Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

The set $S$ is determined by the inequality $\frac{|x+3|-1}{|x|-2} \ge 0$, which holds on the intervals $[-6, -4]$ and $(2, 3]$, while the set $T$ consists of integers satisfying $x^{2-}7|x|+9 \le 0$. By solving $x^{2-}7|x|+9 \le 0$, we find $|x| \in [\frac{7-\sqrt{13}}{2}, \frac{7+\sqrt{13}}{2}]$, which includes the integers $\{-5, -4, -3, -2, 2, 3, 4, 5\}$. Identifying the intersection $S \cap T$ by comparing these two sets confirms that only the integers $-5, -4,$ and $3$ are common to both, resulting in a total of 3 elements.

The sum of squares of the roots is expressed as $(a+b)^{2-}2ab$, which for the given equation calculates to $\frac{(\alpha-6)^2}{9}-2\frac{\alpha+3}{3}=\frac{\alpha^{2-}18\alpha+18}{9}$. For real roots, the discriminant $D=(\alpha-6)^{2-}12(\alpha+3)$ must be non-negative, simplifying to $\alpha^{2-}24\alpha \ge 0$, meaning $\alpha \le 0$ or $\alpha \ge 24$. The function $f(\alpha)=\frac{\alpha^{2-}18\alpha+18}{9}$ has its vertex at $\alpha=9$; since this point lies outside the valid domain for $\alpha$, no real roots exist at the minimum of this parabola.

The roots of $x^{2-}x-4=0$ satisfy the recurrence $P_n = P_{n-1} + 4P_{n-2}$, which can be rearranged to $P_n - P_{n-1} = 4P_{n-2}$. Factoring the numerator $P_{15}P_{16} - P_{14}P_{16} - P_{15}^2 + P_{14}P_{15}$ yields $(P_{16}-P_{15})(P_{15}-P_{14})$. Substituting the recurrence relations $P_{16}-P_{15} = 4P_{14}$ and $P_{15}-P_{14} = 4P_{13}$ into the expression results in rac{(4P_{14})(4P_{13})}{P_{13}P_{14}}, which simplifies to 16.

This sum is a telescoping series where each term takes the form rac{1}{20}(rac{1}{20k-a} - rac{1}{20(k+1)-a}). Summing these nine terms collapses to rac{1}{20}(rac{1}{20-a} - rac{1}{200-a}) = rac{1}{256}. Solving the resulting quadratic equation $(20-a)(200-a) = 2304$ gives $a^2 - 220a + 1696 = 0$, leading to roots of 8 and 212, with the maximum value being 212.

Polynomial factorization reveals that the expression $x^{4-}3x^{3-}2x^{2+}3x+1=0$ can be written as $(x^{2-}1)(x^{2-}3x-1)=0$. This indicates that the roots of the equation are $1$ and $-1$, along with the roots $\alpha$ and $\beta$ of the quadratic $x^{2-}3x-1=0$. The sum of the cubes of all roots is $1^{3+}(-1)^{3+}\alpha^{3+}\beta^3$, which reduces to $\alpha^{3+}\beta^3$. Applying the identity \alpha^{3+}\beta^3 = (\alpha+\beta)^{3-}3\alpha\beta(\alpha+\eta) with $\alpha+\beta=3$ and $\alpha\beta=-1$ yields $3^{3-}3(-1)(3) = 27+9=36$.

Vieta's formulas establish that for the equation $3x^{2+}\lambda x-1=0$, the sum of roots $\alpha+\beta = -\lambda/3$ and their product $\alpha\beta = -1/3$. The given sum of reciprocal squares $\frac{1}{\alpha^2}+\frac{1}{\beta^2}=15$ simplifies to $\frac{(\alpha+\beta)^{2-}2\alpha\beta}{(\alpha\beta)^2}=15$, which leads to $\frac{\lambda^2/9+2/3}{1/9}=15$, yielding $\lambda^2=9$. Substituting this into the expression 6(\alpha^{3+}\beta^3)^2 = 6\[(\alpha+\beta)((\alpha+\beta)^{2-}3\alpha\beta)]^2$results \in$6$[-\frac{\lambda}{3}(\frac{\lambda^2}{9}+1)]^2 = 6(\frac{\lambda^2}{9})(\frac{\lambda^2}{9}+1)^2$. With$\lambda^2=9$, this evaluates to$6(1)(2)^2$, which equals 24.

Factoring the given polynomial reveals it as $(x-1)^2(x+1)(x^{5+}3x-1)=0$, which immediately identifies $x=1$ and $x=-1$ as roots. The term $x^{5+}3x-1$ is a strictly increasing function, as its derivative $5x^{4+}3$ is always positive, ensuring it crosses the x-axis at exactly one point. Consequently, the equation possesses exactly three distinct real roots.

The number of real roots is determined by the behavior of $f(x)=x^{4-}4x+1$. The derivative $f'(x)=4(x^{3-}1)$ shows a single global minimum at $x=1$ where $f(1)=-2$. Since $f(0)=1$ and $f(2)=9$, the function crosses the $x$-axis once in $(0, 1)$ and once in $(1, 2)$, resulting in 2 real roots.

A quadratic polynomial with roots $\alpha$ and $\beta$ is expressed as $f(x) = a(x-\alpha)(x-\beta)$. Given one root is $-1$, we write $f(x) = a(x+1)(x-\beta)$. Evaluating the condition $f(-2) + f(3) = a(2+\beta) + a(4)(3-\beta) = 0$ simplifies to $a(14-3\beta) = 0$, implying $\beta = \frac{14}{3}$. The sum of the roots is thus $-1 + \frac{14}{3} = \frac{11}{3}$.

A repeated root in a quadratic $ax^{2-}2bx+15=0$ implies that its discriminant is zero, giving $b^2=15a$, and the repeated root $\alpha$ is expressed as $\frac{-(-2b)}{2a} = \frac{b}{a} = \frac{15}{b}$. Substituting $\alpha = \frac{15}{b}$ into $x^{2-}2bx+21=0$ results in $(\frac{15}{b})^{2-}2b(\frac{15}{b})+21=0$, which simplifies to $\frac{225}{b^2}-30+21=0$, leading to $b^2=25$. Given the second equation $x^{2-}2bx+21=0$ has roots $\alpha$ and $\beta$, the sum of their squares is $\alpha^{2+}\beta^2 = (\alpha+\beta)^{2-}2\alpha\beta = (2b)^{2-}2(21) = 4(25)-42=58$.

The relationship between roots and coefficients allows for the simplification of the given expression, $(\frac{1}{p}+\frac{1}{q})^{-2} = (\frac{p+q}{pq})^{-2} = (\frac{pq}{p+q})^2$. Given $p+q=3$, we utilize the identity $(p^{2+}q^2)^2 = p^{4+}q^{4+}2p^2q^2$ and $p^{2+}q^2 = (p+q)^{2-}2pq = 9-2pq$. Substituting these into the original equation yields $(9-2pq)^{2-}2(pq)^2=369$, which simplifies to $2p^2q^{2-}36pq-288=0$, or $p^2q^{2-}18pq-144=0$. Factoring this quadratic gives $(pq-24)(pq+6)=0$, and since $pq$ must be $-6$ to satisfy the condition for real roots, the expression becomes $(\frac{-6}{3})^2 = 4$.

Substituting $t=e^x$ transforms the equation into $t^2 - 11t - \frac{45}{t} + 40.5 = 0$. Clearing the denominator gives $t^3 - 11t^2 + 40.5t - 45 = 0$. By Vieta's formulas, the product of the roots $t_1 t_2 t_3 = e^{\alpha_1+\alpha_2+\alpha_3} = 45$. Thus, $\alpha_1+\alpha_2+\alpha_3 = \ln 45$, and $p=45$.

Equating the two rational expressions allows for a comparison of the numerators and denominators. Rearranging the equality to rac{3x^{2-}9x+17}{5x^{2-}7x+19} = rac{x^{2+}3x+10}{3x^{2+}5x+12} reveals that the difference between the numerator and denominator on both sides is the same expression, $2x^{2+}2x+2$. Since setting this difference to zero yields no real roots, the expressions are equal when $5x^{2-}7x+19 = 3x^{2+}5x+12$, which simplifies to $2x^{2-}12x+7=0$. By Vieta's formulas, the sum of the roots is -rac{-12}{2} = 6.

Substituting $u=e^{x}$ turns the equation into $u^{6}-u^{4}-2u^{3}-12u^{2}+u+1=0$, which simplifies to $(u^{3}-1)^{2}-u(u^{3}-1)-12u^{2}=0$. This quadratic in $(u^{3}-1)$ factors into $(u^{3}-1-4u)(u^{3}-1+3u)=0$, leading to the two equations $u^{3}-4u-1=0$ and $u^{3}+3u-1=0$. Analyzing these functions for $u=e^{x}>0$ confirms that each equation contributes exactly one positive real root, resulting in two real solutions for $x$.

The greatest integer function $f(x) = [x]$ evaluates to $n/2$ when $n$ is even and $-(n+1)/2$ when $n$ is odd. Evaluating the terms from $n=8$ to $n=100$ reveals that each even integer $n$ and the subsequent odd integer $n+1$ pair up to sum to zero, specifically when considering the range from $n=9$ onwards. Since the series starts at $n=8$ with the value $4$, the remaining terms cancel out in pairs, leaving a total sum of $4$.

Since $\alpha$ is a common root of both quadratic equations, it must satisfy both, allowing us to equate the expressions by eliminating the quadratic terms. Multiplying the first equation by 3 and subtracting it from the second yields $7\alpha = 21\lambda$, meaning $\alpha = 3\lambda$. Substituting this value into the first quadratic equation leads to $9\lambda^2 - \lambda = 0$, and given $\lambda \neq 0$, we find $\lambda = \frac{1}{9}$ and $\alpha = \frac{1}{3}$. Finally, utilizing Vieta's formulas for each equation, we determine $\beta = 1 - \alpha$ and $\gamma = \frac{10}{3} - \alpha$, which allows us to calculate the value of $\frac{\beta\gamma}{\lambda}$ as 18.

The infinite continued fraction is defined by the recurrence $x=4+\frac{1}{5+\frac{1}{x}}$. Simplifying this leads to $x=4+\frac{x}{5x+1}$, which cross-multiplies to the quadratic $5x^{2}-20x-4=0$. Applying the quadratic formula gives $x=\frac{20 \pm \sqrt{400+80}}{10}=2 \pm \frac{\sqrt{480}}{10}$. Simplifying the radical to $4\sqrt{30}$ results in $x=2 \pm \frac{2}{5}\sqrt{30}$, and taking the positive root yields $2+\frac{2}{5}\sqrt{30}$.

A quadratic expression $ax^{2}+bx+c$ remains strictly positive for all real $x$ if the leading coefficient is positive and the discriminant $D=b^{2}-4ac$ is negative. For the given expression, the discriminant $D=[2(3k-1)]^{2}-4(8k^{2}-7)$ must be less than zero. This simplifies to $4(9k^{2}-6k+1)-32k^{2}+28 < 0$, which reduces to $4k^{2}-24k+32 < 0$, or $(k-2)(k-4) < 0$. The inequality holds for $2 < k < 4$, and the only integer value in this range is $3$.

To solve the equation $e^{4x}-e^{3x}-4e^{2x}-e^{x}+1=0$, divide the entire equation by $e^{2x}$ to obtain $e^{2x}-e^{x}-4-e^{-x}+e^{-2x}=0$. Grouping the terms as $(e^{2x}+e^{-2x})-(e^{x}+e^{-x})-4=0$ and substituting $y=e^{x}+e^{-x}$ transforms the equation into $y^{2}-y-6=0$, which factors to $(y-3)(y+2)=0$. Since $e^{x}+e^{-x} ext{-} ext{ge} 2$ for all real $x$, only $y=3$ is valid, leading to $e^{x}+e^{-x}=3$. This results in a quadratic equation $e^{2x}-3e^{x}+1=0$ with a positive discriminant, confirming there are two distinct real solutions for $x$.

Using the properties of logarithms, specifically $\log_{a^n}b = \frac{1}{n}\log_a b$, the given equation simplifies by expressing all terms with base 2. Combining the logarithmic terms into a single expression leads to the relation $1 + \log_2 \frac{10 \cdot 2^x - 1}{(3+2^x)^2} = 0$, which rearranges to a quadratic equation in terms of $2^x$. Substituting $y = 2^x$ gives $y^2 - 14y + 11 = 0$, where the product of the roots $y_1 y_2 = 2^{x_1} \cdot 2^{x_2} = 2^{x_1+x_2}$ is equal to 11. Taking the logarithm base 2 of both sides directly yields the sum of the roots $x_1 + x_2 = \log_2 11$.