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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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,$A = \sqrt{\sin 	heta}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 -

Similar Questions

The line $L$ is given by $\frac{x}{5} + \frac{y}{b} = 1$ and passes through the point $(13, 32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$. Then the distance between $L$ and $K$ is

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is

If $(x_1, y_1)$ and $(x_2, y_2)$ are two points on the line $x+y+3=0$ such that each of them is at a distance of $\sqrt{5}$ units from the line $x+2y+2=0$,then the value of $|x_1-x_2|$ is:

The distance of the point $(2, 5)$ from the line $3x + y + 4 = 0$,measured parallel to the line $3x - 4y + 8 = 0$,is

The points $(2,3)$ and $(-4,-4/3)$ lie on the opposite sides of the line $L \equiv 5x - 6y + k = 0$,where $k$ is an integer. If the points $(1,2)$ and $(4,5)$ lie on the same side of the line $L = 0$,then the perpendicular distance from the origin to the line $L = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module