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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.
| 51. |
Find the minimum value of the quadratic expression `4x^(2) -3x + 4`.A. `(-55)/(16)`B. `(55)/(16)`C. `(16)/(15)`D. `(161)/(22)` |
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Answer» Correct Answer - B Use the formula of tfind the minimum value. |
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| 52. |
Write the quadratic equation whose roots are `5/2` and `8/3`. |
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Answer» Sum of the roots `= 5/2 + 8/3 = 31/6`. Product of the roots `= 5/3(8/3) = 20/3`. The required equation is, `x^(2)` - (sum of the roots) x + (proudct of the roots) = 0 `rArr x^(2) - (31/6) x + (20)/(3) =0` `rArr 6x^(2) - 31x + 40 = 0` `:.` a quadratic equation with roots `5/2` and `83` is `6x^(2) - 31x + 40 = 0`. |
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| 53. |
(a) Solve `x^(2) - 15x+ 26 = 0` (b) Solve `x+1/x = 5/2`. |
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Answer» (a) First, let us resolve `x^(2) - 15x + 26` into factors. `rArr x^(2) - 15x + 26` `= x^(2) - 13x - 2x + 26` `= x(x-13) - 2(x-13)` `= (x-13)(x-2)`. The given equation, `x^(2) - 15 x +26 = 0` is reduced to `(x-13)(x-2) = 0` `rArr x - 13= 0` (or) `x - 2 = 0` `rArr x = 13` (or) x = 2. `:. x = 2. 13` are the roots of the given equation. (b) `(x^(2) + 1)/(x) = 5/2` `rArr 2x^(2) + 2 = 5x`. `rArr 2x^(2) - 5x + 2 = 0` `rArr 2x^(2) -4x - x + 2 = 0` `rArr 2x(x-2) - 1 (x-2)= 0` `rArr (2x-1)(x-2) = 0` `rArr 2x - 1 = 0` or `x - 2 = 0` `rArr x = 1/2` or `x = 2`. `:. x = 1/2, 2` are the roots of the given equation. |
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| 54. |
If `x` is a real, then the maximum value `(x^2+14x+9)/(x^2+2x+3)``(i)2 (ii)4 (iii)6 (iv) 8`A. 3, 1B. 4, -5C. `0, -oo`D. `oo, -oo` |
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Answer» Correct Answer - B Let `y=(x^(2)+14x+9)/(x^(2)+2x+3)`. Then, `x^(2)(y-1)+2x(y-7) + 3y - 9 = 0` `rArr" "4(y-7)^(2) - 4(y-1)(3y-9) ge 0" "[because "x is real"]` `rArr" "y^(2) + y - 20 le 0 rArr -5 le y le 4` |
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| 55. |
If `alpha, beta` are the roots of `px^(2) + qx + r = 0`, then `alpha^(3) + beta^(3) = "______"`.A. `(3qpr-q^(3))/(p^(3))`B. `(3pqr-3q)/(p^(3))`C. `(pqr - 3q)/(p^(3))`D. `(3pqr -q)/(p^(3))` |
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Answer» Correct Answer - A `alpha + beta = (-q)/(r )` `alpha.beta = (r )/(p)` `alpha. Beta = r/p` `alpha^(3) + beta^(3) = (alpha + beta)^(3) - 3alpha beta (alpha + beta)` `= ((-q)/(p))^(3)- 3(r/p) (-q/p)` `= (-q^(3) + 3 pqr)/(p^(3))` `:. alpha^(3) + beta^(5) = (3pqr - q^(3))/(p^(3))`. |
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| 56. |
If one of the roots of the an equation, `x^(2) - 2x + c =0` is thrice the other, then `c =?`A. `1/2`B. `4/3`C. `-1/2`D. `3/4` |
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Answer» Correct Answer - D Let the root be `alpha` and `3alpha`. |
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| 57. |
If `(x+2)` is a common factor of the expression `x^(2) + ax - 6, x^(2) + bx +2 ` and `kx^(2) - ax - (a+b)`, then `k= "_____"`.A. 2B. 3C. 1D. -2 |
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Answer» Correct Answer - C Substitute `x = - 2` in the first two expression, equated to zero. |
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| 58. |
Solve equation is `x^(2) - 11x +30 = 0` by using formula. |
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Answer» Given equation is `x^(2) - 11 x + 30 = 0`. The roots of `ax^(2) + bx + c = 0`, are `(-b+-sqrt(b^(2) - 4ac))/(2a)` Here, a=1, b = - 11, and c = 30. That is, `x = (-(-11) +-sqrt((-11)^(2) -4(1)(30)))/(2xx1) = (11+-sqrt(121-120))/(2)` `x = (11+-1)/(2)` , `x = (11+1)/(2)` and `(11-1)/(2) rArr 6` and 5. `:.` The roots are 5 and 6. |
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| 59. |
If `alpha, beta` are the roots of the equation `x^(2) - lm + m = 0`, then find the value of `1/(alpha^(2)) + 1/(beta^(2))` in terms of l and m. |
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Answer» Given, `alpha, beta` are the roots of `x^(2) - lx + m = 0` `rArr` Sum of the roots `= alpha + beta = (-(-l))/(1) = L " "(1)` `rArr` Product of the roots `= alpha beta = (m)/(1) = m" "(2)` Now, `1/(alpha^(2)) + 1/(beta^(2)) = (alpha^(2) + beta^(2))/(alpha^(2)beta^(2))` `= ((alpha^(2) + beta^(2))^(2) - 2(alphabeta))/((alphabeta)^(2))` Substituting the values of `alpha + beta` and `alphabeta` in the above equation , we get , `(1)/(alpha^(2)) +(1)/(beta^(2)) = (l^(2)-2m)/(m^(2))` `:.` The value of `1/(alpha^(2)) + 1/(beta^(2)) = (l^(2) - 2m)/(m^(2))`. |
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| 60. |
If `x^2+px+1` is a factor of `ax^3+bx+c` thena) `a^2+c^2=-ab` b) `a^2+c^2ab` c) `a^2-c^2=ab` d) `a^2-c^2=-ab`A. `a^(2) + c^(2) = -ab`B. `a^(2) - c^(2) = -ab`C. `a^(2) - c^(2) = ab`D. none of these |
| Answer» Correct Answer - C | |
| 61. |
If one of the roots of `ax^(2) + bx + c = 0` is thrice that of the other root, then b can beA. `(4ac)/(3)`B. `(16ac)/(9)`C. `4sqrt((ac)/(3))`D. `sqrt((4ac)/(3))` |
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Answer» Correct Answer - C Let the roots be k and 3k. Sum of the roots `= k + 3 k`. `rArr 4 k = (-b)/(a) rArr k = (-b)/(4a)`. Product of the roots `= k xx 3 k` `rArr 3k^(2) = c/a` `rArr 3((-b)/(4a))^(2) = c/a` `rArr (3b^(2))/(16a^(2)) = c/a` `rArr 3b^(2) = 16 ac` `rArr b = +-sqrt((ac)/(3))`. |
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| 62. |
Find the nature of the roots of the equations of given below : (a) `x^(2) - 13x + 11 =0` (b ) `18x^(2) - 14x + 17 = 0` (c ) `9x^(2) - 36x + 36 = 0` (d) `3x^(2) - 5x - 8 = 0` |
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Answer» (a) Given `x^(2) - 13x + 11 = 0` Comparing it with `ax^(2) + bx + c = 0`, we have `a = 1, b = -13` and `c = 11`. Now, `b^(2) - 4ac = (-13)^(2) - 4(1)(11)` `= 169 - 44 = 125 lt 0`. `rArr b^(2) - 4ac gt 0` and is not perfet square. `:.` The roots are distinct and irrational . (b) Given `18x^(2) - 14x + 17 =0` Comparing it with `ax^(2) + bx + c = 0, a = 18, b = -14` and c = 17. Now , `b^(2) - 4ac = (-14)^(2) - 4(18) (17)` `= 196 - 1224` `= - 1028 lt 0` `rArr b^(2)-4ac lt 0`. `:.` The roots are imagniary. (c) Given `9x^(2) - 36x + 36 = 0`. `rArr 9(x^(2) - 4x+4) =0` `rArr x^(2) - 4x + 4 = 0` Comparing the above equation , with `ax^(2) + bx + c= 0, a= 1, b = - 4` and c = 4 Now, `b^(2)- 4ac = (-4)^(2) - 4(1)(4)` `= 16 - 16 = 0` `rArr b^(2) - 4ac = 0` `:.` The roots are equal. (d) Given `3x^(2) - 5x - 8 = 0` Comparing the above equation with `ax^(2) + bx + c = 0`, we getm `a = 3, b = -5` and `c = - 8`. Now, `b^(2)- 4ac = (-5)^(2) - 4(3) (-8)` `= 25+96 = 121 gt 0` , `rArr b^(2) - 4ac lt 0` and is a perfect square. `:.` The roots are rational and distinct. |
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| 63. |
If `x^(2) + ax+ b` and `x^(2) + bx + c` have a common factor `(x-k)`, then `k ="_____"`.A. `(a-b)/(b-c)`B. `(b-c)/(c-a)`C. `(c-b)/(b-a)`D. `(c-b)/(a-b)` |
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Answer» Correct Answer - D `x^(2) + ax + b` and `x^(2) + bx + c` have a common factor `(x-k)`. `rArr k^(2) + ak + b = 0` and `k^(2) + bk + c= 0` `rArr k^(2) + akl + b = k^(2) + bk + c` `ak + b = bk + c` `k = (c-b)/(a-b)`. |
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| 64. |
Maximum value of `(2+12x-3x^(2))/(2x^(2) - 8x + 9)` is `"______"`.A. 14B. 17C. 11D. Cannot be determined |
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Answer» Correct Answer - A For the maximum value of `(2+12x-3x^(2))/(2x^(2) - 8x+9) , 2 +12x-3x^(2)` is , maximum and `2x^(2) - 8x+9` is minimum. The maximum value of `2+12x-3x^(2)` and minimum-value of `2x^(2) - 8x +9` occurs at `x = (-b)/(2a)`, i.e, When `x= 2`, `(2+12x+-3x^(2))/(2x^(2) - 8x + 9) = (2+24-12)/(8-16+9) = 14`. |
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| 65. |
Find the quadratic equation whose roots are 2 times the roots of `x^(2) - 12x - 13 = 0`.A. `x^(2) - 24x - 52 = 0`B. `x^(2) - 24x - 26 = 0`C. `x^(2) - 14x - 15 = 0`D. None of these |
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Answer» Correct Answer - A `f(x) = x^(2) - 12x - 13 = 0` If the roots of `g(x)` are 2 times the roots of `f(x)` , then `g(x) = f(x/2) - 13 = 0` `rArr (x/2)^(2) - 12(x/2) - 13 = 0` `rArr (x^(2))/(4) - (12x)/(2) - 13 = 0` `rArr x^(2) - 24 x - 52 = 0`. |
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| 66. |
If `alpha, beta` are the roots of `ax^(2) + bx + c = 0`, then find the quadratic equation whose roots are `alpha + beta, alpha beta`.A. `ax^(2) + (ab-ac) x - c = 0`B. `ax^(2) + (b-c) x - bc = 0`C. `a^(2)x^(2) + (b-c) x - ac = 0`D. `a^(2)x^(2) + (ab-ac) x - bc = 0` |
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Answer» Correct Answer - D `alpha + beta` are the root of `ax^(2) + bx + c = 0` `rArr alpha + beta = (-b)/(a)` `alpha. Beta = (c )/(a)` Quadratic equation whose roots are `alpha + beta`, and `alphabeta` is `x^(2)- ((-b)/(c) + (c )/(a)) x + (-b)/(a)xx c/a= 0`. `x^(2) + ((b-c)/(a)) x - (bc)/(a^(2)) = 0` `a^(2)x^(2) + (ab-ac) x - bc = 0`. |
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| 67. |
Choose the minimum value of `(2x^(2) + 12x - 3)/(1 + 18x -3x^(2))` from the following options : (a) `(-15)/(29)`, (b) `15/28` , (c ) `(-15)/(28)`, (d) None of these |
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Answer» For the minimum value of `(2x^(2)- 12x + 3)/(1+18x+3x^(2)), 2x^(2) - 12x + 3` is minimum and `1+18x-3x^(2)` is maximum. The minimum value of `2x^(2) -12x +3` occurs at `x = (-b)/(2a) = (-(-12))/(2xx2) = 3`. The maximum value of `1+18x-3x^(2)` occurs at `x = (-b)/(2a) = (-18)/(2xx-3) = 3`. Minimum value of given expresson is `(2(3)^(2) - 12(3) + 3)/(1+18(3) - 3(3)^(2)) = (18-36+3)/(55-27) = (-15)/(28)`. |
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| 68. |
If `9x-3y + z = 0` , then the value of `y/(2x)+ sqrt((x^(2) - 4xz)/(4x^(2)))` (where x,y,z and constants).A. 9B. 2C. 3D. 6 |
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Answer» Correct Answer - C `9x-3y+z= 0` consider ` xa^(2) - ya + z = 0` a quadratic equation in a. Where x, y, z are constants Let `a = 3 rArr (3)^(2) x - 3 y + z = 0` is a quadratic equation in 3. `3 =(-(-y)+sqrt(y^(2)-4.x.z))/(2x)`. `rArr = y/(2x) + sqrt((y^(2) - 4xz)/(4x^(2)))` |
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| 69. |
If the roots of `3x^(2)- 12x +k = 0` are complex, then find the range of k.A. `k lt 22`B. `a lt - 10`C. `k gt 11`D. `k gt 12` |
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Answer» Correct Answer - D Given the roots of the given equation are complex `rArr b^(2) - 4 ac lt 0` `rArr (-12)^(2) - 4(3) k lt 0` `144 - 12 k lt 0` `- 12 k lt - 144` `12 k gt 144` `k lt 12`. |
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| 70. |
If f(x) = `(x^2 -2x + 4)/(x^2+2x+4), x in R` then range of function isA. `[1//3, 3]`B. `(1//3, 3)`C. (3, 3)D. `(-1//3, 3)` |
| Answer» Correct Answer - A | |
| 71. |
If `aA. imaginaryB. realC. one real and imaginaryD. equal and imaginary |
| Answer» Correct Answer - A | |
| 72. |
If `b gt a`, then the equation `(x-a)(x-b)-1=0`, hasA. both roots in [a, b]B. both roots in `(-oo, a)`C. roots in `(-oo, a)` and other in `(b, oo)`D. both roots in `(b, oo)` |
| Answer» Correct Answer - D | |
| 73. |
If a, b, c are real and `x^3-3b^2x+2c^3` is divisible by x -a and x - b, then (a) a =-b=-c (c) a = b = c or a =-2b-_ 2c (b) a = 2b = 2c (d) none of theseA. a = - b = - cB. a = 2b = 2cC. a = b = c or a = - 2b = -2cD. none of these |
| Answer» Correct Answer - C | |
| 74. |
The roots of the quadratic equation `(a + b-2c)x^2+ (2a-b-c) x + (a-2b + c) = 0` areA. `a+b+c and a-b+c`B. `(1)/(2) and a-2b + c`C. `a-2b + c and (1)/(a+b-c)`D. none of these |
| Answer» Correct Answer - D | |
| 75. |
Real roots of equation `x^2 + 5 |x| + 4 = 0` areA. `-1, -4`B. 1, 4C. `-4, 4`D. none of these |
| Answer» Correct Answer - D | |
| 76. |
The real roots of the equation `|x|^3-3x^2+3|x|-2=0` areA. 0, 2B. `+- 1`C. `+- 2`D. 1, 2 |
| Answer» Correct Answer - C | |
| 77. |
If a and b are distinct positive real numbers such that `a, a_(1), a_(2), a_(3), a_(4), a_(5), b` are in A.P. , `a, b_(1), b_(2), b_(3), b_(4), b_(5), b` are in G.P. and `a, c_(1), c_(2), c_(3), c_(4), c_(5), b` are in H.P., then the roots of `a_(3)x^(2)+b_(3)x+c_(3)=0` areA. real and distinctB. real and equalC. imaginaryD. none of these |
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Answer» Correct Answer - C Clearly, `a_(3), b_(3) and c_(3)` are A.M., G.M. and H.M. respectively of a and b. Therefore, `b_(3)^(2)=a_(3) c_(3)` Now, Discriminant `= b_(3)^(2) - 4a_(3) c_(3) = b_(3)^(2) - 4b_(3)^(2) lt 0" "[because a_(3) c_(3) = b_(3)^(2)]` Hence, roots of the given equation are imaginary. |
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| 78. |
If `b_(1),b_(2),b_(3),"….."b_(n)` are positive then the least value of `(b_(1) + b_(2) +b _(3) + "….." + b_(n)) ((1)/(b_(1)) + (1)/(b_(2)) + "….." +(1)/(b_(n)))` isA. `b_(1)b_(2)"…."b_(n)`B. `n^(2) + 1`C. `n(n+1)`D. `n^(2)` |
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Answer» Correct Answer - D (i) Take `b_(1)=b_(2)=b_(03)"……."b_(n) = k` and find the value . (ii) AM `(a_(1),a_(2),"……"a_(n)) ge HM (a_(1),a_(2),"…….",a_(n))` (iii) `(a_(1)+a_(2)+"….."+a_(n)).(n) ge (n)/((1)/(a_(1))+ (1)/(a_(2)) + (1)/(a_(2)))`. |
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| 79. |
If a,b,c are positive real numbers, then the number of real roots of the equation `ax^2+b|x|+c` isA. 2B. 4C. 0D. none of these |
| Answer» Correct Answer - C | |
| 80. |
Total number of integral values of a such that `x^2 + ax + a + 1 = 0` has integral roots is equal to : (A) one 45. (B) two (C) three (D) fourA. oneB. twoC. threeD. four |
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Answer» Correct Answer - B If `x^(2) + ax + a + 1 = 0` has integral roots. Then, ist discriminant must be a perfect square. `therefore" "a^(2) - 4a - 4 = lambda^(2), "where" lambda in Z`. `rArr" "(a-2)^(2)-lambda^(2) = 8` `rArr" "(a-2+lambda)(a-2-lambda)=8` `rArr" "a-2+lambda = 4 and a - 2 - lambda = 2` `or," "a-2+lambda =2 and a-2-lambda = 4` `or," "a-2+lambda=-4 and a-2-lambda = -2` `or," "a-2+lambda = -2 and a-2 - lambda = -4` `rArr" "(a=5, lambda=1)or, (a=5, lambda =-1)or, (a =-1, lambda = -1)or, (a = -1, lambda = 1)` Thus, there are only two integral values of a viz 5 and -1. |
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| 81. |
The number of positive integral roots of `x^(4) + x^(3) - 4 x^(2) + x + 1 = 0`, is |
| Answer» Correct Answer - C | |
| 82. |
If two equation `a_(1) x^(2) + b_(1) x + c_(1) = 0 and, a_(2) x^(2) + b_(2) x + c_(2) = 0` have a common root, then the value of `(a_(1) b_(2) - a_(2) b_(1)) (b_(1) c_(2) - b_(2) c_(1))`, isA. `-(a_(1) c_(2) - a_(2) c_(1))^(2)`B. `(a_(1) a_(2) - c_(1) c_(2))^(2)`C. `(a_(1) c_(1) - a_(2) c_(2))^(2)`D. `(a_(1) c_(2) - c_(1) a_(2))^(2)` |
| Answer» Correct Answer - D | |
| 83. |
solve : `16x^(4)- 28x^(2) - 8 = 0` |
| Answer» Correct Answer - `+-sqrt(2)` | |
| 84. |
If `.^(6)C_(k) + 2* .^(6)C_(k+1) + .^(6)C_(k+2) gt .^(8)C_(3)` then the quadratic equation whose roots are `alpha, beta and alpha^(k-1), beta^(k-1)` haveA. no common rootB. one common rootC. both common rootsD. imaginary roots |
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Answer» Correct Answer - C We have, `.^(6)C_(k) + 2* .^(6)C_(k+1)+.^(6)C_(k+2) gt .^(8)C_(3)` `rArr" "(.^(6)C_(k)+.^(6)C_(k+1))+(.^(6)C_(k+1)+ .^(6)C_(k+2)) gt .^(8)C_(3)` `rArr" ".^(7)C_(k+1)+.^(7)C_(k+2) gt .^(8)C_(3)` `rArr" ".^(8)C_(k+2) gt .^(8)C_(3) rArr k + 2 = 4" "[because .^(8)C_(4) gt .^(8)C_(3)]` `rArr" "k = 2` `therefore" "alpha^(k-1) = alpha and beta^(k-1) = beta` Thus, the quadratic equations having roots `alpha and beta and alpha^(k-1) and beta^(k-1)` are identical. Hence, they have both roots common. |
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| 85. |
The number of integral values of a for which `x^(2) - (a-1) x+3 = 0` has both roots positive and `x^(2) + 3x + 6 - a = 0` has both roots negative is |
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Answer» Correct Answer - B Let `f(x) = x^(2) - (a-1) x + 3 and g(x) = x^(2) + 3x + 6 - a`. If `f(x)=0` has both positive roots, then `f(0) gt 0, (a-1)^(2) - 12 gt 0 and, (a-1)/(2) gt 0` `rArr" "(a-1-2 sqrt(3))(a-1+2 sqrt(3)) gt 0 and a gt 1` `rArr" "a gt 1 + 2 sqrt(3)" "...(i)` If g(x) = 0 has both roots negative, then `g(0) gt 0, -(3)/(2) lt 0 and, 9-4 (6-a)gt 0` `rArr" "6-a gt 0 and 4a - 15 gt 0` `rArr" "a lt 6 and a gt (15)/(4) rArr (15)/(4) lt a lt 6" "...(ii)` From (i) and (ii), we have `1+2 sqrt(3) lt a lt 6 rArr a = 5" "[because "a is an integer"]` Thus, there is only one integral value of a. |
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| 86. |
the values of `a` for which `(a^2-1)x^2+2(a-1)x+2` is positive for all real `x` are.A. `a ge 1`B. `a le 1`C. `a gt - 3`D. `a le - 3 or a ge 1` |
| Answer» Correct Answer - D | |
| 87. |
If `alpha, beta, gamma` are the roots of the equation `x^(3) + x + 1 = 0`, then the value of `alpha^(3) + beta^(3) + gamma^(3)`, is |
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Answer» Correct Answer - C We have, `alpha + beta + gamma = 0 and alpha beta gamma = -1` Now, `alpha + beta + gamma = 0` `rArr" "alpha^(3) + beta^(3) + gamma^(3) = 3 alpha beta gamma` `therefore" "alpha^(3) + beta^(3) + gamma^(3) = 3 xx (-1) = -3" "[because alpha beta gamma = -1]` |
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| 88. |
The value of p for which the difference between the roots of the equation `x^(2) + px + 8 = 0` is 2, areA. `+-2`B. `+-4`C. `+-6`D. `+-8` |
| Answer» Correct Answer - C | |
| 89. |
If `alpha` and `beta` are the roots of the equation `x^2+sqrt(alpha)x+beta=0` then the values of `alpha` and `beta` are -A. `alpha = 1, beta = -1`B. `alpha = 1, beta = -2`C. `alpha = 2, beta = 1`D. `alpha = 2, beta = -2` |
| Answer» Correct Answer - B | |
| 90. |
If `alpha, beta` are roots of the equation `x^(2) + x + 1 = 0`, then the equation whose roots are `(alpha)/(beta) and (beta)/(alpha)`, isA. `x^(2) + x + 1 = 0`B. `x^(2) - x + 1 = 0`C. `x^(2) - x - 1 = 0`D. `x^(2) + x - 1 = 0` |
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Answer» Correct Answer - A We hve, `alpha = omega and beta = omega^(2)` `therefore" "(alpha)/(beta)=omega^(2) and (beta)/(alpha) = omega` Hence, the equation having `(alpha)/(beta)` i.e. `omega^(2) and (beta)/(alpha)` i.e. `omega` as its roots is `x^(2) + x + 1 = 0` |
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| 91. |
If `a, b and c` are in AP and if the equations `(b-c)x^2 + (c -a)x+(a-b)=0 and 2 (c +a)x^2+(b +c)x+(a-b)=0` have a common root, thenA. `a^(2), b^(2), c^(2)` are in A.P.B. `a^(2), c^(2), b^(2)` are in A.P.C. `a^(2), c^(2), b^(2)` are in G.P.D. none of these |
| Answer» Correct Answer - B | |
| 92. |
If p and q are the roots of `x^2 + px + q = 0`, then find p.A. p = 1B. p = 1 or 0C. p = -2D. p = -2 or 0 |
| Answer» Correct Answer - B | |
| 93. |
The value of `a` for which the equation `(1-a^2)x^2+2ax-1=0` has roots belonging to `(0,1)` isA. `a lt (1+ sqrt(5))/(2)`B. `a gt 2`C. `(1+sqrt(5))/(2) lt a lt 2`D. `a gt sqrt(2)` |
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Answer» Correct Answer - B Let `f(x) = (1 -a^(2)) x^(2) + 2ax - 1`. Then, f(x) = 0 will have its roots between 0 and 1, if (i) Discriminant `ge` 0 (ii) 0 and 1 are outside the roots of f(x) = 0 i.e. `(1-a^(2)) f(0) gt 0 and (1-a^(2)) f(1) gt 0` Now, (i) Discriminant `ge` 0 `rArr" "4a^(2) + 4(1-a^(2)) gt 0`, which is always true for all `a in R` (ii) `(1-a^(2)) f(0) gt 0` `rArr" "-(1-a^(2)) gt 0 rArr a^(2) - 1 gt 0 rArr a lt - 1 or, a gt 1" "...(i)` and, `(1-a^(2)) f(1) gt 0` `rArr" "(1-a^(2))(2a - a^(2)) gt 0` `rArr" "a(a-1)(a+1)(a-2) gt 0` `rArr" "a lt - 1 or, a gt 2 or, 0 lt a lt 1" "...(ii)` From (i) and (ii), we get `a lt -1 or, a gt 2`. |
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| 94. |
If `alpha, beta` are the roots of the equation `lambda(x^(2)-x)+x+5=0` and if `lambda_(1) and lambda_(2)` are two values of `lambda` obtained from `(alpha)/(beta)+(beta)/(alpha)=(4)/(5)`, then `(lambda_(1))/(lambda_(2)^(2))+(lambda_(2))/(lambda_(1)^(2))` equalsA. 4192B. 4144C. 4096D. 4048 |
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Answer» Correct Answer - D Since `alpha and beta` are the roots of the equation `lambda x^(2) + x(1-lambda)+5 = 0`. `therefore" "alpha + beta = (lambda -1)/(lambda) and alpha beta =(5)/(lambda)` Now, `(alpha)/(beta)+(beta)/(alpha)=(4)/(5)` `rArr" "5(alpha + beta)^(2) = 14 alpha beta` `5(lambda -1)^(2) = 70 lambda rArr lambda^(2) - 16 lambda + 1 = 0` Since `lambda_(1) and lambda_(2)` are roots of this equation. `therefore" "lambda_(1) + lambda_(2) = 16 and lambda_(1) lambda_(2) = 1` Hence, `(lambda_(1))/(lambda_(2)^(2))+(lambda_(2))/(lambda_(1)^(2))=(lambda_(1)^(3)+lambda_(2)^(3))/((lambda_(1)lambda_(2)))=((lambda_(1)+ lambda_(2))^(3)-3lambda_(1) lambda_(2)(lambda_(1)+lambda_(2)))/((lambda_(1) lambda_(2))^(2))=16^(3) - 3 xx 16 = 4048` |
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| 95. |
If he expression `[m x-1+(1//x)]`is non-negative for all positive real `x ,`then the minimum value of `m`must be`-1//2`b. `0`c. `1//4`d. `1//2`A. `-(1)/(2)`B. 0C. `(1)/(4)`D. `(1)/(2)` |
| Answer» Correct Answer - C | |
| 96. |
Let a, be the roots of the equation `x^2+x+1=0`. The equation whose roots are `alpha^19` and `beta^7` are:A. `x^(2) - x - 1 = 0`B. `x^(2) - x + 1 = 0`C. `x^(2) + x - 1 = 0`D. `x^(2) + x + 1 = 0` |
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Answer» Correct Answer - D correct answer is D |
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| 97. |
Let S denote the set of all real values of a for which the roots of the equation `x^(2) - 2ax + a^(2) - 1 = 0` lie between 5 and 10, then S equalsA. `(-1, 2)`B. (2, 9)C. (4, 9)D. (6, 9) |
| Answer» Correct Answer - D | |
| 98. |
The set of possible values of `lambda` for which `x^2-(lambda^2-5 lambda+5)x+(2 lambda^2-3lambda-4)=0` has roots whose sum and product are both less than 1 isA. `(-1, 5//2)`B. `(1, 4)`C. `[1, 5//2]`D. `(1, 5//2)` |
| Answer» Correct Answer - D | |
| 99. |
The set of values of p for which the roots of the equation `3x^2 +2x + p(p-1) =0` are of opposite signs is:A. `(-oo, 0)`B. (0, 1)C. `(1, oo)`D. `(0, oo)` |
| Answer» Correct Answer - B | |
| 100. |
The maximum number of real roots of the equation `x^(2n) -1 = 0`, isA. 2B. 3C. nD. 2n |
| Answer» Correct Answer - A | |