The minimum value of $f(x) = x^2 + 4x + 5$ is . . . . . . ,where $x \in R$.

Solve the equation $x^{2}+x+\frac{1}{\sqrt{2}}=0$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+3x^2-x-3=0$,then $(1+\alpha^2)(1+\beta^2)(1+\gamma^2) = $

Assertion $(A)$: The maximum value of $-x^2+3x+1$ is $\frac{13}{4}$.
Reason $(R)$: If $a < 0$,the maximum value of $ax^2+bx+c$ exists at $x = -\frac{b}{2a}$.
The correct option among the following is

If the roots of the equation $(a^2 + b^2)x^2 - 2b(a + c)x + (b^2 + c^2) = 0$ are equal,then in which progression are $a, b, c$?

Let $x_1, x_2, x_3 \in \mathbb{R} \setminus \{0\}$,$x_1 + x_2 + x_3 \neq 0$ and $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$. Then $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$ holds good for: