Question

If $2si{n}^{-1}x+3si{n}^{-1}y+4si{n}^{-1}z=\frac{9\pi }{2}$ then $\frac{1}{2x}+\frac{1}{3y}+\frac{1}{4z}$ equals

• A. $\frac{11}{12}$
• B. $\frac{13}{12}$
• C. $\frac{15}{12}$
• D. $\frac{17}{12}$

Answer 1

The correct option is B $\frac{13}{12}$

Given $2{\text{sin}}^{-1}x+3{\text{sin}}^{-1}y+4{\text{sin}}^{-1}z=\frac{9\pi }{2}$

$⟹2{\text{sin}}^{-1}x+3{\text{sin}}^{-1}y+4{\text{sin}}^{-1}z=2\times \frac{\pi }{2}+3\times \frac{\pi }{2}+4\times \frac{\pi }{2}$

As we know that

The maximum value of ${\text{sin}}^{-1}A$ is $\frac{\pi }{2}$

So the only possibility is $x=y=z=1$

$⟹\frac{1}{2x}+\frac{1}{3y}+\frac{1}{4z}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{6+4+3}{12}=\frac{13}{12}$