If the range of the function f(x)= tan -1(3 x2+b x+c) is [0, (π/2)) then

Question

If the range of the function f(x)= tan -1(3 x2+b x+c) is [0, (π/2)) then (A) b 2=3 c (B) b2=4 c (C) b 2=12 c (D) b 2=8 c

Tardigrade Admin · Accepted Answer

Range of f(x)= tan -1(3 x2+b x+c) is [0, (π/2)) if and only if range of g(x)=3 x2+b x+c is [0, ∞). This is possible only when discriminant of the equation 3 x 2+ bx + c =0 is equal to zero. i.e. b2=12c

Q. If the range of the function $f (x) = tan^{- 1} (3 x^{2} + b x + c)$ is $[0, \frac{π}{2})$ then

$b^{2} = 3 c$

$b^{2} = 4 c$

$b^{2} = 12 c$

$b^{2} = 8 c$

Range of $f (x) = tan^{- 1} (3 x^{2} + b x + c)$ is $[0, \frac{π}{2})$ if and only if range of $g (x) = 3 x^{2} + b x + c$ is $[0, \infty)$ .
This is possible only when discriminant of the equation $3 x^{2} + b x + c = 0$ is equal to zero. i.e. $b^{2} = 12 c$

Q. If the range of the function f(x)=tan−1(3x2+bx+c) is [0,2π​) then

b2=3c

b2=4c

b2=12c

b2=8c