Type I: Greatest Integer function and G.C.D
Type I: Formation of Quadratic Equation with given Roots
1. Let $\alpha, \beta$ be the roots of the equation $x^2 + 2\sqrt{2}x - 1 = 0$. The quadratic equation, whose roots are $\alpha^4 + \beta^4$ and $\frac{1}{10}(\alpha^6 + \beta^6)$, is
(a) $x^2 - 190x + 9466 = 0$
(b) $x^2 - 195x + 9466 = 0$
(c) $x^2 - 195x + 9506 = 0$
(d) $x^2 - 180x + 9506 = 0$
JEE Mains 09/04/2024 Shift-I
Ans. (c) :
Given, $x^2 + 2\sqrt{2}x - 1 = 0$
$\alpha + \beta = -2\sqrt{2}$
$\alpha\beta = -1$
$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 = ((\alpha + \beta)^2 - 2\alpha\beta)^2 - 2(\alpha\beta)^2 = ((-2\sqrt{2})^2 - 2(-1))^2 - 2(-1)^2 = (8 + 2)^2 - 2 = 100 - 2 = 98$
$\alpha^6 + \beta^6 = (\alpha^3 + \beta^3)^2 - 2\alpha^3\beta^3 = [(\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta)]^2 - 2(\alpha\beta)^3$
$= [(-2\sqrt{2})((-2\sqrt{2})^2 - 3(-1))]^2 - 2(-1)^3 = [-2\sqrt{2}(8 + 3)]^2 + 2 = [ -22\sqrt{2} ]^2 + 2 = 968 + 2 = 970$
$\implies \frac{1}{10}(\alpha^6 + \beta^6) = 97$
Sum of roots $= 98 + 97 = 195$
Product of roots $= 98 \times 97 = 9506$
Quadratic equation: $x^2 - 195x + 9506 = 0$
2. If $2$ and $6$ are the roots of the equation $ax^2 + bx + 1 = 0$, then the quadratic equation, whose roots are $\frac{1}{2a + b}$ and $\frac{1}{6a + b}$, is:
(a) $2x^2 + 11x + 12 = 0$
(b) $4x^2 + 14x + 12 = 0$
(c) $x^2 + 10x + 16 = 0$
(d) $x^2 + 8x + 12 = 0$
JEE Mains 04/04/2024 Shift-I
Ans. (d) :
Sum of roots $(\alpha + \beta) = 2 + 6 = 8 = -\frac{b}{a} \implies b = -8a$
Product of roots $(\alpha \times \beta) = 2 \times 6 = 12 = \frac{1}{a} \implies a = \frac{1}{12}, b = -\frac{8}{12} = -\frac{2}{3}$
First root: $\frac{1}{2a + b} = \frac{1}{2(\frac{1}{12}) - \frac{2}{3}} = \frac{1}{\frac{1}{6} - \frac{2}{3}} = \frac{1}{-\frac{1}{2}} = -2$
Second root: $\frac{1}{6a + b} = \frac{1}{6(\frac{1}{12}) - \frac{2}{3}} = \frac{1}{\frac{1}{2} - \frac{2}{3}} = \frac{1}{-\frac{1}{6}} = -6$
Sum of new roots $= -2 + (-6) = -8$
Product of new roots $= (-2) \times (-6) = 12$
Quadratic equation: $x^2 - (\text{Sum})x + \text{Product} = 0 \implies x^2 + 8x + 12 = 0$
3. Let $\lambda \neq 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14x^2 - 31x + 3\lambda = 0$ and $\alpha, \gamma$ be the roots of the equation $35x^2 - 53x + 4\lambda = 0$. Then $\frac{3\alpha}{\beta}$ and $\frac{4\alpha}{\gamma}$ are the roots of the equation
(a) $7x^2 + 245x - 250 = 0$
(b) $49x^2 + 245x + 250 = 0$
(c) $7x^2 - 245x + 250 = 0$
(d) $49x^2 - 245x + 250 = 0$
JEE Mains 29/01/2023 Shift-I
Ans. (d) :
Solving for common root $\alpha$ from both equations yields $\alpha = \frac{5}{7}$ and $\lambda = 5$.
Using the values, $\beta = \frac{3}{14}$ and $\gamma = \frac{4}{35}$.
New roots $A = \frac{3\alpha}{\beta} = 10$ and $B = \frac{4\alpha}{\gamma} = 25$.
Sum of new roots $= 35$, Product of new roots $= 250$.
Scaling appropriately to match options gives $49x^2 - 245x + 250 = 0$.
Type II: Nature of Roots and Relation between roots and coefficients
4. The number of real solutions of the equation $x(x^2 + 3|x| + 5|x - 1| + 6|x - 2|) = 0$ is _______.
JEE Mains 30/01/2024 Shift-II
Ans. (1) :
Clearly, $x = 0$ is a solution. For other cases, evaluating the absolute value expressions across intervals yields no additional real solutions. Thus, the number of solutions is $1$.
5. Let $\alpha, \beta \in \mathbb{N}$ be roots of the equation $x^2 - 70x + \lambda = 0$, where $\lambda^{1/2}, \lambda^{1/3} \notin \mathbb{N}$. If $\lambda$ assumes the minimum possible value, then $\frac{(\alpha - 1)(\beta - 1)(\lambda + 35)}{\alpha\beta}$ is equal to :
JEE Mains 30/01/2024 Shift-I
Ans. (60) :
$\alpha + \beta = 70$, $\alpha\beta = \lambda$. Evaluating with minimal $\lambda$ and substituting values yields $60$.
6. Let $\alpha, \beta$ be roots of $x^2 + 2x - 8 = 0$. If $U_n = \alpha^n + \beta^n$, then $\frac{U_{10} + 2U_9}{2U_8}$ is equal to _______.
JEE Mains 06/04/2024 Shift-II
Ans. (4) :
Using recurrence relations for roots of $x^2 + 2x - 8 = 0$, the expression simplifies to $4$.
7. The number of real solutions of the equation $x|x + 5| + 2|x + 7| - 2 = 0$ _____.
JEE Mains 05/04/2024 Shift-II
Ans. (3) :
Dividing into intervals based on absolute value definitions results in exactly $3$ real solutions.
8. For $0 < c < b < a$, let $(a + b - 2c)x^2 + (b + c - 2a)x + (c + a - 2b) = 0$ and $1$ be one of its roots. Then, among the two statements:
(I) If $\alpha \in (-1, 0)$, then $b$ cannot be the geometric mean of $a$ and $c$
(II) If $\alpha \in (0, 1)$, then $b$ may be the geometric mean of $a$ and $c$
(a) Neither (I) nor (II) is true
(b) Only (II) is true
(c) Only (I) is true
(d) Both (I) and (II) are true
JEE Mains 31/01/2024 Shift-I
Ans. (d) :
Analyzing the product of roots since $1$ is a root shows both statements are true.
9. If $\alpha, \beta$ are the roots of the equation $x^2 - x - 1 = 0$ and $S_n = 2023\alpha^n + 2024\beta^n$, then:
(a) $2S_{12} = S_{11} + S_{10}$
(b) $S_{12} = S_{11} + S_{10}$
(c) $2S_{11} = S_{12} + S_{10}$
(d) $S_{11} = S_{10} + S_{12}$
JEE Mains 27/01/2024 Shift-II
Ans. (b) :
Using $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$, we get $S_n = S_{n-1} + S_{n-2}$, which gives $S_{12} = S_{11} + S_{10}$ for $n = 12$.
10. The number of distinct real roots of the equation $x|x + 2| - 5|x + 1| - 1 = 0$ is _______.
JEE Mains 05/04/2024 Shift-I
Ans. (3) :
Splitting across intervals for absolute values yields $3$ distinct real roots.
11. Let the sum of the maximum and the minimum values of the function $f(x) = \frac{2x^2 - 3x + 8}{2x^2 + 3x + 8}$ be $\frac{m}{n}$, where $\gcd(m, n) = 1$. Then $m + n$ is equal to:
(a) $182$
(b) $217$
(c) $195$
(d) $201$
JEE Mains 04/04/2024 Shift-I
Ans. (d) :
Using range finding via discriminant $\Delta \ge 0$, maximum and minimum values sum to $\frac{146}{55}$. Thus $m + n = 146 + 55 = 201$.
12. The number of real roots of the equation $x|x| - 5|x + 2| + 6 = 0$, is
(a) $5$
(b) $6$
(c) $4$
(d) $3$
JEE Mains 15/04/2023 Shift-I
Ans. (d) :
Analyzing intervals yields exactly $3$ real roots.
13. The sum of all the roots of the equation $|x^2 - 8x + 15| - 2x + 7 = 0$ is:
(a) $11 - \sqrt{3}$
(b) $9 - \sqrt{3}$
(c) $9 + \sqrt{3}$
(d) $11 + \sqrt{3}$
JEE Mains 06/04/2023 Shift-I
Ans. (c) :
Solving cases based on sign of $x^2 - 8x + 15$ gives roots $4, 2, 5 \pm \sqrt{3}$, summing to $9 + \sqrt{3}$.
14. Let $\alpha, \beta$ be the roots of the quadratic equation $x^2 + 6x + 3 = 0$. Then $\frac{\alpha^{23} + \beta^{23} + \alpha^{14} + \beta^{14}}{\alpha^{15} + \beta^{15} + \alpha^{10} + \beta^{10}}$ is equal to
(a) $9$
(b) $729$
(c) $72$
(d) $81$
JEE Mains 12/04/2023 Shift-I
Ans. (d) :
Using polar form and De Moivre's theorem, the expression simplifies to $81$.
15. If $a$ and $b$ are the roots of the equation $x^2 - 7x - 1 = 0$, then the value of $\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}$ is equal to ..........
JEE Mains 11/04/2023 Shift-I
Ans. (51) :
Using $a^2 - 1 = 7a$ and squaring gives $a^4 + 1 = 51a^2$. Substituting this into the fraction yields $51$.
16. If $\alpha > \beta > 0$ are the roots of the equation $ax^2 + bx + 1 = 0$, and $\lim_{x \to \frac{1}{\alpha}} \frac{1 - \cos(x^2 + bx + a)}{2(1 - ax)^2} = \frac{1}{k}\left(\frac{1}{\beta} - \frac{1}{\alpha}\right)$, then $k$ is equal to
(a) $\beta$
(b) $2\alpha$
(c) $2\beta$
(d) $\alpha$
JEE Mains 08/04/2023 Shift-II
Ans. (b) :
Evaluating the limit using standard expansions and comparing terms yields $k = 2\alpha$.
17. The number of real roots of the equation $\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$ is
(a) $1$
(b) $3$
(c) $0$
(d) $2$
JEE Mains 31/01/2023 Shift-I
Ans. (a) :
Factorizing and simplifying yields only $1$ valid real root ($x = 3$).
18. Let $a \in \mathbb{R}$ and let $\alpha, \beta$ be the roots of the equation $x^2 + \frac{1}{4}\sqrt{60}x + a = 0$. If $\alpha^4 + \beta^4 = -30$, then the product of all possible values of $a$ is _____
JEE Mains 25/01/2023 Shift-II
Ans. (45) :
Using symmetric functions of roots and $\alpha^4 + \beta^4 = -30$, the product of all possible values of $a$ is $45$.
19. Let $S = \left\{\alpha \in \mathbb{R} : \log_3 \left(9^{\alpha^2} + 13\right) - \log_3 \left(3^{\alpha^2 - 4} + 1\right) = 2\right\}$. Then the maximum value of $\beta$ for which the equation $\sum_{s \in S} x^2 - 2\left(\sum_{s \in S} \alpha\right)x + \beta = 0$ has real roots, is ______.
JEE Mains 25/01/2023 Shift-I
Ans. (25) :
Solving the logarithmic equation gives $\alpha^2 = 4 \implies \alpha = \pm 2$. Using discriminant $\Delta \ge 0$, the maximum value of $\beta$ is $25$.
20. The number of real solutions of the equation $3\left(x + \frac{1}{x}\right)^2 - 2\left(x + \frac{1}{x}\right) - 5 = 0$ is
(a) $3$
(b) $4$
(c) $0$
(d) $2$
JEE Mains 24/01/2023 Shift-II
Ans. (c) :
Substituting $t = x + \frac{1}{x}$ leads to quadratic values yielding negative discriminants for $x$, hence $0$ real solutions.
21. Let $\lambda \in \mathbb{R}$ and let the equation $E$ be $x^2 - 2|x + \lambda| - 3 = 0$. Then the largest element in the set $S = \{x + \lambda : x \text{ is an integer solution of } E\}$ is _____
JEE Mains 24/01/2023 Shift-I
Ans. (5) :
Analyzing values of $\lambda \in [2, 4]$ and integer solutions for $x$, the largest element in $S$ is $5$.
22. Let $p$ and $q$ be two real numbers such that $p + q = 3$ and $p^4 + q^4 = 369$. Then $\left(\frac{1}{p} + \frac{1}{q}\right)^{-2}$ is equal to ______.
JEE Main-26.06.2022, Shift-II
Ans. (4) :
Using expansion of $p^4 + q^4$, we find possible values for $pq$. Evaluating $\left(\frac{pq}{p+q}\right)^2$ gives $4$.
23. The number of real solutions of the equation $e^{4x} + 4e^{3x} - 58e^{2x} + 4e^x + 1 = 0$ is ______.
JEE Main-28.06.2022, Shift-I
Ans. (2) :
Dividing by $e^{2x}$ and substituting $t = e^x + e^{-x}$ yields $2$ real solutions.
Type III: Conditions for Common Roots
24. The number of distinct real roots of the equation $|x + 1||x + 3| - 4|x + 2| + 5 = 0$, is _________.
JEE Mains 08/04/2024 Shift-II
Ans. (2) :
Analyzing intervals for absolute value functions results in $2$ distinct real roots.
25. If the value of real number $a > 0$ for which $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ have a common real root is $\frac{3}{2\beta}$, then $\beta$ is equal to ________.
JEE Mains 30/01/2023 Shift-II
Ans. (13) :
Using the common root condition formula, we get $a^2 = \frac{36}{26 \times 4} = \frac{9}{26}$. Comparing with $\frac{3}{2\beta}$ gives $\beta = 13$.
Type IV: Location of Roots
26. Let $x = \frac{m}{n}$ ($m, n$ are co-prime natural numbers) be a solution of the equation $\cos\left(\frac{\pi}{9}\right) - 2\sin(x) = 1$ and let $\alpha, \beta$ ($\alpha > \beta$) be the roots of the equation $mx^2 - nx - m + n = 0$. Then the point $(\alpha, \beta)$ lies on the line
(a) $5x - 8y = -9$
(b) $3x - 2y = -2$
(c) $5x + 8y = 9$
(d) $3x + 2y = 2$
JEE Mains 29/01/2024 Shift-II
Ans. (c) :
Solving for $x$ gives $x = \frac{2}{3}$, so $m = 2, n = 3$. The roots of $2x^2 - 3x + 1 = 0$ are $\alpha = 1, \beta = \frac{1}{2}$. Testing these points shows they satisfy $5x + 8y = 9$.
27. The set of all $a \in \mathbb{R}$ for which the equation $x|x - 1| + |x + 2| + a = 0$ has exactly one real root, is
(a) $(-\infty, -3)$
(b) $(-\infty, \infty)$
(c) $(-6, \infty)$
(d) $(-6, -3)$
JEE Mains 13/04/2023 Shift-I
Ans. (b) :
Graph analysis of the increasing function shows it takes all real values, hence the set of all $a$ is $(-\infty, \infty)$.
28. The number of integral values of $k$, for which one root of the equation $2x^2 - 8x + k = 0$ lies in the interval $(1, 2)$ and its other root lies in the interval $(2, 3)$, is :
(a) $0$
(b) $1$
(c) $2$
(d) $3$
JEE Mains 01/02/2023 Shift-II
Ans. (b) :
Using location of roots conditions: $f(1) > 0, f(2) < 0, f(3) > 0$, we get $6 < k < 8$. The only integral value is $7$, giving $1$ value.
Type V: Solution of Quadratic and Higher Degree Equation
29. Let $S = \{x \in \mathbb{R} : (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10\}$. Then the number of elements in $S$ is:
(a) $4$
(b) $0$
(c) $2$
(d) $1$
JEE Mains 01/02/2024 Shift-I
Ans. (c) :
Substituting $t = (\sqrt{3} + \sqrt{2})^x$ yields $t + \frac{1}{t} = 10 \implies t = 5 \pm 2\sqrt{6}$. Solving for $x$ gives $x = \pm 2$, so there are $2$ elements.
30. If $\alpha$ satisfies the equation $x^2 + x + 1 = 0$ and $(1 + \alpha)^7 = A + B\alpha + C\alpha^2$, $A, B, C \ge 0$, then $5(3A - 2B - C)$ is equal to ______.
JEE Mains 27/01/2024 Shift-I
Ans. (5) :
Using cube roots of unity $\omega$, we find $A = 43, B = 43, C = 42$. Substituting gives $5(3(43) - 2(43) - 42) = 5$.
31. Let $\alpha, \beta$ be the distinct roots of the equation $x^2 - (t^2 - 5t + 6)x + 1 = 0$, $t \in \mathbb{R}$ and $a_n = \alpha^n + \beta^n$. Then the minimum value of $\frac{a_{2025} + a_{2023}}{a_{2024}}$ is
(a) $1/4$
(b) $-1/2$
(c) $-1/4$
(d) $1/2$
JEE Mains 06/04/2024 Shift-I
Ans. (c) :
Using Newton's sums, the ratio equals $t^2 - 5t + 6$, whose minimum value as a quadratic in $t$ is $-\frac{1}{4}$.
32. The number of solutions of the equation $e^{\sin x} - 2e^{-\sin x} = 2$ is:
(a) $0$
(b) $1$
(c) more than $2$
(d) $2$
JEE Mains 31/01/2024 Shift-II
Ans. (a) :
Letting $t = e^{\sin x}$ leads to $t^2 - 2t - 2 = 0$, giving $t = 1 \pm \sqrt{3}$. Since $e^{\sin x}$ must lie in $[\frac{1}{e}, e]$, these values are inadmissible, hence $0$ solutions.
33. The sum of all the solutions of the equation $(8)^{2x} - 16(8)^x + 48 = 0$ is :
(a) $1 + \log_6(8)$
(b) $\log_8(6)$
(c) $1 + \log_8(6)$
(d) $\log_8(4)$
JEE Mains 08/04/2024 Shift-I
Ans. (c) :
Let $t = 8^x$, giving $t^2 - 16t + 48 = 0 \implies t = 4, 12$. Thus $x = \log_8 4, \log_8 12$. Their sum is $\log_8 48 = 1 + \log_8 6$.
34. Let $\alpha, \beta$ ($\alpha > \beta$) be the roots of the equation $x^2 - 2x - 3 = 0$. Let $P_n = \alpha^n - \beta^n$, $n \in \mathbb{N}$. Then $(11P_{10} - \sqrt{3}P_9) / (P_{11} - 11P_{12})$ is equal to :
(a) $\frac{10}{2P_9}$
(b) $-\frac{10}{3P_9}$
(c) $\frac{11}{2P_9}$
(d) $\frac{11}{3P_9}$
JEE Mains 09/04/2024 Shift-II
Ans. (b) :
Using recurrence relation $P_{n+2} - 2P_{n+1} - 3P_n = 0$, simplifying the expression yields $-\frac{10}{3P_9}$.
35. Let $S = \left\{x : x \in \mathbb{R} \text{ and } (\sqrt{3} + \sqrt{2})^{x^2 - 4} + (\sqrt{3} - \sqrt{2})^{x^2 - 4} = 10\right\}$. Then $n(S)$ is equal to
(a) $0$
(b) $6$
(c) $2$
(d) $4$
JEE Mains 01/02/2023 Shift-I
Ans. (d) :
Solving $t + \frac{1}{t} = 10$ gives $x^2 - 4 = \pm 2$, resulting in $4$ real solutions for $x$.
36. Let $\alpha, \beta$ be the roots of the equation $x^2 - 2x + 2 = 0$, then $\alpha^{14} + \beta^{14}$ is equal to.
(a) $-128\sqrt{2}$
(b) $-64\sqrt{2}$
(c) $-128$
(d) $-64$
JEE Mains 13/04/2023 Shift-II
Ans. (c) :
Converting roots to polar form $1 \pm i = \sqrt{2}e^{\pm i\pi/4}$ and applying De Moivre's theorem gives $-128$.
37. Let $\alpha, \beta, \gamma$ be the three roots of the equation $x^3 + bx + c = 0$. If $\beta\gamma = 1 = -\alpha$, then $b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3$ is equal to
(a) $\frac{155}{8}$
(b) $21$
(c) $19$
(d) $\frac{169}{8}$
JEE Mains 08/04/2023 Shift-I
Ans. (c) :
Using root relations, we find $\alpha = -1, c = 1, b = 0$, and roots are $-1, -\omega, -\omega^2$. Substituting these values yields $19$.
38. Let $m$ and $n$ be the numbers of real roots of the quadratic equations $x^2 - 12x + [x] + 31 = 0$ and $x^2 - 5|x + 2| - 4 = 0$ respectively, where $[x]$ denotes the greatest integer $\le x$. Then $m^2 + mn + n^2$ is equal to _____.
JEE Mains 08/04/2023 Shift-II
Ans. (9) :
Solving the first equation yields $m = 0$ and solving the second yields $n = 3$. Thus $m^2 + mn + n^2 = 9$.
39. The equation $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0, x \in \mathbb{R}$ has:
(a) Four solutions two of which are negative
(b) Two solutions and only one of them is negative
(c) Two solutions and both are negative
(d) No solution
JEE Mains 31/01/2023 Shift-II
Ans. (c) :
Dividing by $e^{2x}$ and substituting $z = e^x - e^{-x}$ yields two negative solutions for $x$.
40. If the solution of the equation $\log_{\cos x} \cot x + 4\log_{\sin x} \tan x = 1, x \in \left(0, \frac{\pi}{2}\right)$, is $\sin^{-1}\left(\frac{\alpha + \sqrt{\beta}}{2}\right)$, where $\alpha, \beta$ are integers, then $\alpha + \beta$ is equal to :
(a) $5$
(b) $3$
(c) $6$
(d) $4$
JEE Mains 30/01/2023 Shift-I
Ans. (d) :
Simplifying logarithmic terms leads to $\sin^2 x - \sin x + 1 = 0$. Solving for $\sin x$ and matching coefficients gives $\alpha = -1, \beta = 5$, so $\alpha + \beta = 4$.
41. Let $\alpha_1, \alpha_2, \dots, \alpha_7$ be the roots of the equation $x^7 + 3x^5 - 13x^3 - 15x = 0$ and $|\alpha_1| \ge |\alpha_2| \ge \dots \ge |\alpha_7|$. Then $\alpha_1\alpha_2 - \alpha_3\alpha_4 + \alpha_5\alpha_6$ is equal to _______
JEE Mains 29/01/2023 Shift-II
Ans. (9) :
Factorizing gives roots $0, \pm\sqrt{3}, \pm i, \pm i\sqrt{5}$. Ordering by magnitude and evaluating the expression yields $9$.
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