JEE MAIN - Mathematics Hindi (2019 - 9th January Morning Slot - No. 6)
यदि $$\cos ^{-1}\left(\frac{2}{3 x}\right)+\cos ^{-1}\left(\frac{3}{4 x}\right)=\frac{\pi}{2}\left(x>\frac{3}{4}\right)$$, तब $$x$$ बराबर है :
$$\frac{\sqrt{145}}{12}$$
$$\frac{\sqrt{145}}{10}$$
$$\frac{\sqrt{146}}{12}$$
$$\frac{\sqrt{145}}{11}$$
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