Basic Algebra

The value of

$$\left((\log_2 9)^2\right)^{\frac{1}{\log_2(\log_2 9)}} \times (\sqrt{7})^{\frac{1}{\log_4 7}}$$

is ______ .

Let $E_1 = \left\{x \in \mathbb{R} : x \ne 1 \text{ and } \frac{x}{x-1} > 0\right\}$ and $E_2 = \left\{x \in E_1 : \sin^{-1}\left(\log_e\left(\frac{x}{x-1}\right)\right) \text{ is a real number}\right\}$.

(Here, the inverse trigonometric function $\sin^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.)

Let $f: E_1 \to \mathbb{R}$ be the function defined by $f(x) = \log_e\left(\frac{x}{x-1}\right)$ and $g: E_2 \to \mathbb{R}$ be the function defined by $g(x) = \sin^{-1}\left(\log_e\left(\frac{x}{x-1}\right)\right)$.

The correct option is:

If $a_4 = 28$, then $p + 2q =$

For any three positive real numbers a, b and c,
$$9(25a^2+b^2)+25(c^2-3ac)=15b(3a+c).$$
Then :

If $(27)^{999}$ is divided by 7, then the remainder is :

If x is a solution of the equation, $\sqrt{2x+1}-\sqrt{2x-1}=1$, $\left(x\ge\dfrac12\right)$, then $\sqrt{4x^2-1}$ is equal to :

Let $S$ be the set of all non-zero real numbers $\alpha$ such that the quadratic equation $\alpha x^2-x+\alpha=0$ has two distinct real roots $x_1$ and $x_2$ satisfying the inequality $|x_1-x_2|<1$. Which of the following intervals is(are) a subset(s) of $S$?

Match List I with List II and select the correct answer using the code given below the lists.

An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is :

The sum of the roots of the equation, $x^2+|2x-3|-4=0$, is :

A number $x$ is chosen at random from the set {1, 2, 3, 4, ....., 100}. Define the event : A = the chosen number $x$ satisfies
$$\frac{(x-10)(x-50)}{(x-30)}\ge0$$
Then P(A) is :

A pack contains $n$ cards numbered from 1 to $n$. Two consecutive numbered cards are removed from the pack and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removed cards is $k$, then $k-20=$

If $3^x=4^{x-1}$, then $x=$

The function $f(x)=2|x|+|x+2|-\big||x+2|-2|x|\big|$ has a local minimum or a local maximum at $x=$

Let $f:\mathbb R\to\mathbb R$ be defined as $f(x)=|x|+|x^2-1|$. The total number of points at which $f$ attains either a local maximum or a local minimum is

The value of $6+\log_{3/2}\left(\dfrac1{3\sqrt2}\sqrt{4-\dfrac1{3\sqrt2}\sqrt{4-\dfrac1{3\sqrt2}\sqrt{4-\dfrac1{3\sqrt2}\cdots}}}\right)$ is

Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p(1)=6$ and $p(3)=2$, then $p'(0)$ is

Let $(x_0,y_0)$ be the solution of the following equations
$$(2x)^{\ln2}=(3y)^{\ln3},\qquad3^{\ln x}=2^{\ln y}.$$
Then $x_0$ is

Let $z=x+iy$ be a complex number where $x$ and $y$ are integers. Then the area of the rectangle whose vertices are the roots of the equation
$$z\bar z^3+\bar zz^3=350$$
is

Match the statements/expressions given in Column I with the values given in Column II.