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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.
| 11201. |
Defination of probability |
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Answer» probability means chances of occurrence\xa0of an event. Probability is favourable outcomes divided by total no. of events |
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| 11202. |
TAN30° IS? |
| Answer» 1/{tex}\\sqrt3{/tex} | |
| 11203. |
State and prove pythagorasTheorem |
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| 11204. |
How can we find relationship between zeros and coefficients of a polynomial |
| Answer» Here the given polynomial f(x)= 2x2+ 5x\xa0-12= 2x2 + 8x - 3x -12= 2x(x + 4) - 3(x + 4)= (x + 4)(2x-3)f(x) = 0, x+4 =0 or 2x-3=0then, x = - 4 or x =\xa0{tex}\\frac{3}{2}{/tex}So, the zeros of\xa0f(x) are -4 and\xa0{tex}\\frac{3}{2}{/tex}Now,Sum of the zeros =\xa0{tex}\\left(-4+\\frac{\\displaystyle3}{\\displaystyle2}\\right)=\\frac{\\displaystyle-5}{\\displaystyle2}=\\frac{\\displaystyle-\\mathrm b}{\\displaystyle\\mathrm a}{/tex},product of the zeros =\xa0{tex}\\text{(-4)×}\\frac{\\displaystyle3}{\\displaystyle2}=\\frac{\\displaystyle-12}{\\displaystyle2}=\\frac{\\displaystyle\\mathrm c\\;}{\\displaystyle\\mathrm a}{/tex}Hence the relation of zeros and coefficients of the polynomial is verified. | |
| 11205. |
The maximum and minimum values of all the 6 trigonometric ratios |
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| 11206. |
Find two irrational number between 3 and 4 ? |
| Answer» Since 3={tex}{\\sqrt 9}{/tex} and 4={tex}{\\sqrt {16}}{/tex} so any number between\xa0{tex}{\\sqrt {10}}{/tex} to\xa0{tex}{\\sqrt {15}}{/tex} is irrational.So,\xa0{tex}{\\sqrt {10}}{/tex} and {tex}{\\sqrt {11}}{/tex} are two irrational numbers between 3 and 4.\xa0 | |
| 11207. |
(X+2)2 = 2x(x2-1) |
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| 11208. |
(-5,7) , (-1,3) |
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| 11209. |
any sample paper regarding 2018 exam |
| Answer» Check Sample Papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 11210. |
x= a sin A and y= b tan A , then show that x*x - y*y=a*a - b*b. |
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| 11211. |
7×2 |
| Answer» 7 x 2= 14 | |
| 11212. |
(2,1),(-1,2)and,(1,0)are the coordinates of three of a parallelogram . Find the fourth vertex |
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| 11213. |
If a and b zeroes of quadratic polynomial is x2-7x+p. Such a2 + b2 = 29 find the value of p |
| Answer» Let a and b be the zeroes of the quadratic polynomial x2\xa0-7x+p.By using relationship between coefficient and zeroes of quadratic polynomial we get\xa0sum of zeroes= a+b={tex}{-(-7)\\over 1} = 7{/tex} -----(1)Product of zeroes = a×b ={tex}{p\\over 1}=p{/tex} -----(2)Given a2 + b2\xa0= 29 -----(3)\xa0from eq.(1)a+b=7on squaring both sides(a+b)2 = 72a2\xa0+ 2ab +b2 =49(a2\xa0+b2)+2ab= 49 29+2p =49 [by using eq.(2) ab=p and eq.(3) a2\xa0+ b2 =29] 2p=49-29 2p=20 p={tex}20\\over 2{/tex} p= 10 | |
| 11214. |
whst is ap |
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| 11215. |
sum of first 14 term of an ap is 1505 and its first term is 10 find its 25 term |
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Answer» Ans is 135 S14=1505 &A=10 we know that Sn=n/2[2a+(n-1)d ] here S14=14/2[2×10 +1(14-1)d ] 1505 = 7(20+13d)1505/7=20+13d215=20+13d215-20=13d195=13dd=195/13d=15Now A25=a+24d = 10+24×15 = 10+360 A25=370 is the answer S14=15051505=n/2 (2a+(n-1)d 1505=14/2 (2 (10)+(14-1)d1505=7 (20+13)d 1505=7 (43)d 1505=301 (d)5=doA25=a+24d 10+24×5 10+120 130 Answer is 120 |
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| 11216. |
Divide 2x^2+3x+1. by 3x-1 |
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| 11217. |
Find the value of k, for which one root of the quadratic equation kx2_14x+8=0 is six times the other |
| Answer» Let one root = {tex}\\alpha{/tex}Other root = 6{tex}\\alpha{/tex}{tex}\\therefore{/tex}Sum of roots =\xa0{tex}\\alpha{/tex}\xa0+ 6{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac{14}{k}{/tex}or,\xa0{tex}7 \\alpha = \\frac { 14 } { k } \\text { or } \\alpha = \\frac { 2 } { k }{/tex} ...........(i)Product of roots\xa0{tex}\\alpha ( 6 \\alpha ) = \\frac { 8 } { k }{/tex}or,\xa0{tex}6 \\alpha ^ { 2 } = \\frac { 8 } { k }{/tex}..........(ii)Solving (i) and (ii),{tex}6 \\left( \\frac { 2 } { k } \\right) ^ { 2 } = \\frac { 8 } { k }{/tex}{tex}6 \\times \\frac { 4 } { k ^ { 2 } } = \\frac { 8 } { k }{/tex}or,\xa0{tex}\\frac { 3 } { k ^ { 2 } } = \\frac { 1 } { k }{/tex}or, 3k = k2or, 3k - k2\xa0= 0k[3-k] = 0{tex}\\therefore{/tex}\xa0k = 0 or k = 3k = 0 is not possibleHence, k = 3 | |
| 11218. |
My exams are starting in 10 days plz suggest any statergy for studying maths plz help |
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Answer» Don\'t be tense just believe in yourself and give the exam Practice, Practice and Practice is the only strategy to gain confidence and have\xa0grip on math.Practice is key to success\xa0\xa0 What are you doing when teacher teaches in the class Easy way is to do practise |
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| 11219. |
What is ordinate? |
| Answer» Ordinate is\xa0a straight line from any point drawn parallel to one coordinate axis and meeting the other, especially a coordinate measured parallel to the vertical. | |
| 11220. |
What is abcissa |
| Answer» X axis is knows abcissa | |
| 11221. |
What is zeroes of polynomial? |
| Answer» 0 | |
| 11222. |
Prove that secA(1-sin A)(secant+tanA)=1 |
| Answer» LHS\xa0{tex}= \\sec A(1 - \\sin A)(\\sec A + \\tan A){/tex}{tex} = (\\sec A - \\sec A \\times \\sin A)(\\sec A + \\tan A){/tex}{tex} = \\left( {\\sec A - \\frac{1}{{\\cos A}} \\times \\sin A} \\right)\\left( {\\sec A + \\tan A} \\right){/tex}\xa0{tex}\\left[ {\\because \\sec A = \\frac{1}{{\\cos A}}} \\right]{/tex}{tex} = \\left( {\\sec A - \\frac{{\\sin A}}{{\\cos A}}} \\right)(\\sec A + \\tan A){/tex}{tex} = (\\sec A - \\tan A)(\\sec A + \\tan A){/tex}\xa0{tex}\\left[ {\\because \\frac{{\\sin A}}{{\\cos A}} = \\tan A} \\right]{/tex}Using identity (a - b)(a + b) = a2 - b2= sec2 - tan2A= 1\xa0{tex}\\left[ {\\because {{\\sec }^2} - {{\\tan }^2}A = 1} \\right]{/tex}= RHSHence proved | |
| 11223. |
What is the formula of trignomentry identies |
| Answer» some of the impotatant trigonometric identities/ functions areconsider a right angled triange ABC rt Ld at B\xa0and LBAC = xsin\xa0x = opposite side/hypotenuse = BC / ACcos x = adjacent side / hypotenuse = AB / ACtan\xa0x = oppo/adjacent = BC / AB | |
| 11224. |
Ap |
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| 11225. |
Prove that 1+((cot^2thetta)/1+codec thetta) = 1/sin thetta |
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| 11226. |
x+y=33x+2y=8 |
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Answer» x + y = 3 => x = 3 - yPutting this value of x in 3x + 2y = 83(3 - y) + 2y = 8=> 9 - 3y + 2y = 8=> -y = -1=> y = 1Putting the value of y in x + y =3x + 1 = 3=> x = 2 y=1. x=2 |
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| 11227. |
Sin-cos=0 find sin^4+cos^4 |
| Answer» Sin θ -cos θ =0Sin θ = cos θ Sin θ /Cos θ =1Tan θ =1 θ =45°Now, (sin θ )^4+(cos θ )^4=(sin45°)^4+(cos45°)^4(1/√2)^4+(1/√2)^4=1/4+1/4=2/4=1/2Hence 1/2 is ans.Hope it will help u. | |
| 11228. |
(SinA-cosesA)(cosA-secA)=tanA+cotA |
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| 11229. |
If x= a sec A + b tan A and y= a tan A + b sec A , then prove that x*x-y*y= a*a-b*b. |
| Answer» L.H.S. =\xa0{tex}{x^2} - {y^2} = {\\left( {a\\sec {\\rm{A}} + b\\tan {\\rm{A}}} \\right)^2} - {\\left( {a\\tan {\\rm{A}} + b\\sec {\\rm{A}}} \\right)^2}{/tex}=\xa0{tex}\\left( {a\\sec {\\rm{A}} + b\\tan {\\rm{A}} + a\\tan {\\rm{A}} + b\\sec {\\rm{A}}} \\right) {\\left( {a\\sec {\\rm{A}} + b\\tan {\\rm{A}} - a\\tan {\\rm{A}} - b\\sec {\\rm{A}}} \\right)}{/tex}=\xa0{tex}\\left[ {a\\left( {\\sec {\\rm{A}} + \\tan {\\rm{A}}} \\right) + b\\left( {\\sec {\\rm{A}} + \\tan {\\rm{A}}} \\right)} \\right]\\left[ {a\\left( {\\sec {\\rm{A}} - \\tan {\\rm{A}}} \\right) + b\\left( { - \\sec {\\rm{A}} + \\tan {\\rm{A}}} \\right)} \\right]{/tex}=\xa0{tex}\\left( {\\sec {\\rm{A}} + \\tan {\\rm{A}}} \\right)\\left( {a + b} \\right)\\left( {\\sec {\\rm{A}} - \\tan {\\rm{A}}} \\right)\\left( {a - b} \\right){/tex}=\xa0{tex}\\left( {{{\\sec }^2}{\\rm{A}} - {{\\tan }^2}{\\rm{A}}} \\right)\\left( {{a^2} - {b^2}} \\right){/tex}=\xa0{tex}{a^2} - {b^2}{/tex}= R.H.S. | |
| 11230. |
23×0 |
| Answer» Answer = 0 | |
| 11231. |
If point p(x,y)is any point on the line joining the points A(a,0) B(0,b) then show that x/a+y/B=1 |
| Answer» If a point (X,Y) lies on a line joining the points A (x1,y1) and B (x2,y2) the equation of the line is given by{tex}\\frac{y-y_{1}}{x-x_{1}}=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}{/tex}Point P(x,y) lies on the line joining the points A (a,0) and B (0,b). So{tex}\\frac{y-0}{x-a}=\\frac{b-0}{0-a}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{y}{x-a}=-\\frac{b}{a}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}a y=-b(x-a){/tex}{tex}\\Rightarrow{/tex}\xa0{tex}a y=-b x+a b{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}a y+b x=a b{/tex}divide both side by ab{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{a y}{a b}+\\frac{b x}{a b}=\\frac{a b}{a b}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{y}{b}+\\frac{x}{a}{/tex}= 1 (proved) | |
| 11232. |
Prove that cube root 6 is not a rational number |
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| 11233. |
What is mean by secant. pls answer this question |
| Answer» It is just a name given to the trigonometric ratio H÷B | |
| 11234. |
Complementary square method |
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| 11235. |
Show that the sequence defined by an=5n-7 is an AP |
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| 11236. |
Find value of a and b for infinitely many solution 2x -3y = 7 (a+b) x -(a+b-3) y =4a+b |
| Answer» Here\xa0{tex}{a_1} = 2,{b_1} = - 3,{c_1} = 7{/tex}\xa0and\xa0{tex}{a_2} = a + b,{b_2} = - \\left( {a + b - 3} \\right),{c_2} = 4a + b{/tex}For infinitely many solutions,\xa0{tex}{{{a_1}} \\over {{a_2}}} = {{{b_1}} \\over {{b_2}}} = {{{c_1}} \\over {{c_2}}}{/tex}=> {tex}{2 \\over {a + b}} = {{ - 3} \\over { - a - b + 3}} = {7 \\over {4a + b}}{/tex}Taking\xa0{tex}{2 \\over {a + b}} = {{ - 3} \\over { - a - b + 3}}{/tex}=> {tex} - 2a - 2b + 6 = - 3a - 3b{/tex}=> {tex}a - b = 6{/tex} .........(i)Again taking\xa0{tex}{2 \\over {a + b}} = {7 \\over {4a + b}}{/tex}=> {tex}8a + 2b = 7a + 7b{/tex}=> {tex}a - 5b = 0{/tex} ..............(ii)On solving eq.(i) and (ii){tex}a = {{15} \\over 2},b = {3 \\over 2}{/tex} | |
| 11237. |
-×-=? |
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| 11238. |
X-a/x-b + x-b/x-a =a/b + b/a solve the quadratic equation |
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Answer» {tex}{{x - a} \\over {x - b}} + {{x - b} \\over {x - a}} = {a \\over b} + {b \\over a}{/tex}=> {tex}{{{x^2} + {a^2} - 2ax + {x^2} + {b^2} - 2bx} \\over {{x^2} - ax - bx + ab}} = {{{a^2} + {b^2}} \\over {ab}}{/tex}=> {tex}{{2{x^2} + {a^2} - 2ax + {b^2} - 2bx} \\over {{x^2} - ax - bx + ab}} = {{{a^2} + {b^2}} \\over {ab}}{/tex}=> {tex}2ab{x^2} + {a^3}b - 2{a^2}bx + a{b^3} - 2a{b^2}x = {a^2}{x^2} - {a^3}x - {a^2}bx + {a^3}b + {b^2}{x^2} - a{b^2}x - {b^3}x + a{b^3}{/tex}=> {tex}2ab{x^2} - {a^2}bx - a{b^2}x = {a^2}{x^2} - {a^3}x + {b^2}{x^2} - {b^3}x{/tex}=> {tex}{a^2}{x^2} - 2ab{x^2} + {b^2}{x^2} - {a^3}x - {b^3}x + {a^2}bx + a{b^2}x = 0{/tex}=> {tex}{x^2}\\left( {{a^2} + {b^2} - 2ab} \\right)+x\\left( { - {a^3} - {b^3} + {a^2}b + a{b^2}} \\right) = 0{/tex}=> {tex}x\\left[ {x\\left( {{a^2} + {b^2} - 2ab} \\right) + \\left( { - {a^3} - {b^3} + {a^2}b + a{b^2}} \\right)} \\right] = 0{/tex}=> {tex}x = 0{/tex}\xa0Or\xa0{tex}x = {{{a^3} + {b^3} - {a^2}b - a{b^2}} \\over {{{\\left( {a - b} \\right)}^2}}}{/tex} x - a /x - b + x - b / x - a = a/b + b/ax2 + a2 - 2ax + x2 + b2 - 2bx /x2 + ab -ax - bx = a2 + b2 /ab |
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| 11239. |
Prove that the line joining the midpoints of any two sides of a triangle is parallel to third side. |
| Answer» According to question it is given that ABC is a triangle in which D and E are the midpoints\xa0of AB and AC respectively.To Prove\xa0{tex}D E \\| B C{/tex}Proof :-\xa0Since D and E are the midpoints of AB and AC respectively, we have AD = DB and AE = EC.{tex}\\therefore \\quad \\frac { A D } { D B } = \\frac { A E } { E C }{/tex}\xa0[each equal to 1].Hence, by the converse of Thales\' theorem,\xa0{tex}D E \\| B C{/tex}. | |
| 11240. |
4sin theta =3 cos theta find the value of 12 sin theta -7 cos theta/8 sin theta +3 cos theta |
| Answer» Given :\xa0{tex}4 sin \\theta = 3 cos \\theta {/tex}=>\xa0{tex}{sin\\theta\\over cos\\theta }= {3\\over 4}{/tex}=>\xa0{tex}tan \\theta = {3\\over 4}{/tex}To find :\xa0{tex}{12sin\\theta - 7cos \\theta\\over 8 sin \\theta+ 3 cos\\theta}{/tex}Divide numerator and denominator by\xa0{tex}cos \\theta {/tex}, we get{tex}12 tan\\theta - 7\\over 8 tan \\theta + 3{/tex}=\xa0{tex}{9-7\\over 6+3}={ 2\\over 9}{/tex} | |
| 11241. |
TanQ+1/tanQ=secQcosecQ |
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Answer» answer tanQ+1 upon tanQ===secQ.cosecQ TanQ+1/tanQ=tanQ×tanQ+1÷tanQ =secQ×secQ÷tanYa =1/cosQ×1/cosQ×cosQ/sinQ =1/cosQ×1/sinQ =secQcosecQ |
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| 11242. |
Tan+1/tan=sec cosec |
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| 11243. |
Proof of phtagorus therom |
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| 11244. |
√5 is rational |
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| 11245. |
If a1x+b1y+c1=0 and a2x+b2x+c2=0 is perpendicular together. So show that a1a2+b1b2=0 |
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| 11246. |
Find the sum of first 1000 +ve integers |
| Answer» (1000*1001)/2=500500 | |
| 11247. |
Solve for thetaCos square theta÷cot square theta - cos square theta |
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| 11248. |
Write the zeroes of tha polynomial x2-3 |
| Answer» X square + alfa + bita + alfa x bita = 0 x square + 2 + -3x square +2-3 | |
| 11249. |
rd sharma exercise solutions 3.7 |
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| 11250. |
If the sum of sequence 2,5,8,11,.........Is 60100,the number of terms is |
| Answer» Here a=2d=3Sn=60100n/2(2a+(n-1)d)=60100n(2*2+(n-1)3)=60100*2n(4+3n-3)=120200n(3n+ | |