Applications Of Derivatives Question 48
Question: At what points of curve $ y=\frac{2}{3}x^{3}+\frac{1}{2}x^{2}, $ the tangent makes equal angle with the axis-
Options:
A) $ ( \frac{1}{2},\frac{5}{24} ) $ and $ ( -1,-\frac{1}{6} ) $
B) $ ( \frac{1}{2},\frac{4}{9} ) $ and $ (-1,0) $
C) $ ( \frac{1}{3},\frac{1}{7} ) $ and $ ( -3,\frac{1}{2} ) $
D) $ ( \frac{1}{3},\frac{4}{47} ) $ and $ ( -1,-\frac{1}{3} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ y=\frac{2}{3}x^{3}+\frac{1}{2}x^{2}\therefore \frac{dy}{dx}=\frac{2}{3}3x^{2}+\frac{1}{2}2x=2x^{2}+x $ Since the tangent makes equal angles with the axes.
$ \therefore \frac{dy}{dx}=\pm 1or2x^{2}+x=\pm 1 $ or
$ 2x^{2}+x-1=0(2x^{2}+x+1=0 $ has no real roots) or $ (2x-1)(x+1)=0 $ i.e., $ x=\frac{1}{2} $ or $ x=-1 $ .
BETA
