MHTCET 22 Apr 2025 Shift 2 Paper

Concept:Use the inverse tangent subtraction identity to simplify the equation and solve separately for each case.Explanation:Let $A = x+\frac{2}{x}$ and $B = \frac{4}{x}$.Then $A-B = x-\frac{2}{x}$.The given equation becomes:$$\tan^{-1}A-\tan^{-1}B=\tan^{-1}(A-B)$$Using the identity $\tan^{-1}A-\tan^{-1}B=\tan^{-1}\left(\frac{A-B}{1+AB}\right)$, we get:$$1+AB = 1+\left(x+\frac{2}{x}\right)\frac{4}{x} = 5+\frac{8}{x^{2}}$$$$\tan^{-1}\left(\frac{x-\frac{2}{x}}{5+\frac{8}{x^{2}}}\right)=\tan^{-1}\left(x-\frac{2}{x}\right)$$Since $\tan^{-1}$ is one-to-one, equate the arguments:$$\frac{x-\frac{2}{x}}{5+\frac{8}{x^{2}}}=x-\frac{2}{x}$$Case 1: $x-\frac{2}{x}=0$$$x^{2}-2=0 \Rightarrow x=\pm\sqrt{2}$$Case 2: $x-\frac{2}{x}\neq0$, then:$$5+\frac{8}{x^{2}}=1 \Rightarrow x^{2}=-2$$This gives no real solution.Hence, the only real solutions are $x=\pm\sqrt{2}$.Answer:There are 2 solutions, i.e. Option B.