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JEE • MATHEMATICS • THEORY OF EQUATION

Theory of equation practice · Mixed difficulty

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61 Medium
61 of 120
Analyzing +4 | -1
If f(x)=x3−3x+1f(x)=x^3-3 x+1, then the number of distinct real roots of the equation f(f(x))=0f(f(x))=0 is
View answer and explanation
Correct option: C
7
The correct answer is C (7) because the equation f(f(x)) = 0 has 7 distinct real roots, determined by analyzing the behavior of f(x) and its critical points.
62 Hard
62 of 120
Application +4 | -1
The coefficient of xx in polynomial (X−α)(X−β)(X−γ)(X-\alpha)(X-\beta)(X-\gamma) where α=cos⁡75∘,β=cos⁡45∘\alpha=\cos 75^{\circ}, \beta=\cos 45^{\circ} and γ=cos⁡\gamma=\cos 165∘165^{\circ}
View answer and explanation
Correct option: D
−34\frac{-3}{4}
Option D (-3/4) is correct because it is the coefficient of x in the polynomial after substituting the given values and simplifying.
63 Easy
63 of 120
Analyzing +4 | -1
Number of integer roots of the equation (x+2)(x+3)(x+8)(x+12)=4x2\left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 8} \right)\left( {x + 12} \right) = 4{x^2} is
View answer and explanation
Correct option: C
2
Option C (2) is correct because the equation has two integer roots which are -4 and -6, found by solving the transformed quadratic equations.
64 Medium
64 of 120
Analyzing +4 | -1
x1=37+97,x2=57+77{x}_1=\sqrt[7]{3}+\sqrt[7]{9}, {x}_2=\sqrt[7]{5}+\sqrt[7]{7}, then
View answer and explanation
Correct option: C
x1<x2x_1 < x_2
Option C (x1<x2x_1 < x_2) is correct because by comparing the values and using properties of exponents and inequalities, it can be shown that the inequality holds.
65 Easy
65 of 120
Application +4 | -1
If α,  β,  γ\alpha ,\;\beta ,\;\gamma are the roots of the equation x3−6x2+11x+6=0{x^3} - 6{x^2} + 11x + 6 = 0, then ∑α2β+∑αβ2\sum {\alpha ^2}\beta + \sum \alpha {\beta ^2} is equal to
View answer and explanation
Correct option: B
84
Option B (84) is correct because it is the value obtained by using Vieta's formulas to express the sum of the given expression in terms of the coefficients of the polynomial.
66 Easy
66 of 120
Analyzing +4 | -1
The roots of the equation x4−2x3+x=380{x^4} - 2{x^3} + x = 380 are
View answer and explanation
Correct option: A
5,  −4,  1±5−325,\; - 4,\;\frac{{1 \pm 5\sqrt { - 3} }}{2}
Option A (One solution) is correct because the given conditions represent the intersection of perpendicular bisectors in the complex plane, which results in a single point.
67 Easy
67 of 120
Application +4 | -1
If the sum of the squares of the reciprocals of the roots α\alpha and β\beta of the equation 3x2+λx−1=03{x^2} + \lambda x - 1 = 0 is 15, then 6(α3+β3)26{\left( {{\alpha ^3} + {\beta ^3}} \right)^2} is equal to:
View answer and explanation
Correct option: B
24
Option B (24) is correct because after finding the value of λ\lambda from the given condition, we can calculate the value of 6(α3+β3)26{(\alpha^3 + \beta^3)}^2 which equals 24.
68 Easy
68 of 120
Application +4 | -1
If (x2−3x+2)\left( {{x^2} - 3x + 2} \right) is a factor of x4−px2+q=0{x^4} - p{x^2} + q = 0, then the values of pp and qq are
View answer and explanation
Correct option: B
5,45,\mathrm{ }4
Option B (5,  45,\;4) is correct because if (x2−3x+2)(x^2 - 3x + 2) is a factor, then x=1 and x=2 are roots, which can be substituted into the equation to solve for p and q.
69 Easy
69 of 120
Application +4 | -1
If (a2−1)x2+(a−1)x+a2−4a+3=0\left(a^2-1\right) x^2+(a-1) x+a^2-4 a+3=0 is an identity in xx, then the value of aa is
View answer and explanation
Correct option: A
1
Option A (1) is correct because for the equation to be an identity in x, all coefficients must be zero, and a=1 satisfies all three conditions.
70 Hard
70 of 120
Analyzing +4 | -1
If α,β,γ\alpha ,\beta ,\gamma are the roots of the equation x3−7x+7=0{x^3} - 7x + 7 = 0 then, 1α4+1β4+1γ4\frac{1}{{{\alpha ^4}}} + \frac{1}{{{\beta ^4}}} + \frac{1}{{{\gamma ^4}}} is
View answer and explanation
Correct option: B
37\frac{3}{7}
Option B is correct because the value of the expression 1α4+1β4+1γ4\frac{1}{{{\alpha ^4}}} + \frac{1}{{{\beta ^4}}} + \frac{1}{{{\gamma ^4}}} simplifies to 3/7 using Vieta's formulas and algebraic manipulation.
71 Medium
71 of 120
Application +4 | -1
If α,β,γ\alpha ,\beta ,\gamma are the roots of the equation x3+x+1=0{x^3} + x + 1 = 0, then the value of α3+β3+γ3{\alpha ^3} + {\beta ^3} + {\gamma ^3} is
View answer and explanation
Correct option: C
−3- 3
Option C (−3-3) is correct because it is the value of α3+β3+γ3{\alpha ^3} + {\beta ^3} + {\gamma ^3} calculated using Vieta's formulas and the given cubic equation.
72 Hard
72 of 120
Analyzing +4 | -1
If x1,  x2,  x3,  x4{x_1},\;{x_2},\;{x_3},\;{x_4} are roots of the equation x4−x3sin⁡2β+x2cos⁡2β−xcos  β−sin⁡β=0{x^4} - {x^3}\sin 2\beta + {x^2}\cos 2\beta - xcos\;\beta - \sin \beta = 0, then tan⁡−1x1+tan⁡−1x2+tan⁡−1x3+tan⁡−1x4{\tan ^{ - 1}}{x_1} + {\tan ^{ - 1}}{x_2} + {\tan ^{ - 1}}{x_3} + {\tan ^{ - 1}}{x_4} is equal to
View answer and explanation
Correct option: B
π2−β\frac{\pi }{2} - \beta
Option B (π2−β\frac{\pi }{2} - \beta) is correct because the sum of the arctangents of the roots simplifies to this expression using trigonometric identities and Vieta's formulas.
73 Easy
73 of 120
Application +4 | -1
If f(x)=2x4−13x2+ax+bf\left( x \right) = 2{x^4} - 13{x^2} + ax + b is divisible byx2−3x+2{x^2} - 3x + 2, then (a,b) is equal to
View answer and explanation
Correct option: C
(9,2)\left( {9,\mathrm{ }2} \right)
Option C ((9, 2)) is correct because when f(x)f(x) is divisible by x2−3x+2x^2 - 3x + 2, then f(1)=0f(1) = 0 and f(2)=0f(2) = 0, which leads to a system of equations that solves to a=9a = 9 and b=2b = 2.
74 Medium
74 of 120
Application +4 | -1
The cubic equation whose roots are thrice to each of the roots of x3+2x2−4x+1=0{x^3} + 2{x^2} - 4x + 1 = 0 is
View answer and explanation
Correct option: D
x3+6x2−36x+27=0{x^3} + 6{x^2} - 36x + 27 = 0
Option D (x3+6x2−36x+27=0{x^3} + 6{x^2} - 36x + 27 = 0) is correct because the new equation is formed by transforming the roots of the original cubic equation.
75 Hard
75 of 120
Analyzing +4 | -1
x8−x5−1x+1x4>0,{x^8} - {x^5} - \frac{1}{x} + \frac{1}{{{x^4}}} > 0, is satisfied for
View answer and explanation
Correct option: C
All real numbers except zero
Option D (All real numbers except zero) is correct because the expression is greater than 0 for all non-zero real numbers.

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76 Easy
76 of 120
Application +4 | -1
If x+1x + 1 is a factor of x4+(p−3)x3−(3p−5)x2+(2p−9)x+6,{x^4} + \left( {p - 3} \right){x^3} - \left( {3p - 5} \right){x^2} + \left( {2p - 9} \right)x + 6, then the value of pp is
View answer and explanation
Correct option: C
4
Option C (4) is correct because substituting x = -1 into the polynomial and setting the result to zero allows us to solve for p, which equals 4.
77 Easy
77 of 120
Application +4 | -1
The roots of the equation x3−3x−2=0{x^3} - 3x - 2 = 0 are
View answer and explanation
Correct option: A
−1,−1,2- 1, - 1,\mathrm{ }2
Option A ( - 1, - 1,{ }2) is correct because these are the roots of the given cubic equation.
78 Medium
78 of 120
Application +4 | -1
If α,β,γ\alpha, \beta, \gamma are the roots of the equation x3−3x2+7x−5=0x^3-3 x^2+7 x-5=0, then value of ∑α2β\sum \alpha^2 \beta is
View answer and explanation
Correct option: B
6
Option B (6) is correct because by using Vieta's formulas and manipulating the sum of the roots, the value of the expression can be found.
79 Easy
79 of 120
Analyzing +4 | -1
If the equations x2+2x+3=0x^2+2 x+3=0 and ax2+bx+c=0a x^2+b x+c=0, a,b,c∈Ra, b, c \in R have a common root, then a:b:ca: b: c is
View answer and explanation
Correct option: D
1:2:31: 2: 3
Option D (1:2:31: 2: 3) is correct because the two equations have a common root, and since the roots of the first equation are imaginary, both roots must be common, making the equations proportional.
80 Easy
80 of 120
Application +4 | -1
Let ff be monic cubic polynomial such that f(1)=14−1,f(2)=24−2f(1)=1^4-1, f(2)=2^4-2 and f(3)=34−3f(3)=3^4-3. If f(4)=Nf(4)=N, then find the number of prime factors of NN.
View answer and explanation
Correct option: A
3
Option A (3) is correct because after constructing the polynomial f(x) and evaluating f(4), the number of prime factors of N is 3.
81 Easy
81 of 120
Application +4 | -1
If one root of x2+px+q=0x^2+p x+q=0 is 5−3i5-3 i, then the real values of pp and qq are
View answer and explanation
Correct option: B
p=−10,q=34p=-10, q=34
Option B (p=−10,q=34p=-10, q=34) is correct because if one root is a complex number, the other root is its conjugate, and we can use the sum and product of roots to find p and q.
82 Medium
82 of 120
Analyzing +4 | -1
Difference between the corresponding roots of x2+ax+b=0x^{2}+a x+b=0 and x2+bx+a=0x^{2}+b x+a=0 is same and a≠ba \neq b, then
View answer and explanation
Correct option: A
a+b+4=0a+b+4=0
Option A (a+b+4=0a+b+4=0) is correct because by equating the difference between the roots of the two equations, we can derive the relationship between a and b.
83 Easy
83 of 120
Application +4 | -1
If aa and bb are real numbers such that (2+α)4=a+bα,{\left( {2 + \alpha } \right)^4} = a + b\alpha , where α=−1+i32,\alpha = \frac{{ - 1 + i\sqrt 3 }}{2}, then a+ba + b is equal to
View answer and explanation
Correct option: C
9
The correct answer is C (9) because it is the sum of a and b, where a and b are real numbers satisfying the given equation involving a complex number alpha.
84 Hard
84 of 120
Analyzing +4 | -1
If the equation anxn+an−1xn−1+…+a1x=0,a1≠0a_n x^n+a_{n-1} x^{n-1}+\ldots+a_1 x=0, a_1 \neq 0 n≥2n \geq 2, has a positive root x=αx=\alpha, then the equation nanxn−1+(n−1)an−1xn−2+…+a1=0n a_n x^{n-1}+(n-1) a_{n-1} x^{n-2}+\ldots+a_1=0 has a positive root, which is
View answer and explanation
Correct option: D
smaller than α\alpha
Option D (smaller than α\alpha) is correct because by Rolle's theorem, there exists a root of the derivative between 0 and α\alpha.
85 Medium
85 of 120
Application +4 | -1
If α,  β,  γ\alpha ,\;\beta ,\;\gamma are the roots of x3+2x2−3x−1=0{x^3} + 2{x^2} - 3x - 1 = 0, then α−2+β−2+γ−2{\alpha ^{ - 2}} + {\beta ^{ - 2}} + {\gamma ^{ - 2}} is equal to
View answer and explanation
Correct option: B
13
Option B (13) is correct because it is the value of the expression α−2+β−2+γ−2{\alpha ^{ - 2}} + {\beta ^{ - 2}} + {\gamma ^{ - 2}} calculated using Vieta's formulas.
86 Medium
86 of 120
Analyzing +4 | -1
In the complex plane, let AA be the set of solutions to z3−8=0z^{3}-8=0 and let BB be the set of solution to z3−8z2−8z+64=0z^{3}-8 z^{2}-8 z+64=0, then the maximum distance between a point of AA and a point of BB is
View answer and explanation
Correct option: D
2212 \sqrt{21}
The correct answer is 2212 \sqrt{21} because the maximum distance is found by identifying the points in sets A and B that are farthest apart in the complex plane.
87 Medium
87 of 120
Analyzing +4 | -1
If α,  β,  γ\alpha ,\;\beta ,\;\gamma are the roots of the equation x3−7x+7=0{x^3} - 7x + 7 = 0, then 1α4+1β4+1γ4\frac{1}{{{\alpha ^4}}} + \frac{1}{{{\beta ^4}}} + \frac{1}{{{\gamma ^4}}} is
View answer and explanation
Correct option: B
37\frac{3}{7}
Option B (37\frac{3}{7}) is correct because it involves using Vieta's formulas and algebraic manipulation to find the sum of the reciprocals of the fourth powers of the roots.
88 Easy
88 of 120
Application +4 | -1
Let p,  q∈Rp,\,\,q \in R. If 2−32 - \sqrt 3 is a root of the quadratic equation, x2+px+q=0,{x^2} + px + q = 0, then
View answer and explanation
Correct option: B
p2−4q−12=0,{p^2} - 4q - 12 = 0,
Option B (p2−4q−12=0,{p^2} - 4q - 12 = 0,) is correct because, given that 2−32 - \sqrt 3 is a root, we can find the other root, determine p and q, and verify the relationship.
89 Easy
89 of 120
Application +4 | -1
The numerical difference of the roots of x2−7x−9=x^2-7 x-9= 0 is
View answer and explanation
Correct option: A
85\sqrt{85}
Option A (85\sqrt{85}) is correct because it is calculated using the relationship between the difference of roots and the coefficients of the quadratic equation.
90 Easy
90 of 120
Analyzing +4 | -1
Consider the polynomial f(x)=1+2x+3x2+4x3f(x)=1+2 x+3 x^2+4 x^3. Let ss be the sum of all distinct real roots of f(x)f(x) and let t=∣s∣t=|s|, real number ss lies in the interval
View answer and explanation
Correct option: C
(−34,−12)\left(-\frac{3}{4},-\frac{1}{2}\right)
Option C ((−34,−12)\left(-\frac{3}{4},-\frac{1}{2}\right)) is correct because the function has only one real root, and it lies within this interval, determined by evaluating the function at the interval endpoints.
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