MHTCET 25 Apr 2025 Shift 1 Paper

Concept:This is a composite function problem where $\mathrm{g}(x)$ is substituted into $\mathrm{f}(x)$, followed by simplification using algebraic identities.Explanation:We need to find $(\mathrm{f} \circ \mathrm{g})(x) = \mathrm{f}(\mathrm{g}(x))$.Given $\mathrm{g}(x) = \frac{3x + x^{3}}{1 + 3x^{2}}$, substitute it into $\mathrm{f}(x) = \log\left(\frac{1+x}{1-x}\right)$.So, $\mathrm{f}(\mathrm{g}(x)) = \log\left(\frac{1 + \frac{3x + x^{3}}{1 + 3x^{2}}}{1 - \frac{3x + x^{3}}{1 + 3x^{2}}}\right)$.Simplify the numerator and denominator by taking the common denominator $1 + 3x^{2}$:$$= \log\left(\frac{1 + 3x^{2} + 3x + x^{3}}{1 + 3x^{2} - 3x - x^{3}}\right)$$Recognize the algebraic identities:$$1 + 3x^{2} + 3x + x^{3} = (1+x)^{3}$$$$1 + 3x^{2} - 3x - x^{3} = (1-x)^{3}$$Therefore,$$= \log\left(\frac{(1+x)^{3}}{(1-x)^{3}}\right) = \log\left(\left(\frac{1+x}{1-x}\right)^{3}\right)$$Using the power rule of logarithms, $\log(a^{n}) = n\log(a)$:$$= 3\log\left(\frac{1+x}{1-x}\right) = 3\mathrm{f}(x)$$Answer:$(\mathrm{fog})(x) = 3\mathrm{f}(x)$Hence, the correct option is B. $3\mathrm{f}(x)$.