Integral Calculus Question 174
Question: The function $ f(x)=\int\limits_{-1}^{x}{t(e^{t}-1)(t-1){{(t-2)}^{3}}{{(t-3)}^{5}}} $ dt has a local minimum at x =
Options:
A) {0}
B) {1, 3}
C) {2}
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \frac{dy}{dx}=f’(x) $
$ \Rightarrow x(e^{x}-1)(x-1){{(x-2)}^{3}}{{(x-3)}^{5}}=0 $ Critical points are 0, 1, 2, 3. Consider change of sign of $ \frac{dy}{dx} $ at $ x=3 $ . $ x<3,\frac{dy}{dx} $ = negative and $ x>3,\frac{dy}{dx}= $ positive Change if from negative to positive, hence minimum at $ x=3 $ . Again minimum and maximum occur alternately.
$ \therefore $ 2nd minimum is at $ x=1 $ .
BETA
