If an integer P is chosen at random in the interval 0 ≤ p ≤ 5, the probability that the roots of the equation x^2 + px -Maths 9th

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Description : An integer is chosen at random from the first two hundred positive integers. What is the probability that the integer chosen is divisible by 6 or 8 ? -Maths 9th

Last Answer : As there are 200 integers, total number of exhaustive, mutually exclusive and equally likely cases, i.e, n(S) = 200 Let A : Event of integer chosen from 1 to 200 being divisible by 6⇒ n(A) = 33 \(\bigg(rac{200}{6}=33rac{1}{3}\ ... (rac{25}{200}\) - \(rac{8}{200}\) = \(rac{50}{200}\) = \(rac{1}{4}\).

Description : Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Last Answer : Let αα and ββ be the roots of the quadratic equation x2+px+q=0x2+px+q=0 Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e., a product of roots ... 1 Now, from Eqs. (ii) and (iii), we get α=−3 and β=−4α=−3 and β=−4 which are correct roots.

Description : If p, q, r are positive and are in A.P., the roots of quadratic equation px^2 + qx + r = 0 are real for : -Maths 9th

Last Answer : Given p,q,r are in A.P. then q=2p+r​.....(1). Now px2+qx+r=0 will have real root then q2−4pr≥0. or, 4(p+r)2​−4pr≥0 or, p2+r2−14pr≥0 or, r2−14rp+49p2≥48p2 or, (r−7p)2≥(43​p)2 or, (pr​−7)2≥(43​)2 [ Since p=0 for the given equation to be quadratic] or, ∣∣∣∣∣​pr​−7∣∣∣∣∣​≥43​.

Description : A computer program selects an integer in the set {k : 1 ≤ k ≤ 10,00,000} at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k = 1 appears in the printout atleast once ? (A) 0.5 (B) 0.704 (C) 0.632121 (D) 0.68

Last Answer : (C) 0.632121

Description : If the difference in the roots of the equation x^2 – px + q = 0 is unity, then which one of the following is correct ? -Maths 9th

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Description : A natural number is chosen at random from amongst the first 300. What is the probability that the number chosen is a multiple of 2 or 3 or 5 ? -Maths 9th

Last Answer : (b) \(rac{11}{15}\)n(S) = 300 Let A : Event of getting a number divisible by 2 B : Event of getting a number divisible by 3 C : Event of getting a number divisible by 5 ∴ A ∩ B : Event of getting a number divisible by ... \(rac{320}{300}\) - \(rac{100}{300}\) = \(rac{220}{300}\) = \(rac{11}{15}\).

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

Description : Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Last Answer : (c) \(rac{1}{10}\)Let S be the sample space.Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon= 6C3 = \(rac{6 imes5 imes4}{3 imes2}\) = 20.Let A : Event that the ... Required probability = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

Description : For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

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Description : For which value of `p` among the following, does the quadratic equation `3x^(2) + px + 1 = 0` have real roots ?

Last Answer : For which value of `p` among the following, does the quadratic equation `3x^(2) + px + 1 = 0` have real roots ?

Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.

Description : When (x^3 – 2x^2 + px – q) is divided by (x^2 – 2x – 3), the remainder is (x – 6), What are the values of p and q respectively ? -Maths 9th

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Description : If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Last Answer : Let f(x)=px3+x2−2x−q Since f(x) is divisible by (x−1) and (x+1) so x=1 and −1 must make f(x)=0. Therefore, p+1−2−q=0, i.e., p−q=1; and −p+1+2−q=0, i.e., p+q=3 Thus p=2 and q=1

Description : If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Last Answer : x3+px+q 3x2+p have common root Let common root =a a3+ap+q=0 a2=3−p​a=3−p​​4p3+27q2=? a3(a2+p)+q=0 ⇒3−p​​[(3−p​)+p]+q=0 3−p​​(3+2p​)=−q ⇒27−p(4p2)​=q2

Description : Let p and q be the roots of the quadratic equation x^2 – (a – 2)x – a – 1 = 0. What is the minimum possible value of p^2 + q^2 ? -Maths 9th

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Description : If a, b are the roots of the equation `x^(2) - px +q = 0`, then find the equation which has `a/b` and `b/a` as its roots.

Last Answer : If a, b are the roots of the equation `x^(2) - px +q = 0`, then find the equation which has `a/b` and `b/a` as its roots.

Description : Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ? -Maths 9th

Last Answer : The chosen consists of players (6 + 5). ∴ Number of ways of selecting 11 players out of 15 players = n(S) = 15C11 Let A : Event of choosing 6 batsmen of 8 batsmen and 5 bowlers of 7 bowlers Then, n(A) = 8C6 ... 7 imes6}{2}\) x \(rac{4 imes3 imes2 imes1}{15 imes14 imes13 imes12}\) = \(rac{28}{65}.\)

Description : If a number x is chosen at random from the numbers -2, -1, 0 1, 2. What is the probability that x 2 < 2? a. 2/5 b. 3/5 c. 1/4 d. 4/7

Last Answer : b. 3/5

Description : If x^2 + mx + n = 0 and x^2 + px + q = 0 have a common root, then the common root is -Maths 9th

Last Answer : Let α be the common root ∴α2+pα+q=0 ...........(1) and α2+qα+p=0 ........ (2) Solving (1) & (2), we get, p2−q2α2​=q−pα​=q−p1​∴α=q−pp2−q2​ and α=1 ⇒q−pp2−q2​=1 ⇒p2−q2=q−p (or) (p2−q2)+(p−q)=0 ⇒(p−q)[p+q+1]=0 ⇒p−q=0 or p+q+1=0

Description : Out of 3n consecutive natural numbers 3 natural numbers are chosen at random without replacement. -Maths 9th

Last Answer : (c) \(rac{3n^2-3n+2}{(3n-1)(3n-2)}\)In 3n consecutive natural numbers, (i) n numbers are of the form 3p (ii) n numbers are of the form 3p + 1 (iii) n numbers are of the form 3p + 2 For the ... ) We can select one number from each set.∴ Favourable number of cases = nC3 + nC3 + nC3 + (nC1 nC1 nC1)

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

Description : If both (x – 2) and (x – 1/2) are factors of px^2 + 5x + r, then: -Maths 9th

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Description : If (x + k) is a common factor of x^2 + px + q and x^2 + lx + m, then the value of k is -Maths 9th

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Description : The equation whose roots are the negatives of the roots of the equation x^7 + 3x^5 + x^3 – x^2 + 7x + 2 = 0 is : -Maths 9th

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Description : A bag contains 7 red and 5 green balls. The probability of drawing all four balls asred balls, when four balls are drawn at random is -Maths 9th

Last Answer : (b) \(rac{7}{99}\)There are (7 + 5) = 12 balls in the bag. 4 balls can be drawn at random from 12 balls in 12C4 ways. ∴ n(S) = 12C4 = \(rac{|\underline{7}}{|\underline3|\underline4}\) = \(rac{7 imes6 imes5}{3 ... ) = 35∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{35}{495}\) = \(rac{7}{99}\).

Description : A bag contains 5 green and 7 red balls, out of which two balls are drawn at random. What is the probability that they are of the same colour ? -Maths 9th

Last Answer : (d) \(rac{31}{66}\)Total number of balls in the bag = 12 (5 Green + 7 Red) Let S be the sample space of drawing 2 balls out of 12 balls.Thenn(S) = 12C2 = \(rac{12 imes11}{2}\) = 66∴ Let A : Event of drawing two red balls⇒ ... \(rac{n(B)}{n(S)}\) = \(rac{21}{66}\) + \(rac{10}{66}\) = \(rac{31}{66}\).

Description : A basket contains 2 blue, 4 red, 3 green and 5 black balls. If 4 balls are picked at random, what is the probability that -Maths 9th

Last Answer : (d) None of theseThe month having 3 days less than 31 days has 28 days, i.e, it is the month of February. P(Choosing February) = \(rac{1}{12}\).

Description : If one of the roots of the equation x^2 + ax + 3 = 0 is 3 and one of the roots of the equation x2 + ax + b = 0 is three -Maths 9th

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Description : If the roots of the equation x^2 – 2ax + a^2 + a – 3 = 0 are real and less than 3, then which one of the following is correct ? -Maths 9th

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Description : If the roots of the equation x^2 + x + 1 = 0 are in the ratio of m : n, then which one of the following relation holds ? -Maths 9th

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Description : What are the roots of the equation 4^x – 3.2^(x + 2) + 32 = 0 ? -Maths 9th

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Description : What are the roots of the quadratic equation a^2 b^2 x^2 – (a^2 + b^2)x + 1 = 0 ? -Maths 9th

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Description : If the roots of the equation a(b – c) x^2 + b(c – a)x + c(a – b) = 0 are equal, then a, b, c are in : -Maths 9th

Last Answer : As we know that for the quadratic equation ax2+bx+c=0, roots will be equal if D=B2−4AC=0 Therefore, for the equation, a(b−c)x2+b(c−a)x+c(a−b)=0 A=a(b−c),B=b(c−a),C=c(a−b) D=0 B2−4AC=0 (b(c−a))2−4(a(b−c))(c(a−b))=0 ⇒ab+bc=2ac Hence a,b and c are in HP.

Description : If the roots of the equation x^3 – ax^2 + bx – c = 0 are three consecutive integers, then what is the smallest possible value of b ? -Maths 9th

Last Answer : Let the roots of the equation x3 – ax2 + bx – c = 0 be (α – 1), α, (α + 1) ∴ S2 = (α – 1)α + α(α + 1) + (α + 1) ( ... ; 1 = b ⇒ 3α2 – 1 = b ∴ Minimum value of b = – 1, when α = 0.

Description : If the equation (a^2 + b^2) x^2 – 2 (ac + bd)x + (c^2 + d^2) = 0 has equal roots, then which one of the following is correct ? -Maths 9th

Last Answer : The given quadratic equation is (a2 + b2)x2 − 2(ac + bd)x + (c2 + d2) = 0. If the roots of given quadratic equation are equal, then its discriminant is zero.

Description : The equation whose roots are twice the roots of the equation x^2 – 3x + 3 = 0 is -Maths 9th

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Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

Description : A coin is tossed 500 times and we get Heads : 285 and tails : 215 times. When a coin is tossed at random, what is the probability of getting a. head? b. tail? -Maths 9th

Last Answer : Given, Total number of events = 500 No. of times heads occur = 285 Probability of getting head when coin is tossed at random = 285/500 = 57/100 No. of times tails occur = 215 Probability of getting tails when coin is tossed at random = 215/500 = 43/100

Description : Two coin are tossed 400 times and we get a. Two Heads : 112 times b. One Head : 160 times c. No Head : 128 times. When two coins are tossed at random, what is the probability of getting a. Two Heads b. One Head c. No Head -Maths 9th

Last Answer : Given, Total number of events = 400 (a) No. of times two heads occur = 112 Probability of getting two heads = 112/400 = 7/25 (b) No. of times one heads occur = 160 Probability of getting one heads = 160/400 = 2/5 (c) No. of times no heads occur = 128 Probability of getting no heads = 128/400 = 8/25

Description : The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Last Answer : There are 7 letters in the word SOCIETY. ∴ Total number of ways of arranging all the 7 letters = n(S) = 7!. When the case of three vowels being together is taken, then the three vowels are considered as one unit, so the ... = 5! 3! ∴ Required probability = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

Description : What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Last Answer : (a) \(rac{1}{25}\) Let us assume S as the sample space in all questions. S means the set denoting the total number of outcomes possible. Let S = {1, 2, 3, , 100} be the sample space. Then, n(S) = 100 Let A : ... ∴Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

Description : From a group of 3 man and 2 women, two person are selected at random. Find the probability that at least one women is selected. -Maths 9th

Last Answer : (b) \(rac{7}{10}\)Total number of ways of selecting 2 persons at random out of 5 persons = 5C2 ∴ n(S) = 5C2 = \(rac{|\underline5}{|\underline3|\underline2}\) = \(rac{5 imes4}{2 imes1}\) = 10Let A : Event of selecting ... = 2 3 + 1 = 7 ∴ Required probability = \(rac{n(A)}{n(S)}\) = \(rac{7}{10}\).

Description : Five horses are in a race. Mr A. Selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is -Maths 9th

Last Answer : (b) \(rac{2}{5}\)As each horse has equal chance of winning the race, Number of ways in which one of the five horses wins the race = 5C1 ∴ n(S) =5C1 = \(rac{|\underline5}{|\underline4|\underline1}\) 5To find the chance ... n(E) = 2C1 = 2 ∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{2}{5}\).

Description : Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that either both are black or both are kings ? -Maths 9th

Last Answer : (b) \(rac{55}{221}\)S : Drawing 2 cards out of 52 cards ⇒ n(S) = 52C2 = \(rac{|\underline{52}}{|\underline{52}|\underline2}\) = \(rac{52 imes51}{2}\) = 1326A : Event of drawing 2 black cards out of 26 black cards⇒ n ... ) + \(rac{6}{1326}\) - \(rac{1}{1326}\) = \(rac{330}{1326}\) = \(rac{55}{221}\).

Description : A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

Last Answer : (d) \(rac{7}{13}\)Here n(S) = 52 Let A, B, C be the events of getting a red card, a diamond and a jack respectively. ∵ There are 26 red cards, 13 diamonds and 4 jacks, n(A) = 26, n(B) = 13, n(C) = 4 ⇒ n(A ∩ B) = ... rac{1}{52}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .

Description : The probability that in the random arrangement of the letters of the word ‘UNIVERSITY’the two I‘s do not come together is -Maths 9th

Last Answer : (b) \(rac{4}{5}\)Let S be the sample space. Then, n(S) = Total number of waysin which the letters of the word UNIVERSITY' can be arranged = \(rac{10!}{2!}\) (∵ There are 2I s) ... ! imes36}{rac{10!}{2!}}\) = \(rac{ ot8! imes36 imes2!}{10 imes9 imes ot8!}\) = \(rac{4}{5}\).