Let f(x) be a cubic polynomial on R which increases in the interval (-∞, 0) ∪(1, ∞) and decreases in the interval (0,1). If f prime(2)=6 and f(2)=2, then the value of tan -1( f (1))+ tan -1( f ((3/2)))+ tan -1( f (0)) cquals

Question

Let f(x) be a cubic polynomial on R which increases in the interval (-∞, 0) ∪(1, ∞) and decreases in the interval (0,1). If f prime(2)=6 and f(2)=2, then the value of tan -1( f (1))+ tan -1( f ((3/2)))+ tan -1( f (0)) cquals (A)  tan -1 2 (B)  cot -1 2 (C) - tan -1 2 (D) - cot -1 2

Tardigrade Admin · Accepted Answer

f prime(x)=k x(x-1) f prime(2)=2 k =6 ∴ k =3 ∴ f prime( x )=3 x 2-3 x ⇒ f ( x )= x 3-(3 x 2/2)+ C f (2)=8-6+ C =2+ C =2 ∴ C =0 ⇒ f ( x )= x 2( x -(3/2)) .

Q. Let $f (x)$ be a cubic polynomial on $R$ which increases in the interval $(- \infty, 0) \cup (1, \infty)$ and decreases in the interval $(0, 1)$ . If $f^{'} (2) = 6$ and $f (2) = 2$ , then the value of $tan^{- 1} (f (1)) + tan^{- 1} (f (\frac{3}{2})) + tan^{- 1} (f (0))$ cquals

$tan^{- 1} 2$

$cot^{- 1} 2$

$- tan^{- 1} 2$

$- cot^{- 1} 2$

$f^{'} (x) = k x (x - 1)$
$f^{'} (2) = 2 k = 6 ∴ k = 3$
$∴ f^{'} (x) = 3 x^{2} - 3 x \Rightarrow f (x) = x^{3} - \frac{3 x ^{2}}{2} + C$
$f (2) = 8 - 6 + C = 2 + C = 2 ∴ C = 0$
$\Rightarrow f (x) = x^{2} (x - \frac{3}{2}) .$

Q. Let f(x) be a cubic polynomial on R which increases in the interval (−∞,0)∪(1,∞) and decreases in the interval (0,1). If f′(2)=6 and f(2)=2, then the value of tan−1(f(1))+tan−1(f(23​))+tan−1(f(0)) cquals

tan−12

cot−12

−tan−12

−cot−12