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(C) For Statement $I$: Given $f(x) = \sec x + 	an x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$.$f'(x) = \sec x 	an x + \sec^2 x = \sec x(	an

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Both Statement $I$ and Statement $II$ are true

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(C) For Statement $I$: Given $f(x) = \sec x + 	an x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$.$f'(x) = \sec x 	an x + \sec^2 x = \sec x(	an x + \sec x)$.Since $\sec x + 	an x = \frac{1+\sin x}{\cos x} = \frac{(\cos(x/2) + \sin(x/2))^2}{\cos^2(x/2) - \sin^2(x/2)} = \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)} = 	an(\frac{\pi}{4} + \frac{x}{2})$.For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$,$\frac{x}{2} \in \left(-\frac{\pi}{4}, \frac{\pi}{4}ight)$,so $\frac{\pi}{4} + \frac{x}{2} \in (0, \frac{\pi}{2})$.In this interval,$\sec x > 0$ and $	an(\frac{\pi}{4} + \frac{x}{2}) > 0$,so $f'(x) > 0$. Thus,$f(x)$ is strictly increasing and one-one.For Statement $II$: Given $f(x) = x^2$ on $[0, \infty)$.If $f(x_1) = f(x_2)$,then $x_1^2 = x_2^2$. Since $x_1, x_2 \geq 0$,this implies $x_1 = x_2$.Thus,$f(x)$ is one-one.Both statements are true.

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