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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.
| 34401. |
Find the sum of the following Aps -37,-33,-29,...,to 12 terms. |
| Answer» Here, {tex}a=-37,\\ d=-33-(-37)=-33+37=4\\ and\\ n=12{/tex}Now we know that ,\xa0Sn={tex}\\frac{n}{2}{/tex}[2a+(n-1)d]Therefore, S12=\xa0{tex}\\frac{{12}}{2}{/tex}[2{tex}\\times{/tex}{tex}(-37)+(12-1)4]{/tex}= {tex}6[-74+44]{/tex}= 6{tex}\\times{/tex}(-30)= -180Therefore, sum of given A.P is -180. | |
| 34402. |
Sin theth-cos theth/sin theta+cos theta |
| Answer» Question is not complete, please send it again with full question. | |
| 34403. |
a=? d=-3 n=18 an=-5 |
| Answer» We know thatan=a+(n-1)d \'an=-5,a=?,n=18,d=-3By putting the value of An,d,n in this formula We get,-5=a+(18-1)-3-5=a+(17*-3)-5=a+(-51)\'a=51-5\'a=46 | |
| 34404. |
The following real numbers have decimal expansion write in the p by q form 43.123456789 |
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Answer» Thank u so much ? Solution :\xa0431234567891000000000 |
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| 34405. |
2x +3ky=55x +2y =6 Find the value of k if the above has unique solution |
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Answer» 2x+ 3ky =55x + 2y =6It is given that it has a unique solution So, a1/ a2 is not equal to b1/b22/5 is not equal to 3k/2Cross multiplying we get,4 = 15k》k=4/15 Solve this problem from elimination method 1st cut the 2x and 5x by multiple |
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| 34406. |
Complete the square of 9x²+7x-2 |
| Answer» 9x2+9x_2x-29x(x+1)_2(x+1)(9x-2)(x+1)9x-2=0/x+1=09x=2/x=-1X=9/2;. X=-1 | |
| 34407. |
If alpha and beta are zeroes of the polynomial f(x)=x^2-x-k such that alpha - beta =9. Find k? |
| Answer» Since\xa0{tex}\\alpha \\text { and } \\beta{/tex}\xa0are the zeroes of the polynomial, then{tex}{/tex}{tex}\\alpha + \\beta = - \\frac { \\text { Coefficient of } x } { \\text { Coefficient of } x ^ { 2 } }{/tex}{tex}{/tex}{tex}{/tex}{tex}\\Rightarrow\\ \\alpha + \\beta = - \\left( \\frac { - 1 } { 1 } \\right) = 1{/tex}.........(i)Given,\xa0{tex}\\alpha - \\beta = 9{/tex}...............(ii)Solving (i) and (ii),\xa0{tex}\\alpha = 5 , \\beta = - 4{/tex}{tex}\\alpha \\beta = \\frac { \\text { Constant term } } { \\text { Coefficient of } x ^ { 2 } }{/tex}{tex}\\Rightarrow\\ \\alpha \\beta = - k{/tex}{tex}\\Rightarrow\\ {/tex}(5)(-4) = -k{tex}\\Rightarrow{/tex}k = 20So,required value of k is 20 | |
| 34408. |
Huda has auctioned plots |
| Answer» | |
| 34409. |
If 1/2 and 1 are zeroes of 2x^4-3x³-3x²-6x-2 , find the other zeroes |
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Answer» 1/2=x, 1=xx-1/2=0 , x-1=0(x-1/2)(x-1)=0=x^2-x-x/2+1/2=x^2-3x/2+1/2=(x^2-3x/2+1/2)/(2x^4-3x^3-3x^2-6x-2)2x^2-4=02x^2=4x^2=2x=+√2,-√2Hence,proved. .... |
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| 34410. |
23*576\\34+589-684(60-43) |
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Answer» 23*576\\34+589-684(60-43)= 13248/34 + 589 - 684(17)= 13248/34 + 589 - 11628= 13248/11005=\xa01.20381644707 13248/34+589-684*17=13248/34+589-11628=13248/11005=1.20381644707 |
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| 34411. |
Genral form of quadric polynomials in alfa bita gama |
| Answer» x^2-(α+β)x+αβ | |
| 34412. |
2 X square + X - 528 middel term splitting |
| Answer» 2x^2+x-528=02x^2+33x-32x-528=0x(2x+33)-16(2x+33)=0x=16,-33/2 | |
| 34413. |
Find the largest three digit number divisible by 12,15,20 and 18 leaving remainder 5 in each case |
| Answer» LCM(12,15,20,18)=180180+5=185The largest 3 digit number=185 | |
| 34414. |
Solve 1352xx +2545x+2325 and find its zero |
| Answer» | |
| 34415. |
SecA+tanA/secA-tanA=(1-tanA)2 (CosA) |
| Answer» | |
| 34416. |
Given that cos a =p/q, find the value of tan a |
| Answer» cos a=p/q{tex}\\begin{array}{l}\\sin\\;a=\\sqrt{1-\\frac{p^2}{q^2}}\\;\\\\=\\frac1q\\sqrt{q^2}-p^2\\\\So\\;\\tan\\;a=\\frac{\\sin\\;a}{\\cos\\;a}=\\frac{\\frac1q\\sqrt{q^2}-p^2}{\\displaystyle\\frac pq}\\\\=\\frac1p\\sqrt{q^2}-p^2\\end{array}{/tex} | |
| 34417. |
The difference between two numbers is 26 and one number is 3times the other.find them |
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Answer» According to second problemsX=3YSubstituting X= 3y 3Y-3=262Y=26Y=13Substituting value of Y in an equation X=3YX=3×13X=39 Let the two numbers X and Y respectively where X is greater than Y According to first problemX-Y=26 |
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| 34418. |
Using prime factorisation find the hcf and lcm of:24,36,40 |
| Answer» {tex}\\begin{array}{l}\\text{24=8×3=2}^3\\times3\\\\\\end{array}{/tex}{tex}\\begin{array}{l}\\text{36=4×9=2}^2\\times3^2\\\\\\end{array}{/tex}{tex} \\begin{array}{l}\\text{40=8×5=2}^3\\times5\\\\\\end{array}{/tex}Therefore, HCF = Product of the smallest power of each common prime factor in the numbers = 22 = 4Therefore, LCM = Product of the greatest power of each prime factor involved in the numbers {tex} \\begin{array}{l}=2^3\\times3^2\\times5=8\\times9\\times5=360\\\\\\end{array}{/tex} | |
| 34419. |
Find the sum of the first 30 terms of an Ap whose nth term is 2+1/2n |
| Answer» Ibwant to know examination patter 2020 | |
| 34420. |
If angle A and B are acute angles such that cosA=cosB,then show that angleA =angleB |
| Answer» Given cos A = cos BHence,\xa0{tex}\\frac{{AC}}{{AB}} = \\frac{{BC}}{{AB}}{/tex}{tex} \\Rightarrow AC = BC{/tex}Since angle opposite to equal sides in a\xa0{tex}\\Delta {/tex}\xa0are equal{tex}\\therefore \\angle B = \\angle A{/tex}Hence proved | |
| 34421. |
For any positive integer n prove that n cube -n is divisible by 6 |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) | |
| 34422. |
Pove that 7 is irrational number |
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Answer» \xa0let us assume that\xa0{tex}\\sqrt 7{/tex}\xa0be a rational number.{tex}\\sqrt { 7 } = \\frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \\neq{/tex}0Squaring both sides, we have{tex}\\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}or,\xa0{tex}a ^ { 2 } = 7 b ^ { 2 }{/tex}--------(i)a2\xa0is divisible by 7.Hence a is divisible by 7..........(ii)Let a = 7c ( where c is any integer)squaring on both sides we get(7c)2\xa0= 7b249c2\xa0= 7b2b2\xa0= 7c2so b2\xa0is divisible by 7hence, b is divisible by 7..........(iii)From equation(ii) and (iii), we have7 is a factor of a and b which is contradicting the fact that a and b are co-primes.Thus, our assumption that\xa0{tex}\\sqrt 7{/tex} is rational number is wrong.Hence,\xa0{tex}\\sqrt 7{/tex}\xa0is an irrational number. Yes |
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| 34423. |
Convert tanA in the term of cotA |
| Answer» For tanA,{tex}\\tan A = \\frac { 1 } { \\cot A }{/tex} | |
| 34424. |
If (x+1) is a factor of 2x3+ax2+2bx+1 , then find the value of a and b given that 2a-3b=4. |
| Answer» Since {tex}(x + 1){/tex} is a factor of {tex}2x^3 + ax^2 + 2bx + 1{/tex}{tex}\\Rightarrow{/tex}{tex}x = -1{/tex} is a zero of {tex}2x^3 + ax^2 + 2bx + 1{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}2(-1)^3 + a(-1)^2 + 2b(-1) + 1 = 0 {/tex}{tex}\\Rightarrow{/tex}\xa0{tex}a - 2b - 1 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0a - 2b = 1 ...(i)Given that {tex}2a - 3b = 4{/tex} ...(ii)Multiplying equation (i) by 2, we get{tex}2a - 4b = 2{/tex} ...(iii)Subtracting equation (iii) from (ii), we getb = 2Substituting b = 2 in equation (i), we havea - 2(2) = 1{tex}\\Rightarrow{/tex}\xa0a - 4 = 1{tex}\\Rightarrow{/tex}\xa0a = 5Hence, a = 5 and b = 2. | |
| 34425. |
Write the decimal representation of 1458 |
| Answer» 18.12 | |
| 34426. |
Word problems on polynomials |
| Answer» | |
| 34427. |
30 example on elimination method |
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Answer» Answers Hey |
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| 34428. |
Find (HCF×LCM) for the numbers 100 and 900 |
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Answer» O hello ....Mr. .....I\'ve already found my answers before u\'r reply.........So don\'t be so oversmart??? Seems you are dumb and have no focus on studies.It is clearly stated in the NCERT book that\xa0HCF x LCM = Product of the\xa0two numbersSo, HCF x LCM = 100 x 900 = 90,000 |
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| 34429. |
How do the sumroot5 |
| Answer» | |
| 34430. |
If Sin+2Cos=1, then prove that 2Sin-Cos=2 |
| Answer» 0 | |
| 34431. |
Evaluate the following. COS (40-A) -SIN (50-A) +COS40+COS50*SIN40+SIN50 |
| Answer» Question is incomplete, please upload it correctly. | |
| 34432. |
(3)-(7) |
| Answer» So silly ? - 4 | |
| 34433. |
The LCM of 2·5,0·5 and 0·175 is (a) 2·5 (b)5 (c) 7·5 (d) 1·75 |
| Answer» 0.5 | |
| 34434. |
If sideof triangle are 16cm,30cm,34cm what is its area |
| Answer» Let In\xa0{tex}\\triangle ABC{/tex}AB=16 cm,BC=30 cm and AC=34 cmAB2+BC2162+302=256+900=1156=342=AC2SO AB2+BC2=AC2HENCE\xa0{tex}\\triangle ABC{/tex}\xa0is a right angle triangleso are of\xa0{tex}\\triangle ABC{/tex}=1/2*AB*BC=1/2*16*30=240 CM2 | |
| 34435. |
(1+tan×tan)(1+cot×cot) |
| Answer» Your question is incorrect.Question.{tex}\\huge \\frac{1+\\tan^2\\theta}{1-\\cot^2\\theta}{/tex}Solution:{tex}\\huge \\frac{1+\\tan^2\\theta}{1-\\cot^2\\theta}{/tex}\xa0{tex}\\huge\\implies \\frac{\\sec^2\\theta}{cosec^2\\theta}\\implies {\\sec^2\\theta}\\times \\frac{1}{cosec^2\\theta}\\space\\space\\space\\space\\space\\space\\space\\space\\space\\space\\begin{cases}\\ \\sec^2\\theta=1+\\tan^2\\theta \\\\cosec^2\\theta=1+\\cot^2\\theta \\\\end{cases}{/tex}{tex}\\huge \\implies{/tex}{tex}\\huge \\frac{1}{\\cos^2\\theta}\\times{\\sin^2\\theta}\\space\\space\\space\\space\\space\\space\\space\\space\\space\\space\\begin{cases}\\ \\sec^2\\theta=\\frac{1}{\\cos^2\\theta} \\\\cosec^2\\theta=\\frac{1}{\\sin^2\\theta} \\\\end{cases}{/tex}{tex}\\huge\\implies \\frac{\\sin^2\\theta}{\\cos^2\\theta}\\implies \\boxed {\\tan^2\\theta}{/tex} | |
| 34436. |
x/a- y/b = a-b ax+by =a2+b2 |
| Answer» {tex}\\frac { x } { a } - \\frac { y } { b } = 0{/tex}{tex} \\Rightarrow x = \\frac { a y } { b }{/tex}....................(i)ax + by = (a2\xa0+ b2) .......................(ii)Substituting (i) in (ii),we get{tex}a \\left( \\frac { a y } { b } \\right) + b y = \\left( a ^ { 2 } + b ^ { 2 } \\right){/tex}{tex}\\Rightarrow \\frac { a ^ { 2 } y } { b } + b y = \\left( a ^ { 2 } + b ^ { 2 } \\right){/tex}{tex}\\Rightarrow a ^ { 2 } y + b ^ { 2 } y = \\left( a ^ { 2 } b + b ^ { 3 } \\right){/tex}{tex}\\Rightarrow y \\left( a ^ { 2 } + b ^ { 2 } \\right) = b \\left( a ^ { 2 } + b ^ { 2 } \\right){/tex}{tex}\\Rightarrow y = b{/tex}Substituting in (i),we get{tex} x = \\frac { a b } { b }{/tex}\xa0⇒ x = a.So, solution of given equation is x = a and y = b. | |
| 34437. |
Rationalise the denominator of five upon four root two+three root three |
| Answer» 4root 2 minus 3root 3 | |
| 34438. |
Express 7180 in decimal notation |
| Answer» The question is incomplete. please reupload it. | |
| 34439. |
What is the biggest number of two digit |
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Answer» 99?? 99.....?? Ovio 99 ?? |
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| 34440. |
Prove root5 is an irrational number |
| Answer» let root 5 be rationalthen it must in the form of p/q [q is not equal to 0][p and q are co-prime]root 5=p/q=> root 5 * q = psquaring on both sides=> 5*q*q = p*p ------> 1p*p is divisible by 5p is divisible by 5p = 5c [c is a positive integer] [squaring on both sides ]p*p = 25c*c --------- > 2sub p*p in 15*q*q = 25*c*cq*q = 5*c*c=> q is divisble by 5thus q and p have a common factor 5there is a contradictionas our assumsion p &q are co prime but it has a common factorso\xa0√5 is an irrational | |
| 34441. |
Who are the students of Tamilnadu ?answer me |
| Answer» | |
| 34442. |
Verify that 3 is a zero of the linear polynomial p (x)=9x - 27 |
| Answer» p(3)=9×3-27 =27-27=0Hence, verified.? | |
| 34443. |
Factorise 2x square - 5 x + 6 |
| Answer» => 2x2 - 5 x + 6(Adding terms 2x and -2x)=>\xa02x2 - 5 x + 6 + 2x - 2x=>\xa02x2 - 5 x -2x + 6 + 2x=>\xa02x2 - 7\xa0x + 6 + 2x=> { (x - 2) (2x - 3) } + 2x (Answer)\xa0 | |
| 34444. |
A+b+c+d=abcd |
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Answer» Hlo Hiii yash .. Hi hlo |
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| 34445. |
Does the standard math paper come from NCERT |
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Answer» Yes board exams paper always comes from ncert book of maths Rd sharma and ouswal No Out of syllabus |
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| 34446. |
What means PT |
| Answer» Practice test. | |
| 34447. |
HOW MANY 2DIGIT NUMBERS ARE DIVISIBLE BY 2 |
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Answer» 45 45 45 |
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| 34448. |
Agar main ncert book ke problemspura karlo toh kya 90 percent mil jayenge ? |
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Answer» Dekh bhai y jaruri nhi ki ncrt ka hi same question aega language change ho skta h digit change ho skta h or 80% ncert s hi puchta h baki 20% side s Number to bhagte - bhagte tumhare pas aayega . |
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| 34449. |
2x+8y=68. Find x |
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Answer» If y=0 ,2x+8(0)=68X=68/2X=34 Where is the second equation? Which is the scecond equation |
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| 34450. |
Board me sabse zyada questions konsi book se atte hai? |
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Answer» Mostly NCERT Ncert Only ncert bas twist krke de dete ha From NCERT book? Ncert ....? NCERT |
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