}{{2\\sin {{{\\rm{B}} - {\\rm{A}}} \\over 2}\\cos {{{\\rm{B}} + {\\rm{A}}} \\over 2}} \\over {\\sin {\\rm{A}}{\\rm{.}}\\sin {\\rm{B}}}} = {{2\\sin {{{\\rm{B}} - {\\rm{A}}} \\over 2}\\sin {{{\\rm{B}} + {\\rm{A}}} \\over 2}} \\over {\\cos {\\rm{A}}{\\rm{.cos B}}}}{/tex}=> {tex}{{\\cos {{{\\rm{B}} + {\\rm{A}}} \\over 2}} \\over {\\sin {{{\\rm{B}} + {\\rm{A}}} \\over 2}}} = {{\\sin {\\rm{A}}{\\rm{.}}\\sin {\\rm{B}}} \\over {\\cos {\\rm{A}}{\\rm{.cos B}}}}{/tex}=> {tex}\\cot {{{\\rm{B}} + {\\rm{A}}} \\over 2} = \\tan {\\rm{A}}{\\rm{.}}\\tan {\\rm{B}}{/tex}=> {tex}\\tan {\\rm{A}}{\\rm{.}}\\tan {\\rm{B = }}\\cot {{{\\rm{B}} + {\\rm{A}}} \\over 2}{/tex}\xa0 | |
| 6512. |
a-(a-b)=a∩b |
| Answer» | |
| 6513. |
(1+cotA+tanA)(sinA-cosA)=secA/cosec2A-cosecA/sec2A\xa0 |
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Answer» I can\'t understand L.H.S.{tex}\\left( {1 + \\cot {\\rm{A}} + \\tan {\\rm{A}}} \\right)\\left( {\\sin {\\rm{A}} - \\cos {\\rm{A}}} \\right){/tex}= {tex}\\left( {1 + {{\\cos {\\rm{A}}} \\over {\\sin {\\rm{A}}}} + {{\\sin {\\rm{A}}} \\over {\\cos {\\rm{A}}}}} \\right)\\left( {\\sin {\\rm{A}} - \\cos {\\rm{A}}} \\right){/tex}= {tex}{{\\left( {\\sin {\\rm{A}} - \\cos {\\rm{A}}} \\right)\\left( {\\sin {\\rm{AcosA}} + {{\\sin }^2}{\\rm{A}} + {{\\cos }^2}{\\rm{A}}} \\right)} \\over {\\sin {\\rm{AcosA}}}}{/tex}= {tex}{{{{\\sin }^3}{\\rm{A}} - {{\\cos }^3}{\\rm{A}}} \\over {\\sin {\\rm{AcosA}}}}{/tex}= {tex}{{{{\\sin }^2}{\\rm{A}}} \\over {{\\rm{cosA}}}} - {{{{\\cos }^2}{\\rm{A}}} \\over {\\sin {\\rm{A}}}}{/tex}= {tex}{1 \\over {{\\rm{cosA}}}}.{{{{\\sin }^2}{\\rm{A}}} \\over 1} - {1 \\over {{\\rm{sinA}}}}.{{{{\\cos }^2}{\\rm{A}}} \\over 1}{/tex}= {tex}{{\\sec {\\rm{A}}} \\over {\\cos e{c^2}{\\rm{A}}}} - {{\\cos ec{\\rm{A}}} \\over {{{\\sec }^2}{\\rm{A}}}}{/tex}= R.H.S. |
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| 6514. |
If f is a real function defined by f(x)=x-1/x+1,then prove that : f(2x)=3f(x)+1/f(x)+3 |
| Answer» | |
| 6515. |
Convert 4 Radian into degree measure |
| Answer» Radian measure =\xa0{tex}{\\pi\\over 180}\\times degree\\ measure{/tex}So,\xa0Degree measure =\xa0{tex}{18\\over \\pi}\\times radian \\ measure{/tex}=> Degree measure =\xa0{tex}{180\\over 22}\\times 7\\times 4{/tex}= 229.09\xa0 | |
| 6516. |
Prove that cos^2A+cos^2(A+120)+cos^2(A-120)=3/2 |
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Answer» Cos5θ=16〖Sin〗^5 θ-20〖Sin〗^3 θ+5cosθ {tex}{\\cos ^2}2{\\rm{A}} + {\\cos ^2}2\\left( {{\\rm{A}} + {{120}^ \\circ }} \\right) + {\\cos ^2}2\\left( {{\\rm{A}} - {{120}^ \\circ }} \\right) = {3 \\over 2}{/tex}We know that{tex}{\\cos ^2}{\\rm{A}} = {{1 + \\cos 2{\\rm{A}}} \\over 2}{/tex} ..............(i){tex}{\\cos ^2}2\\left( {{\\rm{A}} + {{120}^ \\circ }} \\right) = {{1 + \\cos \\left( {{\\rm{2A}} + {{240}^ \\circ }} \\right)} \\over 2}{/tex} ..............(ii)And {tex}{\\cos ^2}2\\left( {{\\rm{A}} - {{120}^ \\circ }} \\right) = {{1 + \\cos \\left( {{\\rm{2A}} - {{240}^ \\circ }} \\right)} \\over 2}{/tex}..............(iii)Adding all these three equations,{tex}{\\cos ^2}2{\\rm{A}} + {\\cos ^2}2\\left( {{\\rm{A}} + {{120}^ \\circ }} \\right) + {\\cos ^2}2\\left( {{\\rm{A}} - {{120}^ \\circ }} \\right){/tex}\xa0= {tex}{3 \\over 2} + {{1 + \\cos 2{\\rm{A}}} \\over 2} + {{1 + \\cos \\left( {{\\rm{2A + }}{{240}^ \\circ }} \\right)} \\over 2} + {{1 + \\cos \\left( {{\\rm{2A}} - {{240}^ \\circ }} \\right)} \\over 2}{/tex}= {tex}{{3 + 1 + \\cos 2{\\rm{A + }}1 + \\cos \\left( {{\\rm{2A + }}{{240}^ \\circ }} \\right) + 1 + \\cos \\left( {{\\rm{2A}} - {{240}^ \\circ }} \\right)} \\over 2}{/tex}= {tex}{{3 + 3 + \\cos 2{\\rm{A}} + \\cos \\left( {{\\rm{2A + }}{{240}^ \\circ }} \\right) + \\cos \\left( {{\\rm{2A}} - {{240}^ \\circ }} \\right)} \\over 2}{/tex}= {tex}{{3 + 3 + \\cos 2{\\rm{A}} + 2\\cos 2{\\rm{A}}.\\cos {{240}^ \\circ }} \\over 2}{/tex}= {tex}{{3 + 3 + \\cos 2{\\rm{A}} - \\cos 2{\\rm{A}}} \\over 2}{/tex}= {tex}{6 \\over 2}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}{\\cos ^2}2{\\rm{A}} + {\\cos ^2}2\\left( {{\\rm{A}} + {{120}^ \\circ }} \\right) + {\\cos ^2}2\\left( {{\\rm{A}} - {{120}^ \\circ }} \\right) = {3 \\over 2}{/tex} |
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| 6517. |
If theta is positive acute angle then solve the equation:--4cos^2 theta -- 4 sin theta =1 |
| Answer» {tex}4{\\cos ^2}\\theta - 4\\sin \\theta = 1{/tex}=> {tex}4\\left( {1 - {{\\sin }^2}\\theta } \\right) - 4\\sin \\theta = 1{/tex}=> {tex}4 - 4{\\sin ^2}\\theta - 4\\sin \\theta = 1{/tex}=> {tex} - 4{\\sin ^2}\\theta - 4\\sin \\theta + 4 - 1 = 0{/tex}=> {tex}4{\\sin ^2}\\theta + 4\\sin \\theta - 3 = 0{/tex}=> {tex}4{\\sin ^2}\\theta + 6\\sin \\theta - 2\\sin \\theta - 3 = 0{/tex}=> {tex}2\\sin \\theta \\left( {2\\sin \\theta + 3} \\right) - 1\\left( {2\\sin \\theta + 3} \\right) = 0{/tex}=> {tex}\\left( {2\\sin \\theta + 3} \\right)\\left( {2\\sin \\theta - 1} \\right) = 0{/tex}=> {tex}2\\sin \\theta + 3 = 0{/tex} or {tex}2\\sin \\theta - 1 = 0{/tex}=> {tex}\\sin \\theta = {{ - 3} \\over 2}{/tex} or {tex}\\sin \\theta = {1 \\over 2}{/tex}But {tex}\\theta {/tex}\xa0is positive, therefore, taking{tex}\\sin \\theta = {1 \\over 2}{/tex}=> {tex}\\sin \\theta = \\sin {30^ \\circ }{/tex}=> {tex}\\theta = {30^ \\circ }{/tex} | |
| 6518. |
convert -3i into polar form. |
| Answer» Let\xa0{tex}-3i= 0-3i=r\\left( cos\\theta +isin\\theta \\right) {/tex}Comparing real and imaginary parts{tex}rcos\\theta =0.......................\\left( i \\right) \\\\ rsin\\theta =-3....................\\left( ii \\right) \\\\ {/tex}Squaring and adding equations (i) and (ii){tex}{ r }^{ 2 }{ cos }^{ 2 }\\theta +{ { r }^{ 2 } }sin^{ 2 }\\theta =9\\\\ \\Rightarrow { r }^{ 2 }=9\\quad \\quad \\quad \\quad \\quad \\quad \\quad \\left( { \\because cos }^{ 2 }\\theta +sin^{ 2 }\\theta =1 \\right) {/tex}{tex}\\Rightarrow r=3{/tex}Substituting\xa0{tex}r=3{/tex}\xa0in equations (i) and (ii), we get\xa0{tex}cos\\theta =0\\quad and\\quad sin\\theta =-1\\\\ {/tex}So\xa0{tex}\\theta {/tex}\xa0lies in fourth quadrant\xa0{tex}\\therefore \\quad \\theta =\\frac { -\\Pi }{ 2 } {/tex} [ in the fourth quarant format of amplitude or principal argument is\xa0{tex}-\\theta {/tex}\xa0(where {tex}\\theta =\\frac { \\Pi }{ 2 } {/tex}\xa0)\xa0]{tex}-3i=3\\left( cos\\left( \\frac { -\\Pi }{ 2 } \\right) +isin\\left( \\frac { -\\Pi }{ 2 } \\right) \\right) \\\\ \\Rightarrow -3i=3\\left( cos\\left( \\frac { \\Pi }{ 2 } \\right) -isin\\left( \\frac { \\Pi }{ 2 } \\right) \\right) \\quad \\quad \\left[ \\because cos\\left( -\\theta \\right) =cos\\theta \\quad ,sin\\left( -\\theta \\right) =-sin\\theta \\right] {/tex}\xa0 | |
| 6519. |
1+x+x2+....+xn=(1-xn-1)/1-x |
| Answer» | |
| 6520. |
Find range and domain of f(x)=x-5 |
| Answer» f(x)=x-5\xa0Domain of a function f is a set of all values of x at which the given function is defind.Here clearly the given function f(x)=x-5 is defind for every real value of x.so domain of f= R (set of real numbers)Range of a function is a set of values of function f(x) at domain of f(x).Range of f= R | |
| 6521. |
3+25 |
| Answer» | |
| 6522. |
Codec A/Cosec A-1 + Codec A/Cosec A+1 = 2+2Tan2A - 2 Sec2A |
| Answer» | |
| 6523. |
Prove that,(cos30° - sin20°)/(cos40°+ cos20°)=(4cos40°cos80°)/√3\u200b\u200b\u200b\u200b\u200b |
| Answer» | |
| 6524. |
A wheel makes 360 revolution in one minute. Through how many radian does it turn in one second?\xa0 |
| Answer» | |
| 6525. |
Evaluate :- 1+i2+i4+i6+.................+i2n |
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Answer» if n is even answer is zeroif n is odd answer is one if n is odd the answer will be zeroif n is even all cancel and answer is 1 We know i\u200b\u200b\u200b\u200b\u200b\u200b2= -1\xa0And\xa0{tex}i^4=(i^2)^2=(-1)^2= 1{/tex}So all the terms having i will be cancelled out. We ll get answer as 1\xa0 |
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| 6526. |
If sum of two numbers is C and their quotient is p/q ,then find number.\xa0 |
| Answer» Let the two numbers are a and b.Then A.T.Q.a+b=C ------(1) and {tex}{a\\over b}={p\\over q}{/tex}\xa0----(2)From eq(2){tex}a={p\\over q}×b{/tex}\xa0---------(3)Substituting this value in eq.(1) we get{tex}{pb\\over q}+b=C{/tex}{tex}{pb+bq\\over q}=C{/tex}pb+bq=C×qb(p+q)=Cq b={tex}{Cq\\over p+q}{/tex}On substituting value of b in eq.(3) we geta={tex}{p\\over q}×{Cq\\over p+q}={pC\\over p+q}{/tex}\xa0 | |
| 6527. |
If a/b=2/3 and b/c=4/5,then find value of a+b/b+c. |
| Answer» Mehod 1:-\xa0In this method first of all we find value of a and c from given equations in variable b. After that we find a+b and b+c and then divide them.Given\xa0{tex}{a\\over b} ={2\\over 3}{/tex}\xa0and\xa0{tex}{b\\over c}={4\\over 5}{/tex}{tex}=> a={2\\over 3}b{/tex}\xa0and\xa0{tex}{c\\over b}={5\\over 4}{/tex}\xa0{tex}=> c={5\\over 4}b{/tex}Now a+b={tex}{2b\\over 3}+b={2b+3b\\over 3}={5b\\over 3}{/tex}and b+c={tex}b+{5b\\over 4}={4b+5b\\over 4}={9b\\over 4}{/tex}So\xa0{tex}{a+b\\over b+c}={{5b\\over 3}\\over {9b\\over 4}}={5b\\over 3}×{4\\over 9b}{/tex}={tex}{20\\over27}{/tex}Method 2:-Given\xa0{tex}{a\\over b}={2\\over 3}{/tex}\xa0and\xa0{tex}{b\\over c}={4\\over 5}{/tex}Now\xa0{tex}{a+b\\over b+c}={{a+b\\over b}\\over {b+c\\over b}}{/tex} [on dividing numerator and denominator by b] ={tex}{{a\\over b}+{b\\over b}}\\over{{b\\over b} +{c\\over b}}{/tex}={tex}{{a\\over b}}+1\\over{1+{c\\over b}}{/tex}={tex}{{2\\over 3}}+1\\over {1+{5\\over4}}{/tex}={tex}{2+3\\over 3}\\over {4+5\\over 4}{/tex}={tex}{5\\over 3}\\over {9\\over 4}{/tex}={tex}{5\\over 3}×{4\\over 9}{/tex}={tex}20\\over 27{/tex} [by substituting the values of a/b and c/b] | |
| 6528. |
Using factor theorem, show that a-b, b-c, c-a are the factors of a(b2-c2) +b(c2-a2)\xa0+c(a2-b2). |
| Answer» If a - b is a factor of given expression, then a - b = 0 => a = bPutting a = b, in the given expression, we getb(b2-c2) +b(c2- b2) +c(b2- b2)= b3 - bc2 + bc2 - b3 + c(0)= 0Therefore, (a - b) is a factor of given expression.Again if (b - c) is a factor of given expression, thenPutting b - c = 0 => b = c in the given expression, we geta(c2 - c2) +c(c2 - a2) + c(a2 - c2)= a(0) + c3 - ca2 + ca2 - c3\xa0= 0Therefore, (b - c) is a factor of given expression.Again if (c - a) is a factor of given expression, thenPutting c - a = 0 => c = a in the given expression, we geta(b2 - a2) + b(a2 - a2) + a(a2 - b2)= ab2 - a3 + b(0) + a3 - ab2= 0Therefore, (c - a) is a factor of given expression | |
| 6529. |
Factorise I) 1+x4+x8II) x4\xa0\xa0+ 4 |
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Answer» ||)\xa0{tex}{x^4} + 4 = {x^2} + {2^2} = {\\left( {x + 2} \\right)^2} - 2 \\times x \\times 2{/tex} [Since\xa0{tex}{a^2} + {b^2} = {\\left( {a + b} \\right)^2} - 2ab{/tex}]=\xa0{tex}{\\left( {x + 2} \\right)^2} - 4x{/tex}=\xa0{tex}{\\left( {x + 2} \\right)^2} - {\\left( {2\\sqrt x } \\right)^2}{/tex}=\xa0{tex}\\left( {x + 2 - 2\\sqrt x } \\right)\\left( {x + 2 + 2\\sqrt x } \\right){/tex}=\xa0{tex}\\left( {x + 2\\sqrt x + 2} \\right)\\left( {x - 2\\sqrt x + 2} \\right){/tex}\xa0\xa0 |)\xa01+x4+x8 = x8 + x4 + 1Adding and subtracting 2(x4)(1){tex}{\\left( {{x^4}} \\right)^2} + 2\\left( {{x^4}} \\right)\\left( 1 \\right) + {\\left( 1 \\right)^2} - 2\\left( {{x^4}} \\right) + {x^4} = {\\left( {{x^4} + 1} \\right)^2} - {x^4}{/tex}=\xa0{tex}\\left( {{x^4} + 1 + {x^2}} \\right)\\left( {{x^4} + 1 - {x^2}} \\right) = \\left( {{x^4} + {x^2} + 1} \\right)\\left( {{x^4} - {x^2} + 1} \\right){/tex}=\xa0{tex}\\left[ {{{\\left( {{x^2}} \\right)}^2} + 2\\left( {{x^2}} \\right)\\left( 1 \\right) + 1 - 2\\left( {{x^2}} \\right)\\left( 1 \\right) + {x^2}} \\right]\\left( {{x^4} - {x^2} + 1} \\right){/tex}=\xa0{tex}\\left[ {{{\\left( {{x^2} + 1} \\right)}^2} - {x^2}} \\right]\\left( {{x^4} - {x^2} + 1} \\right){/tex}=\xa0{tex}\\left( {{x^2} + 1 + x} \\right)\\left( {{x^2} + 1 - x} \\right)\\left( {{x^4} - {x^2} + 1} \\right){/tex}=\xa0{tex}\\left( {{x^2} + x + 1} \\right)\\left( {{x^2} - x + 1} \\right)\\left( {{x^4} - {x^2} + 1} \\right){/tex}\xa0\xa0 |
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| 6530. |
A wheel makes 360 revolution in one minute. Through how many radians does it turn in one second?\xa0 |
| Answer» In 60sec = 360revolutions\xa0 1 sec =6 revolution\xa01 revolution =2 pie6 revolution = 12 pie | |
| 6531. |
-2+5i |
| Answer» | |
| 6532. |
Is a complex no. Come under -infinity to +infinity |
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| 6533. |
Sin(A-B)/cosA.cosB + sin (B-C)/cosB.cosC +sin (C-A)/cosC.cosA =0 |
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Answer» {tex}{{\\sin \\left( {{\\rm{A}} - {\\rm{B}}} \\right)} \\over {\\cos {\\rm{A}}\\cos {\\rm{B}}}} + {{\\sin \\left( {{\\rm{B}} - {\\rm{C}}} \\right)} \\over {\\cos {\\rm{B}}\\cos {\\rm{C}}}} + {{\\sin \\left( {{\\rm{C}} - {\\rm{A}}} \\right)} \\over {\\cos {\\rm{C}}\\cos {\\rm{A}}}}{/tex}=\xa0{tex}{{\\sin {\\rm{A}}\\cos {\\rm{B}} - \\cos {\\rm{A}}\\sin {\\rm{B}}} \\over {\\cos {\\rm{A}}\\cos {\\rm{B}}}} + {{\\sin {\\rm{B}}\\cos {\\rm{C}} - \\cos {\\rm{B}}\\sin {\\rm{C}}} \\over {\\cos {\\rm{B}}\\cos {\\rm{C}}}} + {{\\sin {\\rm{C}}\\cos {\\rm{A}} - \\cos {\\rm{C}}\\sin {\\rm{A}}} \\over {\\cos {\\rm{C}}\\cos {\\rm{A}}}}{/tex}=\xa0{tex}\\eqalign{ & {{\\sin {\\rm{A}}\\cos {\\rm{B}}\\cos {\\rm{C}} - \\cos {\\rm{A}}\\sin {\\rm{B}}\\cos {\\rm{C}} + \\cos {\\rm{A}}\\sin {\\rm{B}}\\cos {\\rm{C}} - \\cos {\\rm{A}}\\cos {\\rm{B}}\\sin {\\rm{C}} + \\sin {\\rm{C}}\\cos {\\rm{A}}\\cos {\\rm{B}} - \\cos {\\rm{B}}\\cos {\\rm{C}}\\sin {\\rm{A}}} \\over {\\cos {\\rm{A}}\\cos {\\rm{B}}\\cos {\\rm{C}}}} \\cr & \\cr} {/tex}=\xa0{tex}{0 \\over {\\cos {\\rm{A}}\\cos {\\rm{B}}\\cos {\\rm{C}}}}{/tex}= 0Hence proved Prove 0 |
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| 6534. |
Lim-xtans 1 (x^4_3x^2+2÷x^3 _5x^2+3x+1 |
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| 6535. |
Show that the four points (0,0),(1,1),(5,—5)and(6,—4)are concyclic |
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| 6536. |
CosA/a + cosB/b + cosC/c = a^2 + b^2 + c^2 |
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| 6537. |
prove by mathmetical induction n2+n is even |
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| 6538. |
Binomial theorems(4x/5 -5/2x) |
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| 6539. |
TanA.+TanB+TanC=TanA.TanB.TanC prove that |
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| 6540. |
What is the value of tan 15° |
| Answer» We can find its value as follows:-tan15°= tan(45°-30°) = {tex}{tan 45° - tan30°\\over 1+tan45°tan30°}{/tex} = {tex}{1- {1\\over {\\sqrt 3}}}\\over1+1×{1\\over {\\sqrt 3}}{/tex} = {tex}{{{\\sqrt 3}-1}\\over {\\sqrt 3}}\\over {{{\\sqrt 3}+1}\\over {\\sqrt 3}}{/tex} = {tex}{\\sqrt 3}-1\\over {\\sqrt 3}+1{/tex} | |
| 6541. |
Cos20.cos40.cos60.cos80=1/16 |
| Answer» =(1/8)cos80°+(1/8)cos100°+(1/8)cos60°=(1/8)(cos80°+cos100°)+(1/8)×(1/2)=(1/8)[{2cos(80°+100°)/2}{cos(80°-100°)/2}]+(1/16)=(1/8)(2cos90°cos10°)+(1/16)=0+(1/16) [cos90°=0]=1/16 (proved) | |
| 6542. |
Prove that cos 20°cos40°.cos60°.cos80°=1÷16 |
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| 6543. |
Lim(x^3+1/x+1)X>0 |
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| 6544. |
Tan15°+cot15°=1 |
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| 6545. |
Cot x =3/4, x lies in third quadrat |
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| 6546. |
if ab+b-a+1=0 find possible integral solution of a and b |
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| 6547. |
Can we get excellent marks in trigonometry in maths.........!!!!! |
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| 6548. |
How can i download the solutions of the book accountancy by dk goel of class 11 |
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| 6549. |
What the value of the third quadrant Sine value |
| Answer» In third quadrant sine is negative\xa0 | |
| 6550. |
If f (x)=2x^2+1 . Find f (1/x) and f (root x) |
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