Q: For x ∈ R , the number of real roots of the equation 3 x 2 - 4 x 2 - 1 + x - 1 = 0 is

x 2 - 1 = x 2 - 1 , x 2 ≥ 1 - x 2 - 1 x 2 < 1 Case 1 x 2 - 1 < 0 ⇒ x ∈ ( - 1 , 1 ) 3 x 2 + 4 x 2 - 1 + x - 1 = 0 7 x 2 + x - 5 = 0 x = - 1 ± 1 + 140 2 × 7 = - 1 + 141 14 , - 1 - 141 14 We know, 141 ≈ 11 . 5 So, both roots lie in the interval - 1 , 1 Case 2 x 2 - 1 ≥ 0 ⇒ x ∈ ( - ∞ , - 1 ] ∪ [ 1 , ∞ ) 3 x

Q: Suppose a , b denote the distinct real roots of the quadratic polynomial x 2 + 20 x − 2020 and suppose c , d denote the distinct complex roots of the quadratic polynomial x 2 - 20

Finding the sum and product of roots of the two given quadratic equations i.e. a + b = - 20 & a b = - 2020 . . . . . . . . . . . i c + d = 20 & c d = 2020 . . . . . . . . . . . i i Now, a c a − c + a d a − d + b c b − c + b d b − d = a 2 c + d − a c 2 + d 2 + b 2 c + d − b c 2 + d 2 = a 2 + b 2 c + d − a + b c 2 + d 2 = a + b 2 − 2 a b c + d − a + b c + d 2 − 2 c d = - 20 2 &

Q: Let α and β be the roots of x 2 - x - 1 = 0 , with α > β . For all positive integers n , define a n = α n - β n α - β , n ≥ 1 b 1 = 1 and b n

Given, α and β are roots of x 2 - x - 1 = 0 a k + 2 - a k = α k + 2 - β k + 2 - α k - β k α - β = α k + 2 - α k - β k + 2 - β k α - β = α k α 2 - 1 - β k β 2 - 1 α - β = a k + 1 ⇒ a k + 2 - a k = a k + 1 ⇒ a k + 2 - a k + 1 = a k ∑ a k k = 1 n = a k + 2 - a 2 = a k + 2 - α 2 - β 2 α - β = a k + 2 - α + β =

Question 1

Match each entry in List-I to the correct entry in List-II and choose the correct option. List-I List-II (P) If and are the distinct roots of the equation x^2 + x + 1 = 0, then the quadratic equation

Accepted Answer

For (P): Since and are roots of x^2 + x + 1 = 0, we have ^2 + + 1 = 0 + 1 = - ^2 and ^3 = 1. 1 ( +1)^ 2026 = 1 (- ^2)^ 2026 = 1 ^ 4052 Since 4052 = 3 1350 + 2, ^ 4052 = ( ^3)^ 1350 ^2 = ^2. Thus, 1 ^2 = ^3 ^2 = . Similarly, 1 ( +1)^ 2026 = . The quadratic equation with roots and is the original equation x^2 + x + 1 = 0. So, (P) (1). For (Q): 1 ( +1)^ 2027 = 1 (- ^2)^ 2027 = - 1 ^ 4054 Since 4054 = 3 1351 + 1, ^ 4054 = . Thus, - 1 = - . Similarly, the other root is - . Sum of roots = - - = -( + )

Question 2

Let a, b, c be positive integers in arithmetic progression such that the equation ax^2 + bx + c = 0 has only integer solutions. Then which of the following statements is (are) TRUE ?

Accepted Answer

Given a, b, c are in arithmetic progression, we have 2b = a + c. Let the integer roots of the equation ax^2 + bx + c = 0 be and . Sum of the roots is + = - b a and product of the roots is = c a . Dividing the arithmetic progression condition by a, we get: 2b a = 1 + c a Substituting the sum and product of the roots: -2( + ) = 1 + + 2 + 2 + 1 = 0 Adding 3 to both sides to factorize: + 2 + 2 + 4 = 3 ( + 2)( + 2) = 3 Since and are integers, ( + 2) and ( + 2) must be integer factors of 3. The possib

Question 3

Let f(x)=x^4+a x^3+b x^2+c be a polynomial with real coefficients such that f(1)=-9. Suppose that i 3 is a root of the equation 4 x^3+3 a x^2+2 b x=0, where i= -1 . If _1, _2, _3, and _4 are all the r

Accepted Answer

f(1)=1+a+b+c=-9 a+b+c=-10.....(1) 4 x^3+3 a x^2+2 b x=0 roots are 3 i,- 3 i, 0 4 x^2+3 a x+2 b=0 < ^ 3i _ - 3i a =0 ~\&~ 2 ~b 4 =( 3 i )(- 3 i ) b =6 use a , b in (1) c =-16 array ll & f ( x )= x ^4+6 x ^2-16=0 \ & ( x ^2+8 ) ( x ^2-2 )=0 \ & x = 8 i , 2 array | _1 |^2+ | _2 |^2+ | _3 |^2+ | _4 |^2=20

Question 4

For x &#8712; R , the number of real roots of the equation  3 x 2 - 4 x 2 - 1 + x - 1 = 0  is

Accepted Answer

x 2 - 1 = x 2 - 1 , &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; &#160; x 2 &#8805; 1 - x 2 - 1 &#160; &#160; &#160; &#160; &#160; x 2 &#60; 1 Case 1 x 2 - 1 &#60; 0 &#8658; x &#8712; ( - 1 , 1 ) 3 x 2 + 4 x 2 - 1 + x - 1 = 0 7 x 2 + x - 5 = 0 x = - 1 &#177; 1 + 140 2 &#215; 7 = - 1 + 141 14 , - 1 - 141 14 We know,  141 &#8776; 11 . 5 So, both roots lie in the interval  - 1 , 1 Case 2 x 2 - 1 &#8805; 0 &#8658; x &#8712; ( - &#8734; , - 1 ] &#8746; [ 1 , &#8734; ) 3 x

Question 5

Suppose a , &#160; b denote the distinct real roots of the quadratic polynomial x 2 + 20 x &#8722; 2020 and suppose c , &#160; d denote the distinct complex roots of the quadratic polynomial x 2 - 20

Accepted Answer

Finding the sum and product of roots of the two given quadratic equations i.e. a + b = - 20 &#160; &#38; &#160; a b = - 2020 &#160; . . . . . . . . . . . i c + d = 20 &#160; &#38; &#160; c d = 2020 &#160; . . . . . . . . . . . i i Now,   a c a &#8722; c + a d a &#8722; d + b c b &#8722; c + b d b &#8722; d = a 2 c + d &#8722; a c 2 + d 2 + b 2 c + d &#8722; b c 2 + d 2 = a 2 + b 2 c + d &#8722; a + b c 2 + d 2 = a + b 2 &#8722; 2 a b c + d &#8722; a + b c + d 2 &#8722; 2 c d = - 20 2 &

Question 6

Let &#945; and &#946; be the roots of x 2 - x - 1 = 0 , with &#945; &#62; &#946; . For all positive integers n , define a n = &#945; n - &#946; n &#945; - &#946; , &#8201; n &#8805; 1 b 1 = 1 and b n

Accepted Answer

Given,  &#945;  and  &#946;  are roots of x 2 - x - 1 = &#160; 0 a k + 2 - a k = &#945; k + 2 - &#946; k + 2 - &#945; k - &#946; k &#945; - &#946; = &#945; k + 2 - &#945; k - &#946; k + 2 - &#946; k &#945; - &#946; = &#945; k &#945; 2 - 1 - &#946; k &#946; 2 - 1 &#945; - &#946; = a k + 1 &#8658; a k + 2 - a k = a k + 1 &#160; &#8658; a k + 2 - a k + 1 = a k &#8721; a k k = 1 n = a k + 2 - a 2 &#160; = a k + 2 - &#945; 2 - &#946; 2 &#945; - &#946; = a k + 2 - &#945; + &#946; =

Quadratic Equation — Previous Year Questions & Solutions