If f ( x )= x 3+ px 2+ qx + r , p , q , r ∈ R is monotonically decreasing in largest possible interval (-2,2) then the value of ( p - q /2) is equal to

Question

If f ( x )= x 3+ px 2+ qx + r , p , q , r ∈ R is monotonically decreasing in largest possible interval (-2,2) then the value of ( p - q /2) is equal to (A) 0 (B) 2 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

f prime( x )=3 x 2+2 Px + q < 0 =3(x2-4)

Q. If $f (x) = x^{3} + p x^{2} + q x + r, p, q, r \in R$ is monotonically decreasing in largest possible interval $(- 2, 2)$ then the value of $\frac{p - q}{2}$ is equal to

0

2

4

6

$f^{'} (x) = 3 x^{2} + 2 P x + q < 0$
$= 3 (x^{2} - 4)$

Q. If f(x)=x3+px2+qx+r,p,q,r∈R is monotonically decreasing in largest possible interval (−2,2) then the value of 2p−q​ is equal to

0

2

4

6

f′(x)=3x2+2Px+q<0 =3(x2−4)

Q. If $f (x) = x^{3} + p x^{2} + q x + r, p, q, r \in R$ is monotonically decreasing in largest possible interval $(- 2, 2)$ then the value of $\frac{p - q}{2}$ is equal to

$f^{'} (x) = 3 x^{2} + 2 P x + q < 0$
$= 3 (x^{2} - 4)$