arthham

JEE Advanced 2019 · Integrals

Let f:R→Rf: R \to R be given by f(x)=(x−1)(x−2)(x−5)f(x)=(x-1)(x-2)(x-5). Define F(x)=∫0xf(t) dt,x>0F(x)=\int_0^x f(t)\,dt, x>0. Then which of the following options is/are correct?

Answers: A. FF has a local maximum at x=2x=2 · B. FF has a local minimum at x=1x=1 · D. F(x)<0F(x)<0 for all x∈(0,5)x \in (0,5)

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Written solution

Let f: R → R be given by f(x)=(x-1)(x-2)(x-5). Define F(x)=∫₀ˣ f(t) dt, x>0. Then which of the following options is/are correct?

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

F(x)=∫0xf(t) dt,x>0F(x)=\int_0^x f(t)\,dt,\quad x>0

To find: correct statement(s) about F\text{To find: correct statement(s) about } F

F′(x)=f(x)F'(x)=f(x) f(x)=0⇒x=1, 2, 5 are critical points of Ff(x)=0 \Rightarrow x=1,\ 2,\ 5 \text{ are critical points of } F f<0 (0,1);f>0 (1,2);f<0 (2,5);f>0 (5,∞)f<0\ (0,1);\quad f>0\ (1,2);\quad f<0\ (2,5);\quad f>0\ (5,\infty) x=1: min⁡;x=2: max⁡;x=5: min⁡x=1:\ \min;\quad x=2:\ \max;\quad x=5:\ \min F(x)=x44−8x33+17x22−10xF(x)=\dfrac{x^4}{4}-\dfrac{8x^3}{3}+\dfrac{17x^2}{2}-10x F(2)=4−643+34−20=−103<0F(2)=4-\dfrac{64}{3}+34-20=-\dfrac{10}{3}<0 F(0)=0, F(2)=−103⇒F(x)<0 ∀x∈(0,5)F(0)=0,\ F(2)=-\tfrac{10}{3} \Rightarrow F(x)<0\ \forall x\in(0,5) (1) true, (2) true, (3) false, (4) true\text{(1) true, (2) true, (3) false, (4) true} options (1), (2), (4)\boxed{\text{options (1), (2), (4)}}

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