Applications Of Derivatives Question 405
Question: The abscissa of the points of the curve $ y=x^{3} $ in the interval
[-2, 2], where the slope of the tangents can be obtained by mean value theorem for the interval [-2, 2], are [MP PET 1993]
Options:
A) $ \pm \frac{2}{\sqrt{3}} $
B) $ \pm \sqrt{3} $
C) $ \pm \frac{\sqrt{3}}{2} $
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
Given that equation of curve $ y=x^{3}=f(x) $
So $ f(2)=8 $ and $ f(-2)=-8 $
Now $ f’(x)=3x^{2}\Rightarrow f’(x)=\frac{f(2)-f(-2)}{2-(-2)} $
therefore $ \frac{8-(-8)}{4}=3x^{2};\therefore x=\pm \frac{2}{\sqrt{3}} $ .
BETA
