Applications Of Derivatives Question 226
Question: At what points of the curve $ y=\frac{2}{3}x^{3}+\frac{1}{2}x^{2}, $ tangent makes the equal angle with axis
[UPSEAT 1999]
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:
$ y=\frac{2}{3}x^{3}+\frac{1}{2}x^{2} $
therefore $ \frac{dy}{dx}=2x^{2}+x $ …………..(i)
Now tangent makes equal angle with axis
$ y=45^{o} $ or $ -45^{o} $
$ \frac{dy}{dx}=\tan (\pm 45^{o})=\pm \tan (45^{o})=\pm 1 $
From equation (i), $ 2x^{2}+x=1 $ (taking +ve sign)
therefore $ 2x^{2}+x-1=0 $
therefore $ (2x-1),(x+1)=0 $
\ $ x=\frac{1}{2},-1 $
From the given curve, when $ x=\frac{1}{2} $ , $ y=\frac{2}{3}.\frac{1}{8}+\frac{1}{2}.\frac{1}{4}=\frac{1}{12}+\frac{1}{8}=\frac{5}{24} $ and when $ x=-1 $ ,
$ y=\frac{2}{3}(-1)+\frac{1}{2}.1 $ = $ -\frac{2}{3}+\frac{1}{2}=-\frac{1}{6} $
Therefore, required points are $ ( \frac{1}{2},,\frac{5}{24} ) $ and $ ( -1,,-\frac{1}{6} ) $ .
BETA
