Let f(b) be the minimum value of the expression y=x2-2x+(b3 - 3 b2 + 4)∀ x∈ R. Then, the maximum value of f(b) as b varies from 0 to 4 is

Question

Let f(b) be the minimum value of the expression y=x2-2x+(b3 - 3 b2 + 4)∀ x∈ R. Then, the maximum value of f(b) as b varies from 0 to 4 is (A) 20 (B) 19 (C) 63 (D) 64

Tardigrade Admin · Accepted Answer

y=(x - 1)2+b3-3b2+3 ∴ y min=f(b)=b3-3b2+3 Now, f' (b) = 3 b2 - 6 b =3b(b - 2) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-ip93apqbqt2yjs0y.jpg" /> Also, f(0)=3 f(2)=-1 f(4)=19 ∴ max f (b)=19

Q. Let $f (b)$ be the minimum value of the expression $y = x^{2} - 2 x + (b^{3} - 3 b^{2} + 4) \forall x \in R .$ Then, the maximum value of $f (b)$ as $b$ varies from $0$ to $4$ is

20

19

63

64

$y = (x - 1)^{2} + b^{3} - 3 b^{2} + 3$
$∴ y_{m i n} = f (b) = b^{3} - 3 b^{2} + 3$
Now, $f^{^{'}} (b) = 3 b^{2} - 6 b$
$= 3 b (b - 2)$

Also, $f (0) = 3$
$f (2) = - 1$
$f (4) = 19$
$∴ max f (b) = 19$

Q. Let f(b) be the minimum value of the expression y=x2−2x+(b3−3b2+4)∀x∈R. Then, the maximum value of f(b) as b varies from 0 to 4 is

20

19

63

64

y=(x−1)2+b3−3b2+3 ∴ymin​=f(b)=b3−3b2+3 Now, f′(b)=3b2−6b =3b(b−2) Also, f(0)=3 f(2)=−1 f(4)=19 ∴maxf(b)=19

Q. Let $f (b)$ be the minimum value of the expression $y = x^{2} - 2 x + (b^{3} - 3 b^{2} + 4) \forall x \in R .$ Then, the maximum value of $f (b)$ as $b$ varies from $0$ to $4$ is

$y = (x - 1)^{2} + b^{3} - 3 b^{2} + 3$
$∴ y_{m i n} = f (b) = b^{3} - 3 b^{2} + 3$
Now, $f^{^{'}} (b) = 3 b^{2} - 6 b$
$= 3 b (b - 2)$

Also, $f (0) = 3$
$f (2) = - 1$
$f (4) = 19$
$∴ max f (b) = 19$