# NIELIT 2019 Feb Scientist D - Section D: 18

1 vote
111 views

If $x^{a}=y^{b}=z^{c}$ and $y^{2}=zx$ then the value of $\frac{1}{a} + \frac{1}{c}$ is  :

1. $\frac{b}{2}$
2. $\frac{c}{2}$
3. $\frac{2}{b}$
4. $\frac{2}{a}$

recategorized

$Given$ : $x^{a}$$=$$y^{b}$$=$$z^{c}$ & $y^{2}=zx$

$Let$ $x^{a}$$=$$y^{b}$$=$$z^{c}$$=$$k$

$\Rightarrow$ $x=k^{\frac{1}{a}}$

$\Rightarrow$ $y=k^{\frac{1}{b}}$

$\Rightarrow$ $z=k^{\frac{1}{c}}$

$\because$ $y^{2}=zx$

$\therefore$ $(k^{\frac{1}{b}})^2$ $=$ $(k^{\frac{1}{c}})*$ $(k^{\frac{1}{b}})$

$\therefore$ $\frac{1}{a}$$+$$\frac{1}{b}$$=$$\frac{2}{b}$

$\textrm{Option C}$
3.5k points 4 10 63

## Related questions

1
88 views
Find all the polynomials with real coefficients $P\left(x \right)$ such that $P\left(x^{2}+x+1 \right)$ divides $P\left(x^{3}-1 \right)$. $ax^{n}$ $ax^{n+2}$ $ax$ $2ax$
2
74 views
The roots of the equation $x^{2/3}+x^{1/3}-2=0$ are : $1, -8$ $-1, -2$ $\frac{2}{3}, \frac{1}{3}$ $-2, -7$
$\left [\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}} \right ]$ equal to : $1$ $0$ $\frac{8}{1-x^{8}}$ $\frac{16}{1-x^{16}}$
If $t^{2}-4t+1=0$, then the value of $\left[t^{3}+1/t^{3} \right]$ is : $44$ $48$ $52$ $64$
If $a^{x}=b$, $b^{y}=c$ and $c^{z}=a$, then the value of $xyz$ is : $0$ $1$ $\frac{1}{3}$ $\frac{1}{2}$