Applications Of Derivatives Question 411
Question: The distance of the point on $ y=x^{4}+3x^{2}+2x $ which is nearest to the line $ y=2x-1 $ is
Options:
A) $ \frac{2}{\sqrt{5}} $
B) $ \sqrt{5} $
C) $ \frac{1}{\sqrt{5}} $
D) $ 5\sqrt{5} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ y=x^{4}+3x^{2}+2x\therefore \frac{dy}{dx}=4x^{3}+6x+2 $ Point on curve which is nearest to the line $ y=2x-1 $ is the point where tangent to curve is parallel to given line.
Therefore, $ 4x^{3}+6x+2=2 $ or $ 2x^{3}+3x=0 $ or $ x=0,y=0 $ .
Therefore, point on the curve at the least distance from the line $ y=2x-1 $ is (0, 0).
Distance of this point from line is $ \frac{1}{\sqrt{5}} $ .
BETA
