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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.
| 38901. |
To draw the graph of a quadratic polynomial and observe |
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| 38902. |
Find k if x = 8+yz |
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Answer» It is wrong question Sorry wrong question |
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| 38903. |
Find the zeros of the polynomial 2x4-11x3+7x2+13x given that two of its zeros are (3+√2)and (3-√2) |
| Answer» PlZ go to 11 class it\'s interesting love story | |
| 38904. |
Solve by using quadratic formula: 4x^2+2(b-3a)x-3ab =0 |
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| 38905. |
Find the maximum number of zeroes that the polynomial p(x) = x^4 + 3x^2 - 7 can have? |
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Answer» This answer is correct or not 4 zeroes |
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| 38906. |
explain by paythagoras |
| Answer» Phythagoras theoram says that in any right angled triangle , the square of hypotenuse is equal to the sum of squares of perpendicular and base | |
| 38907. |
a+b the whole cube formula |
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Answer» Thank you a³+b³+3ab(a+b) |
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| 38908. |
(Bc+ca-ab)abc |
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| 38909. |
Prove n³-n is divisible by 6 where n is any natural number |
| 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) | |
| 38910. |
Perimeter of circle |
| Answer» 2πr | |
| 38911. |
Find the number of an A.P. is equal to 3 times its 6th term.if its 9th term is 19 find the A.P. |
| Answer» Plz say me the answer | |
| 38912. |
If tanA=root2-1,show that sinAcosA=root2/4 |
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Answer» Hint :::: find hypotenuse and put the value in sinA.cosA according to formula How i can i send By wriiting is too long And i can\'t send u pic |
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| 38913. |
2 x square - 5 X Y + 2 y square |
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| 38914. |
[sin30°+tan45°-cosec60°]÷[sec30°+cos60°+cot45°] |
| Answer» Put the values and divide | |
| 38915. |
Trigonometry means |
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Answer» Measure of Three angels of triangles Knoledge of three angles Dimaag Ka dahi -(दिमाग का दही) |
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| 38916. |
Find all the zeroes of the polynomial p(x) = x^4 + 3x^2 - 7. |
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| 38917. |
What is zero of the polynomial |
| Answer» Instead to X substitute value we get zero then it is called zero of | |
| 38918. |
Find the zeros of x2-root 5 |
| Answer» X2=root5X=4rootof5 | |
| 38919. |
Chapter 10 ex10. 2 9 ques |
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| 38920. |
How many terms of the AP 27 24 21... should be taken to get the sum zero |
| Answer» a +(n-1)d= 027+ (n-1)-3=027-3n+3=030=3nn=10 | |
| 38921. |
If P= x^2 -2x -15/x^2 +7x+12 and Q= x^2 -4x -77/x^2 +11x +28 .Find P÷Q |
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| 38922. |
Find the quadratic polynomial whose sum and product of zeros are 1/2a and 1/2a |
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| 38923. |
Find the value of all trionometeric ratios |
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| 38924. |
(a/b)-(b/y)=0 ; [a( b^2)]/x + [(a^2)b]/y = a^2 + b^2 Solve...... |
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| 38925. |
Which book we should prefer for maths for extra questions?? |
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Answer» You can buy examidea for extra question.This is avilable in all subject. You have to give priority first to science and after that R.S Agrawal Comprehensive is bst I think u should more focus on ncert at first Lakhmir Singh First prefer NCERT.... In Secondary u can use... RD Sharma, Rs Agrwal, ncert exampler or Xam Idea RD sharma and RS Agrawal both were best You can choose anyone Rs aggarwal Rd Sharma |
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| 38926. |
Find the hcf of 95and 190 |
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Answer» 95 Hcf = 95 only As 95×2=190 Using Euclid division lemma —a=bq+r190=95×2+0 Your HCF is 95 2 |
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| 38927. |
Express 23150. as product of prime factors. |
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| 38928. |
Find the HCF of (2cube×3square × 5),(2square× 3cube× 5square) and 2ke upar 4×3×5ke upar cube×7) |
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| 38929. |
If sin A+B=1 and sin A-B=1/2 or 0 |
| Answer» Sin 90 degree is 1and sin 30 degree 1/2 so this implies A+B=90 degree and A-B=30 degree then by 90-30=60 SO A=60,and 60-30=30 SO B=30 | |
| 38930. |
how to draw graph of an qudratic |
| Answer» You dont know | |
| 38931. |
Sin2A =2sinA when A =? |
| Answer» 0° | |
| 38932. |
Prove that 7-2*3 is an irrational number. |
| Answer» \xa0let us assume that\xa0{tex}\\sqrt 3{/tex}\xa0be a rational number.{tex}\\sqrt { 3 } = \\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 } = 3 b ^ { 2 }{/tex}--------(i)a2\xa0is divisible by 3.Hence a is divisible by 3..........(ii)Let a = 3c ( where c is any integer)squaring on both sides we get(3c)2\xa0= 3b29c2\xa0= 3b2b2\xa0= 3c2so b2\xa0is divisible by 3hence, b is divisible by 3..........(iii)From equation(ii) and (iii), we have3 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 3{/tex} is rational number is wrong.Hence,\xa0{tex}\\sqrt 3{/tex}\xa0is an irrational number.Let us assume that 7 - 2{tex}\\sqrt 3{/tex}\xa0is a rational number7 -\xa02{tex}\\sqrt 3{/tex}\xa0=\xa0{tex}\\frac { p } { q }{/tex} (q\xa0{tex}\\neq{/tex}0 and p and q are co-primes){tex}\\style{font-family:Arial}{\\begin{array}{l}7-2\\sqrt3=\\frac pq\\\\-2\\sqrt3=\\frac pq-7\\\\2\\sqrt3=7-\\frac pq\\\\2\\sqrt3=\\frac{7q-p}q\\\\\\sqrt3=\\frac{7q-p}{2q}\\end{array}}{/tex}Here 7q-p\xa0and 2q both are integers, hence\xa0{tex}\\sqrt 3{/tex} is a rational number.But this contradicts the fact that\xa0{tex}\\sqrt 3{/tex}\xa0is an irrational number.This contradict is due to our assumption that\xa0{tex}\\style{font-family:Arial}{7-2\\sqrt3}{/tex}\xa0is rational.Hence, 7 -\xa02{tex}\\sqrt3{/tex}\xa0is irrational. | |
| 38933. |
2x square - (1+2root2)+2 |
| Answer» Mere question ka answer to do | |
| 38934. |
Best book for 10th |
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Answer» Yes ncert good book for 10th Yes Ncert Ncert |
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| 38935. |
Give me time table for score 1st in school board 10 |
| Answer» If you will make your own time table you can score 1st becoz everyone knows their own routine | |
| 38936. |
(tan65°)÷(cot25°) |
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Answer» 1 1 |
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| 38937. |
Is there exercises of chapters |
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| 38938. |
What do you mean by trigonmetri ratios |
| Answer» Trigonometry ratios are of six types sin,cos,tan,cosine,sec,cot.These are the angle created or the form of waves generated on xy curve | |
| 38939. |
Find the zeroes of the polynomial x^3-13x^2+52x-60 |
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| 38940. |
find a quadratic polynomial whose zeroes are -3and 4 |
| Answer» Zeroes are -3 & 4Factors are x+3 & x-4(x+3)(x-4)= p(x)x^2-4x+3x-12=p(x)x^2-x-12=p(x)The required polynomial is x^2-x-12 | |
| 38941. |
Prove that (a+b)2 = a2 +b2 +2ab |
| Answer» (a+b)^2= (a+b)(a+b) =a^2 + ab+ ab+ b^2 =a^2+ 2ab+ b^2 | |
| 38942. |
Prove that : ((Root3+1)to the power 2x)+ ((root3-1)to the power 2x)=2to the power 3x |
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| 38943. |
Find the zeroes of the polynomial 2xsquare-9 |
| Answer» 2x^2-92x^2=92x=√92x=+3or-3X=3/2andx=-3/2 | |
| 38944. |
Which is the smallest composite number |
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Answer» 4 4 |
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| 38945. |
Four add four |
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Answer» 8 8 |
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| 38946. |
by using the method of completing the square show that the 4x^2+3x+5=0 has no real roots |
| Answer» We have,4x2 + 3x + 5 = 0Divide whole equation by 4{tex} \\Rightarrow \\quad x ^ { 2 } + \\frac { 3 } { 4 } x + \\frac { 5 } { 4 } = 0{/tex}{tex}\\Rightarrow \\quad x ^ { 2 } + 2 \\left( \\frac { 3 } { 8 } x \\right) = - \\frac { 5 } { 4 }{/tex}Add ( half of coefficient of x )2 both sides{tex}\\Rightarrow \\quad x ^ { 2 } + 2 \\left( \\frac { 3 } { 8 } \\right) x + \\left( \\frac { 3 } { 8 } \\right) ^ { 2 } = \\left( \\frac { 3 } { 8 } \\right) ^ { 2 } - \\frac { 5 } { 4 }{/tex}{tex}\\Rightarrow \\quad \\left( x + \\frac { 3 } { 8 } \\right) ^ { 2 } = - \\frac { 71 } { 64 }{/tex}{tex}\\left( x + \\frac { 3 } { 8 } \\right) ^ { 2 }{/tex} cannot be negative for any real value of x. Hence, the given equation has no real roots. | |
| 38947. |
If the sum of squares of the zeroes of polynomial 6x2+x+k is 25/36. Find the value of k ? |
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| 38948. |
Can (x-7) be the remainder on division of polynomial p(x) by (7x+2)?justify your answer |
| Answer» 37 | |
| 38949. |
If alpha and beta are the zeroes of polynmial x^2-5x+k such that alpha-beta=1,then find k |
| Answer» K is 5 | |
| 38950. |
Solve 71x+37y=253 and 37x+71y=287 by elimination method |
| Answer» x=2, y=3 | |