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EEJ MAIN Mathematics QUESTION #6961
Question 1

If $\cos^{-1}\!\left(\dfrac{2}{3x}\right)+\cos^{-1}\!\left(\dfrac{3}{4x}\right)=\dfrac{\pi}{2}$ for $x>\dfrac{3}{4}$, find $x$.

  • $\dfrac{\sqrt{145}}{12}$✔️
  • $\dfrac{\sqrt{145}}{10}$
  • $\dfrac{\sqrt{146}}{12}$
  • $\dfrac{\sqrt{145}}{11}$
Correct Answer Explanation

$\cos^{-1}A+\cos^{-1}B=\pi/2 \Rightarrow \cos^{-1}A=\pi/2-\cos^{-1}B=\sin^{-1}B$, i.e., $A=\sin(\cos^{-1}B)=\sqrt{1-B^2}$.

$\dfrac{2}{3x}=\sqrt{1-\dfrac{9}{16x^2}}$

Square: $\dfrac{4}{9x^2}=1-\dfrac{9}{16x^2}$

Multiply by $144x^2$: $64=144x^2-81 \Rightarrow 144x^2=145 \Rightarrow x=\dfrac{\sqrt{145}}{12}$