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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.
| 16901. |
Obtain all others zeroes of polynomial 2x4+3x3-15x2-24x-8 |
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| 16902. |
Prove that 1/√2 irrational |
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Answer» If you have RD Sharma..refer to page 1.50 Irrational |
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| 16903. |
For any positive integer n ,prove that n^3-n is divisible by 6 |
| Answer» yes dear. | |
| 16904. |
X- y +=0 |
| Answer» ?? | |
| 16905. |
The arithematic mean of set of variable a,a+d,a+2d,a+3d..........a+2nd is |
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Answer» Please give explanation a+nd |
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| 16906. |
Cosec 85-sin 56 |
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| 16907. |
Prove that 2+5√3 is an irtational number, given that √3 is an irrational number |
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Answer» Let 2+5√3 is an rational no.2+5√3 = a/b5√3 = a/b - 2So, 5√3 is equal to a/b - 2 And we know that a/b - 2 is an rational numberBut it is a fact that √3 is an irrational no.So, therefore, 2+5√3 is an irrational no. If possible, let us suppose that 2 + 5{tex}\\sqrt{3}{/tex} is\xa0a rational numberThen,\xa0we can write\xa0{tex}2+5\\sqrt{3}=\\frac{p}{q} {/tex}( Where p and q are coprime){tex}\\implies 5\\sqrt{3}=\\frac{p}{q}-2{/tex}{tex}\\implies 5\\sqrt{3}=\\frac{p-2q}{q}{/tex}{tex}\\implies \\sqrt{3}=\\frac{p-2q}{5q} {/tex}{tex}\\implies \\sqrt{3}=\\frac{integer}{integer} {/tex}\xa0(Since p and q are integers){tex}\\implies \\sqrt{3} {/tex}\xa0is rational numberwhich is a contradiction to the given fact that {tex}\\sqrt{3}{/tex}\xa0is irrational.{tex}\\therefore{/tex}\xa02 + 5{tex}\\sqrt{3}{/tex}\xa0cannot be rationalHence, 2 + 5{tex}\\sqrt{3}{/tex}\xa0is irrational. |
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| 16908. |
Express 429 as product of primes. |
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Answer» product of prime :--429 =3×13×11. The product of the prime factors of 429 is:- 429 = 3 × 13 × 11 |
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| 16909. |
If the HCF of 55 and 99 is expressible in the form55m -99, then find the value of m. |
| Answer» HCF of 55 and 99 is\xa011.Thus, 55m - 99 = 11 11(5m - 9) = 11 5m - 9 = 1 5m = 9 + 1 = 10 thus, m = 2. | |
| 16910. |
For what least value \'n\' a natural number is divisuble by 8 |
| Answer» (24)1\xa0= 24{tex}24 \\over 8{/tex}\xa0= 3Hence, the least value of n is 1. | |
| 16911. |
Is solving last year question paper and ncert book is enough for class 10 |
| Answer» YES, THAT IS ENOUGH | |
| 16912. |
If HCF of 144 and 180 is expressed in the form of 13m-3 find the value of m |
| Answer» The hcf of 180 and 144 is 36.13m-3=hcf . 13m-3=3613m=36+3.13m=39m=39÷13=3m=3 | |
| 16913. |
The sum of a rational and an irratioal number is |
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Answer» irrational no. Irrational number 1 |
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| 16914. |
2x +1=? |
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Answer» x=-1/2. X=-1/2 Sry ...by mstke the msg was posted Hola |
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| 16915. |
Find k so that 15, k-1 are in AP |
| Answer» less number of terms. give at least 3 terms. | |
| 16916. |
Angle M=Angle N and PM÷MQ=PN÷NR.prove that PQR is an isosceles triangle. |
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| 16917. |
Pm parallel to ab and pn parallel to ad.prove thatbm bycm is equal to dn by cn |
| Answer» Incomplete data given | |
| 16918. |
Show that there is no positive interger n for which √n-1 +√n+1 is rational |
| Answer» Let us assume that there is a positive integer n for {tex}\\sqrt{n-1}+\\sqrt{n+1}{/tex}which is rational and equal to {tex}\\frac pq{/tex}, where p and q are positive integers and (q\xa0{tex}\\neq{/tex}\xa00).{tex}\\sqrt { n - 1 } + \\sqrt { n + 1 } = \\frac { p } { q }{/tex}......(i)or,\xa0{tex}\\frac { q } { p } = \\frac { 1 } { \\sqrt { n - 1 } + \\sqrt { n + 1 } }{/tex}on multiplication of numerator and denominator by\xa0{tex}\\sqrt{n-1}-\\sqrt{n+1}{/tex}\xa0we get{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( \\sqrt { n - 1 } + \\sqrt { n + 1 } ) ( \\sqrt { n - 1 } - \\sqrt { n + 1 } ) }{/tex}{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( n - 1 ) - ( n + 1 ) } = \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { - 2 }{/tex}or,\xa0{tex}\\sqrt { n + 1 } - \\sqrt { n - 1 } = \\frac { 2 q } { p }{/tex} ........(ii)On adding (i) and (ii), we get{tex}2 \\sqrt { n + 1 } = \\frac { p } { q } + \\frac { 2 q } { p } = \\frac { p ^ { 2 } + 2 q ^ { 2 } } { p q }{/tex}{tex}\\sqrt{n+1}\\;=\\frac{p^2+2q^2}{2pq}{/tex}...............(iii)From (i) and (ii),{tex}\\style{font-family:Arial}{\\sqrt{n-1}\\;=\\frac{p^2-2q^2}{2pq}}{/tex}........(iv)In RHS of (iii) and (iv)\xa0{tex}\\frac{p^2+2q^2}{2pq}\\;and\\;\\frac{\\displaystyle p^2-2q^2}{\\displaystyle2pq}\\;are\\;rational\\;number\\;because\\;p\\;and\\;q\\;are\\;positive\\;integers{/tex}But it is possible only when (n + 1) and (n - 1) both are perfect squares.Now n+1-(n-1)=n+1-n+1=2Hence they differ by 2 and two perfect squares never differ by 2.So both (n + 1) and (n -1 ) cannot be perfect squares. Hence there is no positive integer n for which\xa0{tex}\\style{font-family:Arial}{\\sqrt{n-1\\;}+\\sqrt{n+1}}{/tex} is rational | |
| 16919. |
Cos 45/sec 30+cosec 30 |
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Answer» root6 +8/4 cos45°=1/√2sec30°=2/√3csc30°=2Then,(1/√2)/(2/√3)+21/√2 * 2/√3 +2√2/√3 + 21/√3/2 + 1/1/21/(√3+1)/22/(√3+2)2(2-√3)/(2+√3)(2-√3)(4-2√3)/(4-3)4-2√3 |
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| 16920. |
Find the largest number divides 70 and 125 leaving remainder 5 in both cases |
| Answer» 5 | |
| 16921. |
What is cubic polynomial |
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Answer» Those polynomial having highest /maximum degree 3 is called cubic polynomial. Which have highest degree as 3 A cubic polynomial is a polynomial which has highest degree 3 which have three roots. 3 no. degree |
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| 16922. |
X/7+y/3=5 and X/6-y/9=6 |
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| 16923. |
अल्फा बीटा गामा ऑफ द पॉलिनॉमियल एक्स क्यूब प्लस 3 स्क्वायर माइनस 5 एक्स प्लस 1 फाइंड द वैल्यू ऑफ |
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| 16924. |
APs2,4,8,16,.... |
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Answer» no, because its common different is different. no ,because its comman factor is different. no, it is not an ap because the common difference is not same. 4-2 -- 2, 8-4 -- 4. No |
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| 16925. |
Write a quadratic polynomial, the sum and product of whose zeroes ate 3 and -2 respectively |
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Answer» x square + 3x -2 sum -- 3product -- -2 quadratic polynomial -- k[xsquare - (sum)x+product](where k is any constant)quadraticpolynomial-- xsquare-3x-2 X2 +3x -2 |
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| 16926. |
Find arithmetic progression whose third term is 16 and the seventh term exceeds the 5 th term by 12 |
| Answer» a3 = 16{tex} \\Rightarrow {/tex}\xa0a + 2d = 16 ..... (i)a7 = a5 + 12{tex} \\Rightarrow {/tex}\xa0a + 6d = a + 4d + 12{tex} \\Rightarrow {/tex}\xa02d = 12{tex} \\Rightarrow {/tex}\xa0d = 6Put the value of d in eq. (i){tex}a + 2 \\times 6 = 16{/tex}{tex} \\Rightarrow {/tex}\xa0a = 16 - 12{tex} \\Rightarrow {/tex}\xa0a = 44, 10, 16.... | |
| 16927. |
Pie is a rational or irrational number |
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Answer» irrational number Irrational number irrational Irrational Irrational number Irrational |
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| 16928. |
What is the probability that an ordinary year has 53 tuesdays |
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| 16929. |
in the equation ax^2+bx+c=0, if (a+b+c)=0, show that the roots of the equation are 1 acd c/a |
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| 16930. |
If cos theta equals to 4 find the value of 4 cos theta minus sin theta upon 2 cos theta + sin theta |
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| 16931. |
What is trigonometry ratio? |
| Answer» There are six trigonometry ratio 1.sin2.Cos 3.Tan 4.Cot 5.Sec 6.Cose | |
| 16932. |
prove that 5 minus 3 by 7 root 3 is an irrational number |
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| 16933. |
What is perpendicular bisector of a triangle |
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| 16934. |
If 12 is added to a natural number, it becomes 160 times of its reciprocal. Find the number. |
| Answer» let natural number be xa to q x+12 -- 160 *1/x(-- is in the place of equals to sign)x+12-160/x -- 0 (take lcm and solve it)xsquare+12x-160 --0(solve it by middle term splitting factorization)x square+20x-8x-160 -- 0x(x+20)-8(x+20)(x+20)(x-8)so the possiblities of natural number is 8 and -20 | |
| 16935. |
Solve 4x^2+4bx-(a^2-b^2)=0 by completing square method. |
| Answer» We have the following equation,4x2 + 4bx - (a2 - b2) = 0Now,4x2 + 4bx - (a2 - b2) = 0{tex}\\Rightarrow{/tex}\xa04x2 - 2(a - b)x + 2(a + b)x - (a2 - b2) = 0{tex}\\Rightarrow{/tex} 2x[2x - (a - b)] + (a + b)[2x - (a - b)] = 0{tex}\\Rightarrow{/tex} [2x - (a - b)] [2x + (a + b)] = 0{tex}\\Rightarrow{/tex} 2x - (a - b) = 0 or 2x + (a + b) = 0{tex}\\Rightarrow{/tex}\xa02x = a - b or 2x = -a - b{tex} \\Rightarrow x = \\frac{{a - b}}{2}{/tex} or {tex}x = \\frac{{ - a - b}}{2}{/tex} | |
| 16936. |
Solve 4x^2 + 4bx-(a^2-b^2)=0 |
| Answer» We have the following equation,4x2 + 4bx - (a2 - b2) = 0Now,4x2 + 4bx - (a2 - b2) = 0{tex}\\Rightarrow{/tex}\xa04x2 - 2(a - b)x + 2(a + b)x - (a2 - b2) = 0{tex}\\Rightarrow{/tex} 2x[2x - (a - b)] + (a + b)[2x - (a - b)] = 0{tex}\\Rightarrow{/tex} [2x - (a - b)] [2x + (a + b)] = 0{tex}\\Rightarrow{/tex} 2x - (a - b) = 0 or 2x + (a + b) = 0{tex}\\Rightarrow{/tex}\xa02x = a - b or 2x = -a - b{tex} \\Rightarrow x = \\frac{{a - b}}{2}{/tex} or {tex}x = \\frac{{ - a - b}}{2}{/tex} | |
| 16937. |
Prove that one out of three consecutive positive integers is divisible by 3. |
| Answer» Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.By Euclid’s division lemma, we havea = bq + r; 0 ≤ r < bFor a = n and b = 3, we haven = 3q + r ...(i)Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.Putting r = 0 in (i), we get{tex}n = 3q{/tex}∴ n is divisible by 3.{tex}n + 1 = 3q + 1{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 2{/tex}∴ n + 2 is not divisible by 3.Putting r = 1 in (i), we get{tex}n = 3q + 1{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 2{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}∴ n + 2 is divisible by 3.Putting r = 2 in (i), we get{tex}n = 3q + 2{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}∴ n + 1 is divisible by 3.{tex}n + 2 = 3q + 4{/tex}∴ n + 2 is not divisible by 3.Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3. | |
| 16938. |
Prove that the product of two consecutive positive integers is divisible by 2. |
| Answer» Let n-1 and n be consecutive positive integers,Let P be their productThen {tex}\\style{font-family:Arial}{\\style{font-size:12px}{\\mathrm n(\\mathrm n-1)=\\mathrm n^2-1\\;.................(1)}}{/tex}We know that any positive integers is of the form 2q or 2q + 1, where q is a positive integerCase I: When n = 2q, thenP=n2\xa0- n = (2q)2\xa0- 2q = 4q2\xa0- 2q = 2q(2q - 1)-----(2)Case II: When n = 2q + 1, thenP=n2\xa0- n = (2q + 1)2\xa0- (2q + 1)= 4q2\xa0+ 4q + 1 - 2q - 1= 4q2\xa0+ 2qP = 2q(2q + 1)..........(3)From (2) and (3) we conclude thatThe product of n-1 and n is divisible by 2 | |
| 16939. |
Find the middle term of an AP 213 205 197 .......37 |
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Answer» an equals to a + (n-1)d37 equals to 213+(n-1)-837-213 equals to(n-1)-8-176/-8 equals to n-122+1equals to n23 equals to n(23 is odd)so middle term equals to n+1/223+1/2 equals to12so 12th term is middle terman equal to a+(n-1)da12 equal to 213+(12-1)-8a12 equal to 213-88a12 equal to 125(sorry for equals to sign) a is 213, d is 205-213(-8), an is 37an equals to |
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| 16940. |
Given that root 2 is an irrational number ,then prove that 5+3root2 is an irrational number . |
| Answer» To Prove: 5+3{tex}\\sqrt{2}{/tex}\xa0is irrational numberProof: If possible let us assume 5 + 3{tex}\\sqrt{2}{/tex}\xa0is a rational number.{tex}\\implies {/tex}\xa05 + 3{tex}\\sqrt{2}{/tex}\xa0=\xa0{tex}\\frac{p}{q}{/tex}\xa0where q\xa0{tex}\\ne{/tex}\xa00 and p and q are coprime integers.{tex}\\implies{/tex}{tex}3\\sqrt{2}=\\frac{p}{q}-5 {/tex}{tex}\\implies{/tex}{tex}3\\sqrt{2}=\\frac{p-5q}{q}{/tex}{tex}\\implies\\sqrt{2}=\\frac{p-5q}{3q}{/tex}{tex}\\implies\\sqrt{2}=\\frac{integer}{integer}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt{2}{/tex} is\xa0a rational number.This contradicts the given fact that\xa0{tex}\\sqrt{2}{/tex}\xa0is irrational.Hence 5 + 3{tex}\\sqrt{2}{/tex} is an irrational number. | |
| 16941. |
Show that 6n can not end with 0 or 5 any natural number |
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Answer» It cannot end because it has 2×3 as its prime factors. If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5Prime factorisation of 6n = (2 ×3)nIt can be observed that 5 is not in the prime factorisation of 6n.Hence, for any value of n, 6n will not be divisible by 5.Therefore, 6n cannot end with the digit 0 or 5 for any natural number n. |
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| 16942. |
Find the mode Class interval 1-3 3-5 5-7 7-9 9-11Frequency 14 16 4 4 2 |
| Answer» 3 - 5 is the modal class since it has the highest frequency. | |
| 16943. |
How to find middle term of an AP 6 13 20 .....216 |
| Answer» Here,{tex} a = 6, l =216, d = 13 - 6 = 7{/tex}Let the number of terms be n{tex}l = a + ( n - 1) d{/tex}216 = {tex}6 + (n -1)(7){/tex}216 -6 = {tex}7(n - 1){/tex}{tex}7(n - 1){/tex}= 210{tex}n - 1{/tex}=\xa0{tex}\\frac { 210 } { 7 }{/tex}= 30{tex}n{/tex} = 30 + 1 = 31The middle term will be =\xa0{tex}\\frac { 31 + 1 } { 2 }{/tex}= 16th term{tex}\\therefore{/tex}a16\xa0= 6 +(16 - 1)(7){tex}= 6 + 15 \\times 7{/tex}= 6 + 105= 111Middle term will be 111. | |
| 16944. |
Prove that the product of three consecutive positive integer is divisible by 6 |
| Answer» Let three consecutive numbers be x, (x + 1) and (x + 2)Let x = 6q + r 0 {tex}\\leq r < 6{/tex}{tex}\\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}{tex}\\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}if x = 6q then which is divisible by 6{tex}\\text { if } x = 6 q + 1{/tex}{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}{tex}= 6 ( 3 q + 1 ) \\cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}which is divisible by 6if x = 6q + 2{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}{tex}= 6 ( 2 q + 1 ) \\cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}Which is divisible by 6{tex}\\text { if } x = 6 q + 3{/tex}{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}which is divisible by 6{tex}\\text { if } x = 6 q + 4{/tex}{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}which is divisible by 6if x = 6q + 5{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}which is divisible by 6{tex}\\therefore {/tex}\xa0the product of any three natural numbers is divisible by 6. | |
| 16945. |
Under what condition the polynomial ax*2+bx+c does have any other real zero |
| Answer» As\xa0given in the book, you didn\'t have been read it yet\xa0Answer: When\xa0{tex}\\huge b^2-4ac<0{/tex} | |
| 16946. |
Plz answer my ques |
| Answer» What is the question | |
| 16947. |
If the polynomial x 4 - 3 x square + 3 X + 42 is divisible by X + 3 then find the value of p |
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| 16948. |
Find k if sum of zeros of polynomial x²-(k+6)x+2(2k-1) is half their product |
| Answer» I think K = 7.A/q, k + 6 = 1/2 × 4k - 2 k + 6 = 2k - 1 k - 2k = - 1- 6 K = 7. | |
| 16949. |
Find the zeroes of the polynomial 2x^3+x^2-5x+2 |
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| 16950. |
Find the zeroes of f (x)=3x^2+11x-4 |
| Answer» 3x^2+11x-43x^2+12x-1x-43x(x+4)-1(x+4)(3x-1)(x+4)x= 1/3 , -4 | |