Find the equation of tangent towhich has slope 2.a)2x – y = 1b)N...
y = 1/(x-3)
dy/dx = d/dx(x-3)-1
dy/dx = (-1) (x-3)(-1-1) . d(x-3)/dx
dy/dx = -(x-3)(-2)
dy/dx = - 1/(x-3)2
Given, slope = 2, dy/dx = 2
- 1/(x-3)2 = 2
⇒ -1 = 2(x-3)2
2(x-3)2 = -1
(x-3)2 = -½
We know that square of any number is always positive So, (x-3)2 > 0
(x-3)2 = -½ is not possible
No tangent to the curve has slope 2.
View all questions of this testFind the equation of tangent towhich has slope 2.a)2x – y = 1b)N...
Y=1/(x-3)^2
dy/dx=(-1/(x-3)^2)
given slope=2
-1/(x-3)^2 =2
-1/2=(x-3)^2
negative number is not equal to square. so no tangent
Find the equation of tangent towhich has slope 2.a)2x – y = 1b)N...
y = 1/(x-3)
dy/dx = d/dx(x-3)-1
dy/dx = (-1) (x-3)(-1-1) . d(x-3)/dx
dy/dx = -(x-3)(-2)
dy/dx = - 1/(x-3)2
Given, slope = 2, dy/dx = 2
- 1/(x-3)2 = 2
⇒ -1 = 2(x-3)2
2(x-3)2 = -1
(x-3)2 = -½
We know that square of any number is always positive So, (x-3)2 > 0
(x-3)2 = -½ is not possible
No tangent to the curve has slope 2.