Let y=fx represent a parabola with focus -1/2,0 and directrix y=-1/2. Then S=x∈ℝ:tan^(-1)√fx+sin^(-1)√(fx+1)=(π)/2

Question

Let y=fx represent a parabola with focus -1/2,0 and directrix y=-1/2. Then S=x∈ℝ:tan^(-1)√fx+sin^(-1)√(fx+1)=(π)/2:

MARKS Content Team · Accepted Answer

Correct answer: (A) contains exactly two elements. Equation of parabola with focus (-1/2, 0) and directrix y=-1/2 is: x+12^2=y+1/4 ⇒y=x^2+x Put f(x)=y=x+x^2 in S, tan^(-1)√(xx+1)+sin^(-1)√(x^2+x+1)=(π)/2 ⇒0≤x^2+x+1≤1 ⇒x^2+x≤0 …1 Also, x^2+x≥0 …2 By 1 & 2, we have x^2+x=0⇒x=0,-1 S contains 2 element.

Let $y = f (x)$ represent a parabola with focus $(- \frac{1}{2}, 0)$ and directrix $y = - \frac{1}{2}$ . Then $S = \{x \in ℝ : \tan^{- 1} (\sqrt{f (x)} + \sin^{- 1} (\sqrt{f (x) + 1})) = \frac{π}{2}\} :$

Solution

Relevant Questions

Let y=fx represent a parabola with focus -12,0 and directrix y=-12. Then S=x∈ℝ:tan-1fx+sin-1fx+1=π2:

Solution

Relevant Questions

Let $y = f (x)$ represent a parabola with focus $(- \frac{1}{2}, 0)$ and directrix $y = - \frac{1}{2}$ . Then $S = \{x \in ℝ : \tan^{- 1} (\sqrt{f (x)} + \sin^{- 1} (\sqrt{f (x) + 1})) = \frac{π}{2}\} :$