Quadratic Equation And Inequalities – JEE Main MathematicsPractice Questions & PYQs

Given the quadratic equation x^{2} + (20)^{rac{1}{4}} x + (5)^{rac{1}{2}} = 0, the sum of roots is ext{-} ext{\alpha} + ext{-} ext{\beta} = -(20)^{rac{1}{4}} and the product is $ext{-} ext{\alpha} ext{-} ext{\beta} = ext{sqrt}{5}$. Squaring the sum yields $ext{-} ext{\alpha}^{2} + ext{-} ext{\beta}^{2} + 2 ext{-} ext{\alpha} ext{-} ext{\beta} = ext{sqrt}{20}$, and since $2 ext{-} ext{\alpha} ext{-} ext{\beta} = 2 ext{sqrt}{5} = ext{sqrt}{20}$, it follows that $ext{-} ext{\alpha}^{2} + ext{-} ext{\beta}^{2} = 0$, implying $ext{-} ext{\alpha}^{2} = - ext{-} ext{\beta}^{2}$. Squaring this relationship gives $ext{-} ext{\alpha}^{4} = ext{-} ext{\beta}^{4}$, and substituting the product $ext{-} ext{\alpha}^{2} ext{-} ext{\beta}^{2} = ext{sqrt}{5}^{2} = 5$ into $- ext{-} ext{\beta}^{4} = 5$ shows that $ext{-} ext{\beta}^{4} = -5$ and $ext{-} ext{\alpha}^{4} = 5$. Thus, $ext{-} ext{\alpha}^{8} + ext{-} ext{\beta}^{8} = 5^{2} + (-5)^{2} = 50$.

The value $a^{4}+b^{4}+c^{4}$ is calculated using elementary symmetric polynomials. Starting with $a^{2}+b^{2}+c^{2} = (a+b+c)^{2}-2(ab+bc+ca) = 1^{2}-2(2) = -3$, and noting that $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} = (ab+bc+ca)^{2}-2abc(a+b+c) = 2^{2}-2(3)(1) = -2$, the final expression is derived from $(a^{2}+b^{2}+c^{2})^{2}-2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$. Substituting the values gives $(-3)^{2}-2(-2) = 9+4=13$.

To determine the values of $k$ for which the equation has no real roots, rearrange the given expression into the standard quadratic form $2x^{2}-(6+k)x+(3k+4)=0$. The condition for no real roots is that the discriminant $D = b^{2}-4ac < 0$, which yields $(6+k)^{2}-8(3k+4) < 0$. Simplifying this inequality results in $k^{2}-12k+4 < 0$, identifying the range as $6-4 ext{-} ext{sqrt}{2} < k < 6+4 ext{-} ext{sqrt}{2}$, or approximately $0.35 < k < 11.65$. The integral values of $k$ are therefore $1, 2, 3, ext{...}, 11$, and their sum is 66.

By substituting $y = 3x^2 + 4x + 2$, the given expression transforms into a simplified quadratic in $y$ of the form $(y+1)^2 - (k+1)y(y+1) + ky^2 = 0$. Simplifying this leads to $y(1-k) = -1$, or $y = \frac{1}{k-1}$. Substituting this result back into the expression for $y$ gives the quadratic equation $3x^2 + 4x + (2 - \frac{1}{k-1}) = 0$, which must have a non-negative discriminant for real roots to exist. Applying the condition $D = b^2 - 4ac \geq 0$ and solving the resulting inequality for $k$ identifies the valid interval $(1, \frac{5}{2}]$.

The greatest integer property $[x+n]=[x]+n$ for any integer $n$ reduces $[e^{x}+1]$ to $[e^{x}]+1$, allowing the equation to be rewritten as [e^{x}]^{2}+\[e^{x}]$-2=0$. Solving for$[e^{x}]$gives the roots$1$and$-2$, but because$e^{x}$is strictly positive, the only valid condition is$[e^{x}]=1$. This implies$1 \le e^{x} < 2$, which simplifies to$0 \le x < \log_{e} 2$.

The value of $\csc 18^\circ$ is derived from the trigonometric identity $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$, which simplifies to $\csc 18^\circ = \sqrt{5} + 1$. By setting $x = \sqrt{5} + 1$, we isolate the radical term as $x - 1 = \sqrt{5}$ and square both sides of the equation. This yields $(x-1)^2 = 5$, which expands to $x^2 - 2x + 1 = 5$, resulting in the quadratic equation $x^2 - 2x - 4 = 0$ that is satisfied by this value.

Evaluating equations involving absolute values requires splitting the domain to redefine the modulus function based on the sign of $x-5$. For the equation $(x+1)^{2}+|x-5|=\frac{27}{4}$, analyzing the case $x \ge 5$ yields no valid solutions, while the case $x < 5$ leads to $(x+1)^{2}-(x-5)=\frac{27}{4}$. This simplifies to the quadratic equation $4x^{2}+4x-3=0$, which produces two distinct roots, $0.5$ and $-1.5$, both of which are valid within the domain $x < 5$.

The continued fraction can be represented by the recursive relation $x=3+\frac{1}{4+\frac{1}{x}}$. Simplifying the right side results in $x=3+\frac{x}{4x+1}$, which rearranges to the quadratic equation $4x^{2}-12x-3=0$. Solving for $x$ using the quadratic formula $x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ with coefficients $a=4, b=-12, c=-3$ yields $x=\frac{12 \pm \sqrt{144+48}}{8}=1.5 \pm \sqrt{3}$. Since the expression must converge to a positive value, the result is $1.5+\sqrt{3}$.

The equation $x^{2}-|x|-12=0$ can be rewritten using the property $x^{2}=|x|^{2}$ as $|x|^{2}-|x|-12=0$. Treating this as a quadratic in terms of $|x|$, we factor the expression to $(|x|-4)(|x|+3)=0$. Since the absolute value of any real number must be non-negative, $|x|=-3$ yields no solutions, leaving $|x|=4$ as the only possibility, which provides two real roots: $x=4$ and $x=-4$.

A quadratic root $x$ of $x^{2}-6x-2=0$ satisfies the identity $x^{2}-2=6x$. Multiplying this equality by $x^{8}$ yields $x^{10}-2x^{8}=6x^{9}$, a property that holds for both roots $\alpha$ and $\beta$. Substituting this into the fraction $\frac{(\alpha^{10}-2\alpha^{8})-(\beta^{10}-2\beta^{8})}{3(\alpha^{9}-\beta^{9})}$ reduces the numerator to $6\alpha^{9}-6\beta^{9}$, or $6(\alpha^{9}-\beta^{9})$. Dividing by $3(\alpha^{9}-\beta^{9})$ results in $2$.

Defining a polynomial $g(x) = xf(x)+2$ allows us to account for the roots at $k=2, 3, 4, 5$, where $kf(k)+2=0$. Because $f(x)$ has degree 3, $g(x)$ is a degree 4 polynomial of the form $g(x) = c(x-2)(x-3)(x-4)(x-5)$. Evaluating $g(0) = 0f(0)+2 = 2$ gives $2 = c(-2)(-3)(-4)(-5)$, so $c = 1/60$. Computing $10f(10)+2 = g(10) = \frac{1}{60}(8)(7)(6)(5) = 28$ yields $10f(10) = 26$, and consequently, $52-10f(10) = 26$.

The roots of the quadratic equation $x^{2}+5\sqrt{2}x+10=0$ satisfy the property $\alpha+5\sqrt{2}=-10/\alpha$, which allows simplifying expressions involving powers of $\alpha$ and $\beta$. By factoring the numerator $P_{17}(\alpha^{20}-\beta^{20}+5\sqrt{2}(\alpha^{19}-\beta^{19}))$ as $P_{17}(\alpha^{19}(\alpha+5\sqrt{2})-\beta^{19}(\beta+5\sqrt{2}))$, it becomes $-10P_{17}(\alpha^{18}-\beta^{18})$, which is $-10P_{17}P_{18}$. A similar reduction applied to the denominator $P_{18}(\alpha^{19}-\beta^{19}+5\sqrt{2}(\alpha^{18}-\beta^{18}))$ yields the same result, confirming the ratio is $1$.

A quadratic equation with roots $p$ and $q$ can be expressed as $x^{2}-(p+q)x+pq=0$. Given the identity $p^{4}+q^{4}=((p+q)^{2}-2pq)^{2}-2p^{2}q^{2}=272$ and $p+q=2$, substitution leads to $(4-2pq)^{2}-2p^{2}q^{2}=272$, which simplifies to $p^{2}q^{2}-8pq-128=0$. Solving this quadratic for $pq$ yields $16$ (since $pq$ must be positive for positive roots $p$ and $q$), resulting in the equation $x^{2}-2x+16=0$.

Since $\alpha$ and $\beta$ are roots of $x^{2}-x-1=0$, they satisfy the relations $\alpha^{2}=\alpha+1$ and $\beta^{2}=\beta+1$. Multiplying these by $\alpha^{n-1}$ and $\beta^{n-1}$ and adding them yields the linear recurrence $p_{n+1}=p_{n}+p_{n-1}$, where $p_{n}=\alpha^{n}+\beta^{n}$. Given $p_{n+1}=29$ and $p_{n-1}=11$, we find $29=p_{n}+11$, which gives $p_{n}=18$. The value of $p_{n}^{2}$ is thus $18^{2}=324$.

By expressing the function $8^{2 ext{sin} 3x} ext{-} ext{cdot} 4^{4 ext{cos} 3x}$ as $2^{6 ext{sin} 3x+8 ext{cos} 3x}$, we recognize that the exponent ranges between $-10$ and $10$. Consequently, the maximum value $ext{-} ext{\alpha}$ is $2^{10}$ and the minimum value $ext{-} ext{\beta}$ is $2^{-10}$. Calculating the roots ext{-} ext{\alpha}^{rac{1}{5}}=2^{2}=4 and ext{-} ext{\beta}^{rac{1}{5}}=2^{-2}=rac{1}{4} allows us to form the quadratic equation x^{2}-(4+rac{1}{4})x+(4 ext{-} ext{cdot} rac{1}{4})=0, which is $x^{2}-4.25x+1=0$. Multiplying by 8 gives $8x^{2}-34x+8=0$, identifying $b=-34$ and $c=8$, which makes $c-b=42$.

The cubic polynomial $x^{3}-2x^{2}+2x-1=0$ can be factored as $(x-1)(x^{2}-x+1)=0$, giving the roots $1$ and $\frac{1 \pm i\sqrt{3}}{2}$. In polar form, the complex roots are $e^{i\pi/3}$ and $e^{-i\pi/3}$, which both have a period of 6. Raising these roots to the 162nd power yields $1^{162} + (e^{i\pi/3})^{162} + (e^{-i\pi/3})^{162}$, which simplifies to $1 + e^{i54\pi} + e^{-i54\pi}$, resulting in a final sum of $1+1+1=3$.

If $\alpha$ is a root, the equation $\alpha^{2}+3^{\frac{1}{4}}\alpha+3^{\frac{1}{2}}=0$ implies $\alpha^{2}+\sqrt{3}=-3^{\frac{1}{4}}\alpha$. Squaring both sides yields $\alpha^{4}+2\sqrt{3}\alpha^{2}+3=\sqrt{3}\alpha^{2}$, which simplifies to $\alpha^{4}+\sqrt{3}\alpha^{2}+3=0$. Multiplying this by $\alpha^{4}-3$ results in $\alpha^{12}-27=0$, or $\alpha^{12}=3^{3}$. Substituting $\alpha^{12}=27$ into $\alpha^{96}(\alpha^{12}-1)+\beta^{96}(\beta^{12}-1)$ gives $3^{24}(27-1)+3^{24}(27-1)$, which equals $26\times 3^{24}+26\times 3^{24}=52\times 3^{24}$.

The condition that if $\alpha$ is a root, then $\alpha^2 - 2$ is also a root, implies that the set of roots $\{\alpha, \beta\}$ must be invariant under the mapping $f(x) = x^2 - 2$. By examining the cases where the roots are equal, such as $\alpha = \beta$, we obtain the pairs $(2, 1)$ and $(-4, 4)$ from the fixed points of the function. When the roots are distinct, we consider the scenarios where the roots are permuted or mapped under $f(x)$, leading to systems of equations for the coefficients $a$ and $b$ through Vieta's relations. Solving these cases yields the remaining four pairs $(-1, -2)$, $(1, -1)$, $(0, -4)$, and $(0, -1)$, confirming that there are 6 such pairs in total.

A quadratic equation $ax^2 - 2bx + 5 = 0$ with a repeated root $\alpha$ satisfies $\alpha = -(-2b)/(2a) = b/a$ and the discriminant equals zero, leading to $4b^2 - 20a = 0$, or $b^2 = 5a$. Substituting $\alpha = b/a$ into the second equation $x^2 - 2bx - 10 = 0$ yields $(b/a)^2 - 2b(b/a) - 10 = 0$, which simplifies to $b^2/a^2 - 2b^2/a - 10 = 0$. Using $b^2 = 5a$, we find $5/a - 10 - 10 = 0$, giving $a = 1/4$ and $\alpha^2 = 20$. With $\alpha\beta = -10$, we obtain $\beta^2 = 100/\alpha^2 = 5$, resulting in $\alpha^2 + \beta^2 = 25$.

Dividing the equation $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$ by $e^{2x}$ transforms it into $(e^{2x} + e^{-2x}) + (e^x + e^{-x}) - 4 = 0$. Setting $t = e^x + e^{-x}$, where $t \ge 2$, we rewrite the expression as $(t^2 - 2) + t - 4 = 0$, or $t^2 + t - 6 = 0$. Factoring gives $(t+3)(t-2) = 0$, and since $t$ must be at least 2, we have $t=2$. Solving $e^x + e^{-x} = 2$ leads to $(e^x - 1)^2 = 0$, which yields the single real root $x = 0$.