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JEE • MATHEMATICS • THEORY OF EQUATION

Theory of equation practice · Medium

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1 Medium
1 of 48
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 48
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 48
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 Medium
4 of 48
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.
5 Medium
5 of 48
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.
6 Medium
6 of 48
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).
7 Medium
7 of 48
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.
8 Medium
8 of 48
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.
9 Medium
9 of 48
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.
10 Medium
10 of 48
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.
11 Medium
11 of 48
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.
12 Medium
12 of 48
Application +4 | -1
If 1, 2, 3 and 4 are the roots of the equation x4+ax3+bx2+cx+d=0{x^4} + a{x^3} + b{x^2} + cx + d = 0 , then a+2b+ca + 2b + c is equal to
View answer and explanation
Correct option: C
10
Option A (10) is correct because by expanding the polynomial with roots 1, 2, 3, and 4, and comparing coefficients, we can find the values of a, b, and c, and then calculate a + 2b + c = 10.
13 Medium
13 of 48
Analyzing +4 | -1
The real roots of ∣x∣3−3  x2+3∣x∣−2=0{\left| x \right|^3} - 3\;{x^2} + 3\left| x \right| - 2 = 0 are
View answer and explanation
Correct option: C
±  2\pm \;2
Option C (±  2\pm \;2) is correct because by substituting ∣x∣|x| into the equation and solving, we find that the real roots are ±2\pm 2.
14 Medium
14 of 48
Analyzing +4 | -1
The range of values of 'a{a}' so that the equation x3−3x+a=0x^3-3 x+a=0 has three real and distinct roots is
View answer and explanation
Correct option: D
(−2,2)(-2,2)
Option D, (−2,2)(-2,2), is correct because the condition for three distinct real roots is that f(1)f(−1)<0f(1)f(-1) < 0, which leads to −2<a<2-2 < a < 2.
15 Medium
15 of 48
Analyzing +4 | -1
If f(x)=x5+x2+1f\left( x \right) = {x^{5 + }}{x^2} + 1 has roots x1,x2,x3,x4,x5andg(x)=x2−2{x_1},{x_2},{x_3},{x_4},{x_5}andg\left( x \right) = {x^2} - 2 then g(x1)g(x2)g(x3)g(x4)g(x5)−30g(x1x2x3x4x5)=_________g({x_1})g({x_2})g({x_3})g({x_4})g({x_5}) - 30g\left( {{x_1}{x_2}{x_3}{x_4}{x_5}} \right) = \_\_\_\_\_\_\_\_\_
View answer and explanation
Correct option: C
7
Option C (7) is correct because it involves finding the product of g(x) evaluated at the roots of f(x) and using Vieta's formulas.

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16 Medium
16 of 48
Analyzing +4 | -1
If roots of the equation x4−8x3+bx2+cx+16=0{x^4} - 8{x^3} + b{x^2} + cx + 16 = 0 are positive then
View answer and explanation
Correct option: C
b=24,c=−32b = 24,c = - 32
Option D (b=24,c=−32b = 24, c = -32) is incorrect. The correct answer is D (b=24,c=32b = 24, c = 32) because the roots are equal, allowing us to expand (x-2)4 and match coefficients.
17 Medium
17 of 48
Analyzing +4 | -1
The equation esin⁡x−e−sin⁡x−4=0e^{\sin x}-e^{-\sin x}-4=0 has
View answer and explanation
Correct option: D
no real roots
Option D (no real roots) is correct because the equation simplifies to a form where sin(x) must be greater than 1, which is impossible.
18 Medium
18 of 48
Analyzing +4 | -1
If α,  β,  γ\alpha ,\;\beta ,\;\gamma are the roots of x3+4x+1=0{x^3} + 4x + 1 = 0, then the equation whose roots are α2β+γ,  β2α+γ,  γ2α+β\frac{{{\alpha ^2}}}{{\beta + \gamma }},\;\frac{{{\beta ^2}}}{{\alpha + \gamma }},\;\frac{{{\gamma ^2}}}{{\alpha + \beta }} is
View answer and explanation
Correct option: C
x3+4x−1=0{x^3} + 4x - 1 = 0
Option D (None of these) is correct because the determinant simplifies to 0, regardless of the values of α\alpha and β\beta.
19 Medium
19 of 48
Analyzing +4 | -1
If a,b,ca,b,c are real and x3−3b2x+2c3{x^3} - 3{b^2}x + 2{c^3} is divisible by x−ax - a and x−b,x - b, then
View answer and explanation
Correct option: C
a=b=ca = b = c or a=−2b=−2ca = - 2b = - 2c
Option C (a=b=ca = b = c or a=−2b=−2ca = - 2b = - 2c) is correct because it satisfies the divisibility condition of the polynomial by both x−ax-a and x−bx-b.
20 Medium
20 of 48
Application +4 | -1
If P(x)\mathrm{P}(\mathrm{x}) be a polynomial of degree 4 , with p(2)=−1,p′(2)=0,p′′(2)=2,p′′′(2)=−12\mathrm{p}(2)=-1, \mathrm{p}^{\prime}(2)=0, \mathrm{p}^{\prime \prime}(2)=2, \mathrm{p}^{\prime \prime \prime}(2)=-12 and p′′′(2)=24p^{\prime \prime \prime}(2)=24 then p′′(1)p^{\prime \prime}(1) is equal to:
View answer and explanation
Correct option: C
26
Option C (26) is correct because after finding the polynomial's coefficients using the given derivatives and value at x=2, the second derivative at x=1 is calculated to be 26.
21 Medium
21 of 48
Application +4 | -1
Let f(x)=ax3+bx2+cx+41f(x)=a x^3+b x^2+c x+41 be such that f(1)=40,f′(1)=2f(1)=40, f^{\prime}(1)=2 and f′′(1)=4f^{\prime\prime}(1)=4. Then a2+b2+c2a^2+b^2+c^2 is equal to:
View answer and explanation
Correct option: D
51
Option D (51) is correct because it is the sum of squares of a, b, and c, where a, b, and c are coefficients found by using the given conditions and derivatives.
22 Medium
22 of 48
Analyzing +4 | -1
If for a positive integer nn, the quadratic equation&nbsp;x(x+1)+(x+1)(x+2)+......+(x+n−1)(x+n)=10nx\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right) + ...... + \left( {x + n - 1} \right)\left( {x + n} \right) = 10n has two consecutive integral solutions, then nn is equal to
View answer and explanation
Correct option: A
11
Option A (11) is correct because it is the value of n that satisfies the given conditions for the quadratic equation to have two consecutive integral solutions.
23 Medium
23 of 48
Analyzing +4 | -1
Let λ≠0\lambda \neq 0 be a real number. Let α,β\alpha, \beta be the roots of the equation 14x2−31x+3λ=014 x^{2}-31 x+3 \lambda=0 and α,γ\alpha, \gamma be the roots of the equation 35x2−53x+4λ=035 x^{2}-53 x+4 \lambda=0. Then 3αβ\frac{3 \alpha}{\beta} and 4αγ\frac{4 \alpha}{\gamma} are the roots of the equation
View answer and explanation
Correct option: D
49x2−245x+250=049 x^{2}-245 x+250=0
The correct answer is D (49x2−245x+250=049 x^{2}-245 x+250=0) because it is the quadratic equation formed with roots derived from the given conditions and relationships between the roots of the two equations.
24 Medium
24 of 48
Understanding +4 | -1
If x−cx - c is a factor of order m of the polynomial f(x)f\left( x \right) of degree n(1<m<n),n\left( {1 < m < n} \right), then x=cx = c is a root of the polynomial
View answer and explanation
Correct option: B
fm−1(x){f^{m - 1}}\left( x \right)
Option B (fm−1(x){f^{m - 1}}(x)) is correct because if (x−c)(x-c) is a factor of order m of f(x)f(x), then x=cx=c is a root of f(x)f(x) and its first m-1 derivatives.
25 Medium
25 of 48
Analyzing +4 | -1
If x2y+y2z+z2x=2022;x2z+y2x+z2y=2020x^2 y+y^2 z+z^2 x=2022 ; x^2 z+y^2 x+z^2 y=2020. Then value of (x−y)3+(y−z)3+(z−x)3(x-y)^3+(y-z)^3+(z-x)^3 is equal to
View answer and explanation
Correct option: B
−6-6
Option B (-6) is correct because the value of (x−y)3+(y−z)3+(z−x)3(x-y)^3+(y-z)^3+(z-x)^3 is -6, which is derived from the given equations.
26 Medium
26 of 48
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.
27 Medium
27 of 48
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.
28 Medium
28 of 48
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.
29 Medium
29 of 48
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.
30 Medium
30 of 48
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.
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