Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be given by $\mathrm{f}(\mathrm{x})=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-5)$. Define $\mathrm{F}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}, \mathrm{x}>0$. Then which of the following options is/are correct ?

(A) F has a local minimum at $\mathrm{x}=1$; (B) F has a local maximum at $\mathrm{x}=2$; (C) $\mathrm{F}(\mathrm{x}) \neq 0$ for all $\mathrm{x} \in(0,5)$

(D) F has two local maxima and one local minimum in $(0, \infty)$

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Application of Derivatives

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Let

f : R \to R

be given by

f (x) = (x - 1) (x - 2) (x - 5)

. Define

F (x) = \int_{0}^{x} f (t) dt, x > 0

. Then which of the following options is/are correct ?

(A)F has a local minimum at $x = 1$ ✓ Correct
(B)F has a local maximum at $x = 2$ ✓ Correct
(C) $F (x) \neq = 0$ for all $x \in (0, 5)$ ✓ Correct
(D)F has two local maxima and one local minimum in $(0, \infty)$

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

Solution

Given

f (x) = (x - 1) (x - 2) (x - 5)

, we have

F^{'} (x) = f (x) = (x - 1) (x - 2) (x - 5)

for

x > 0

. The critical points are at

x = 1, 2, 5

. By analyzing the sign of

F^{'} (x)

, we see that

F (x)

has local minima at

x = 1

and

x = 5

, and a local maximum at

x = 2

. Expanding

f (x) = x^{3} - 8 x^{2} + 17 x - 10

and integrating:

F (x) = \frac{x ^{4}}{4} - \frac{8 x ^{3}}{3} + \frac{17 x ^{2}}{2} - 10 x

. From the graph of

y = F (x)

, we can verify that

F (x) \neq = 0

for all

x \in (0, 5)

. Therefore, options (A), (B), and (C) are correct.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Application of Derivatives

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