Algebra PYQ — Page 5

For the functions $f(\theta) = \alpha\tan^2\theta + \beta\cot^2\theta$, and $g(\theta) = \alpha\sin^2\theta + \beta\cos^2\theta$, $\alpha > \beta > 0$, let $\min_{0 < \theta < \pi/2}f(\theta) = \max_{0 < \theta < \pi}g(\theta)$. If the first term of a G.P. is $\left(\dfrac{\alpha}{2\beta}\right)$, its common ratio is $\left(\dfrac{2\beta}{\alpha}\right)$ and the sum of its first $10$ terms is $\dfrac{m}{n}$, $\gcd(m, n) = 1$, then $m + n$ is equal to _______.

If the set of all solutions of $|x^2 + x - 9| = |x| + |x^2 - 9|$ is $[\alpha, \beta] \cup [\gamma, \infty)$, then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to:

The positive integer n, for which the solutions of the equation $x(x+2)+(x+2)(x+4)+\cdots+(x+2 n-2)(x+2 n)=\frac{8 n}{3}$ are two consecutive even integers, is :

For the function $f:[1,\infty) \rightarrow [1,\infty)$ defined by $f(x)=(x-1)^4+1$, among the two statements: (I) The set $S=\{x \in [1,\infty): f(x)=f^{-1}(x)\}$ contains exactly two elements, and (II) The set $S=\{x \in [1,\infty): f(x)=f^{-1}(x+1)\}$ is an empty set,

The sum $1 + \dfrac{1}{2}(1^2 + 2^2) + \dfrac{1}{3}(1^2 + 2^2 + 3^2) + \ldots$ upto 10 terms is equal to :

Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - ax + 2 = 0$. Let the sets $\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of hyperbolas}\} = (\alpha, \beta)$, and $\{a \in \mathbb{R} : e_1 \text{ and } e_2 \text{ are the eccentricities of an ellipse and a hyperbola, respectively}\} = (\gamma, \infty)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:

Let one root of the quadratic equation in $x$: $(k^2 - 15k + 27)x^2 + 9(k-1)x + 18 = 0$ be twice the other. Then the length of the latus rectum of the parabola $y^2 = 6kx$ is equal to:

If the quadratic equation $(\lambda+2)x^2-3\lambda x+4\lambda=0$, $\lambda \neq -2$, has two positive roots, then the number of possible integral values of $\lambda$ is:

Let $a, b \in \mathbb{C}$. Let $\alpha, \beta$ be the roots of the equation $x^2 + ax + b = 0$. If $\beta - \alpha = \sqrt{11}$ and $\beta^2 - \alpha^2 = 3i\sqrt{11}$, then $(\beta^3 - \alpha^3)^2$ is equal to:

If $\alpha, \beta$, where $\alpha<\beta$, are the roots of the equation $\lambda x^{2}-(\lambda+3) x+3=0$ such that $\frac{1}{\alpha}-\frac{1}{\beta}=\frac{1}{3}$, then the sum of all possible values of $\lambda$ is

Let $\alpha$ and $\beta$ be the roots of the equation $x^{2}+2 a x+(3 a+10)=0$ such that $\alpha<1<\beta$. Then the set of all possible values of $a$ is :

Let $z=(1+i)(1+2 i)(1+3 i) \ldots(1+n i)$, where $i=\sqrt{-1}$. If $|z|^{2}=44200$, then $n$ is equal to $\_\_\_\_$

If $z=\frac{\sqrt{3}}{2}+\frac{i}{2}, i=\sqrt{-1}$, then $\left(z^{201}-i\right)^{8}$ is equal to

Let $\alpha=\frac{-1+i \sqrt{3}}{2}$ and $\beta=\frac{-1-i \sqrt{3}}{2}, i=\sqrt{-1}$. If $(7-7 \alpha+9 \beta)^{20}+(9+7 \alpha-7 \beta)^{20}+(-7+9 \alpha+7 \beta)^{20}+(14+7 \alpha+7 \beta)^{20}=m^{10}$, then $m$ is $\_\_\_\_\_$

Two players $A$ and $B$ play a series of games of badminton. The player, who wins $5$ games first, wins the series. Assuming that no game ends in a draw, the number of ways, in which player $A$ wins the series is __________.

Let $A=\{(a,b,c): a,b,c \text{ are non-negative integers and } a+b+2c=22\}$. Then $n(A)$ is equal to:

The number of 4 -letter words, with or without meaning, which can be formed using the letters PQRPQRSTUVP, is $\_\_\_\_$.

Let S denote the set of 4-digit numbers $a b c d$ such that $a>b>c>d$ and P denote the set of 5 -digit numbers having product of its digits equal to 20. Then $n(\mathrm{~S})+n(\mathrm{P})$ is equal to $\_\_\_\_$

Let $z_1, z_2 \in \mathbb{C}$ be the distinct solutions of the equation $z^2 + 4z - (1 + 12i) = 0$. Then $|z_1|^2 + |z_2|^2$ is equal to :

If $g(x)=3 x^{2}+2 x-3, f(0)=-3$ and $4 g(f(x))=3 x^{2}-32 x+72$, then $f(g(2))$ is equal to:

Let $A_1, A_2, A_3, \ldots, A_{39}$ be $39$ arithmetic means between the numbers $59$ and $159$. Then the mean of $A_{25}, A_{28}, A_{31}$ and $A_{36}$ is equal to :

If the sum of the first $10$ terms of the series $\dfrac{1}{1 + 1^4 \times 4} + \dfrac{2}{1 + 2^4 \times 4} + \dfrac{3}{1 + 3^4 \times 4} + \dfrac{4}{1 + 4^4 \times 4} + \ldots$ is $\dfrac{m}{n}$, $\gcd(m, n) = 1$, then $m + n$ is equal to :

$\displaystyle\sum_{n=1}^{10} \left( \dfrac{528}{n(n+1)(n+2)} \right)$ is equal to:

Let the sum of the first $n$ terms of an A.P. be $3n^2 + 5n$. Then the sum of squares of the first $10$ terms of the A.P. is: