Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

The equation $x |x|-5|x+2|+6=0$ is solved by analyzing three distinct intervals based on the absolute value definitions. For $x \ge 0$, the equation simplifies to $x^{2-}5x-4=0$, yielding the valid root $x=\frac{5+\sqrt{41}}{2}$. In the interval $-2 \le x < 0$, the equation becomes $x^{2+}5x+4=0$, which gives $x=-1$ as a valid root while discarding $x=-4$. For the final interval $x < -2$, the equation simplifies to $x^{2-}5x-16=0$, providing the root $x=\frac{5-\sqrt{89}}{2}$, which satisfies the interval condition, resulting in a total of three distinct real roots.

For the first equation $x^{2-}12x+ ext{lfloor} x ext{rfloor}+31=0$, rewrite it as $x^{2-}11x+31= ext{lbrace} x ext{rbrace}$. Since the fractional part satisfies $0 ext{leq} ext{lbrace} x ext{rbrace} < 1$, the expression $x^{2-}11x+31$ must fall within this range, constraining $x$ to the interval $(5, 6)$ where $ext{lfloor} x ext{rfloor} = 5$. Substituting this integer value back into the original equation leads to $x^{2-}12x+36=0$, which yields $x=6$, a value outside the necessary domain, thus confirming $m=0$. For the second equation $x^{2-}5|x+2|-4=0$, treating the cases for $x ext{geq} -2$ and $x < -2$ reveals the roots $ext{lbrace} -2, 7, -3 ext{rbrace}$, meaning $n=3$, so the final result is $0^2 + 0(3) + 3^2 = 9$.

The greatest integer function property relies on algebraic factorization to isolate the fractional part. By rearranging the terms to $(x-1)(x-3) = ext{floor}(x)(x-1)$, we identify two cases: either $x=1$ or $x-3 = ext{floor}(x)$. The second case implies the fractional part of $x$, denoted as $ext{curlybraces}(x)$, must be 3, which is impossible since $0 ext{leq} ext{curlybraces}(x) < 1$. Therefore, $x=1$ is the unique solution.

Given the quadratic $x^2 - 7x - 1 = 0$, define $S_n = a^n + b^n$. According to the recurrence relation derived from the characteristic equation, $S_n - 7S_{n-1} - S_{n-2} = 0$, which implies $S_n = 7S_{n-1} + S_{n-2}$. Evaluating the expression rac{S_{21} + S_{17}}{S_{19}} involves substituting $S_{21} = 7S_{20} + S_{19}$ and $S_{17} = S_{19} - 7S_{18}$ to rewrite the numerator as $7S_{20} + 2S_{19} - 7S_{18}$. Since $S_{20} - S_{18} = 7S_{19}$, the numerator becomes $7(7S_{19}) + 2S_{19} = 51S_{19}$, which simplifies to the final value of 51.

The equation involves absolute values and is solved by completing the square, yielding $(|x|-1)^2 + | ext{\lambda}-3| = 1$. Since both squared terms must be non-negative and sum to 1, we test integer values for $|x|$ in the set ${0, 1, 2}$, which determine the possible values for $ext{\lambda}$. Evaluating $x + ext{\lambda}$ for every valid integer pair $(x, ext{\lambda})$ derived from the equation produces the maximum possible value of 5.

An absolute value equation of the form $|f(x)| = g(x)$ requires $g(x) \geq 0$ and is solved by considering the cases $f(x) = g(x)$ and $f(x) = -g(x)$. Solving the first case $x^{2-}8x+15 = 2x-7$ leads to $x^{2-}10x+22=0$, with roots $5 \pm \sqrt{3}$, where only $5+\sqrt{3}$ satisfies the condition $2x-7 \geq 0$. The second case $x^{2-}8x+15 = -2x+7$ simplifies to $(x-4)(x-2)=0$, where $x=4$ is the only valid solution since $x=2$ fails the non-negativity constraint. Summing the valid roots $5+\sqrt{3}$ and $4$ results in $9+\sqrt{3}$.

For set $A$, the condition $ext{lfloor} x+3 ext{rfloor} + ext{lfloor} x+4 ext{rfloor} ext{leq} 3$ simplifies to $2 ext{lfloor} x ext{rfloor} + 7 ext{leq} 3$, or $ext{lfloor} x ext{rfloor} ext{leq} -2$, which corresponds to the interval $(- ext{infty}, -1)$. For set $B$, evaluate the geometric series sum as rac{0.3}{1 - 0.1} = rac{1}{3}, simplifying the inequality to $3^x (1/3)^{x-3} < 3^{-3x}$, or $3^{x-(x-3)} < 3^{-3x}$. This leads to $3^3 < 3^{-3x}$, implying $3 < -3x$, which simplifies to $x < -1$, demonstrating that set $A$ is equal to set $B$.

Recognizing the reciprocal relationship between the bases allows the substitution $t = (\sqrt{3}+\sqrt{2})^{x^{2-}4}$, which transforms the equation into $t + t^{-1} = 10$. Solving the quadratic $t^2 - 10t + 1 = 0$ gives $t = 5 \pm 2\sqrt{6}$, and observing that $5 \pm 2\sqrt{6} = (\sqrt{3} \pm \sqrt{2})^2$ enables equating the exponents $x^{2-}4$ to $2$ and $-2$. These yield $x^2=6$ and $x^2=2$, providing a total of four distinct real solutions for $x$.

Applying Vieta's formulas to $14x^{2-}31x+3\lambda=0$ and $35x^{2-}53x+4\lambda=0$ gives the relationships $\alpha+\beta=\frac{31}{14}$, $\alpha\beta=\frac{3\lambda}{14}$, $\alpha+\gamma=\frac{53}{35}$, and $\alpha\gamma=\frac{4\lambda}{35}$. The ratio $\frac{\beta}{\gamma} = \frac{\alpha\beta}{\alpha\gamma} = \frac{15}{8}$, and the difference $\beta-\gamma = \frac{7}{10}$, which leads to the values $\gamma=\frac{4}{5}$, \beta=rac{3}{2}, $\alpha=\frac{5}{7}$, and $\lambda=5$. The new roots are $\frac{10}{7}$ and $\frac{25}{7}$, giving the sum $5$ and product $\frac{250}{49}$, which forms the quadratic equation $49x^{2-}245x+250=0$.

For a quadratic equation with roots $\alpha, \beta$, the power sum $\alpha^{4+}\beta^4$ is equivalent to $(\alpha^{2+}\beta^2)^{2-}2(\alpha\beta)^2$, where $\alpha^{2+}\beta^2 = (\alpha+\beta)^{2-}2\alpha\beta$. Using $\alpha+\beta=-60^{1/4}$ and $\alpha\beta=a$, we obtain $\alpha^{2+}\beta^2 = \sqrt{60}-2a$. Substituting this into the given condition $(\sqrt{60}-2a)^{2-}2a^2=-30$ results in $60-4a\sqrt{60}+4a^{2-}2a^2=-30$, or $2a^{2-}4\sqrt{60}a+90=0$. Dividing by 2 yields $a^{2-}2\sqrt{60}a+45=0$, and the product of the possible values for $a$ is 45.

Since $ext{\alpha}, ext{\beta}, ext{\gamma}$ are roots of $x^{3+}bx+c=0$ and the product of roots $ext{\alpha} ext{\beta} ext{\gamma} = -c$, substituting $ext{\alpha}=-1$ and $ext{\beta} ext{\gamma}=1$ yields $(-1)(1) = -c$, giving $c=1$. Plugging $ext{\alpha}=-1$ into the polynomial $x^{3+}bx+1=0$ leads to $-1-b+1=0$, which implies $b=0$. The equation simplifies to $x^{3+}1=0$, with roots $ext{\alpha}=-1$, $ext{\beta}=- ext{\omega}$, and $ext{\gamma}=- ext{\omega}^2$. Calculating the expression $b^{3+}2c^{3-}3 ext{\alpha}^{3-}6 ext{\beta}^{3-}8 ext{\gamma}^3$ with these values results in $0 + 2(1) - 3(-1)^3 - 6(- ext{\omega})^3 - 8(- ext{\omega}^2)^3 = 2+3+6+8 = 19$.

Given $\alpha$ is a root of $1+x^{2+}x^4=0$, we know $\alpha^6=1$ and $\alpha^2 \neq 1$, making $\alpha^2$ a primitive cube root of unity $\omega$. Using modular arithmetic for the powers, $\alpha^{1011}=(\alpha^6)^{168}\alpha^3=\alpha^3=\omega\alpha$, and $\alpha^{2022}=(\alpha^6)^{337}=1$, and $\alpha^{3033}=(\alpha^6)^{505}\alpha^3=\omega\alpha$. Substituting these into the original expression results in $\omega\alpha+1-\omega\alpha=1$.

Dividing the equation by $e^{2x}$ transforms it into $(e^{2x}+e^{-2x}) + 4(e^x+e^{-x}) - 58 = 0$. Substituting $t = e^x + e^{-x}$ gives $t^2 - 2 + 4t - 58 = 0$, which simplifies to $(t+10)(t-6)=0$. Since $e^x + e^{-x} \ge 2$, we take $t=6$, leading to $e^{2x}-6e^x+1=0$. This quadratic in $e^x$ has two distinct positive roots, providing two real solutions for $x$.

The sum of the squares of the roots $\alpha^{2+}\beta^2$ is related to the coefficients of the quadratic equation by the identity $\alpha^{2+}\beta^2=(\alpha+\beta)^{2-}2\alpha\beta$. Using Vieta's relations for $x^{2+}(3-a)x+1-2a=0$, we have $\alpha+\beta=a-3$ and $\alpha\beta=1-2a$. Substituting these gives $(a-3)^{2-}2(1-2a)=a^{2-}6a+9-2+4a=a^{2-}2a+7$. This expression can be rewritten as $(a-1)^{2+}6$, which reaches a minimum value of 6 when $a=1$.

The roots of the equation $x^{4+}x^{3+}x^{2+}x+1=0$ are the 5th roots of unity excluding $x=1$, because the equation is equivalent to $\frac{x^{5-}1}{x-1}=0$ for $x \neq 1$. The sum of the $p$-th powers of the $n$-th roots of unity is 0 if $n$ does not divide $p$, and $n$ if it does. Here, $p=2021$ and $n=5$, so the sum of all 5th roots of unity is 0, which implies $1^{2021}+\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021}=0$. Rearranging this sum yields $-1$.

The shared root condition implies that the common root is $4$, which when substituted into the equation $(p^{2+}q^2)x^{2-}2q(p+r)x+q^{2+}r^2=0$ yields the identity $(4p-q)^{2+}(4q-r)^2=0$. This requires the relations $q=4p$ and $r=4q=16p$ to hold. Substituting these values into the ratio $\frac{q^{2+}r^2}{p^2}$ provides $\frac{(4p)^{2+}(16p)^2}{p^2}$, which simplifies to $16+256=272$.

Applying Rolle's theorem to $f(x)=x^{7-}7x-2$ helps determine the number of real roots by analyzing its derivative, $f'(x)=7(x^{6-}1)$. The function has critical points at $x=1$ and $x=-1$, where $f(1)=-8$ and $f(-1)=4$. Since the function is increasing on $(-\infty , -1)$, decreasing on $(-1, 1)$, and increasing on $(1, \infty )$, the Intermediate Value Theorem confirms one root in each interval, resulting in exactly $3$ distinct real roots.

For $S_1$, the quadratic $x^{2+}3x+5$ is always positive, so the inequality simplifies to $\frac{x+2}{-(x-1)(x-2)} \ge 0$, which is $\frac{x+2}{(x-1)(x-2)} \le 0$. Testing intervals gives $S_1=(-\infty , -2] \cup (1, 2)$. For $S_2$, substituting $t=3^x$ into $t^{2-}3t-9t+27 \le 0$ yields $(t-3)(t-9) \le 0$, resulting in $3 \le 3^x \le 9$, or $1 \le x \le 2$. The union $S_1 \cup S_2$ is $(-\infty , -2] \cup (1, 2) \cup [1, 2]$, which simplifies to $(-\infty , -2] \cup [1, 2]$.

Absolute value inequalities define intervals $A = (-3, 1)$ and B = (-\infty , -1] \cup \[3, ∞)$. Evaluating the set operations,$A-B$is$(-3, 1) \cap ((-1, 3)) = (-1, 1)$, while$B-A$is$B \cap A^c = (-∞, -3]$ \cup [3, \infty )$. Since the expression$R-(-3, 1)$equates to$(-\infty , -3] \cup [1, \infty )$, it does not match the actual set$B-A$.

Substituting $t=e^x$ transforms the given equation into $(t^{2-}4)(6t^{2-}5t+1)=0$. Factoring this leads to $(t-2)(t+2)(2t-1)(3t-1)=0$, indicating $t=2, -2, 1/2, 1/3$. Since the exponential term $e^x$ must be positive for real $x$, the value $t=-2$ is discarded, leaving solutions $x = ext{ln } 2, x = - ext{ln } 2,$ and $x = - ext{ln } 3$. Summing these real roots results in $ext{ln } 2 - ext{ln } 2 - ext{ln } 3$, which simplifies to $- ext{ln } 3$.