MCQ
The solution of the equation ${\log _7}{\log _5}$ $(\sqrt {{x^2} + 5 + x} ) = 0$
- A$x = 2$
- B$x = 3$
- ✓$x = 4$
- D$x = - 2$
$ \Rightarrow $ ${({x^2} + 5 + x)^{1/2}} = 5$
$ \Rightarrow $ $({x^2} + x + 5) = 25$$ \Rightarrow $ ${x^2} + x - 20 = 0$
$ \Rightarrow $ $ (x - 4)\,(x + 5) = 0$ $ \Rightarrow $ $ x = \,4,\, - 5$ ==> $x = 4$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Statement $I$ : For any two non-zero complex numbers $\mathrm{z}_1, \mathrm{z}_2$
$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ and
Statement $II$ : If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three distinct complex numbers and a, b, c are three positive real numbers such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$, then
$\frac{\mathrm{a}^2}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^2}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^2}{\mathrm{x}-\mathrm{y}}=1$
Between the above two statements,