Let f(x)=∫0x(t-1)(t-2)2 dt. If f(x) ≥ k for all x and for some k, then the set of exhaustive values of k is

Question

Let f(x)=∫0x(t-1)(t-2)2 dt. If f(x) ≥ k for all x and for some k, then the set of exhaustive values of k is (A) (0 , ∈ fty) (B) (0 , 2) (C) (1 , ∈ fty) (D) (-∞,-(17/12)]

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-fk4hqcrt9q4w.png" /> x=1 is a point of global minima of f(x) Minimum value of f(x)= displaystyle ∫ 01 (t - 1)(t - 2)2dt = displaystyle ∫ 01 (t3 - 5 t2 + 8 t - 4)dt =(- 17/12)

Q. Let $f (x) = \int_{0}^{x} (t - 1) (t - 2)^{2} d t$ . If $f (x) \geq k$ for all $x$ and for some $k,$ then the set of exhaustive values of $k$ is

$(0, \in f t y)$

$(0, 2)$

$(1, \in f t y)$

$(- \infty, - \frac{17}{12}]$

$x = 1$ is a point of global minima of $f (x)$
Minimum value of
$f (x) = \int_{0}^{1} (t - 1) (t - 2)^{2} d t$
$= \int_{0}^{1} (t^{3} - 5 t^{2} + 8 t - 4) d t$
$= \frac{- 17}{12}$

Q. Let f(x)=∫0x​(t−1)(t−2)2dt. If f(x)≥k for all x and for some k, then the set of exhaustive values of k is

(0,∈fty)

(0,2)

(1,∈fty)

(−∞,−1217​]