Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

Substituting $t = 8^x$ into the given equation transforms it into the quadratic equation $t^2 - 16t + 48 = 0$. Factoring this expression yields $(t - 4)(t - 12) = 0$, giving the solutions $t = 4$ and $t = 12$. These correspond to $8^x = 4$ and $8^x = 12$, which imply $x = \log_8 4$ and $x = \log_8 12$. The sum of these values is $\log_8 4 + \log_8 12 = \log_8 48$, which can be expanded to $\log_8(8 \cdot 6) = \log_8 8 + \log_8 6$, resulting in $1 + \log_8 6$.

Use a substitution method to convert the given expression into a solvable quadratic form. Letting $t = (\sqrt{3}+\sqrt{2})^x$ and noting that $(\sqrt{3}-\sqrt{2}) = 1/(\sqrt{3}+\sqrt{2})$, the equation becomes $t+1/t=10$, which rearranges to $t^{2-}10t+1=0$. Solving for $t$ gives $5 \pm 2\sqrt{6}$, which are equivalent to $(\sqrt{3} \pm \sqrt{2})^2$. Since $t = (\sqrt{3}+\sqrt{2})^x$, matching the bases confirms that $x=2$ and $x=-2$ are the only solutions, totaling 2 elements in the set.

We analyze the equation $x|x+5|+2|x+7|-2=0$ across different intervals. For $x \geq -5$, the equation becomes $x^{2+}7x+12=0$, yielding the solutions $x=-3$ and $x=-4$. For the interval $-7 < x < -5$, the equation becomes $-x^2 - 3x + 12 = 0$, where the only root falling within the specified interval is $x = \frac{-3 - \sqrt{57}}{2}$. In the final interval $x \leq -7$, the equation simplifies to $x^2 + 7x + 16 = 0$, which has a negative discriminant and therefore no real solutions. Counting the valid roots, we find a total of 3 real solutions.

The equation x (x^{2+}3|x|+5|x-1|+6|x-2|) = 0 is satisfied if either x=0 or the parenthetical expression is zero. For any x ≠ 0, the expression x^{2+}3|x|+5|x-1|+6|x-2| is strictly positive because x^2 > 0 and the absolute value terms are non-negative, meaning the sum cannot be zero. Thus, only x=0 is a valid solution, resulting in exactly one solution.

The roots $\alpha$ and $\beta$ of the quadratic equation $x^2 + \sqrt{2}x - 8 = 0$ satisfy the relation $\alpha^2 + \sqrt{2}\alpha = 8$. Multiplying this by $\alpha^8$ gives $\alpha^{10} + \sqrt{2}\alpha^9 = 8\alpha^8$, and applying the same logic to $\beta$ yields $\beta^{10} + \sqrt{2}\beta^9 = 8\beta^8$. Summing these two equations provides $U_{10} + \sqrt{2}U_9 = 8(U_8)$. Substituting this into the given expression $\frac{U_{10} + \sqrt{2}U_9}{2U_8}$ simplifies to $\frac{8U_8}{2U_8}$, which equals 4.

Expand and complete the square for the expression to isolate the variables, revealing that $(a^{2+}b^2)x^{2-}2b(a+c)x+(b^{2+}c^2)=0$ is equivalent to $(ax-b)^{2+}(bx-c)^2=0$. For this to hold in the context of a triangle, the sides must satisfy the triangle inequality, and the bounds for $x$ are derived as $\alpha=\frac{\sqrt{5}-1}{2}$ and $\beta=\frac{\sqrt{5}+1}{2}$. Substituting these values into $12(\alpha^{2+}\beta^2)$ results in $12(\frac{6-2\sqrt{5}}{4} + \frac{6+2\sqrt{5}}{4})$, which simplifies to $12(3)=36$.

Apply the relationship between the coefficients and the roots of a quadratic equation, where for $ax^{2+}bx+1=0$ with roots $r_1=2$ and $r_2=6$, the equation can be written as $a(x-2)(x-6)=0$. Expanding this gives $ax^{2-}8ax+12a=0$, and by comparing coefficients, we find $12a=1$ or $a=1/12$, and $b=-8a=-2/3$. Calculating the new roots $\frac{1}{2a+b}$ and $\frac{1}{6a+b}$ results in $1/(1/6 - 2/3) = -2$ and $1/(1/2 - 2/3) = -6$, respectively. Using these roots, the required quadratic equation is $(x+2)(x+6)=0$, which simplifies to $x^{2+}8x+12=0$.

To determine the number of real roots for the given equation $e^{8x}-e^{6x}-3e^{4x}-e^{2x}+1=0$, introduce the substitution $t = e^{2x}$, where $t > 0$. This transforms the equation into $t^4 - t^3 - 3t^2 - t + 1 = 0$, which can be divided by $t^2$ to obtain $(t^2 + 1/t^2) - (t + 1/t) - 3 = 0$. By setting $u = t + 1/t$, the equation becomes $u^2 - u - 5 = 0$, yielding roots u = rac{1 eq ext{sqrt}{21}}{2}. Given that $u$ must satisfy $|u| ext{geq} 2$ for $t$ to be a positive real number, only the root u = rac{1 + ext{sqrt}{21}}{2} ext{approx} 2.79 produces valid real values for $t$, resulting \in two distinct real solutions for $x$.

Substituting $t=x+\frac{1}{x}$ into the equation gives $3(t^{2-}2)-2t+5=0$ since $x^{2+}\frac{1}{x^2}=(x+\frac{1}{x})^{2-}2$. This simplifies to $3t^{2-}2t-1=0$, which factors as $(3t+1)(t-1)=0$, resulting in $t=1$ or $t=-\frac{1}{3}$. Because real solutions for $x$ require $|t| \ge 2$, neither root provides a valid real solution for $x$, meaning there are no real solutions.

We seek pairs of natural numbers (x, y) such that x^2 - 2^y = 2023, or x^2 - 2023 = 2^y. If y ≥ 3, then 2^y is divisible by 8, implying x^2 ≡ 2023 ≡ 7 (mod 8), which is impossible for any square. Testing y=1 gives x^2 = 2025, so x=45, and y=2 yields x^2 = 2027, which is not a perfect square, so the only solution is (45, 1) resulting in x+y=46.

The location of roots for a quadratic function $f(x)$ is determined by the sign of the function at interval endpoints, where $f(x_1) \cdot f(x_2) < 0$ indicates exactly one root lies between $x_1$ and $x_2$. For this equation, the condition for one root in $(1,2)$ is $f(1) \cdot f(2) < 0$, resulting in $(k-6)(k-8) < 0$, which restricts $k$ to the range $(6,8)$. Similarly, the condition for one root in $(2,3)$ is $f(2) \cdot f(3) < 0$, which leads to the same range for $k$. As $k$ must be an integer, the only valid value is 7.

The expression $\sum_{n=1}^{120} [\sqrt{n}]$ is evaluated by grouping terms where $[\sqrt{n}]$ remains constant. For each integer $k$, the value $[\sqrt{n}]=k$ holds for $(k+1)^{2-}k^2=2k+1$ values of $n$. Summing these across the range up to $k=10$, we calculate $\sum_{k=1}^{10} k(2k+1)$, which expands to $2\sum_{k=1}^{10} k^2 + \sum_{k=1}^{10} k$. Applying the standard summation formulas gives $2 \cdot \frac{10 \cdot 11 \cdot 21}{6} + \frac{10 \cdot 11}{2} = 770 + 55 = 825$.

Factoring the equation as $x(x^{6+}3x^{4-}13x^{2-}15)=0$ and substituting $t=x^2$ leads to the cubic $t^{3+}3t^{2-}13t-15=0$, which factors as $(t-3)(t+1)(t+5)=0$. The roots are $x=0$ and $x^2=3, -1, -5$, giving $x = \pm\sqrt{3}, \pm i, \pm\sqrt{5}i$. Arranging these by magnitude, the root pairs are $\pm\sqrt{5}i$, $\pm\sqrt{3}$, and $\pm i$, which yields the product values $5, -3, 1$ respectively. Calculating the expression $5 - (-3) + 1$ results in 9.

If $\alpha$ is a common root of the two equations, then $\alpha^{2-}5a\alpha+1=0$ and $\alpha^{2-}a\alpha-5=0$ must both be satisfied. Subtracting the second equation from the first yields $-4a\alpha+6=0$, which implies $\alpha = \frac{6}{4a} = \frac{3}{2a}$. Substituting this expression into the second equation gives $(\frac{3}{2a})^2 - a(\frac{3}{2a}) - 5 = 0$, simplifying to $\frac{9}{4a^2} - \frac{3}{2} - 5 = 0$. This leads to $\frac{9}{4a^2} = \frac{13}{2}$, resulting in $a^2 = \frac{9}{26}$, so $a = \frac{3}{\sqrt{26}}$, which identifies $\beta = 13$.

The equation is valid only where all radicands are non-negative, imposing the domain condition $x \leq -3$ or $x \geq 3$. Factoring the terms under the square roots leads to $\sqrt{x-3} (\sqrt{x-1} + \sqrt{x+3}) = \sqrt{2(2x-1)(x-3)}$, which confirms $x=3$ is a solution. Dividing by $\sqrt{x-3}$ for $x > 3$ results in the equation $\sqrt{x-1} + \sqrt{x+3} = \sqrt{4x-2}$, which yields $x=7/6$ after squaring, but this value lies outside the established domain. Consequently, only one real root exists.

For x^2 - x - 1 = 0, the roots α and β satisfy α^2 = α+1 and β^2 = β+1. Multiplying these by α^{n-2} and β^{n-2} gives α^n = α^{n-1} + α^{n-2} and β^n = β^{n-1} + β^{n-2}, allowing us to express S_n = 2023 α^n + 2024 β^n as S_n = S_{n-1} + S_{n-2}. Setting n=12 yields S_{12} = S_{11} + S_{10}.

The function $f(x)=x|x-1|+|x+2|$ is continuous and strictly increasing over the domain of real numbers, as its derivative is positive in every defined interval. Because $f(x)$ ranges from $-\infty$ to $\infty$ as $x$ spans the real line, the equation $f(x)=-a$ will intersect the function at exactly one point for any real value of $a$. Consequently, the set of all such $a$ encompasses the entire set of real numbers, $(-\infty , \infty )$.

The roots of the quadratic equation $x^2 - \sqrt{2} x + 2 = 0$ can be determined using the quadratic formula, resulting in $x = \frac{\sqrt{2} \pm i\sqrt{6}}{2} = \sqrt{2} \left( \frac{1 \pm i \sqrt{3}}{2} \right)$. Expressing these roots in polar form, we have $\alpha = -\sqrt{2} \omega$ and $\beta = -\sqrt{2} \omega^2$, where $\omega = e^{i 2\pi/3}$. Raising these to the power of 14, we obtain $\alpha^{14} + \beta^{14} = 2^7 (\omega^{14} + \omega^{28}) = 128(\omega^2 + \omega)$, which equals $-128$ because $\omega^2 + \omega = -1$.

This equation is a reciprocal type, solved by dividing through by $e^{2x}$ to group terms as $(e^x)^2 + (e^{-x})^2 + 8(e^x - e^{-x}) + 13 = 0$. Substituting $z = e^x - e^{-x}$ transforms the expression into $z^2 + 8z + 15 = 0$, which yields $z = -3$ and $z = -5$. For each resulting equation $e^x - e^{-x} = z$, the positive root $t=e^x$ satisfies $0 < t < 1$, ensuring that $x = \ln t$ is negative in both cases.

For the roots $\alpha, \beta$ of $x^{2+}\sqrt{6} x+3=0$, we can express them as $\sqrt{3} e^{\pm i 3\pi/4}$. The sum of powers $\alpha^n+\beta^n$ follows the form $2(\sqrt{3})^n \cos\left(\frac{3n\pi}{4}\right)$. Substituting this into the given ratio, the terms with $\cos(42\pi/4)$ and $\cos(30\pi/4)$ become zero, simplifying the expression to $\frac{2(\sqrt{3})^{23}\cos(69\pi/4)}{2(\sqrt{3})^{15}\cos(45\pi/4)}$. Given that $\cos(69\pi/4) = \cos(45\pi/4) = -1/\sqrt{2}$, the ratio reduces to $(\sqrt{3})^{23-15} = (\sqrt{3})^8 = 81$.