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JEE • MATHEMATICS • THEORY OF EQUATION

Theory of equation practice · Hard

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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.

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1 Hard
1 of 9
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.
2 Hard
2 of 9
Analyzing +4 | -1
If x2−2r  arx+r=0;r=1,  2,  3{x^2} - 2r\;{a_r}x + r = 0;r = 1,\;2,\;3 are three quadratic equations of which each pair has exactly one root common, then the number of solutions of the triplet (a1,a2,a3)\left( {{a_1},{a_2},{a_3}} \right) is
View answer and explanation
Correct option: B
2
Option B is correct because there are two possible sets of triplets (a1, a2, a3) that satisfy the given conditions of the quadratic equations having exactly one root in common.
3 Hard
3 of 9
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.
4 Hard
4 of 9
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.
5 Hard
5 of 9
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
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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.
6 Hard
6 of 9
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.
7 Hard
7 of 9
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
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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.
8 Hard
8 of 9
Analyzing +4 | -1
Let Pn(x)=1+2x+3x2+…+(n+1)xn{P_n}\left( x \right) = 1 + 2x + 3{x^2} + \ldots + \left( {n + 1} \right){x^n} be a polynomial such that nn is even. Then, the number of real roots of Pn(x),{P_n}\left( x \right), is
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Correct option: A
0
Option A (0) is correct because the polynomial has no real roots when n is even.
9 Hard
9 of 9
Analyzing +4 | -1
The number of real roots of the equation 1+a1x+a2x2+…+anxn=0,1 + {a_1}x + {a_2}{x^2} + \ldots + {a_n}{x^n} = 0, where ∣x∣<13  and  ∣an∣<2,\left| x \right| < \frac{1}{3}\;\mathrm{and\;}\left| {{a_n}} \right| < 2, is
View answer and explanation
Correct option: C
0 for any n∈Nn \in N
Option C (0 for any n∈Nn \in N) is correct because the given equation has no real roots for any n, as the expression is always positive within the specified range.
Revision notes and key takeaways

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Key Takeaways

  • ✓ This set contains 9 single-correct-answer MCQs on THEORY OF EQUATION.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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