Practice question on quadratic equations for IITJEE , NDA And Air force page no. 3 (Set 3)

quadratic equations

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Question no. 31

If α and β are the distinct roots of x² -x+1 =0 ,then the value of α¹⁰¹ + β¹⁰⁷ is

looks_one 1

looks_two -1

looks_3 2

looks_4 3

Option looks_one 1

Solution:
x² -x+1 =0
β = -ω , α =-ω²
α¹⁰¹ + β¹⁰⁷ = (-1)(ω²)¹⁰¹ + (-1)(ω)¹⁰⁷
= -(ω)²⁰² -(ω)¹⁰⁷
= -(ω³)⁶⁷ω -(ω³)³⁵ω²
= -ω -ω² = 1

Question no. 32

the sum of all real valuesof x satisfying the equation (x² -5x+ 5)^{x2 +4x- 60}=1 is

looks_one 5

looks_two 3

looks_3 -4

looks_4 6

Option looks_two 3

Solution :
(x² -5x+ 5)^{x2 +4x- 60}=1
Case 1 : x² +4x- 60 = 0 ⇒ x= 6 ,-10
Case 2: x² -5x+ 5 = 1
x² -5x+ 4 = 0 ⇒ x =1 , 4
Case 3 : x² -5x+ 5 = -1
x² -5x+ 6 = 0 ⇒ x =2 , 3
At x= 2 , x² + 4x - 60 = -48 , it means x = 2 can be a solution
At x= 3 , x² + 4x - 60 = -45 , odd power will give -1 , therefore 3 can't be solution
Sum or roots = 6 -10 +1 + 4 + 2 = 3

Question no. 33

If α and β are the distinct roots of x² -6x-2 =0, a_n = αⁿ -βⁿ for n>=1, then the value of (a₁₀-2a₈)/2a₉

looks_one 3

looks_two -3

looks_3 6

looks_4 -6

Option looks_one 3

Solution :
x² -6x-2 =0
α² -6α-2 =0 ⇒ α² = 6α + 2
β² -6β-2 =0 ⇒ β² = 6β + 2

Question no. 34

If the equation x² +2x+3 =0 and ax² +bx+c =0 , a, b,c ​∈ R have a common root, then a:b:c

looks_one 1:2:3

looks_two 3:2:1

looks_3 1:3:2

looks_4 3:1:2

Option looks_one 1:2:3

Solution :
x² +2x+3 =0
D = 4 - 4*3 = -8
D < 0 . So both the roots are complex
therefore both the roots are same of both equation
therefore a= 1, b= 2, c= 3

Question no. 35

the equation e^sinx -e^-sinx-4=0 has

looks_one exactly four real roots

looks_two infinity number of real roots

looks_3 no real roots

looks_4 exactly one real root

Option looks_3 no real roots

Solution :
e^sinx -e^-sinx-4=0
e^sinx = Y
Y - 1/Y -4 = 0
Y² -1 - 4Y = 0
Y² - 4Y - 1 = 0 ⇒ Y = 2 ± √5
e^sinx = 2 + √5 = 4.23
e^sinx = 2- √5 < 0
but e^-1 ≤e^sinx ≤ e¹

Question no. 36

Sachin and rahul attemptd to solve quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots(4,3).Rahul made a mistake in writing coefficient of x to get roots(3,2), the correct roots of quadratic equation re:

looks_one -1,-3

looks_two 6,1

looks_3 4,3

looks_4 -6,-1

Option looks_two 6,1

Solution :
Roots when constant term is wrong : (3 ,4) , but coefficient of x is correct , So sum of roots(will be corrected) is 7
Similarly roots when coefficient of x is wrong : (3,2) , but constant is not wrong , product of roots will be 3*2 = 6
only 6,1 satisfy both the equation

Question no. 37

If α and β are the distinct roots of x² -x+1 =0, then α²⁰⁰⁹ +β²⁰⁰⁹

looks_one -2

looks_two 1

looks_3 2

looks_4 -1

Option looks_two 1

Solution :
x² -x+1 =0
α = -ω ,β =-ω²
α²⁰⁰⁹ +β²⁰⁰⁹ = (-ω)²⁰¹⁹ + (-ω²)²⁰¹⁹
= (-1)(ω³)⁶⁶⁹ω² + (-1)(ω³)¹³³⁹ω
= -(ω + ω²) = 1

Question no. 38

If α ≠ β and α²=5α -3 , β²=5β -3 , then the equation having the roots β/α and α/β is

looks_one 3x² -19x+ 3=0

looks_two 3x² -19x- 3=0

looks_3 3x² +19x- 3=0

looks_4 3x² -19x+ 3=0

Option looks_one 3x² -19x+ 3=0

Solution :
α²=5α -3 , β²=5β -3
α + β = 5 , α.β = 3
Equation with roots β/α and α/β
x² -(β/α + α/β)x + 1 = 0
α.βx² - ( α² + β² ) + α.β = 0
α² + β² = (α + β)² -2α.β = 25 -6 = 19
α.βx² - ( α² + β² ) + α.β = 3x² -19x + 3 =0

Question no. 39

Product of real roots of the equation t²x²+|x|+9=0

looks_one always positive

looks_two always negative

looks_3 does not exist

looks_4 none of these

Option looks_3 does not exist

Solution :
a > 0 , D = 1 - 36t² , it will be less than zero
real roots does not exist

Question no. 40

If p and q are the roots of equation x² +px+q=0 , then

looks_one p= 1, q= -2

looks_two p=0 , q= 1

looks_3 p= -2, q=0

looks_4 p= -2, q= 1

Option looks_one p= 1, q= -2

Solution :
-p = p+q (Sum of roots)
q = -2p
q =pq (product of roots)
q(1-p) = 0
p = 1 , q = -2

Question no. 41

Let Arithmetic mean and geometric mean of two numbers are 9 and 4 respectively. these two numbers are roots of equation

looks_one x² -18x +16=0

looks_two x² -18x -16=0

looks_3 x² +18x +16=0

looks_4 x² +18x -16=0

Option looks_one x² -18x +16=0

Solution :
(a+b)/2 = 9 ⇒ a+b = 18
√ab = 4 ⇒ ab = 16
x² - (a+b)x + ab = 0
x² - 18x + 16 = 0

Question no. 42

If (1-p) is the root of equation x² +px+(1-p)=0, then its roots are

looks_one 0,1

looks_two -1,2

looks_3 0,-1

looks_4 1,-1

Option looks_3 0,-1

Solution :
(1-p) is the root of equation x² +px+(1-p)=0
(1-p)² + p(1-p) + (1-p) = 0
(1-p)(1-p + p + 1) = 0 ⇒ p = 1
x² + x = 0 ⇒ x= 0 , -1

Question no. 43

If the root of equation x² -bx+c=0, be two consecutive integers, then b²-4c

looks_one -2

looks_two 3

looks_3 2

looks_4 1

Option looks_4 1

Solution :
x² -bx+c=0
a, a+1 are roots of given equation
b = 2a +1 (sum of roots)
c = a(a+1) (products of roots)
b² - 4c = (2a+1)² - 4a(a+1)
= 4a² + 1 + 4a - 4a² -4a = 1

Question no. 44

If both the roots of equation x² -2kx +k² +k-5=0 are less than 5, then k lies in the interval

looks_one (5,6]

looks_two (6,∞)

looks_3 (-∞,4)

looks_4 [4,5]

Option looks_3 (-∞,4)

Solution :
x² -2kx +k² +k-5=0
a > 0 therefore f(5) > 0
25 -2k(5) + k² + k - 5 > 0
k² -9k + 20 > 0
(k-4)(k-5) > 0
k =(-∞,4)U(5,+∞)

Question no. 45

If the root of equation x² +px+q=0 are tan30^o and tan15^o respectively, then the value of 2+p -q

looks_one 1

looks_two 2

looks_3 3

looks_4 0

Option looks_one 1

Solution:
p = -(tan30^o + tan15^o)
q = tan30^o.tan15^o
tan45^o = (tan30^o + tan15^o)/(1- tan30^o.tan15^o )
1 = -p/(1-q)
1-q = -p ⇒ p-q = -1
2 +p-q = 2 -1 = 1

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