The length of the longest interval in which the function f(x)=x3-3a2x+4 is decreasing is (∀ a 0)

Question

The length of the longest interval in which the function f(x)=x3-3a2x+4 is decreasing is (∀ a > 0) (A) a (B) 2a (C) 3a (D) 4a

Tardigrade Admin · Accepted Answer

f' (x) = 3 x2 - 3 a2 =3(x - a)(x + a) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-v35dotodwmr1kavf.jpg" /> ∴ f(x) decreases in x∈ [- a , a] i.e. length =2a

Q. The length of the longest interval in which the function $f (x) = x^{3} - 3 a^{2} x + 4$ is decreasing is $(\forall a > 0)$

$a$

$2 a$

$3 a$

$4 a$

$f^{^{'}} (x) = 3 x^{2} - 3 a^{2}$
$= 3 (x - a) (x + a)$

$∴ f (x)$ decreases in $x \in [- a, a]$
i.e. length $= 2 a$

Q. The length of the longest interval in which the function f(x)=x3−3a2x+4 is decreasing is (∀a>0)

a

2a

3a

4a

f′(x)=3x2−3a2 =3(x−a)(x+a) ∴f(x) decreases in x∈[−a,a] i.e. length =2a

Q. The length of the longest interval in which the function $f (x) = x^{3} - 3 a^{2} x + 4$ is decreasing is $(\forall a > 0)$

$a$

$2 a$

$3 a$

$4 a$

$f^{^{'}} (x) = 3 x^{2} - 3 a^{2}$
$= 3 (x - a) (x + a)$

$∴ f (x)$ decreases in $x \in [- a, a]$
i.e. length $= 2 a$