JEE MATHEMATICS MCQ: STRAIGHT LINES

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Analyzing +4 | -1

A\left(\sin \alpha, \frac{1}{\sqrt{2}}\right)

and

B\left(\frac{1}{\sqrt{2}}, \cos \alpha\right),-\pi \leq \alpha \leq \pi

, are two points on the same side of the line

x-y=0

, then

\alpha

belongs to the interval

View answer and explanation

Correct option: A

\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)

Option A (

\left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, \frac{3 \pi}{4}\right)

) is correct because it represents the interval where both points A and B lie on the same side of the line x-y=0.

More Questions from STRAIGHT LINES

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x-3 y=p, a x+2 y=q

a x+y=r

\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\}

y = 2\;x + 4

6x = 3y + 5

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

(3,8)

x+3 y=7

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0