Let b be a nonzero real number. Suppose f: R | vedclass

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(D) Given the differential equation $f^{\prime}(x) = \frac{f(x)}{b^2+x^2}$.
Separating variables,we get $\frac{f^{\prime}(x)}{f(x)} = \frac{1}{b^2+x^

Vedclass · Accepted Answer

$A, C$

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(D) Given the differential equation $f^{\prime}(x) = \frac{f(x)}{b^2+x^2}$.
Separating variables,we get $\frac{f^{\prime}(x)}{f(x)} = \frac{1}{b^2+x^2}$.
Integrating both sides with respect to $x$,we have $\int \frac{f^{\prime}(x)}{f(x)} dx = \int \frac{1}{b^2+x^2} dx$.
This yields $\ln|f(x)| = \frac{1}{b} 	an^{-1}\left(\frac{x}{b}ight) + C$.
Using the initial condition $f(0)=1$,we get $\ln(1) = \frac{1}{b} 	an^{-1}(0) + C$,which implies $C=0$.
Thus,$f(x) = e^{\frac{1}{b} 	an^{-1}(\frac{x}{b})}$.
Now,consider $f(x)f(-x) = e^{\frac{1}{b} 	an^{-1}(\frac{x}{b})} \cdot e^{\frac{1}{b} 	an^{-1}(\frac{-x}{b})} = e^{\frac{1}{b} (	an^{-1}(\frac{x}{b}) - 	an^{-1}(\frac{x}{b}))} = e^0 = 1$. So,$(C)$ is true.
For $b>0$,$f^{\prime}(x) = \frac{f(x)}{b^2+x^2} > 0$ since $f(x) = e^{\dots} > 0$ and $b^2+x^2 > 0$. Thus,$f$ is an increasing function. So,$(A)$ is true.
Therefore,the correct statements are $(A)$ and $(C)$.

Column $I$	Column $II$
$(A)$ Interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y^{\prime}+y=0$	$(p)$ $(-\frac{\pi}{2}, \frac{\pi}{2})$
$(B)$ Interval containing the value of the integral $\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) dx$	$(q)$ $(0, \frac{\pi}{2})$
$(C)$ Interval in which at least one of the points of local maximum of $\cos^2 x+\sin x$ lies	$(r)$ $(\frac{\pi}{8}, \frac{5\pi}{4})$
$(D)$ Interval in which $\tan^{-1}(\sin x+\cos x)$ is increasing	$(s)$ $(0, \frac{\pi}{8})$
	$(t)$ $(-\pi, \pi)$

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