If $f(x)=\int_{0}^{x} e^{t^{2}}(t-2)(t-3) d t$ for all $x \in(0, \infty)$, then

(A) $f$ has a local maximum at $x=2$; (B) $f$ is decreasing on $(2,3)$; (C) there exists some $c \in(0, \infty)$ such that $f^{\prime \prime}(c)=0$; (D) $f$ has a local minimum at $x=3$

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Application of Derivatives

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f (x) = \int_{0}^{x} e^{t^{2}} (t - 2) (t - 3) d t

for all

x \in (0, \infty)

, then

(A) $f$ has a local maximum at $x = 2$ ✓ Correct
(B) $f$ is decreasing on $(2, 3)$ ✓ Correct
(C)there exists some $c \in (0, \infty)$ such that $f^{''} (c) = 0$ ✓ Correct
(D) $f$ has a local minimum at $x = 3$ ✓ Correct

Answer: (A), (B), (C), (D)

Solution

By the Fundamental Theorem of Calculus:

f^{'} (x) = e^{x^{2}} (x - 2) (x - 3)

Since

e^{x^{2}} > 0

always, the sign of

f^{'} (x)

depends on

(x - 2) (x - 3)

For $x < 2$ : $f^{'} (x) > 0$ (increasing)
For $2 < x < 3$ : $f^{'} (x) < 0$ (decreasing)
For $x > 3$ : $f^{'} (x) > 0$ (increasing)

Therefore:

$f$ has a local maximum at $x = 2$ ✓ (A)
$f$ is decreasing on $(2, 3)$ ✓ (B)
$f$ has a local minimum at $x = 3$ ✓ (D)

For (C):

f^{'} (x) = 0

x = 2

and

x = 3

. By Rolle's theorem, there exists

c \in (2, 3)

such that

f^{''} (c) = 0

✓ (C)

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