MCQ
If $11^x=3^y=99^z=k$, then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$
- ✓$\frac{2}{z}-\frac{1}{y}$
- B$\frac{2}{z}+\frac{1}{y}$
- C$-\frac{1}{y}$
- D$0$
✓
Answer
Correct option: A.
$\frac{2}{z}-\frac{1}{y}$
(a)
Let $11^x=3^y=99^2=k$. Then, $k^{1 / x}=11, k^{1 / y}=3$ and $k^{1 / 2}=99$.
Now, $\quad 99=11 \times 3^2$
$\Rightarrow \quad k^{1 / z}=k^{1 / x} \times\left(k^{1 / y}\right)^2 \Rightarrow k^{1 / z}=k^{1 / x+2 / y} \Rightarrow \frac{1}{z}=\frac{1}{x}+\frac{2}{y} \Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{z}-\frac{1}{y} \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{z}-\frac{1}{y}$
Let $11^x=3^y=99^2=k$. Then, $k^{1 / x}=11, k^{1 / y}=3$ and $k^{1 / 2}=99$.
Now, $\quad 99=11 \times 3^2$
$\Rightarrow \quad k^{1 / z}=k^{1 / x} \times\left(k^{1 / y}\right)^2 \Rightarrow k^{1 / z}=k^{1 / x+2 / y} \Rightarrow \frac{1}{z}=\frac{1}{x}+\frac{2}{y} \Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{z}-\frac{1}{y} \Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{z}-\frac{1}{y}$
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