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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.
| 37151. |
Express 23150 as product of its prime factors |
| Answer» Do with algorithm method | |
| 37152. |
What is a composite, prime, and co-prime number ? |
| Answer» Composite Numbers -A composite number is a positive integer that has a factor other than 1 and itself. For instance, 12 is a composite number because it is a positive integer and it has factors other than 1 and itself. 2, 3, 4, 6 are its factors other than 1 and itself. All even numbers greater than 2 are composite numbers. 2 is a prime number. The smallest composite number is 4.Prime numbers:Numbers which have only two factors 1 and the number itself are called prime numbers.E.g 2 ,3 ,5, 7 ,11 ,13, 17 and so on are all prime numbers.Of the first 100 numbers 25 are prime numbers.Only 2 is an even prime number all prime numbers are odd.Co prime numbers:Numbers which do not have any common factor between them other than one are called coprime numbers.Two natural numbers( not necessarily Prime) are coprime if their HCF is 1.For e.g\xa0(1,2) , (1,3), (3,4), (3,10), (3,8), (5,6), (17,23)even two composite numbers can also be Prime. for e.g(16,25), (84,65) | |
| 37153. |
/5_3 and -/5_3 |
| Answer» | |
| 37154. |
Find the value of k if a+b=1/2ab and polynomial is x^2-6x+4k-2-kx .a is alpha and b is betta. |
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| 37155. |
Find the value of k, for which the following will represent coincident lines. 2x +ky+8y=14 |
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| 37156. |
Rd sharma exercise 2.1ques |
| Answer» | |
| 37157. |
If both a and b are rational number then a and b from the following 3-√5/3+2√5=a√5- b,are |
| Answer» LHS = ( 3 - √5 )/(3 + 2√5 )= [(3-√5)(2√5-3)]/[(2√5+3)(2√5-3)]= [6√5 - 9 -10 + 3√5 ]/[ ( 2√5 )² - 3² ]= ( 9√5 - 19 )/( 20 - 9 )= ( 9√5 - 19 )/11= ( 9/11 )√5 - 19/11 ---( 1 )= a√5 - b ----------( 2 ) [ RHS ]from ( 1 ) and ( 2 ) ,a = 9/11 ,b = 19/11 | |
| 37158. |
9y+11x=26x×y=60 |
| Answer» | |
| 37159. |
If HCF of 65 and 117 is expressible in the form of 65n-117 , find n |
| Answer» 117 = 13 {tex}\\times{/tex}\xa03 {tex}\\times{/tex}\xa0365 = 13 {tex}\\times{/tex}\xa05HCF (117, 65) = 13LCM(117,65) = 13 {tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa03 {tex}\\times{/tex}\xa03 = 585Here is given that:{tex}HCF =65m-117 {/tex}{tex}13=65m-117{/tex}{tex}65m=130{/tex}m =\xa0{tex}\\frac { 130 } { 6 5 } ={/tex}2 | |
| 37160. |
Cosec^3 30× cos60×tan^45×sin^90×sec^40×cot30 |
| Answer» We know that, cosec30°=2, cos60°=(1/2), tan45°=1=sin90°, sec45°=√2 & cot30°=√3,putting these values in the given expression, we get:-{tex}\\cos e{c^3}30^\\circ \\cos 60^\\circ {\\tan ^3}45^\\circ {\\sin ^2}90^\\circ {\\sec ^2}45^\\circ \\cot 30^\\circ {/tex}{tex} = {(2)^3} \\times {\\left( {\\frac{1}{2}} \\right)} \\times {(1)^3} \\times {(1)^2} \\times {\\left( {\\sqrt 2 } \\right)^2} \\times \\sqrt 3 {/tex}{tex} = 8 \\times \\frac{1}{2} \\times 1 \\times 1 \\times 2 \\times \\sqrt 3 {/tex}{tex} = 8\\sqrt 3 {/tex} | |
| 37161. |
Find HCF of 276& 1238 by prime factorization |
| Answer» | |
| 37162. |
x-y+z=4x+y+z=22x+y-3z=0 |
| Answer» {tex}A = \\left[ {\\begin{array}{*{20}{c}} 1&{ - 1}&1 \\\\ 2&1&{ - 3} \\\\ 1&1&1 \\end{array}} \\right]{/tex}{tex}\\left| A \\right| = \\left[ {\\begin{array}{*{20}{c}} 1&{ - 1}&1 \\\\ 2&1&{ - 3} \\\\ 1&1&1 \\end{array}} \\right]{/tex}{tex} = 10 \\ne 0{/tex}Here,A11 = 4, A12 = -5, A13 = 1A21 = 2, A22 = 0, A23 = -2A31 = 2, A32 = 5, A33 = 3{tex}adjA = \\left[ {\\begin{array}{*{20}{c}} 4&2&2 \\\\ { - 5}&0&5 \\\\ 1&{ - 2}&3 \\end{array}} \\right]{/tex}{tex}{A^{ - 1}} = \\frac{1}{{\\left| A \\right|}}\\left( {adjA} \\right){/tex}{tex}= \\frac{1}{{10}}\\left[ {\\begin{array}{*{20}{c}} 4&2&2 \\\\ { - 5}&0&5 \\\\ 1&{ - 2}&3 \\end{array}} \\right]{/tex}System of equation can be written isX = A-1B{tex} = \\frac{1}{{10}}\\left[ {\\begin{array}{*{20}{c}} 4&2&2 \\\\ { - 5}&0&5 \\\\ 1&{ - 2}&3 \\end{array}} \\right]\\left[ {\\begin{array}{*{20}{c}} 4 \\\\ 0 \\\\ 2 \\end{array}} \\right]{/tex}{tex}\\left[ {\\begin{array}{*{20}{c}} x \\\\ y \\\\ z \\end{array}} \\right] = \\left[ {\\begin{array}{*{20}{c}} 2 \\\\ { - 1} \\\\ 1 \\end{array}} \\right]{/tex}x = 2, y = -1, z = 1 | |
| 37163. |
(a-b+c)2 |
| Answer» | |
| 37164. |
Use Euclid\'s division leema to show square of any positive integer are either in form 4q ,4q+1 |
| Answer» Let a = 4q + r, when r = 0, 1, 2 and 3{tex}\\therefore{/tex}Numbers are 4q, 4q + 1, 4q + 2 and 4q + 3{tex}( a ) ^ { 2 } = ( 4 q ) ^ { 2 } = 16 q ^ { 2 } = 4 ( 4 q ) ^ { 2 } = 4 m{/tex}{tex}( a ) ^ { 2 } = ( 4 q + 1 ) ^ { 2 } = 16 q ^ { 2 } + 8 q + 1 = 4 \\left( 4 q ^ { 2 } + 2 q \\right) + 1 = 4 m + 1{/tex}{tex}( a ) ^ { 2 } = ( 4 q + 2 ) ^ { 2 } = 16 q ^ { 2 } + 16 q + 4 = 4 \\left( 4 q ^ { 2 } + 4 q + 1 \\right) = 4 m{/tex}{tex}( a ) ^ { 2 } = ( 4 q + 3 ) ^ { 2 } = 16 q ^ { 2 } + 24 q + 9 = 4 \\left( 4 q ^ { 2 } + 6 q + 2 \\right) + 1 = 4 m + 1{/tex}{tex}\\therefore {/tex}\xa0the square of any +ve integer is of the form 4q or 4q + 1\xa0 | |
| 37165. |
Find the probability of (a+b)^2=a^2+b^2 |
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Answer» Its 0 1 |
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| 37166. |
√2 and√5 between 4 rational no. |
| Answer» 1.616661661.818818881.99199919999 | |
| 37167. |
2x+y=63x-y=14Solve by elimination method |
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Answer» 2x+y+3x-y=14+65y=20Y=42x+y=62x+4=62x=6/4X=6/4*2X=3/4 2x+y=63x-y=14_________-X = 20By subing the value of x in 1 2×20+y=6Y=46 |
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| 37168. |
X-9=2/3 |
| Answer» X=29/3=9.66(approx.) | |
| 37169. |
How many is better for study?? |
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Answer» Hours does not depends for study . Its how much you learn or understand during your study time. We have to focus on our studies. It depends on you. That how much you can concentrate in studies. How much you can study. It depends on your choice.From my point of view. It doesn\'t matter how long u study ..... instead it matters how effectively u study !!!! sorry my question is that How many hour is better for study?? Sorry, not able to understand yur ques..!! |
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| 37170. |
Loss |
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Answer» Kiska loss bhaiya?? Hlo What loss?? ...? Kiska loss... |
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| 37171. |
Solve for x and y 1\\(3x+y)+1\\(3x-y)=3\\4 |
| Answer» | |
| 37172. |
Find the value of k in 2x + 3y=7 and (k-1)x + (k+2)y=3k |
| Answer» {tex}2x + 3y = 7{/tex}{tex}(k - 1) x + (k + 2)y = 3k{/tex}These are of the form{tex}a_1x + b_1y + c_1\xa0= 0 ,\\ a_2x + b_2y + c_2\xa0= 0{/tex}where,{tex}a_1= 2\xa0,\\ b_1= 3,\\ c_1\xa0= -7,{/tex}{tex}a_2=k - 1\xa0\\ ,b_2= k + 2 ,\\ c_2\xa0= -3k {/tex} for infinitely many solutions, we must have{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}This hold only when{tex}\\frac { 2 } { k - 1 } = \\frac { 3 } { k + 2 } = \\frac { - 7 } { - 3 k }{/tex}{tex}\\frac { 2 } { k - 1 } = \\frac { 3 } { k + 2 } = \\frac { 7 } { 3 k }{/tex}Now the following cases arises:Case I:{tex}\\frac { 2 } { k - 1 } = \\frac { 3 } { k + 2 }{/tex}{tex}\\Rightarrow{/tex}2(k + 2) = 3(k -1){tex}\\Rightarrow{/tex}2k + 4= 3k - 3{tex}\\Rightarrow{/tex}\xa0k = 7Case II:{tex}\\frac { 3 } { k + 2 } = \\frac { 7 } { 3 k }{/tex}{tex}\\Rightarrow{/tex}7(k + 2) = 9k{tex}\\Rightarrow{/tex}7k + 14= 9k{tex}\\Rightarrow{/tex}\xa0k = 7Case III:{tex}\\frac { 2 } { k - 1 } = \\frac { 7 } { 3 k }{/tex}{tex}\\Rightarrow{/tex}\xa07k - 7 = 6k{tex}\\Rightarrow{/tex}\xa0k = 7For k = 7, there are infinitely many solutions of the given system\xa0of equations. | |
| 37173. |
4/3 {sin square 59 -cot ^31}- 2/3 sin 90+3 tan^56 tan^2 34=x/3 |
| Answer» 21/3. Ha xnxn | |
| 37174. |
Find largest no. which when divides 705 & 1053 leaving remainder 9 in each case |
| Answer» | |
| 37175. |
Cot-1 (√1+sinx+√1-sinx/√1+sinx-√1-sinx) |
| Answer» | |
| 37176. |
Show that 3 under root 6 is an irrational |
| Answer» Let 3 under root 6 be a rational no.3 under root 6 = P/Q where p and q are co-prime numbers and Q is not equal to 03 under root 6= PQUnder root6=P/3QRoot 6 is irrational no So p/3q is also irrational no So there is contradiction3 under root 6 is irrational number | |
| 37177. |
9x^-3x-2=0 |
| Answer» We have,9x2 - 3x - 2 = 0{tex}\\Rightarrow{/tex}\xa09x2 - 6x + 3x - 2 = 0{tex}\\Rightarrow{/tex}\xa03x(3x - 2) + 1(3x - 2) = 0{tex}\\Rightarrow{/tex}\xa0(3x - 2)(3x + 1) = 0{tex}\\Rightarrow{/tex}\xa03x - 2 = 0 or, 3x + 1 = 0\xa0{tex}\\Rightarrow \\quad x = \\frac { 2 } { 3 } \\text { or, } x = - \\frac { 1 } { 3 }{/tex}Thus,\xa0{tex}x = \\frac { 2 } { 3 } \\text { and } x = - \\frac { 1 } { 3 }{/tex} are the required roots of the given equation. | |
| 37178. |
3√5x×x+25x-10√5=0 |
| Answer» | |
| 37179. |
What is cumulative frequency |
| Answer» It is just adding of numbers | |
| 37180. |
How to find class mark in statistics |
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Answer» Upper limit +lower limit ,and ➗ by 2 lower limit subtrated by upper limit, then divided by 2 |
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| 37181. |
Trigo trigonometric ratio of complementary angle |
| Answer» Trigonometric Ratios Of Complementary AnglesWe know complementary angles are pair of angles whose sum is 90°Like 40°, 50°, 60°, 30°, 20°, 70°, 15°, 75° ; etc,Formulae:sin (90° – θ) = cos θ, cot (90° – θ) = tanθcos (90° – θ) = sin θ, sec (90° – θ) = cosec θtan (90° – θ) = cot θ, cosec (90° – θ) = sec θ | |
| 37182. |
Given that HCF(306,657)is 9,find lcm |
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Answer» HCF × LCM= product of two no.s9× LCM= 306×657LCM= 22338 HCF(a & b)×LCM (a & b)=a×b9 × LCM(a & b) =306 × 6579 × LCM(a & b ) =201042LCM(a & b ) = 201042 ________ 9LCM(a & b)= 22338 2238 |
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| 37183. |
2x +3y = 7: (a-b)x + (a+b)y = 3a + b - 2 |
| Answer» 2x + 3y - 7 = 0{tex}a _ { 1 } = 2 , b _ { 1 } = 3 , c _ { 1 } = - 7{/tex}\xa0a(x + y) - b(x - y) = 3a + b - 2ax +ay -b x + by = 3a + b - 2(a - b)x + (a + b)y - (3a + b - 2) = 0{tex}a _ { 2 } = a - b , b _ { 2 } = a + b , c _ { 2 } = - ( 3 a + b - 2 ){/tex}\xa0for infinite many solutions{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}\xa0or,\xa0{tex}\\frac { 2 } { a - b } = \\frac { 3 } { a + b } = \\frac { - 7 } {- ( 3 a + b - 2 ) }{/tex}\xa0{tex}\\frac { 2 } { a - b } = \\frac { 7 } { 3 a + b - 2 }{/tex}\xa02(3a + b -2 ) = 7(a-b)6a + 2b - 4 = 7a - 7b7a- 7b - 6a - 2b + 4 = 0a - 9b = - 4 ....(i){tex}\\frac { 3 } { a + b } = \\frac { 7 } { 3 a + b - 2 }{/tex}\xa03(3a + b - 2) = 7(a + b )9a + 3b - 6 = 7a + 7b9a + 3b - 7a - 7b = 62a - 4 b = 6a - 2 b = 3 .......(ii)Subtracting eqn. (i) from (ii){tex} \\Rightarrow {/tex}\xa0b = 1On putting the value of b in eqn. (i),we geta = 5Hence, a = 5, b = 1. | |
| 37184. |
4x^2 -4x - 3 find the zeros |
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Answer» 4x2 - 4x - 3= 4x2 - 6x + 2x - 3= 2x ( 2x - 3) + 1 (2x - 3)= (2x - 3) ( 2x + 1)2x - 3 = 0 or 2x + 1 = 02x = 3 or 2x = -1x = 3/2 or x = -1/2 First put the values and check it 6 and -2 are the zeros . 3/2 and -1/2 |
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| 37185. |
Class 10 math 2019 solved question paper set 2 |
| Answer» Check the papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 37186. |
If secA+tanA=p then show that psquarep-1÷psquare+1=simA |
| Answer» secA + tanA = p....................(1)We know that,sec²A - tan²A=1or, (secA + tanA)(secA-tanA)=1or, p(secA-tanA)=1or, secA-tanA=1/p .............(2)Adding (1) and (2) we get,2secA=p+1/por, secA=(p²+1)/2p∴, cosA=1/secA=2p/(p²+1)∴, sinA=√(1-cos²A)=√{1-4p²/(p²+1)²=√{(p²+1)²-4p²}/(p²+1)²=√(p⁴+2p²+1-4p²)/(p²+1)=√(p²-1)²/(p²+1)=(p²-1)/(p²+1) (Proved) | |
| 37187. |
Show that there is no positive integer \'n\'dor which√n-1 +√n+1 is a rational |
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Answer» I don\'t thik so it is a question I dont no |
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| 37188. |
I want 100 differen type of questions of chapter 1 with solutions |
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Answer» My suggestion is to clear up all the examples along with exercise questions of both NCERT and RD SHARMA for better clearance all ur doubts. If u solve it by urself it is easy to understand If u didn\'t get 100 questions solve all questions to of both book u will definitely got ur answer Solve all examples of ncert as well as r.s agarwal questions |
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| 37189. |
Proof of sin0=0 |
| Answer» Draw a circle of radius one unit.Through the centre draw the y- axis and the x-axis.Mark the coordinates of the points where the axes intersect the circle.Starting on the x-axis at 0 degrees ,The coordinates are (1,0).Move anticlockwise along the circle to the point (0,1) which corresponds to 90 degrees.Continue to the point (-1,0) which is 180 degrees.Again move to the point (0,-1) which is 270 degrees.And finally back to (1,0) which is 360 degrees.For 0 degrees, the x-co-ordinate represents the cosine of 0,and is equal to 1. And the y -co-ordinate which is zero is the sine of 0. | |
| 37190. |
Sec(6) theetha=tan(6)thetha+ 3tan(2)thetha(sec(2)thetha+1 |
| Answer» \xa0We have to prove that :-\xa0{tex} \\Rightarrow {\\sec ^6}\\theta = {\\tan ^6}\\theta +3{\\tan ^2}\\theta {\\sec ^2}\\theta + 1{/tex}{tex} \\Rightarrow {\\sec ^6}\\theta - {\\tan ^6}\\theta = 3{\\tan ^2}\\theta {\\sec ^2}\\theta + 1{/tex}Now, LHS\xa0{tex} = {\\sec ^6}\\theta - {\\tan ^6}\\theta {/tex}{tex} = {\\left( {{{\\sec }^2}\\theta } \\right)^3} - {({\\tan ^2}\\theta )^3}{/tex}{tex}= \\left( {{{\\sec }^2}\\theta - {{\\tan }^2}\\theta } \\right)\\left[ {{{({{\\sec }^2}\\theta )}^2} + {{\\sec }^2}\\theta {{\\tan }^2}\\theta + {{({{\\tan }^2}\\theta )}^2}} \\right]{/tex}{Since, a3\xa0- b3\xa0= (a - b )(a2\xa0- ab + b2\xa0)}{tex} = 1\\left[ {{{\\sec }^4}\\theta + {{\\sec }^2}\\theta {{\\tan }^2}\\theta + {{\\tan }^4}\\theta } \\right]{/tex}\xa0{tex}\\left[ {\\because {{\\sec }^2}\\theta - {{\\tan }^2}\\theta = 1} \\right]{/tex}{tex} = {\\sec ^4}\\theta + {\\tan ^4}\\theta + {\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}{tex} = {({\\sec ^2}\\theta )^2} + {({\\tan ^2}\\theta )^2} + {\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}Adding and subtracting\xa0{tex}2{\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}{tex} = {({\\sec ^2}\\theta )^2} + {({\\tan ^2}\\theta )^2} - 2{\\sec ^2}\\theta {\\tan ^2}\\theta + 2{\\sec ^2}\\theta {\\tan^2}\\theta+ {\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}{tex} = {({\\sec ^2}\\theta - {\\tan ^2}\\theta )^2} + 3{\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}\xa0{tex}\\left[ {\\because {{(a - b)}^2} = {a^2} + {b^2} - 2ab} \\right]{/tex}{tex} = 1 + 3{\\sec ^2}\\theta {\\tan ^2}\\theta {/tex}\xa0{tex}\\left[ {\\because {{\\sec }^2}\\theta - {{\\tan }^2}\\theta = 1} \\right]{/tex}= RHSHence proved | |
| 37191. |
X-2y+3=o |
| Answer» It can be represented only graphically There are many solutions for the questionEx:- (X,Y)= (1,2) | |
| 37192. |
How to do direct method in statistics |
| Answer» Mean=fixi/fi | |
| 37193. |
(2+/2) 3 |
| Answer» 6+3√2 | |
| 37194. |
f(x)=12x4+324x,g(x)=36x³+90x²-54x find hcf and lcm |
| Answer» | |
| 37195. |
X+y=3 and 3y-y=2 solve this question by elimination methode |
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Answer» The equation is wrong Thnx The variable should be x in second equation |
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| 37196. |
If 3 is zero of polynomial X ^ 2 x+k find the value of k |
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Answer» Second method X=3 Put the value of x in equation and get your abswer I have two method to solve this question 1. If x = 3 So, x-3=0 X^2 + x +k=0 divided by x - 3 =0 X^2+x+k=0 Put x=3(3)^2+3+k=0 9+3+k=0 12+k=0 K=-12 |
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| 37197. |
Find cubic polynomial whose zeroes are 2,1,1 |
| Answer» Let\xa0{tex}\\alpha,\\mathrm\\beta\\;\\mathrm{and}\\;\\mathrm\\gamma{/tex}\xa0be the zeroes of the given polynomial.\xa0Then, we have\xa0{tex}\\alpha{/tex}\xa0= 2,\xa0{tex}\\beta{/tex}\xa0= 1\xa0and\xa0{tex}\\gamma{/tex}\xa0= 1\xa0Hence{tex}\\alpha + \\beta + \\gamma{/tex}\xa0= 2\xa0+ 1\xa0+\xa01\xa0= 4 ...............(1){tex}\\alpha \\beta + \\beta \\gamma + \\gamma \\alpha{/tex}\xa0= 2(1) + 1(1) + 1(2)\xa0= 2\xa0+ 1\xa0+ 2\xa0= 5 ................(2){tex}\\alpha \\beta \\gamma{/tex}\xa0= 2(1)(1) = 2 .............(3)Now, a cubic polynomial whose zeros are {tex}\\alpha , \\beta{/tex}\xa0and\xa0{tex}\\mathrm\\gamma{/tex}\xa0is equal top(x) = x3\xa0-\xa0{tex}( \\alpha + \\beta + \\gamma ) x ^ { 2 } + ( \\alpha \\beta + \\beta y + \\gamma \\alpha ) x - \\alpha \\beta \\gamma{/tex}On substituting values from (1),(2) and (3) we get{tex}\\mathrm p(\\mathrm x)=\\mathrm x^3-(4)\\mathrm x^2+(5)\\mathrm x-(2){/tex}= x3\xa0- 4x2\xa0+\xa05x - 2 | |
| 37198. |
In the given figure if pos similar roq prove that ps is parallel to qr |
| Answer» Given: In figure, {tex}\\triangle {/tex} POS {tex} \\sim {/tex}{tex}\\triangle {/tex}ROQTo prove: PS {tex}\\parallel{/tex}QRProof: {tex}\\triangle {/tex} POS {tex} \\sim {/tex}{tex}\\triangle {/tex}ROQ ........[Given]{tex}\\therefore {/tex}{tex}\\angle{/tex} PSO ={tex}\\angle{/tex} RQO .........[ {tex}\\because {/tex} corresponding angle of two similar triangles are equal]But these form a pair of alternate angles{tex}\\therefore {/tex} PS {tex}\\parallel{/tex}QR | |
| 37199. |
So that the square of any positive integer cannot be of the form 5q+2 or 5q+3 dor any integer q ? |
| Answer» Let n be any positive integer. Applying Euclids division lemma with divisor = 5, we get{tex}\\style{font-family:Arial}{\\begin{array}{l}n=5q+1,5q+2,5q+3\\;and\\;5q+4\\;\\\\\\end{array}}{/tex}Now (5q)2 = 25q2 = 5m, where m = 5q2, which is an integer;{tex}\\style{font-family:Arial}{\\begin{array}{l}(5q\\;+\\;1)^{\\;2}\\;=\\;25q^2\\;+\\;10q\\;+\\;1\\;=\\;5(5q^2\\;+\\;2q)\\;+\\;1\\;=\\;5m\\;+\\;1\\\\where\\;m\\;=\\;5q^2\\;+\\;2q,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;2)^2\\;=\\;25q^2\\;+\\;20q\\;+\\;4\\;=\\;5(5q^2\\;+\\;4q)\\;+\\;4\\;=\\;5m\\;+\\;4,\\\\\\;where\\;m\\;=\\;5q^2\\;+\\;4q,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;3)^{\\;2}\\;=\\;25q^2\\;+\\;30q\\;+\\;9\\;=\\;5(5q^2\\;+\\;6q+\\;1)\\;+\\;4\\;=\\;5m\\;+\\;4,\\\\\\;where\\;m\\;=\\;5q^2\\;+\\;6q\\;+\\;1,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;4)^2\\;=\\;25q^2\\;+\\;40q\\;+\\;16\\;=\\;5(5q^2\\;+\\;8q\\;+\\;3)\\;+\\;1\\;=\\;5m\\;+\\;1,\\;\\\\where\\;m\\;=\\;5q^2\\;+\\;8q\\;+\\;3,\\;which\\;is\\;an\\;integer\\\\\\end{array}}{/tex}Thus, the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m.It follows that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m. | |
| 37200. |
2x square +x-528=0 |
| Answer» 2x2 + x - 528= 02x2 + 33x - 32x - 528 = 0x(2x + 33)- 16(2x+33)= 0(x - 16)(2x + 33)Zeros are x = 16 and x= -33/2 | |