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(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cos x$ and $Q = \cos x$. The In

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all of the above

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(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cos x$ and $Q = \cos x$. The Integrating Factor $(I.F.)$ is given by $e^{\int P \, dx} = e^{\int \cos x \, dx} = e^{\sin x}$. The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + C$. Substituting the values: $y \cdot e^{\sin x} = \int \cos x \cdot e^{\sin x} \, dx + C$. Let $u = \sin x$,then $du = \cos x \, dx$. The integral becomes $\int e^u \, du = e^u = e^{\sin x}$. So,$y \cdot e^{\sin x} = e^{\sin x} + C$. Dividing by $e^{\sin x}$,we get $y = 1 + C e^{-\sin x}$. Since the curve passes through $(0, 1)$,we substitute $x = 0$ and $y = 1$: $1 = 1 + C e^{-\sin 0} \implies 1 = 1 + C(1) \implies C = 0$. Thus,the function is $y = 1$. Since $y = 1$ is a constant function,it is also periodic (with any period $T > 0$) and it is continuous and differentiable for all $x \in \mathbb{R}$. Therefore,all the given statements are correct.

The graph of the function $y = f(x)$ passing through the point $(0, 1)$ and satisfying the differential equation $\frac{dy}{dx} + y \cos x = \cos x$ is such that

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