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JEE • MATHEMATICS • THEORY OF EQUATION

Theory of equation practice · Mixed difficulty

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Showing 1-30 of 120 questions
1 Medium
1 of 120
Understanding +4 | -1
The maximum number of real roots of the equation x2n−1=0{x^{2n}} - 1 = 0 is
View answer and explanation
Correct option: A
2
Option A (2) is correct because the equation x(2n) - 1 = 0 has two real roots, which are 1 and -1.
2 Medium
2 of 120
Analyzing +4 | -1
The number of roots of the equation 1x+1(1−x2)=3512\frac{1}{x}+\frac{1}{\sqrt{\left(1-x^2\right)}}=\frac{35}{12} is
View answer and explanation
Correct option: D
3
Option D (3) is correct because the equation has three roots, found by substitution and solving the resulting equations.
3 Medium
3 of 120
Analyzing +4 | -1
How many real solutions does the equation x7+14x5+16x3+30x−560=0x^7+14 x^5+16 x^3+30 x-560=0 have?
View answer and explanation
Correct option: A
1
Option A (1) is correct because the derivative of the given function is always positive, indicating that the function is strictly increasing and thus has only one real root.
4 Easy
4 of 120
Analyzing +4 | -1
If x1,x2,x3,…,xnx_1, x_2, x_3, \ldots, x_n are the roots of the equation xn+ax+b=0x^n+a x+b=0, the value of (x1−x2)(x1−x3)(x1−x4)…(x1−xn)\left(x_1-x_2\right)\left(x_1-x_3\right)\left(x_1-x_4\right) \ldots\left(x_1-x_n\right) is
View answer and explanation
Correct option: C
n(x1)n−1+an\left(x_1\right)^{n-1}+a
Option C (n(x1)n−1+an(x_1)^{n-1}+a) is correct because it is the result of differentiating the polynomial and evaluating it at one of its roots.
5 Easy
5 of 120
Application +4 | -1
The polynomials P(x)=kx3+3x2−3P(x)=k x^3+3 x^2-3 and Q(x)=2x3−5x+kQ(x)=2 x^3-5 x+k, when divided by (x−4)(x-4) leave the same remainder, then kk is equal to:
View answer and explanation
Correct option: B
1
Option B (1) is correct because when the polynomials are divided by (x-4), setting the remainders equal and solving for k results in k=1.
6 Easy
6 of 120
Analyzing +4 | -1
If (x−1)3{\left( {x - 1} \right)^3} is a factor of x4+ax3+bx2+cx−1,{x^4} + a{x^3} + b{x^2} + cx - 1, then the other factor is
View answer and explanation
Correct option: B
x+1x + 1
Option B (x+1x + 1) is correct because by using the given factor and polynomial division or by finding the roots, we can determine the remaining factor.
7 Easy
7 of 120
Analyzing +4 | -1
The number of positive integral solutions of x4−y4=3789108x^4-y^4=3789108 is
View answer and explanation
Correct option: A
0
Option A (0) is correct because the given equation has no positive integral solutions since the left side is always divisible by 8, while the right side is not.
8 Medium
8 of 120
Analyzing +4 | -1
Let α,β\alpha, \beta be the roots of the equation px2+qx+r=0p x^2+q x+r=0, p≠0p \neq 0. If p,q,rp, q, r are in AP\mathrm{AP} and 1α+1β=4\frac{1}{\alpha}+\frac{1}{\beta}=4, the value of ∣α−β∣|\alpha-\beta|, is
View answer and explanation
Correct option: B
2139\frac{2 \sqrt{13}}{9}
Option B (2139\frac{{2 \sqrt{{13}}}}{{9}}) is correct because it is the result of calculating the difference between the roots of the quadratic equation given the conditions that p, q, r are in AP and 1α+1β=4\frac{{1}}{{\alpha}} + \frac{{1}}{{\beta}} = 4.
9 Medium
9 of 120
Analyzing +4 | -1
If α\alpha is only real root of the equation x3+cos⁡1⋅x2+sin⁡1⋅x+1=0x^3+\cos 1 \cdot x^2+\sin 1 \cdot x+1=0, then tan⁡−1α+tan⁡−11α\tan ^{-1} \alpha+\tan ^{-1} \frac{1}{\alpha} is equal to
View answer and explanation
Correct option: C
−π2-\frac{\pi}{2}
Option C (−π2\frac{-\pi}{2}) is correct because the real root is negative, and the sum of the arctangents of a number and its reciprocal is −π2-\frac{\pi}{2} when the number is negative.
10 Medium
10 of 120
Analyzing +4 | -1
Let α,β,γ\alpha, \beta, \gamma be the three roots of the equation x3+bxx^{3}+b x +c=0+c=0. If βγ=1=−α\beta \gamma=1=-\alpha, then b3+2c3−3α3−6β3−b^{3}+2 c^{3}-3 \alpha^{3}-6 \beta^{3}- 8γ38 \gamma^{3} is equal to
View answer and explanation
Correct option: D
19
Option A (e+2e+1−log⁡e(e+1)\frac{e+2}{e+1}-\log _{e}(e+1)) is correct because after solving the integral and applying the given limit condition, the value of I(1) is found to be e+2e+1−log⁡e(e+1)\frac{e+2}{e+1}-\log _{e}(e+1).
11 Easy
11 of 120
Application +4 | -1
If α,β,γ\alpha ,\beta ,\gamma are the roots of x3−2x2+3x−4=0{x^3} - 2{x^2} + 3x - 4 = 0, then the value of α2β2+β2γ2+γ2α2{\alpha ^2}{\beta ^2} + {\beta ^2}{\gamma ^2} + {\gamma ^2}{\alpha ^2} is
View answer and explanation
Correct option: A
−7- 7
Option A (−7- 7) is correct because it is the value of the expression calculated using Vieta's formulas.
12 Medium
12 of 120
Analyzing +4 | -1
Let f(x)={x}[x];g(x)=ax2f(x)=\{x\}[x] ; g(x)=a x^{2} the sum of all real solutions of equation satisfying f(x)=g(x)f(x)=g(x) is 420 . (Where aa is (+ve) rational number), then aa is equal to (Where [.] and {.} represents greatest integer function and fractional part function respectively)
View answer and explanation
Correct option: B
29900\frac{29}{900}
Option B (29900\frac{29}{900}) is correct because it is the value of 'a' that satisfies the given equation after considering the sum of real solutions.
13 Medium
13 of 120
Analyzing +4 | -1
If x2+px+1{x^2} + px + 1 is a factor of the expression ax3+bx+ca{x^3} + bx + c, then
View answer and explanation
Correct option: C
a2−c2=ab{a^2} - {c^2} = ab
Option C (a2−c2=ab{a^2} - {c^2} = ab) is correct because by equating coefficients after assuming ax3+bx+ca{x^3} + bx + c is divisible by x2+px+1{x^2} + px + 1, we arrive at the relationship a2−c2=ab{a^2} - {c^2} = ab.
14 Medium
14 of 120
Analyzing +4 | -1
The sum of all the real roots of the equation (e2x−4)(6e2x−5ex+1)=0({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0 is
View answer and explanation
Correct option: B
−log⁡e3- {\log _e}3
Option B (−log⁡e3- \log_e 3) is correct because it is the sum of the real roots obtained by solving the given equation.
15 Easy
15 of 120
Application +4 | -1
If the roots of the equation 4x3−12x2+11x+k=04{x^3} - 12{x^2} + 11x + k = 0 are in arithmetic progression, then kk is equal to
View answer and explanation
Correct option: A
−3- 3
Option D (None of these) is correct because the determinant simplifies to x2, which is not among the given options.

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16 Easy
16 of 120
Application +4 | -1
The sum of all the real values of xx satisfying the equation 2(x−1)(x2+5x−50)=12^{(\mathrm{x}-1)\left(\mathrm{x}^{2}+5 \mathrm{x}-50\right)}=1 is :
View answer and explanation
Correct option: C
-4
Option C (-4) is correct because the sum of all real values of x that satisfy the equation is -4, which is obtained by solving the equation and summing the real roots.
17 Easy
17 of 120
Application +4 | -1
Let α,β\alpha, \beta be the roots of the quadratic equation x2+px+p3=0(p≠0)x^2+p x+p^3=0(p \neq 0). If (α,β)(\alpha, \beta) is a point on the parabola y2=xy^2=x, then the roots of the quadratic equation are
View answer and explanation
Correct option: A
4,−24,-2
Option A (4, -2) is correct because it satisfies both the quadratic equation and the parabola equation.
18 Easy
18 of 120
Application +4 | -1
If α,β,γ\alpha ,\beta ,\gamma are the roots of the equation x3+4x−1=0{x^3} + 4x - 1 = 0, then (α+β)−1+(β+γ)−1+(γ+α)−1{\left( {\alpha + \beta } \right)^{ - 1}} + {\left( {\beta + \gamma } \right)^{ - 1}} + {\left( {\gamma + \alpha } \right)^{ - 1}} is equal to
View answer and explanation
Correct option: C
−4- 4
Option C (-4) is correct because it is the result of simplifying the expression using Vieta's formulas for the sum and product of roots of the given cubic equation.
19 Medium
19 of 120
Application +4 | -1
Let α,β,δ\alpha ,\beta ,\delta be the roots of x4+x2+1=0.{x^4} + {x^2} + 1 = 0. Then, the equation whose roots are α2,β2,γ2,δ2{\alpha ^2},{\beta ^2},{\gamma ^2},{\delta ^2} is
View answer and explanation
Correct option: B
(x2+x+1)2=0{\left( {{x^2} + x + 1} \right)^2} = 0
The correct answer is B ((x2+x+1)2=0{\left( {{x^2} + x + 1} \right)^2} = 0) because substituting y=x2y = x^2 transforms the equation and squaring the result gives the desired equation.
20 Hard
20 of 120
Analyzing +4 | -1
If λ∈R\lambda \in R is such that the sum of the cubes of the roots of the equation, x2+(2−λ)x+(10−λ)=0x^{2}+(2-\lambda) x+(10-\lambda)=0 is minimum, then the magnitude of the difference of the roots of this equation is
View answer and explanation
Correct option: B
252 \sqrt{5}
Option B (252 \sqrt{5}) is correct because the magnitude of the difference of the roots is calculated by minimizing the sum of the cubes of the roots.
21 Easy
21 of 120
Analyzing +4 | -1
Number of real roots of the equation x+x−(1−x)=1\sqrt{x}+\sqrt{x-\sqrt{(1-x)}}=1 is
View answer and explanation
Correct option: B
1
The correct answer is B (1) because after simplification and squaring, only one real root satisfies the original equation.
22 Easy
22 of 120
Application +4 | -1
The equation log⁡3(3x−8)=2−x\log _3\left(3^x-8\right)=2-x has the solution
View answer and explanation
Correct option: B
x=2\mathrm{x}=2
Option B (x=2) is correct because it is the only solution that satisfies the given logarithmic equation after simplification and solving the resulting exponential equation.
23 Easy
23 of 120
Application +4 | -1
If α,β\alpha ,\beta and γ\gamma are the roots of x3+8=0,{x^3} + 8 = 0, then the equation whose roots are α2,β2{\alpha ^2},{\beta ^2} and γ2{\gamma ^2} is
View answer and explanation
Correct option: D
x3−64=0{x^3} - 64 = 0
The correct answer is D (x3−64=0{x^3} - 64 = 0) because it is derived by substituting x=yx = \sqrt{y} into the original equation and simplifying.
24 Easy
24 of 120
Analyzing +4 | -1
The real number kk for which the equation, 2x3+3x+k=02{x^3} + 3x + k = 0 has two distinct real roots in [0, 1]
View answer and explanation
Correct option: D
does not exist
Option D (does not exist) is correct because the derivative of the function is always positive, meaning the function is strictly increasing and can have at most one real root.
25 Medium
25 of 120
Application +4 | -1
If f(x)=2x3+mx2−13x+nf\left( x \right) = 2{x^3} + m{x^2} - 13x + n and 2, 3 are roots of the equation f(x)=0f\left( x \right) = 0, then the values of mm and nn are
View answer and explanation
Correct option: B
−5,30- 5,30
Option A (−5,−30- 5, - 30) is incorrect. Option B (−5,30- 5,30) is correct because substituting the roots 2 and 3 into the equation f(x)=0f(x) = 0 yields two equations that can be solved for mm and nn, resulting in m=−5m = -5 and n=30n = 30.
26 Easy
26 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 by using Vieta's formulas and manipulating the expressions, we can find the value of the given expression.
27 Easy
27 of 120
Analyzing +4 | -1
Let α,β,γ\alpha, \beta, \gamma be the roots of the equation (x−a)(x−b)(x−c)=d,d≠0(x-a)(x-b)(x-c)=d, d \neq 0, the roots of the equation (x−α)(x−β)(x−γ)+d=0(x-\alpha)(x-\beta)(x-\gamma)+d=0 are
View answer and explanation
Correct option: C
a,b,ca, b, c
Option C (a,b,ca, b, c) is correct because by substituting the roots of the first equation into a modified second equation, we find that a,b,a, b, and cc are the roots.
28 Easy
28 of 120
Analyzing +4 | -1
Maximum number of real solution for the equation axn+x2+bx+c=0a x^n+x^2+b x+c=0, where a,b,c∈Ra, b, c \in R and nn is an even positive number, is
View answer and explanation
Correct option: D
infinite
Option D (infinite) is correct because by choosing specific values for a, b, and c, the equation can have infinite solutions.
29 Easy
29 of 120
Application +4 | -1
If (x+1)(x+1) is factor of x3+kx2−3x+k+2x^3+k x^2-3 x+k+2, then kk is equal to:
View answer and explanation
Correct option: C
−2-2
Option C is correct because if (x+1) is a factor, then substituting x = -1 into the polynomial results in 0, which solves to k = -2.
30 Easy
30 of 120
Analyzing +4 | -1
The number of pairs (x,y)(x, y) which will satisfy the equation x2−xy+y2=4(x+y−4)x^2-x y+y^2=4(x+y-4), is
View answer and explanation
Correct option: A
1
Option A (1) is correct because by completing the square and analyzing the discriminant, we find only one pair (4,4) satisfies the equation.
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