JEE Advanced 2019 · Paper 2 · Mathematics Q7

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MathematicsMultiple correct+4 / -1Question 43 of 54

Let $f:\mathbb R\to\mathbb R$ be given by $f(x) = (x-1)(x-2)(x-5)$ . Define

$F(x) = \int_0^xf(t)\,dt,\quad x > 0.$

Then which of the following options is/are correct?

F

has a local minimum at

x = 1

F

has a local maximum at

x = 2

F

has two local maxima and one local minimum in

(0,\infty)

F(x) \ne 0

for all

x\in(0,5)

Answer and worked solution

Official answer: A, B, D

$F'=f=(x-1)(x-2)(x-5)$ . Its sign is negative on $(0,1)$ , positive on $(1,2)$ , negative on $(2,5)$ and positive beyond 5.

Extrema. $F$ has minima at 1 and 5 and a maximum at 2.

(A) is true.
(B) is true.
(C) is false: there are two minima and one maximum.

(D) $F(x)=\frac{x^4}4-\frac{8x^3}3+\frac{17x^2}2-10x$ . It starts at 0 and is decreasing near 0. Its local maximum on $(0,5)$ is $F(2)=-\frac{10}3<0$ , so $F<0$ throughout $(0,5)$ . True.

(A), (B), (D)

Topics:Fundamental theorem of calculus Maxima and minima

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JEE Advanced 2019 · Paper 2, Mathematics question 7