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InterviewSolution
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.
| 37301. |
Express sec A in term of cot A |
| Answer» We have to express the trigonometric ratios sin A, sec A\xa0and tan A\xa0in terms of cot A.For sec A,By using identity {tex} \\sec ^ { 2 } A - \\tan ^ { 2 } A = 1{/tex}{tex} \\Rightarrow \\sec ^ { 2 } A = 1 + \\tan ^ { 2 } A{/tex}{tex} \\Rightarrow \\sec ^ { 2 } A = 1 + \\frac { 1 } { \\cot ^ { 2 } A } = \\frac { \\cot ^ { 2 } A + 1 } { \\cot ^ { 2 } A }{/tex}{tex} \\Rightarrow \\sec ^ { 2 } A = \\frac { 1 + \\cot ^ { 2 } A } { \\cot ^ { 2 } A }{/tex}{tex} \\Rightarrow \\sec A = \\frac { \\sqrt { 1 + \\cot ^ { 2 } A } } { \\cot A }{/tex} | |
| 37302. |
Describe the politics of ur grandparents time ..... How is ....? |
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| 37303. |
Describe about the previous price of ur grandparents time ... |
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| 37304. |
Describe the previous food habits of ur grandparents ... |
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| 37305. |
Find the value of k so that tha lines 2x-3y=9 and kx-9y=18 will be parallel |
| Answer» K,6 | |
| 37306. |
Find the value of \'a\' so that the point (3,9) lies on the line represented by 2x-2y=5 |
| Answer» Since the point (3, a) lies on the line 2x - 3y = 5, we have2\xa0{tex}\\times{/tex}\xa03 - 3\xa0{tex}\\times{/tex}\xa0a = 5{tex}\\Rightarrow{/tex}\xa06 - 3a = 5{tex}\\Rightarrow{/tex}\xa03a = 1{tex}\\Rightarrow{/tex}\xa0{tex}a = \\frac { 1 } { 3 }{/tex} | |
| 37307. |
If the point A(1,2) ,B(4,3), C(6,6) are three points of a parallelogram ABCD then find fourth point |
| Answer» Let coordinates of D be\xa0{tex}( \\alpha , \\beta ){/tex}\xa0P is mid-point of AC and BD.{tex}\\therefore \\quad \\left( \\frac { \\alpha + 4 } { 2 } , \\frac { \\beta + 3 } { 2 } \\right){/tex}{tex}= \\left( \\frac { 1 + 6 } { 2 } , \\frac { 2 + 6 } { 2 } \\right){/tex}{tex}\\Rightarrow \\frac { \\alpha + 4 } { 2 }{/tex}{tex}= \\frac { 7 } { 2 } ; \\frac { \\beta + 3 } { 2 } = \\frac { 8 } { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\alpha + 4 = 7 ; \\beta + 3 = 8{/tex}{tex}\\Rightarrow \\quad \\alpha = 3 ; \\beta = 5{/tex}{tex}\\therefore{/tex}\xa0Coordinates of D are (3, 5). | |
| 37308. |
By method of Completing square x2+4x+10=0 |
| Answer» x2+4x+10=0x2+4x=-10Multiply by 4a=4*1=44x2+16x=-40adding b2=(-4)2=164x2+16x+16=16-40(2x)2+2*(2x)(4)+42=-24{tex}\\begin{array}{l}(2x+4)^2=4\\times-6\\\\2x+4=\\pm\\sqrt{4\\times-6}\\\\2x=-4\\pm2\\sqrt{-6}\\\\x=-2\\pm\\sqrt{-6}\\\\\\\\\\\\\\end{array}{/tex}\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0 | |
| 37309. |
Find all zeroes of polynomial 4x4 - 20x3 + 5x - 6 , if two of its zeroes are 2and 3 |
| Answer» 16-60+5×2-6-44+4=-40 | |
| 37310. |
What is the difference between the zeroes and the roots?PLEASE |
| Answer» They both are used to denote the solution of the polynomial. You can either use roots or zeroes . Both are same. | |
| 37311. |
proof that triangle abc angle b 90 and ab is 8 cm and ac is 6 cm. bc 5 cm |
| Answer» Not possible , as AB2\xa0+ BC2\xa0\xa0= 82\xa0+ 52\xa0 => 64\xa0+ 25\xa0 => 89\xa0which is not equals to 62\xa0i.e. 36.If angle B is 90 degrees then its opposite side will be AC and also it will be the hypotenuse, and AC = 6cm\xa0.But in a right-angled triangle the hypotenuse is the most longest side.\xa0and here in the given situation the hypotenuse is smaller than one of the sides given.Thus , this condition is not possible. | |
| 37312. |
What is siddhacharyas rule |
| Answer» Is it siddhacharyas rule or shree dharacharyas rule? Confirm it fast so I can told you the right one.... | |
| 37313. |
7429 lcm kese nikalw |
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Answer» By factorisation method 23×17×19 7429 LCM |
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| 37314. |
31038 upon cot 52 - 4 sec 5 upon cosec 85 |
| Answer» Question is not clear, please resend the correct question.\xa0 | |
| 37315. |
If 3cot a =4 chech |
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| 37316. |
(1-sin²A)sec²A=1 |
| Answer» (1-sin²A)sec²A=1LHS =(1-sin²A)sec²A=cos2A*1/cos2A=1=RHS | |
| 37317. |
Write the degree of zero polynomial |
| Answer» Can\'t be defined. | |
| 37318. |
Solve the equation:0.5(x-0.4)/0.35-0.6(x-2.17)=x+6.1 |
| Answer» 1st u multiple both side by 10 then equation will be easy and when do not solve questions of u please say me | |
| 37319. |
Solve x and y 152x-378y =-74,-378x+152y=-604 |
| Answer» The given equations are{tex}152x - 378y = -74{/tex} ...(i){tex}-378x\xa0+ 152y = -604{/tex}. ... (ii)Clearly, the coefficients of x\xa0and y in one equation are interchanged in the other.Adding (i) and (ii), we get{tex}(152-378)x\xa0+ (-378 +152) y = -(74 + 604){/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(-226)x\xa0+ (-226)y = -678{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(-226)(x + y) = -678{/tex}{tex}\\Rightarrow \\quad ( x + y ) = \\frac { - 678 } { - 226 } \\Rightarrow x + y = 3{/tex}. ...(iii)Subtracting (ii) from (i), we get{tex}(152 + 378) x + (-378 - 152)y = (-74 + 604){/tex}{tex}\\Rightarrow{/tex}530x\xa0- 530y = 530{tex}\\Rightarrow{/tex}\xa0530(x - y) = 530 {tex}\\Rightarrow{/tex}\xa0x\xa0- y = 1. ... (iv)Adding (iii) and (iv), we get 2x = 4 {tex}\\Rightarrow{/tex}\xa0x = 2.Subtracting (iv) from (iii), we get2y = 2 {tex}\\Rightarrow{/tex} y = 1.Hence, x = 2 and y = 1 | |
| 37320. |
Find the middle term of the A.P:6,13,20.....216 |
| Answer» Ap:6 13 20.....216\xa0a=6. d=7. an=216an=a+(n-1)d216=6+(n-1)7On solving\xa0n=31\xa0So middle term is 16th term\xa0a16=a+15d =6+15×7 =6+105=111 | |
| 37321. |
Solve the following pair of linear equations by the substitution and cross multiplication method |
| Answer» Bhai pair hi nhi h | |
| 37322. |
If 3cos`0 - 4sin`0 = 2cos`0 + sin`0 , then find value of tan`0 |
| Answer» 3cos\'0-4sin\'0=2cos\'0+sin\'03cos\'0-2cos\'0=5sin\'0cos\'0=5sin\'0cos\'0÷sin\'0=5cot\'0=51÷tan\'0=5tan\'0=1÷5 | |
| 37323. |
Show that square of an odd positive integer can be of the form 6q+1 or 6q+3 for some integer \'q\' |
| Answer» We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5 for some integer m.So, an odd positive integer x is of the form 6m +1 or 6m + 3.CASE I When x = 6m + 1: In this case,x2 = (6m + 1)2 = 36m2 + 12m + 1 = 6 (6m2 + 2m) + 1 = 6q + 1, where q = 6m2 + mCASE II When x = 6m + 3: In this case,x2 = (6m + 3)2 = 36m2 + 36m + 9 = (36m2 + 36m + 6) + 3= 6 (6m2 + 6m + 1) + 3 = 6q + 3, where q = 6m2 + 6m + 3. | |
| 37324. |
Divide 39 into two parts such that their product is 324.A question from quadratic equation. |
| Answer» {tex}\\large Solution:{/tex}\t\xa0{tex}\\large Let \\space \'a\'\\space\\space and\\space \'b\'\\space be\\space the\\space two\\space numbers.{/tex}{tex}\\large Thus,\\begin{cases} \\ \\\\ \\ \\\\ \\ \\end{cases}{/tex}\xa0{tex}\\Large a+b=39 \\dots (i)\\\\\\Large a\\times b=324\\dots(ii){/tex}{tex}\\large Putting \\space b=39-a\\space in\\space the\\space eq.\\space(ii)\\space,we\\space get{/tex}{tex}\\Large \\implies a\\times b=324\\\\\\Large \\implies a\\times (39-a)=324\\\\\\Large \\implies39a-a^2=324\\\\\\Large \\therefore a^2-39a+324=0\\\\ \\space\\\\\\Large \\therefore a=12,\\space 27.{/tex}\xa0{tex}\\huge a{/tex}{tex}\\huge b{/tex}{tex}\\huge a+b{/tex}{tex}\\huge a\\times b{/tex}{tex}\\huge 12{/tex}{tex}\\huge 27{/tex}{tex}\\huge 39{/tex}\xa0{tex}\\huge 324{/tex}\xa0{tex}\\huge 27{/tex}\xa0{tex}\\huge 12{/tex}\xa0{tex}\\huge 39{/tex}\xa0{tex}\\huge 324{/tex}\xa0\t | |
| 37325. |
Show that cube of any positive integer is of form of 4 m 4m+1 4m+8 |
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Answer» Let a be the positive integer and b = 4.Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.{tex}(4q)^3\\;=\\;64q^3\\;=\\;4(16q^3){/tex}= 4m, where m is some integer.{tex}(4q+1)^3\\;=\\;64q^3+48q^2+12q+1=4(\\;16q^3+12q^2+3q)+1{/tex}= 4m + 1, where m is some integer.{tex}(4q+2)^3\\;=\\;64q^3+96q^2+48q+8=4(\\;16q^3+24q^2+12q+2){/tex}= 4m, where m is some integer.{tex}(4q+3)^3\\;=\\;64q^3+144q^2+108q+27{/tex}=4×(16q3+36q2+27q+6)+3= 4m + 3, where m is some integer.Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m. Thank you so much |
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| 37326. |
If the points (2,1) and(1,-2) are equidistant from the point (x,y),show that x+3y=0 |
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| 37327. |
1/7x+1/6y=3, 1/2x-1/3y=5 |
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Answer» 6X+7Y=126......(eq. 1)3X-2Y=30..........(eq. 2)For making l.h.s. sameMultiply eq (2) with 2,We will get 6X-4Y=60Mark it as 3 Solve 1 and 3 We will getX=14Y=6 Take lcm and solve |
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| 37328. |
12(6) |
| Answer» 72 is the answer | |
| 37329. |
If one zero of the polynomial (a^2)x^2 +13x+6a is reciprocal of the other , find the value of a |
| Answer» 6 | |
| 37330. |
Find the value of k for which the roots are real and equal1) kx(x-2root5)+10=0 |
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| 37331. |
1st term syllabus of class 10 maths |
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Answer» 1,2,3&4th chapter No thanks PolynomialsLinear eqAPTrianglesStatisticsQuadratic eqns |
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| 37332. |
Find the value of K for which the roots are real and equal1) (2K+1)x^2+2(K+3)x+(K+5)=0 |
| Answer» We have, {tex}(2k+1)x^2+2(k+3)x+(k+5)=0{/tex}Here, a=(2k+1),b=2(k+3), c=(k+5){tex}\\therefore D=b^2-4ac=4(k+3)^2-4(2k+1)(k+5){/tex}{tex}=4k^2+36+24k-8k^2-40k-4k-20=-4k^2-20k+16{/tex}{tex}=-4(k^2+5k-4){/tex}For the equation having equal roots,{tex}D=0 \\implies k^2+5k-4=0{/tex}{tex}k = {-5 \\pm \\sqrt{5^2-4(1)(-4)} \\over 2(1)}={-5 \\pm \\sqrt{41} \\over 2}{/tex} | |
| 37333. |
If x-p is a factor of both x2-ax+b and x2-cx+d then prove p=d-b÷c-a( note that the x2 is x square) |
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| 37334. |
Solve x:root 2x +5 +x = 132 |
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| 37335. |
Sim15°cos75°+cos15°sin75°Tan5°tan30°tan55°tan85° |
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| 37336. |
All formula of chapter 2 |
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| 37337. |
2/X+3/y =135/X-4/y=-2 |
| Answer» Putting {tex}\\frac 1x{/tex}= u and {tex}\\frac 1y{/tex}\xa0= v, the given equations become{tex}2u+ 3v\xa0= 13{/tex} .... (i){tex}5u - 4v = -2{/tex} .......(ii){tex}\\text{Multiplying (i) by 4 and (ii) by 3 and adding the results, we get}{/tex}{tex}8u\xa0+ 15u\xa0= 52 - 6{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}23u = 46{/tex}{tex}\\Rightarrow \\quad u = \\frac { 46 } { 23 } = 2{/tex}{tex}\\text{Putting u\xa0= 2 in (i), we get}{/tex}(2 {tex}\\times {/tex}\xa02) + 3v\xa0= 13 {tex}\\Rightarrow{/tex}\xa03v\xa0= 13 - 4 = 9 {tex}\\Rightarrow{/tex}\xa0v\xa0= 3.Now,u\xa0= 2{tex}\\Rightarrow \\frac { 1 } { x } = 2 \\Rightarrow 2 x = 1 \\Rightarrow x = \\frac { 1 } { 2 }{/tex}And, v\xa0= 3 {tex}\\Rightarrow \\frac { 1 } { y } = 3 \\Rightarrow 3 y = 1 \\Rightarrow y = \\frac { 1 } { 3 }{/tex}Hence,\xa0{tex}x = \\frac { 1 } { 2 } \\text { and } y = \\frac { 1 } { 3 }{/tex} | |
| 37338. |
Find the quadratic polynomial whose zeroes are 3+ √5 and 3_√5 |
| Answer» This also seems like you didn\'t learn chapter 4 in a better way.{tex}\\large{Question: \\space Zeroes=3+\\sqrt5 \\space and \\space 3-\\sqrt 5,\\space find\\space the \\space quadratic\\space equation?}{/tex}{tex}\\large Solution:{/tex}\t{tex}\\large {For\\space any\\space quadratic\\space equation,\\space x^2+x\\times(\\alpha+\\beta)+(\\alpha\\times\\beta)=0\\space }{/tex}{tex}\\large {Here,\\space let\\space\\space\\alpha=3+\\sqrt5 \\space\\space\\space\\space,\\space\\space\\space\\space\\beta=3-\\sqrt5.\\space }{/tex}{tex}\\large {Thus,\\space\\space\\space\\alpha+\\beta=3+\\sqrt5 +3-\\sqrt5 = 6\\space\\space\\space\\space,\\space\\space and\\space\\space\\alpha\\times\\beta=(3+\\sqrt5)(3-\\sqrt5)=3^2-\\sqrt5 ^2=9-5=4.\\space }{/tex}{tex}\\large {\\therefore\\space x^2+x\\times(\\alpha+\\beta)+(\\alpha\\times\\beta)=0\\space }\\\\\\large {\\implies\\space x^2+x\\times(6)+(4)=0\\space }\\\\\\large {\\implies\\boxed{\\space x^2+6x+4=0}\\space }\\\\{/tex}\tNow after seeing this, don\'t you think this is a very very simple thing to do. | |
| 37339. |
Can 72 and 20 be the L.C.M and H.C.F of two numbers? Write down the reason. |
| Answer» Question : Can 72 and 20 be the L.C.M and H.C.F of two numbers? Write down the reason.This is a very pathetic question to ask. It really shows that you are not studying well.Solution:\xa0{tex}\\large{\\implies \\boxed{LCM\\times HCF=a\\times b}}{/tex}{tex}\\large{\\implies{72\\times20=a\\times b}}{/tex}{tex}\\large{\\implies{1440=a\\times b}}{/tex}Now, 1440 can be multiple of any two numbers.{tex}\\large{\\implies{\\space\\space1\\times1440=1440 }}\\\\\\large{\\implies{\\space\\space2\\times720\\space\\space=1440 }}\\\\\\large{\\implies{\\space\\space3\\times480\\space\\space=1440 }}\\\\\\large{\\implies{\\space\\space4\\times360\\space\\space=1440}}\\\\\\large{\\implies{\\space\\space5\\times288\\space\\space=1440}}\\\\\\large{\\implies{\\space\\space6\\times240\\space\\space=1440}}\\\\\\large{\\implies{\\space\\space8\\times180\\space\\space=1440}}\\\\\\large{\\implies{\\space\\space9\\times160\\space\\space=1440 }}\\\\\\large{\\implies{10\\times144\\space\\space=1440 }}\\\\\\large{\\implies{12\\times120\\space\\space=1440 }}\\\\\\large{\\implies{15\\times96\\space\\space\\space\\space=1440 }}\\\\\\large{\\implies{16\\times90\\space\\space\\space\\space=1440 }}\\\\\\large{\\implies{18\\times80\\space\\space\\space\\space=1440 }}\\\\\\large{\\implies{20\\times72\\space\\space\\space\\space=1440 }}\\\\\\large{\\implies{24\\times60\\space\\space\\space\\space=1440 }}\\\\\\large{\\implies{30\\times48\\space\\space\\space\\space=1440 }}\\\\\\large{\\implies{32\\times45\\space\\space\\space\\space=1440 }}\\\\\\large{\\implies{36\\times40\\space\\space\\space\\space=1440 }}\\\\{/tex} | |
| 37340. |
What is quardatic equation |
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Answer» Quadratic equeation is those equation when the power of equation is two QUADRATIC EQUATIONSThe polynomial of degree two is called quadratic polynomial and equation corresponding to a quadratic polynomial P(x) is called a quadratic equation in variable x.Thus, P(x) = ax2\xa0+ bx + c =0, a ≠ 0, a, b, c ∈ R is known as the standard form of quadratic equation.There are two types of quadratic equation.(i)\xa0Complete quadratic equation :\xa0The equation ax2\xa0+ bx + c 0 where a ≠ 0, b ≠ 0,c ≠ 0(ii)\xa0Pure quadratic equation :\xa0An equation in the form of ax2\xa0= 0, a ≠ 0, b = 0, c = 0 A equation whose maximum power is 2 called quadratice equation |
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| 37341. |
Find the largest number that will divide 398 436 and 542 leaving remainders 7 11 and 15 respectively |
| Answer» The required number = HCF ( 398-7),(436-11),(542-15)= HCF (391),(425),(527)= 17. | |
| 37342. |
Form the cubic polynomial whose zeros are minus 3 minus 1 and 2 |
| Answer» x^3 -(-3)x^2 + (-1)x - 2= x^3 + 3x^2 - x - 2 | |
| 37343. |
If the product of the zeros of the polynomial ax²-6x-6 is 4 find the value of a |
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Answer» What is value of a Product of zeroes =4 |
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| 37344. |
3x÷2 -5y÷3 = -2 |
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| 37345. |
Draw the graph of if(x)=x^2-2x-8 2)f(x)=2x^2 |
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| 37346. |
If if the LCM of 65 and 117 is expressible in the form 65 M - 117 then find the value of m |
| Answer» h. c. f. of 65 & 117 =65=5*13117=3*3*13h. c. f. = 13.A. T. Q. ,H. C. F=65m-11713=65m-11713+117=65m130=65m130/65=m2=m. ,m=2L. C. M=65=5*13117=3*3*13L. C. M= 3*3*5*13585(ans).it will help | |
| 37347. |
Has the rational number 441/2^2 x5^7 x7^2 a terminating or a non-terminating decimal representation |
| Answer» 441 / ( 2^2 * 5^7 * 7^2 )= ( 3^2 * 7^2 ) / ( 2^2 * 5^7 * 7^2 )= 3^2 / ( 2^2 * 5^7 )Since the denominator of this rational number is in the form of 2^ m * 5^n,so it will have terminating decimal representation. | |
| 37348. |
sin 45\'+cos45 |
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Answer» 1 1 |
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| 37349. |
Difference between zeroes and roots of quadratice equatoin |
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Answer» There are no difference between zeroes and roots Zeroes is the other name of roots. Both are the same roots are related with equation eg 2x+3y=0 . zeros are related with factorization eg 2x+3y+3 Zero is for a function and root is for an equation. The zero of the function f(x)=x+1 is -1. The root of the equation x+1=0 is -1. |
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| 37350. |
If two zeros of a polynomial X^3-3x^2+2 are 1+√3 and 1-√3, then find the third zero |
| Answer» use solved example of chapter 2 for reference | |