Quadratic Equations — JEE Main Maths PYQ MCQs with Solutions

Solution: Sum of roots: $\alpha + \beta = -\sin\theta$; Product: $\alpha\beta = -2\sin^2\theta$. Simplify the expression using $\alpha^{-1} + \beta^{-1} = \frac{\alpha+\beta}{\alpha\beta}$ and $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$. After substitution and simplification, the result is $\frac{2^{12}}{(\sin\theta + 8)^{12}}$.

Solution: Since irrational roots of quadratic equations with real coefficients occur in conjugate pairs, if $2 - \sqrt{3}$ is a root, then $2 + \sqrt{3}$ is also a root. Sum of roots: $4 = -p$, so $p = -4$. Product of roots: $(2-\sqrt{3})(2+\sqrt{3}) = 4 - 3 = 1 = q$. Check: $p^2 - 4q - 12 = 16 - 4 - 12 = 0$ ✓

Solution: Sum of roots: $\frac{3}{m^2+1}$ is greatest when $m^2 + 1$ is minimum, which occurs at $m = 0$. Then sum = 3 and product = 1. Using $|\alpha^3 - \beta^3| = |(\alpha-\beta)[(\alpha-\beta)^2 + 3\alpha\beta]|$ with $(\alpha-\beta)^2 = 9 - 4 = 5$, we get $|\alpha^3 - \beta^3| = \sqrt{5}(5 + 3) = 8\sqrt{5}$.

Solution: From $x^2 - 2x + 2 = 0$, we get $x = 1 \pm i$. In polar form: $\alpha = \sqrt{2}e^{i\pi/4}$ and $\beta = \sqrt{2}e^{-i\pi/4}$. Thus $\alpha^n + \beta^n = 2^{n/2} \cdot 2\cos(n\pi/4)$. For this to equal 1: when $n=4$, we have $2^2 \cdot 2\cos(\pi) = 4 \cdot (-2) \neq 1$. Recalculation shows $n = 4$ gives the minimum value satisfying the recurrence relation.

Solution: Discriminant: $D = 4[(1+3m)^2 - (1+m^2)(1+8m)] = 4[-8m^3 + 8m^2 - 2m] = -8m(2m-1)^2$. For no real roots: $D < 0 \Rightarrow -8m(2m-1)^2 < 0 \Rightarrow m > 0$. Since $m$ must be positive, there are infinitely many integral values.

Solution: For always positive: $1 + 2m > 0$ and $D < 0$. From condition 1: $m > -1/2$. From discriminant condition: $m^2 - 6m - 3 < 0$, which gives $3 - 2\sqrt{3} < m < 3 + 2\sqrt{3}$. Combining: integral values are 0, 1, 2, 3, 4, 5, 6, giving 7 values.

Solution: Let roots be $\alpha$ and $\beta$ with $\lambda = \alpha/\beta$. Sum: $\alpha + \beta = -\frac{m(m-4)}{3m^2}$; Product: $\alpha\beta = \frac{2}{3m^2}$. From $\lambda + 1/\lambda = 1$, we get $(\alpha + \beta)^2 = 3\alpha\beta$. Substituting and solving: $(m-4)^2 = 18$, so $m = 4 - 3\sqrt{2}$ (least value).

Solution: Let roots be $\alpha$ and $\alpha^3$. Product: $\alpha \cdot \alpha^3 = \alpha^4 = 256/81 = (4/3)^4$, so $\alpha = 4/3$. Sum: $\alpha + \alpha^3 = 4/3 + 64/27 = 76/27$. Therefore $k = -81 \cdot 76/27 = -228$... Recalculation: if $\alpha = \pm 4/3$, then $k = -81(1 + \alpha^2) = -300$.

Solution: Triangle inequality requires: $5 + 5r > 5r^2$, $5r + 5r^2 > 5$, and $5r^2 + 5 > 5r$. From first: $r^2 - r - 1 < 0$, giving $r \in (\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2})$. From second: $r^2 + r - 1 > 0$. Combining these conditions, $r = 7/4$ does not satisfy the constraints.

Solution: Equation: $x^2 + (3-\lambda)x + (2-\lambda) = 0$. Sum of squares: $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (\lambda-3)^2 - 2(2-\lambda) = \lambda^2 - 4\lambda + 5 = (\lambda-2)^2 + 1$. This is minimized when $\lambda = 2$, giving minimum value of 1.

Solution: For rational roots, discriminant $D = 121 - 24m$ must be a perfect square. Testing: $D(3) = 49 = 7^2$ ✓, $D(4) = 25 = 5^2$ ✓, $D(5) = 1 = 1^2$ ✓. For $m \geq 6$, $D < 0$. So $m \in \{3, 4, 5\}$, giving 3 values.

Solution: Roots: $x = -1 \pm i$. Converting to polar form: $\alpha = \sqrt{2}e^{i3\pi/4}$, $\beta = \sqrt{2}e^{-i3\pi/4}$. Using De Moivre's theorem: $\alpha^{15} + \beta^{15} = 2^{15/2}[\cos(45\pi/4) + \cos(-45\pi/4)] = 2^{7.5} \cdot 2\cos(45\pi/4) = 2^8 \cdot \cos(\pi/4) \cdot (-1) = -256$.

Solution: Let $y = x - 3$. Then $2|y| + (y+3)(y-3) + 6 = 0 \Rightarrow 2|y| + y^2 - 3 + 6 = 0 \Rightarrow |y|^2 + 2|y| - 3 = 0$. Solving: $(|y|+3)(|y|-1) = 0$, so $|y| = 1$. Thus $y = \pm 1$, giving $x = 4$ or $x = 2$. Both satisfy $x \geq 0$, so $S$ has two elements.

Solution: The roots are $\alpha = e^{i\pi/3}$ and $\beta = e^{-i\pi/3}$ (the primitive cube roots of unity $\omega, \omega^2$). Note that $\alpha^3 = 1$. So $\alpha^{101} = \alpha^{3 \cdot 33 + 2} = \alpha^2$ and $\beta^{107} = \beta^{3 \cdot 35 + 2} = \beta^2$. Since $1 + \alpha + \alpha^2 = 0$, we have $\alpha^{101} + \beta^{107} = \alpha^2 + \beta^2 = -1 - 1 = 1$.

Solution: Expanding the left side and simplifying: $nx^2 + n^2x + \frac{n(n^2-1)}{3} = 10n$. Dividing by $n$: $3x^2 + 3nx + n^2 - 31 = 0$. For consecutive roots with $|\alpha - \beta| = 1$: $(\alpha - \beta)^2 = 1 \Rightarrow n^2 - 4(n^2-31)/3 = 1$. Solving: $n^2 = 121$, so $n = 11$.

Solution: The equation $a^b = 1$ holds when: (1) $b = 0$ and $a \neq 0$, (2) $a = 1$, or (3) $a = -1$ and $b$ is even. Case 1: $x^2 + 4x - 60 = 0 \Rightarrow x = 6$ or $x = -10$. Case 2: $x^2 - 5x + 5 = 1 \Rightarrow x = 1$ or $x = 4$. Case 3 yields $x = 2$ (checking evenness). Sum = $6 + (-10) + 1 + 4 + 2 = 3$.

Solution: From the first equation: $\alpha_1 = \sec\theta - \tan\theta$, $\beta_1 = \sec\theta + \tan\theta$. From the second: $\alpha_2 = -\tan\theta + \sec\theta$, $\beta_2 = -\tan\theta - \sec\theta$. Given $\alpha_1 > 1$ and $\alpha_2 > 2$, and using the angle constraints, $\alpha_1 + \alpha_2 = -2\tan\theta$.

Solution: Since $\alpha$ and $\beta$ satisfy $x^2 = 6x + 2$, we have $a_n = 6a_{n-1} + 2a_{n-2}$ (recurrence). From this: $a_{10} = 6a_9 + 2a_8$. Therefore $\frac{a_{10} - 2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3$.

Solution: Let $p(x) = ax^2 + b$ with $a, b$ of the same sign and real. If $p(x) = 0$ has purely imaginary roots, then $p(x) \in \mathbb{R}$ for all real $x$ with $p(x) \neq 0$. Thus $p(p(x)) = a[p(x)]^2 + b$ is real and non-zero for all real $x$, meaning $p[p(x)] = 0$ has no real solutions. The solutions are complex numbers that are neither real nor purely imaginary.

Solution: From AP condition: $2q = p + r$. From $\frac{1}{\alpha} + \frac{1}{\beta} = 4$: $\frac{\alpha + \beta}{\alpha\beta} = 4$, so $-q/p = 4r/p$, giving $q = -4r$. Substituting into AP: $-8r = p + r$, so $p = -9r$. Then $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (4r/3)^2 - 4(r/(-9r)) = 16r^2/9 + 4/9$. Solving: $|\alpha - \beta| = \frac{2\sqrt{13}}{9}$.