Application of Derivatives 4 Question 32

34. If $f(x)=2-e^{x-1}, \quad 1<x \leq 2$ and $g(x)=\int _0^{x} f(t) d t$,

$$ x-e, \quad 2<x \leq 3 $$

$x \in[1,3]$, then

(2006, 3M)

(a) $g(x)$ has local maxima at $x=1+\log _e 2$ and local minima at $x=e$

(b) $f(x)$ has local maxima at $x=1$ and local minima at $x=2$

(c) $g(x)$ has no local minima

(d) $f(x)$ has no local maxima

Show Answer

Answer:

Correct Answer: 34. (b, d)

Solution:

  1. Given,

$g^{\prime \prime}\left(1+\log _e 2\right)=-e^{\log _e 2}<0, g(x)$ has a local maximum.

Also, at $x=e$,

$g^{\prime \prime}(e)=1>0, g(x)$ has a local minima.

$\because f(x)$ is discontinuous at $x=1$, then we get local maxima at $x=1$ and local minima at $x=2$.

Hence, (a) and (b) are correct answers.



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक