Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

For the roots of a quadratic equation with integer coefficients to be rational, its discriminant must be a perfect square. Calculating the discriminant for $6x^2 - 11x + \alpha = 0$ gives $D = 121 - 24\alpha$, which must be a perfect square. Testing positive integer values $\alpha \le 5$, we find that $\alpha=3$ gives $49$, $\alpha=4$ gives $25$, and $\alpha=5$ gives $1$, all of which are perfect squares, resulting in 3 possible values.

A quadratic expression is strictly positive for all real values if its leading coefficient is positive and its discriminant is negative. For $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$, this requires $1 + 2m > 0$, giving $m > -0.5$. The discriminant condition $D < 0$ simplifies to $[-2(1 + 3m)]^2 - 4(1 + 2m)(4(1 + m)) < 0$, which results in $4(1 + 6m + 9m^2) - 16(1 + 3m + 2m^2) < 0$. This inequality reduces to $m^2 - 6m - 3 < 0$. Solving the quadratic $m^2 - 6m - 3 = 0$ gives $m = 3 eq \sqrt{12}$, approximately $-0.46$ and $6.46$. Combining these conditions, the interval is $3 - \sqrt{12} < m < 3 + \sqrt{12}$, and the integral values within this range are $0, 1, 2, 3, 4, 5, 6$, totaling 7 values.

The location of roots for a quadratic $f(x)$ depends on the signs of the function values at the boundaries of the given intervals. For one root to lie in $(0,2)$ and the other in $(2,3)$, the value at $x=2$ must be on the opposite side of the origin relative to the values at $x=0$ and $x=3$. Evaluating $f(0) = c-4$, $f(2) = c-24$, and $f(3) = 4c-49$, the condition $f(0)f(2) < 0$ yields $4 < c < 24$, while $f(2)f(3) < 0$ yields $12.25 < c < 24$. Combining these, the integer values for $c$ are the set {13, 14, ..., 23}, which contains exactly 11 elements.

The absolute value requires analyzing the equation in two cases based on the domain of $\sqrt{x}$. For $\sqrt{x} \geq 2$, the equation becomes $(\sqrt{x}-2) + x - 4\sqrt{x} + 2 = 0$, simplifying to $x-3\sqrt{x}=0$, which yields $\sqrt{x}=3$ and thus $x=9$. For $\sqrt{x} < 2$, the equation becomes $(2-\sqrt{x}) + x - 4\sqrt{x} + 2 = 0$, leading to $x-5\sqrt{x}+4=0$, which factors as $(\sqrt{x}-4)(\sqrt{x}-1)=0$ to give $\sqrt{x}=1$, so $x=1$. The sum of these valid solutions is $9+1=10$.

A quadratic equation has no real roots when its discriminant $D=b^{2-}4ac$ is strictly less than zero. Calculating the discriminant for $(1+m^2)x^{2-}2(1+3m)x+(1+8m)=0$ gives $[2(1+3m)]^2 - 4(1+m^2)(1+8m) < 0$, which simplifies to $4(1+6m+9m^2) - 4(1+8m+m^{2+}8m^3) < 0$. Algebraic reduction yields $8m^{3-}8m^{2+}2m > 0$, or $2m(2m-1)^2 > 0$. Since $(2m-1)^2$ is positive for all real $m$ except $0.5$, the condition requires $m>0$ and $m \neq 0.5$, which allows for infinitely many positive integral values for $m$.

Let the roots be $\alpha$ and $\alpha^3$. According to Vieta's formulas, their product is $\alpha \cdot \alpha^3 = \alpha^4 = 256/81$, which gives $\alpha^2 = 16/9$ and $\alpha = \pm 4/3$. The sum of the roots is $\alpha + \alpha^3 = -k/81$, which we can write as $\alpha(1 + \alpha^2) = -k/81$. Substituting $\alpha = \pm 4/3$ and $\alpha^2 = 16/9$, we get $\pm (4/3)(1 + 16/9) = \pm (4/3)(25/9) = \pm 100/27 = -k/81$. Multiplying by $-81$ leads to $k = \mp 300$, and given the options, $-300$ is the valid solution.

Given that $\alpha, \beta, \gamma$ are consecutive terms in a geometric progression, we express them as $\alpha, \alpha r, \alpha r^2$, allowing the quadratic equation $\alpha x^2 + 2\beta x + \gamma = 0$ to be written as $\alpha(x^2 + 2rx + r^2) = 0$, which factors into $\alpha(x+r)^2 = 0$. This identifies the common root as $x = -r$, and since it must satisfy $x^2 + x - 1 = 0$, substituting yields $r^2 - r - 1 = 0$, implying $r^2 = r+1$. Finally, computing $\alpha(\beta + \gamma)$ gives $\alpha(\alpha r + \alpha r^2) = \alpha^2 r(1+r) = \alpha^2 r(r^2) = (\alpha r)(\alpha r^2)$, which is equal to $\beta \gamma$.

When both roots of a quadratic equation lie within a closed interval, the discriminant must be non-negative, the function values at the interval boundaries must satisfy specific sign conditions, and the vertex must reside within the interval. For $x^2 - mx + 4 = 0$ with roots in $[1, 5]$, we require $m^2 - 16 > 0$, $f(1) \ge 0$, $f(5) \ge 0$, and the vertex $m/2$ to be in $[1, 5]$. These constraints lead to the intersection $m \in (4, 5]$, which is not represented among the provided choices, making none of the options correct.

For a quadratic equation $ax^2 + bx + c = 0$, the roots $\alpha$ and $\beta$ satisfy $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$. Given the ratio $\lambda = \alpha/\beta$, the condition $\lambda + 1/\lambda = 1$ is equivalent to $(\alpha^2 + \beta^2)/(\alpha\beta) = 1$, which simplifies to $(\alpha + \beta)^2 = 3\alpha\beta$. Substituting the sum $\alpha + \beta = -(m - 4)/(3m)$ and product $\alpha\beta = 2/(3m^2)$, we get $((m - 4)/(3m))^2 = 3(2/(3m^2))$. This simplifies to $(m - 4)^2 = 18$, so $m - 4 = \pm 3\sqrt{2}$. Solving for $m$, the least value is $4 - 3\sqrt{2}$.

We can rewrite the inequality as $2^{\sqrt{(\sin x - 1)^2 + 4}} \le 2^{2\sin^2 y}$. Since the minimum value of the expression $\sqrt{(\sin x - 1)^2 + 4}$ is $\sqrt{4} = 2$, occurring when $\sin x = 1$, the minimum value of the left-hand side is $2^2 = 4$. The right-hand side reaches a maximum value of $4^{\sin^2 y} = 2^{2\sin^2 y} = 4$ when $\sin^2 y = 1$. For the inequality to hold, both sides must equal 4, requiring $\sin x = 1$ and $\sin^2 y = 1$, which satisfies the condition $\sin x = |\sin y| = 1$.

The sum of the roots of a quadratic equation $ax^{2+}bx+c=0$ is given by $-\frac{b}{a}$, meaning the sum for $(m^{2+}1)x^{2-}3x+(m^{2+}1)^2=0$ is $\frac{3}{m^{2+}1}$. This value reaches its maximum when the denominator $m^{2+}1$ is minimized, which occurs at $m=0$, resulting in the equation $x^{2-}3x+1=0$. The absolute difference of cubes $|\alpha^{3-}\beta^3|$ is calculated using the identity $|\alpha-\beta| \cdot |(\alpha+\beta)^{2-}\alpha\beta|$. Given $\alpha+\beta=3$ and $\alpha\beta=1$, we find $|\alpha-\beta| = \sqrt{(\alpha+\beta)^{2-}4\alpha\beta} = \sqrt{9-4} = \sqrt{5}$, yielding an absolute difference of $\sqrt{5} \cdot (9-1) = 8\sqrt{5}$.

For the quadratic equation $3x^2 - 10x - 25 = 0$ with roots $\tan A$ and $\tan B$, Vieta's formulas provide the sum and product of the roots as $\tan A + \tan B = \frac{10}{3}$ and $\tan A \tan B = -\frac{25}{3}$. Applying the tangent addition identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, we obtain $\tan(A+B) = \frac{10/3}{1 - (-25/3)} = \frac{10/3}{28/3} = \frac{5}{14}$.

Dividing the given expression $3\sin^2(A+B) - 10\sin(A+B)\cos(A+B) - 25\cos^2(A+B)$ by $\cos^2(A+B)$ results in $\cos^2(A+B) [3\tan^2(A+B) - 10\tan(A+B) - 25]$. Substituting $\tan(A+B) = \frac{5}{14}$ and noting that $\cos^2(A+B) = \frac{1}{1 + \tan^2(A+B)} = \frac{1}{1 + (5/14)^2} = \frac{196}{221}$, the expression becomes $\frac{196}{221} [3(\frac{25}{196}) - 10(\frac{5}{14}) - 25]$. Simplifying the term inside the brackets gives $\frac{75}{196} - \frac{700}{196} - \frac{4900}{196} = -\frac{5525}{196}$, and multiplying this by $\frac{196}{221}$ results in $-\frac{5525}{221}$, which evaluates to $-25$.

Given $5 \cos A + 3 = 0$, we find $\cos A = -\frac{3}{5}$, which implies $\sin A = \frac{4}{5}$ and thus $\tan A = -\frac{4}{3}$ and $\sec A = -\frac{5}{3}$. Solving the quadratic equation $9x^2 + 27x + 20 = 0$ using the quadratic formula yields roots $\frac{-27 \pm \sqrt{729-720}}{18}$, which simplifies to $-\frac{4}{3}$ and $-\frac{5}{3}$. Comparing these results confirms the roots are $\tan A$ and $\sec A$.

The sum of the cubes of the roots $\alpha$ and $\beta$ of the equation $x^2 + (2-\lambda)x + (10-\lambda) = 0$ is given by $S = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$, where Vieta's formulas provide $\alpha + \beta = \lambda - 2$ and $\alpha\beta = 10 - \lambda$. Substituting these into the expression for $S$ yields $S = (\lambda - 2)^3 - 3(10 - \lambda)(\lambda - 2) = (\lambda - 2)(\lambda^2 - 4\lambda + 4 - 30 + 3\lambda) = \lambda^3 - 3\lambda^2 - 24\lambda + 52$. Setting the derivative $S' = 3\lambda^2 - 6\lambda - 24$ to zero results in $\lambda^2 - 2\lambda - 8 = 0$, with solutions $\lambda = 4$ and $\lambda = -2$. The second derivative $S'' = 6\lambda - 6$ is positive for $\lambda = 4$, identifying it as the value producing the minimum sum. The magnitude of the difference of the roots is $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \sqrt{(\lambda - 2)^2 - 4(10 - \lambda)} = \sqrt{\lambda^2 - 36}$. Substituting $\lambda = 4$ results in $\sqrt{|16 - 36|} = \sqrt{20} = 2\sqrt{5}$.

A quadratic expression $f(x)$ with a known root at $-1$ can be written in factored form as $a(x+1)(x-r)$, where $r$ is the other root. Substituting this form into the given equation $f(1) + f(2) = 0$, we obtain $a(1+1)(1-r) + a(2+1)(2-r) = 0$. Simplifying this leads to $2(1-r) + 3(2-r) = 0$, which results in $8 - 5r = 0$, yielding $r = \frac{8}{5}$.

Substituting $t = \sqrt{x}$ into the equation, with the constraint $t \ge 0$, allows us to rewrite the expression as $2|t-3| + t^2 - 6t + 6 = 0$. Since the absolute value function changes behavior at $t=3$, splitting the domain into two intervals reveals the potential values for $t$.

When $t \ge 3$, the absolute value term becomes $2(t-3)$, simplifying the equation to $2t - 6 + t^2 - 6t + 6 = 0$, or $t^2 - 4t = 0$. This factors to $t(t-4) = 0$, giving $t=0$ and $t=4$. Because our condition requires $t \ge 3$, only $t=4$ is valid, resulting in $x = 16$.

When $t < 3$, the absolute value term becomes $2(3-t)$, changing the equation to $6 - 2t + t^2 - 6t + 6 = 0$, which simplifies to $t^2 - 8t + 12 = 0$. Factoring the quadratic as $(t-6)(t-2) = 0$ provides solutions $t=6$ and $t=2$. With the constraint $t < 3$, we must discard $t=6$ and keep $t=2$, which corresponds to $x = 4$. Bringing these findings together, the set $S$ consists of $4$ and $16$, establishing that it contains exactly two elements.

The sum of roots being zero is the key condition for roots to be equal in magnitude but opposite in sign. For the equation $\frac{2x+p+q}{(x+p)(x+q)} = \frac{1}{r}$, we expand to $x^2 + (p+q-2r)x + (pq-rp-rq) = 0$. Setting the coefficient of $x$ to zero gives $r = \frac{p+q}{2}$. Substituting this into the constant term, the equation becomes $x^2 = rp + rq - pq$, which simplifies to $x^2 = \frac{p^{2+}q^2}{2}$. Since the roots are $\pm x$, their squares are both $x^2$, so the sum is $2x^2 = p^{2+}q^2$.

A quadratic polynomial $p(x) = ax^{2+}bx+c$ is defined by three conditions. Given $p(0)=1$, we find $c=1$. Applying the remainder theorem for $x-1$ and $x+1$, we get $a+b+1=4$ and $a-b+1=6$. Solving these linear equations yields $a=4$ and $b=-1$, resulting in $p(x)=4x^{2-}x+1$. Evaluating at $x=-2$, we calculate $p(-2) = 4(-2)^2 - (-2) + 1 = 16+2+1=19$.

The series expands to $n x^2 + n^2 x + \frac{n(n^{2-}1)}{3} = 10n$, which simplifies to $x^2 + nx + \frac{n^{2-}31}{3} = 0$ after dividing by $n$. For consecutive integral roots $\alpha$ and $\alpha+1$, their sum is $2\alpha+1 = -n$ and their product is $\alpha(\alpha+1) = \frac{n^{2-}31}{3}$. Using these relations to eliminate $\alpha$ leads to the equation $n^2=121$, confirming that $n=11$.

Since the base is 2, the exponent must be zero for the expression to equal 1. Factoring the equation $(x-1)(x^{2+}5x-50) = 0$ yields $(x-1)(x+10)(x-5) = 0$. The roots are $x=1$, $x=-10$, and $x=5$. Their sum is $1 - 10 + 5 = -4$.