Question 1

Consider the quadratic equation (n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0, n R . Let be the minimum value of the product of its roots and be the maximum value of the sum of its roots. Then the sum

Accepted Answer

The given quadratic equation is (n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0. Let k = n^2 - 2n + 2 = (n-1)^2 + 1. Since (n-1)^2 0 for all n R , the minimum value of k is 1 at n=1. The product of the roots is given by P = (n^2 - 2n + 2)^2 n^2 - 2n + 2 = n^2 - 2n + 2 = k. The minimum value of the product of the roots is = (k) = 1. The sum of the roots is given by S = 3 n^2 - 2n + 2 = 3 k . The maximum value of the sum of the roots occurs when k is minimum. Thus, = ( 3 k ) = 3 1 = 3. We are given

Question 2

Let one root of the quadratic equation in x: (k^2 - 15k + 27)x^2 + 9(k-1)x + 18 = 0 be twice the other. Then the length of the latus rectum of the parabola y^2 = 6kx is equal to:

Accepted Answer

Let the roots of the given quadratic equation be and 2 . Sum of the roots: + 2 = -9(k-1) k^2 - 15k + 27 3 = -9(k-1) k^2 - 15k + 27 = -3(k-1) k^2 - 15k + 27 Product of the roots: 2 = 18 k^2 - 15k + 27 2 ^2 = 18 k^2 - 15k + 27 ^2 = 9 k^2 - 15k + 27 Substituting the value of from the sum into the product equation: ( -3(k-1) k^2 - 15k + 27 )^2 = 9 k^2 - 15k + 27 9(k-1)^2 (k^2 - 15k + 27)^2 = 9 k^2 - 15k + 27 Since k^2 - 15k + 27 0, we can simplify to: (k-1)^2 = k^2 - 15k + 27 k^2 - 2k + 1 = k^2 - 15

Question 3

Let e_1 and e_2 be two distinct roots of the equation x^2 - ax + 2 = 0. Let the sets \ a R : e_1 and e_2 are the eccentricities of hyperbolas \ = ( , ), and \ a R : e_1 and e_2 are the eccentricities

Accepted Answer

The given quadratic equation is x^2 - ax + 2 = 0 with roots e_1 and e_2. The sum of the roots is e_1 + e_2 = a and the product is e_1 e_2 = 2. For the roots to be real and distinct, the discriminant must be positive: = a^2 - 8 > 0 a (- , -2 2 ) (2 2 , ) Since e_1 and e_2 represent eccentricities, they must be positive. Given e_1 e_2 = 2 > 0, both roots have the same sign. For them to be positive, their sum must be positive, so a > 0. Thus, a > 2 2 . For e_1 and e_2 to be the eccentricities of hy

Question 4

Let , be the roots of the equation x^2 - x + p = 0 and , be the roots the equation x^2 - 4x + q = 0; p, q Z . If , , , are in G.P., then |p + q| equals :

Accepted Answer

Let = a, = ar, = ar^2, = ar^3 be the terms of the G.P. From the given quadratic equations, the sum of the roots are: + = 1 a(1 + r) = 1 + = 4 ar^2(1 + r) = 4 Dividing the second equation by the first equation: ar^2(1 + r) a(1 + r) = 4 1 r^2 = 4 r = 2 or r = -2 If r = 2, then a(1 + 2) = 1 a = 1 3 . The product of the roots of the first equation is p = = a^2 r = ( 1 3 )^2 2 = 2 9 . Since p Z , r = 2 is rejected. If r = -2, then a(1 - 2) = 1 a = -1. The product of the roots of the first equation is

Question 5

Let a, b C . Let , be the roots of the equation x^2 + ax + b = 0. If - = 11 and ^2 - ^2 = 3i 11 , then ( ^3 - ^3)^2 is equal to:

Accepted Answer

Given - = 11 and ^2 - ^2 = 3i 11 + = ^2 - ^2 - = 3i 11 11 = 3i Using the identity ( - )^2 = ( + )^2 - 4 ( 11 )^2 = (3i)^2 - 4 11 = -9 - 4 4 = -20 = -5 Now, ^3 - ^3 = ( - )( ^2 + + ^2) ^3 - ^3 = ( - )(( + )^2 - ) ^3 - ^3 = 11 ((3i)^2 - (-5)) ^3 - ^3 = 11 (-9 + 5) = -4 11 Squaring both sides, we get: ( ^3 - ^3)^2 = (-4 11 )^2 = 16 11 = 176 Answer: 176

Question 6

Let A, B, where A, B (- 2 , 2 ), be the roots of the quadratic equation x^2 - 2x - 5 = 0. Then 20 ^2 ( A+B 2 ) is equal to:

Accepted Answer

Given that A and B are the roots of the quadratic equation x^2 - 2x - 5 = 0. Sum of the roots: A + B = 2 Product of the roots: A B = -5 Using the tangent addition formula: (A+B) = A + B 1 - A B = 2 1 - (-5) = 2 6 = 1 3 Since A, B (- 2 , 2 ) and their product A B = -5 Since (A+B) = 1 3 > 0, the angle A+B must lie in the first quadrant, i.e., A+B (0, 2 ). Thus, (A+B) is positive. (A+B) = 1 1 + ^2(A+B) = 1 1 + ( 1 3 )^2 = 3 10 We need to find the value of 20 ^2 ( A+B 2 ). Using the half-angle ident

Question 7

If =1 and =1+i 2 , where i= -1 are two roots of the equation x^3+ax^2+bx+c=0, a,b,c R , then _ -1 ^ 1 (x^3+ax^2+bx+c)dx is equal to:

Accepted Answer

Since a, b, c R , the complex roots of the equation x^3+ax^2+bx+c=0 must occur in conjugate pairs. Given roots are = 1 and = 1+i 2 . The third root must be = 1-i 2 . The polynomial is given by: x^3+ax^2+bx+c = (x-1)(x-(1+i 2 ))(x-(1-i 2 )) = (x-1)((x-1)^2 - (i 2 )^2) = (x-1)(x^2 - 2x + 1 + 2) = (x-1)(x^2 - 2x + 3) = x^3 - 3x^2 + 5x - 3 We need to evaluate the integral: I = _ -1 ^ 1 (x^3 - 3x^2 + 5x - 3) dx Using the property of definite integrals _ -a ^ a f(x) dx = 0 for odd functions and 2 _ 0

Question 8

If the quadratic equation ( +2)x^2-3 x+4 =0, -2, has two positive roots, then the number of possible integral values of is:

Accepted Answer

For the quadratic equation ( +2)x^2-3 x+4 =0 to have two positive roots, the following three conditions must be satisfied: 1. Discriminant D 0 (-3 )^2 - 4( +2)(4 ) 0 9 ^2 - 16 ^2 - 32 0 -7 ^2 - 32 0 7 ^2 + 32 0 (7 + 32) 0 [- 32 7 , 0 ] 2. Sum of roots > 0 + = 3 +2 > 0 (- , -2) (0, ) 3. Product of roots > 0 = 4 +2 > 0 (- , -2) (0, ) Taking the intersection of all three conditions, we get: [- 32 7 , -2 ) Since - 32 7 -4.57, the possible integral values of in this interval are -4 and -3. Thus, ther

Question 9

Let , be the roots of the equation x^2 - 3x + r = 0, and 2 , 2 be the roots of the equation x^2 + 3x + r = 0. If the roots of the equation x^2 + 6x = m are 2 + + 2r and - 2 - r 2 , then m is equal to:

Accepted Answer

From the first equation x^2 - 3x + r = 0, the sum and product of the roots are: + = 3 = r From the second equation x^2 + 3x + r = 0, the sum and product of the roots are: 2 + 2 = -3 ( 2 )(2 ) = r = r We have a system of linear equations in terms of and : + = 3 + 4 = -6 Subtracting the first equation from the second gives: 3 = -9 = -3 Substituting = -3 into + = 3 gives: = 6 Now, we can find r: r = = 6 (-3) = -18 The roots of the third equation x^2 + 6x - m = 0 are given as 2 + + 2r and - 2 - r 2

Question 10

Let a, b, c \ 1, 2, 3, 4\ . If the probability, that ax^2 + 2 2 \,bx + c > 0 for all x R , is m n , (m, n) = 1, then m + n is equal to _______.

Accepted Answer

For the quadratic expression ax^2 + 2 2 bx + c > 0 for all x R , the leading coefficient must be positive and the discriminant must be negative. Since a \ 1, 2, 3, 4\ , the condition a > 0 is always satisfied. The discriminant condition is: D (2 2 b)^2 - 4ac 8b^2 - 4ac 2b^2 The total number of possible triplets (a, b, c) is 4 4 4 = 64. Now, we find the number of favorable outcomes by checking the possible values of b \ 1, 2, 3, 4\ : Case 1: b = 1 We need 2(1)^2 2. The total number of pairs (a, c

Question 11

Let S = \ x^ 3 +a x^ 2 +b x+c: a, b, c ~N . and .a, b, c 20 \ be a set of polynomials. Then the number of polynomials in S, which are divisible by x^ 2 +2, is

Accepted Answer

If x^2 + 2 divides x^3 + ax^2 + bx + c, write x^3 + ax^2 + bx + c = (x^2 + 2)(x + k). Expanding: x^3 + kx^2 + 2x + 2k. Comparing coefficients: a = k, b = 2, c = 2a. Constraints: a N , a 20, and c = 2a 20 a 10. So a \ 1, 2, 3, ..., 10\ , giving 10 polynomials.

Question 12

If , , where < , are the roots of the equation x^ 2 -( +3) x+3=0 such that 1 - 1 = 1 3 , then the sum of all possible values of is

Accepted Answer

+ = +3 , = 3 . 1 - 1 = - = 1 3 , so - = 3 = 1 . ( - )^2 = ( + )^2 - 4 : 1 ^2 = ( +3)^2 ^2 - 12 1 = ( +3)^2 - 12 = ^2 - 6 + 9 ^2 - 6 + 8 = 0 = 2 or = 4. Both values satisfy < (verified). Sum = 2 + 4 = 6.

Question 13

The smallest positive integral value of a, for which all the roots of x^ 4 -a x^ 2 +9=0 are real and distinct, is equal to

Accepted Answer

Let t = x^2, so the equation becomes t^2 - at + 9 = 0. For all four roots of the original equation to be real and distinct, both roots of this quadratic must be positive and distinct. Discriminant > 0: a^2 - 36 > 0 |a| > 6. Since a > 0, we need a > 6. Product of roots = 9 > 0 and sum of roots = a > 0, so both roots are positive. The smallest positive integer satisfying a > 6 is a = 7.

Question 14

The sum of all the real solutions of the equation _ (x+3) (6 x^ 2 +28 x+30 )=5-2 _ (6 x+10) (x^ 2 +6 x+9 ) is equal to

Accepted Answer

Note 6x^2+28x+30 = 2(3x+5)(x+3), x^2+6x+9 = (x+3)^2, 6x+10 = 2(3x+5). Let a = _ (x+3) [2(3x+5)]. Then LHS = a + 1 and _ 2(3x+5) (x+3)^2 = 2 a . Equation becomes a + 1 = 5 - 4 a , i.e., a^2 - 4a + 4 = 0 a = 2. So _ (x+3) [2(3x+5)] = 2 (x+3)^2 = 6x+10 x^2 = 1 x = 1 or x = -1. Both satisfy domain conditions (bases > 0, 1; arguments > 0). Sum = 1 + (-1) = 0.

Question 15

A building construction work can be completed by two masons A and B together in 22.5 days. Mason A alone can complete the construction work in 24 days less than mason B alone. Then mason A alone will

Accepted Answer

Let A take a days and B take a + 24 days. Together: 1 a + 1 a+24 = 1 22.5 = 2 45 . 2a+24 a(a+24) = 2 45 45(2a+24) = 2a(a+24) 90a + 1080 = 2a^2 + 48a a^2 - 21a - 540 = 0 (a-36)(a+15) = 0 a = 36. Mason A alone completes the work in 36 days.

Question 16

If and ( < ) are the roots of the equation (-2+ 3 )(| x -3|)+(x-6 x )+(9-2 3 )=0, x 0, then + is equal to :

Accepted Answer

Let t = x 0. The equation becomes (-2+ 3 )|t-3| + (t^2 - 6t) + (9 - 2 3 ) = 0. Since t^2 - 6t + 9 = (t-3)^2, we get (t-3)^2 + ( 3 -2)|t-3| - 2 3 = 0. Let u = |t-3| 0: u^2 + ( 3 -2)u - 2 3 = 0. Discriminant = ( 3 -2)^2 + 8 3 = 7 + 4 3 = (2+ 3 )^2. u = (2- 3 ) (2+ 3 ) 2 , giving u = 2 or u = - 3 (rejected). |t-3| = 2 t = 5 or t = 1, so x = 25 or x = 1. = 1, = 25. / + = 25 + 25 = 5 + 5 = 10.

Question 17

Let , be the roots of the quadratic equation 12 x^ 2 -20 x+3 =0, Z . If 1 2 | - | 3 2 , then the sum of all possible values of is :

Accepted Answer

For 12x^2-20x+3 =0: + = 5 3 , = 4 . | - |= 25-9 3 Given 1 2 25-9 3 3 2 9 4 25-9 81 4 From left: 91 36 2.53 From right: 19 36 0.53 Since Z : \ 1,2\ . Sum =3.

Question 18

Let and be the roots of the equation x^ 2 +2 a x+(3 a+10)=0 such that <1< . Then the set of all possible values of a is :

Accepted Answer

Let f(x) = x^2 + 2ax + (3a + 10). For f(1) = 1 + 2a + 3a + 10 = 5a + 11 a When f(1) a (- , - 11 5 ).

Question 19

The sum of all the roots of the equation (x-1)^ 2 -5|x-1|+6=0, is :

Accepted Answer

Let t = |x-1| where t 0. Equation becomes t^2 - 5t + 6 = 0. (t-2)(t-3) = 0 gives t = 2 or t = 3. For |x-1| = 2: x = 3 or x = -1. For |x-1| = 3: x = 4 or x = -2. Sum of roots = 3 + (-1) + 4 + (-2) = 4.

Quadratic Equation — Previous Year Questions & Solutions