Let f(theta) be the distance of the line (sqr | vedclass

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(D) The distance $d$ of a line $Ax + By + C = 0$ from the origin $(0, 0)$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.Here,$A = \sqrt{\sin 	heta}$

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$\left[ \frac{1}{2^{1/4}}, 1 \right]$

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(D) The distance $d$ of a line $Ax + By + C = 0$ from the origin $(0, 0)$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.Here,$A = \sqrt{\sin 	heta}$,$B = \sqrt{\cos 	heta}$,and $C = 1$.Thus,$f(	heta) = \frac{1}{\sqrt{\sin 	heta + \cos 	heta}}$.For the expression to be defined,$\sin 	heta \ge 0$ and $\cos 	heta \ge 0$,which implies $	heta \in [0, \pi/2]$.Let $g(	heta) = \sin 	heta + \cos 	heta = \sqrt{2} \sin(	heta + \pi/4)$.For $	heta \in [0, \pi/2]$,the range of $g(	heta)$ is $[1, \sqrt{2}]$.Since $f(	heta) = \frac{1}{\sqrt{g(	heta)}}$,we evaluate the range of $f(	heta)$ based on the range of $g(	heta)$.When $g(	heta) = 1$,$f(	heta) = 1/\sqrt{1} = 1$.When $g(	heta) = \sqrt{2}$,$f(	heta) = 1/\sqrt{\sqrt{2}} = 1/2^{1/4}$.Thus,the range of $f(	heta)$ is $[1/2^{1/4}, 1]$.

Let $f(\theta)$ be the distance of the line $(\sqrt{\sin \theta})x + (\sqrt{\cos \theta})y + 1 = 0$ from the origin. Then the range of $f(\theta)$ is -

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