The acute angles between the curves y=2x2-x and texty text2= textx at (0,0) and (1,1) are α and β respectively, then

Question

The acute angles between the curves y=2x2-x and texty text2= textx at (0,0) and (1,1) are α and β respectively, then (A) α -β =0 (B) α +β =0 (C) α >β  (D) α < β

Tardigrade Admin · Accepted Answer

y=2x2-x ⇒ (d y/d x)=4x-1 Let, m1=4x-1 y2=x ⇒ 2y(d y/d x)=1 Let, m2=(d y/d x)=(1/2 y) At P(0,0),m1=-1,m2 arrow not defined Angle α =(π /4) At Q(1,1),m1=3,m2=(1/2) β =(t a n)- 1((3 - (1/2)/1 + (3)2))=(π /4)

Q. The acute angles between the curves $y = 2 x^{2} - x$ and $y^{2} = x$ at $(0, 0)$ and $(1, 1)$ are $α$ and $β$ respectively, then

$α - β = 0$

$α + β = 0$

$α > β$

$α < β$

Q. The acute angles between the curves y=2x2−x and y2=x at (0,0) and (1,1) are α and β respectively, then

α−β=0

α+β=0

α>β

α<β

y=2x2−x ⇒dxdy​=4x−1 Let, m1​=4x−1 y2=x ⇒2ydxdy​=1 Let, m2​=dxdy​=2y1​ At P(0,0),m1​=−1,m2​→ not defined Angle α=4π​ At Q(1,1),m1​=3,m2​=21​ β=(tan)−1(1+23​3−21​​)=4π​

Q. The acute angles between the curves $y = 2 x^{2} - x$ and $y^{2} = x$ at $(0, 0)$ and $(1, 1)$ are $α$ and $β$ respectively, then

$α - β = 0$

$α + β = 0$

$α > β$

$α < β$