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
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Answer:
Correct Answer: 34. (b, d)
Solution:
- 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.