Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

For roots in a geometric progression, let them be represented as $\alpha, \alpha r, \alpha r^{2}, \alpha r^{3}$. Summing the roots for the first equation yields $\alpha(1+r)=3$, and for the second equation $\alpha r^{2}(1+r)=6$, revealing $r^{2}=2$. Expressing the products $p = \alpha^{2}r$ and $q = \alpha^{2}r^{5}$ allows the ratio $\frac{2q+p}{2q-p}$ to be simplified to $\frac{2r^{4}+1}{2r^{4}-1}$. With $r^{2}=2$ and thus $r^{4}=4$, the ratio is calculated as $\frac{2(4)+1}{2(4)-1} = 9/7$.

For a quadratic equation to have real roots, its discriminant must be non-negative. Applying this to $2x^2 + (a - 10)x + (\frac{33}{2} - 2a) = 0$, we calculate $D = (a - 10)^2 - 8(\frac{33}{2} - 2a) \ge 0$. This expands to $a^2 - 20a + 100 - 132 + 16a \ge 0$, simplifying to $a^2 - 4a - 32 \ge 0$. Factoring the inequality as $(a-8)(a+4) \ge 0$ identifies the solution range as $a \in (-\infty , -4] \cup [8, \infty )$, making 8 the least positive value.

The greatest integer function satisfies the property $[x+n] = [x] + n$ for any integer $n$. Applying this to the equation $[x]^{2}+2[x+2]-7=0$ yields $[x]^{2}+2([x]+2)-7=0$, which simplifies to the quadratic $[x]^{2}+2[x]-3=0$. Factoring this expression gives $([x]+3)([x]-1)=0$, implying that $[x]$ must be either $1$ or $-3$. Since $[x]=1$ corresponds to $x ext{ \in } [1, 2)$ and $[x]=-3$ corresponds to $x ext{ \in } [-3, -2)$, there are infinitely many real values of $x$ that satisfy the equation.

For any quadratic equation $ax^{2+}bx+c=0$ with roots $\alpha$ and \eta, the power sums $S_n = \alpha^n+\beta^n$ satisfy the recurrence relation $a S_{n+2} + b S_{n+1} + c S_n = 0$. Substituting the coefficients from $5x^{2+}6x-2=0$, we obtain $5 S_{n+2} + 6 S_{n+1} - 2 S_n = 0$. By setting $n=4$, the equation becomes $5 S_6 + 6 S_5 - 2 S_4 = 0$, which directly simplifies to $5 S_6 + 6 S_5 = 2 S_4$.

A quadratic polynomial with a root at $3$ can be expressed as $f(x)=a(x-3)(x-\alpha)$, where $\alpha$ is the other root. Evaluating at the given points, $f(-1)=a(-4)(-1-\alpha)=4a(1+\alpha)$ and $f(2)=a(-1)(2-\alpha)=a(\alpha-2)$. Setting $f(-1)+f(2)=0$ results in $a(4+4\alpha+\alpha-2)=a(2+5\alpha)=0$. Since $a$ cannot be zero, $5\alpha=-2$, giving $\alpha=-0.4$, which falls within the interval $(-1, 0)$.

Substituting $t = 3^x$ for $t > 0$ yields the equation $t^2 - t + 2 = |t-1| + |t-2|$. For $t < 1$, the equation becomes $t^2 - t + 2 = 3 - 2t$, which simplifies to $t^2 + t - 1 = 0$, giving $t = \frac{-1 + \sqrt{5}}{2}$ as the only positive root. For $1 \le t < 2$, the equation reduces to $t^2 - t + 1 = 0$, which has no real roots. For $t \ge 2$, the equation becomes $t^2 - 3t + 5 = 0$, which also lacks real roots, confirming that only one value of $t$ exists, making the set of real roots a singleton.

Starting with the characteristic equation $x^2 - x - 1 = 0$, we observe that any root $x$ satisfies $x^2 = x + 1$. By multiplying this relation by $x^{k-2}$, we derive the recurrence relation $p_k = p_{k-1} + p_{k-2}$ for the power sums $p_k = \alpha^k + \beta^k$. Calculating the initial values with $p_1 = \alpha + \beta = 1$ and $p_2 = (\alpha+\beta)^2 - 2\alpha\beta = 1^2 - 2(-1) = 3$, we find the subsequent values as $p_3 = 4$, $p_4 = 7$, and $p_5 = 11$. Testing the given options, we find that $p_5 = p_2 \cdot p_3$ results in $11 = 3 \cdot 4$, which is an incorrect statement.

For a quadratic function $f(x)=ax^{2+}bx+c$ to have exactly one root in the interval $(0, 1)$, either $f(0) ext{ and } f(1)$ must have opposite signs or one root must lie on the boundary of the interval. Evaluating $f(0)=2$ and $f(1)=\lambda^{2-}4\lambda+3$, the condition $f(0)f(1)<0$ leads to $2(\lambda-1)(\lambda-3)<0$, which restricts $\lambda$ to the interval $(1, 3)$. Checking the endpoints, $\lambda=1$ results in $2(x-1)^2=0$ (no root in $(0, 1)$), while $\lambda=3$ yields $10x^{2-}12x+2=0$ with roots $1$ and $1/5$; since $1/5$ lies within $(0, 1)$, the value $\lambda=3$ is included, resulting in the interval $(1, 3]$.

Using Vieta's formulas, the sum of roots $\alpha+\beta=3/7$ and the product $\alpha\beta=-2/7$ are obtained from the given equation. The expression $\frac{\alpha}{1-\alpha^{2}}+\frac{\beta}{1-\beta^{2}}$ simplifies to $\frac{\alpha+\beta-\alpha\beta(\alpha+\beta)}{1-(\alpha+\beta)^{2}+2\alpha\beta+(\alpha\beta)^{2}}$. Substituting the values $3/7$ and $-2/7$ results in a numerator of $27/49$ and a denominator of $16/49$, which gives a final value of $27/16$.

Given roots $\alpha, \beta$ for $x^{2+}px+2=0$, we have $\alpha+\beta=-p$ and $\alpha\beta=2$. For the second equation $2x^{2+}2qx+1=0$ with roots $1/\alpha, 1/\beta$, the product $1/(\alpha\beta)=1/2$ is consistent and the sum $1/\alpha+1/\beta = -q$ implies $(\alpha+\beta)/(\alpha\beta) = -p/2 = -q$, so $p=2q$. Expanding the expression $(\alpha-1/\alpha)(\beta-1/\beta)(\alpha+1/\beta)(\beta+1/\alpha)$ gives $[(\alpha^{2-}1)(\beta^{2-}1)(\alpha\beta+1)^2] / (\alpha\beta)^2$. Substituting $\alpha^2=-p\alpha-2$ and $\beta^2=-p\beta-2$, the numerator becomes $(-p\alpha-3)(-p\beta-3)(3)^2 = 9(p^2\alpha\beta + 3p(\alpha+\beta) + 9) = 9(2p^2 - 3p^2 + 9) = 9(9-p^2)$, which when divided by $(\alpha\beta)^2=4$ yields $\frac{9}{4}(9-p^2)$.

Root properties of $4x^{2}+2x-1=0$ define the relationship between roots $\alpha$ and $\beta$ via $\alpha+\beta=-\frac{2}{4}=-0.5$. Since $\alpha$ is a root, it satisfies $4\alpha^{2}+2\alpha=1$, which is equivalent to $2\alpha^{2}+\alpha=0.5$. Substituting $0.5=2\alpha^{2}+\alpha$ into the sum relation $\beta=-0.5-\alpha$ gives $\beta=-(2\alpha^{2}+\alpha)-\alpha$, which simplifies to $-2\alpha^{2}-2\alpha$ or $-2\alpha(\alpha+1)$.

For the quadratic equation $x^{2}-64x+256=0$, Vieta's formulas give $\alpha+\beta=64$ and $\alpha\beta=256$. The expression $\left(\frac{\alpha^{3}}{\beta^{5}}\right)^{1/8}+\left(\frac{\beta^{3}}{\alpha^{5}}\right)^{1/8}$ simplifies to $\frac{\alpha+\beta}{(\alpha\beta)^{5/8}}$. Since $256=2^{8}$, the denominator becomes $(2^{8})^{5/8} = 2^{5} = 32$, and the final expression evaluates to $64/32 = 2$.

Vieta's formulas for the first equation $x^{2}-x+2\lambda=0$ give $\alpha+\beta=1$ and $\alpha\beta=2\lambda$, while for the second equation $3x^{2}-10x+27\lambda=0$ they give $\alpha+\gamma=10/3$ and $\alpha\gamma=9\lambda$. Subtracting the sums yields $\gamma-\beta=7/3$, and relating the product equations shows $\gamma/\beta = 4.5$. Solving these simultaneous equations for $\beta$ and $\gamma$ yields $\beta=2/3$ and $\gamma=3$, which implies $\lambda=1/9$, leading to $\frac{\beta\gamma}{\lambda} = \frac{(2/3)(3)}{1/9} = 18$.

The value $\alpha = \frac{-1 + i\sqrt{3}}{2}$ represents the primitive cube root of unity $\omega$, where $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. The sum $a = (1+\omega)\sum_{k=0}^{100} \omega^{2k}$ simplifies to $(1+\omega)\frac{1-\omega^{202}}{1-\omega^2}$, and since $\omega^{202} = \omega$, this equals $(1+\omega)\frac{1-\omega}{1-\omega^2} = 1$. The sum $b = \sum_{k=0}^{100} \omega^{3k}$ is simply $\sum_{k=0}^{100} 1$, which equals $101$. The quadratic equation with roots $1$ and $101$ is given by $(x-1)(x-101) = x^2 - 102x + 101 = 0$.

By substituting $t=|x|$, the equation transforms into the quadratic form $9t^{2}-18t+5=0$, which factors into $(3t-5)(3t-1)=0$. This leads to $|x|=5/3$ or $|x|=1/3$, meaning the roots are $\pm 5/3$ and $\pm 1/3$. The product of these four roots is calculated as $(5/3)(-5/3)(1/3)(-1/3) = 25/81$.

Factoring the quadratic equation $x^2 \sin \theta - x(\sin \theta \cos \theta + 1) + \cos \theta = 0$ reveals roots $x = \cos \theta$ and $x = 1/\sin \theta$, or $\csc \theta$. Given the constraint $0 < \theta < 45^{\circ}$, $\cos \theta < 1 < \csc \theta$, so $\alpha = \cos \theta$ and $\beta = \csc \theta$. The series becomes $\sum_{n=0}^{\infty } (\cos \theta)^n + \sum_{n=0}^{\infty } (-\sin \theta)^n$. Using the formula for an infinite geometric series $\sum r^n = 1/(1 - r)$, we obtain $1/(1 - \cos \theta) + 1/(1 - (-\sin \theta))$, which simplifies to $1/(1 - \cos \theta) + 1/(1 + \sin \theta)$.

Letting $u = 2^x$, the equation transforms into $5 + |u - 1| = u^2 - 2u$. In the case $u \ge 1$, the equation becomes $u - 1 = u^2 - 2u - 5$, simplifying to $u^2 - 3u - 4 = 0$, which yields the valid root $u=4$ corresponding to $x=2$. In the case $u < 1$, the equation becomes $-(u - 1) = u^2 - 2u - 5$, or $u^2 - u - 6 = 0$, leading to $u=3$ or $u=-2$, neither of which satisfy the condition $u < 1$ for an exponential function $2^x$. Thus, there is only one real root.

Vieta's formulas express the sum and product of the roots as $\alpha+\beta=-\sin\theta$ and $\alpha\beta=-2\sin\theta$. Simplifying the expression $\frac{\alpha^{12}+\beta^{12}}{(\alpha^{-12}+\beta^{-12})(\alpha-\beta)^{24}}$ involves multiplying the numerator and denominator by $(\alpha\beta)^{12}$, which transforms the term $\frac{\alpha^{12}+\beta^{12}}{\alpha^{-12}+\beta^{-12}}$ into $(\alpha\beta)^{12}$. Consequently, the entire expression simplifies to $\frac{(\alpha\beta)^{12}}{(\alpha-\beta)^{24}}$, which is $\left(\frac{\alpha\beta}{(\alpha-\beta)^2}\right)^{12}$. Utilizing the identity $(\alpha-\beta)^2=(\alpha+\beta)^{2-}4\alpha\beta$, we compute $(\alpha-\beta)^2=\sin^2\theta+8\sin\theta$, leading to the final form $\frac{(-2\sin\theta)^{12}}{(\sin^2\theta+8\sin\theta)^{12}}$, which reduces to $\frac{2^{12}}{(\sin\theta+8)^{12}}$.

The sum of the squares of the roots $\alpha^2 + \beta^2$ is given by $(\alpha + \beta)^2 - 2\alpha\beta$. For the equation $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$, Vieta's formulas provide $\alpha + \beta = \lambda - 3$ and $\alpha\beta = 2 - \lambda$. Substituting these into the expression results in $(\lambda - 3)^2 - 2(2 - \lambda) = \lambda^2 - 6\lambda + 9 - 4 + 2\lambda = \lambda^2 - 4\lambda + 5$. Completing the square gives $(\lambda - 2)^2 + 1$, which reaches a minimum value of 1 when $\lambda = 2$.

For a quadratic equation with rational coefficients, if $2-\sqrt{3}$ is a root, then its conjugate $2+\sqrt{3}$ must also be a root. The sum of these roots is $(2-\sqrt{3})+(2+\sqrt{3})=4$, which corresponds to $-p$, giving $p=-4$. Their product is $(2-\sqrt{3})(2+\sqrt{3})=4-3=1$, which corresponds to $q=1$. Substituting these values into the expression $p^{2-}4q-12$ results in $(-4)^{2-}4(1)-12 = 16-4-12=0$.