Skip to main content
Back to THEORY OF EQUATION

Your Progress

0 / 63 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • THEORY OF EQUATION

Theory of equation practice · Easy

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →
Showing 31-60 of 63 questions
31 Easy
31 of 63
Analyzing +4 | -1
The sum of all real values of xx satisfying the equation (x2−5x+5)x2+4x−60=1\left(x^2-5 x+5\right)^{x^2+4 x-60}=1 is
View answer and explanation
Correct option: C
3
Option C (3) is the correct answer because it represents the sum of all real values of x that satisfy the given equation, considering all possible cases where the equation equals 1.
32 Easy
32 of 63
Application +4 | -1
The real values of pp and qq such that the equation x2+px+q=0x^2+p x+q=0 has 2+i32+i \sqrt{3} as a root, are
View answer and explanation
Correct option: A
p=−4,q=7p=-4, q=7
Option A (p=−4,q=7p=-4, q=7) 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.
33 Easy
33 of 63
Application +4 | -1
The roots of the equation x4−8x2−9=0{x^4} - 8{x^2} - 9 = 0 are
View answer and explanation
Correct option: B
±  3,  ±  i\pm \;3,\; \pm \;i
Option B (±  3,  ±  i\pm \;3,\; \pm \;i) is correct because these are the roots obtained by factoring the given quartic equation.
34 Easy
34 of 63
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.
35 Easy
35 of 63
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.
36 Easy
36 of 63
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.
37 Easy
37 of 63
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.
38 Easy
38 of 63
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.
39 Easy
39 of 63
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.
40 Easy
40 of 63
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.
41 Easy
41 of 63
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.
42 Easy
42 of 63
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.
43 Easy
43 of 63
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.
44 Easy
44 of 63
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.
45 Easy
45 of 63
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.

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
46 Easy
46 of 63
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.
47 Easy
47 of 63
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.
48 Easy
48 of 63
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.
49 Easy
49 of 63
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.
50 Easy
50 of 63
Application +4 | -1
The quadratic equation with rational coefficients, one of whose roots is 12+5\frac{1}{2+\sqrt{5}}, is
View answer and explanation
Correct option: A
x2+4x−1=0x^2+4 x-1=0
Option A (x2+4x−1=0x^2+4 x-1=0) is correct because it is the quadratic equation with rational coefficients having 12+5\frac{1}{2+\sqrt{5}} as one of its roots.
51 Easy
51 of 63
Application +4 | -1
If α,β\alpha ,\beta and γ\gamma are the roots of equation x3−3x2+x+5=0{x^3} - 3{x^2} + x + 5 = 0, then y=∑α2+αβγy = \sum {\alpha ^2} + \alpha \beta \gamma satisfies the equation
View answer and explanation
Correct option: B
y3−y2−y−2=0{y^3} - {y^2} - y - 2 = 0
Option B (y3−y2−y−2=0{y^3} - {y^2} - y - 2 = 0) is correct because the value of y is 2, which satisfies the given equation.
52 Easy
52 of 63
Application +4 | -1
If the roots of the equation 8x3−14x2+7x−1=08{x^3} - 14{x^2} + 7x - 1 = 0 are in GP, then the roots are
View answer and explanation
Correct option: A
1,12,141,\frac{1}{2},\frac{1}{4}
Option A (1,12,141,\frac{1}{2},\frac{1}{4}) is correct because the roots are in GP and their product is 18\frac{1}{8}, allowing us to find the common ratio.
53 Easy
53 of 63
Analyzing +4 | -1
Let α,β\alpha, \beta be the roots of the equation x2−x+p=0x^2-x+p=0 and γ,δ\gamma, \delta be the roots of the equation x2−4x+q=0x^2-4 x+q=0. If α,β,γ\alpha, \beta, \gamma and δ\delta are in GP, the integral values of pp and qq respectively, are
View answer and explanation
Correct option: A
−2,−32-2,-32
Option A (-2, -32) is correct because the values of p and q are found by using the properties of GP and the roots of quadratic equations.
54 Easy
54 of 63
Analyzing +4 | -1
The roots α,β  and γ\alpha ,\beta \,\,\mathrm{and}\,\gamma of an equation x3−3  a  x2+3  b  x−c=0{x^3} - 3\;a\;{x^2} + 3\;b\;x - c = 0 are in H.P. Then,
View answer and explanation
Correct option: D
β=cb\beta = \frac{c}{b}
Option D (β=cb\beta = \frac{c}{b}) is correct because the harmonic mean is the reciprocal of the arithmetic mean of the reciprocals, leading to the given relationship.
55 Easy
55 of 63
Application +4 | -1
If α,β\alpha, \beta are the roots of the equation x2−3x+5=0x^2-3 x+5=0 and γ\gamma, δ\delta are the roots of the equation x2+5x−3=0x^2+5 x-3=0, then the equation whose roots are αγ+βδ\alpha \gamma+\beta \delta and αδ+βγ\alpha \delta+\beta \gamma, is
View answer and explanation
Correct option: D
x2+15x+158=0x^2+15 x+158=0
Option D x2+15x+158=0x^2+15 x+158=0 is correct because it is the quadratic equation formed using the sum and product of the roots αγ+βδ\alpha \gamma + \beta \delta and αδ+βγ\alpha \delta + \beta \gamma.
56 Easy
56 of 63
Analyzing +4 | -1
If a∈Ra \in R and the equation
−3(x−[x])2+2(x−[x])+a2=0-3(x-[x])^2+2(x-[x])+a^2=0 (where [⋅][\cdot] denotes the greatest integer function) has no integral solution, then all possible values of aa lie in the interval
View answer and explanation
Correct option: C
(−1,0)∪(0,1)(-1,0) \cup(0,1)
Option C ((-1,0) ∪(0,1)) is correct because it ensures that the fractional part of x is non-zero and the equation has no integral solutions, leading to the given interval for a.
57 Easy
57 of 63
Analyzing +4 | -1
If 2+3i2+3 i is one of the roots of the equation 2x3−9x2+kx−132 x^{3}-9 x^{2}+k x-13 =0,k∈R=0, \mathrm{k} \in \mathrm{R}, then the real root of this equation :
View answer and explanation
Correct option: B
exists and is equal to 12\frac{1}{2}.
Option B exists and is equal to 12\frac{1}{2} is correct because the product of the roots can be used to find the real root.
58 Easy
58 of 63
Analyzing +4 | -1
Let α(a)\alpha(a) and β(a)\beta(a) be the roots of the equation ((1+a)3−1)x2+((1+a)−1)x+((1+a)6−1)=0(\sqrt[3]{(1+a)}-1) x^2+(\sqrt{(1+a)}-1) x+(\sqrt[6]{(1+a)}-1)=0 where a>−1a>-1, then lim⁡a→0+α(a)\lim\limits_{a \to {0^ + }} \alpha (a) and lim⁡a→0+β(a)\lim\limits_{a \to {0^ + }} \beta (a), are
View answer and explanation
Correct option: B
(−12)\left(-\frac{1}{2}\right) and (−1)(-1)
Option B ((−12)\left(-\frac{1}{2}\right) and (−1)(-1)) is correct because after applying L'Hopital's rule and simplifying, the roots of the resulting quadratic equation are -1/2 and -1.
59 Easy
59 of 63
Application +4 | -1
If α,β∈C\alpha ,\beta \in C are the distinct roots of the distinct roots of the equation x2−x+1=0,{x^2} - x + 1 = 0, then α101+β107{\alpha ^{101}} + {\beta ^{107}} is equal to
View answer and explanation
Correct option: C
1
Option A (Atleast 1000) is correct because the number of arrangements is calculated to be 1080, which is greater than 1000.
60 Easy
60 of 63
Application +4 | -1
If x2+x+1x^2+x+1 is a factor of ax3+bx2+cx+da x^3+b x^2+c x+d, the real root of ax3+bx2+cx+d=0a x^3+b x^2+c x+d=0 is
View answer and explanation
Correct option: A
−da-\frac{d}{a}
Option A (-d/a) is correct because if x2+x+1x^2+x+1 is a factor, then the remaining factor must be linear, and the real root is found by considering the product of the roots.
Revision notes and key takeaways

Review THEORY OF EQUATION in your textbook or class notes, then use this set to check your understanding.

Showing 31-60 of 63 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - THEORY OF EQUATION MCQs (All) 120 MCQs → ENGINEERING MATHEMATICS - THEORY OF EQUATION MCQs (Hard) 9 MCQs → ENGINEERING MATHEMATICS - THEORY OF EQUATION MCQs (Medium) 48 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → INDEFINITE INTEGRATION 1179 MCQs
📚 Switch Set (EASY)
Chapter Sets
ALL 120 MCQs EASY 63 MCQs HARD 9 MCQs MEDIUM 48 MCQs
0 /63
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →