If the function f(x)=x3+2px2+27x+16 is strictly increasing for all x∈R, then the range of P is

f'(x)>0⇒3x2+4px+27>0Δ<0⇒p2−814<0⇒p∈(−92,92)

If the function f(x)=x3+2px2+27x+16 is strictly increasing for all x∈R, then the range of P is

f'(x)>0⇒3x2+4px+27>0Δ<0⇒p2−814<0⇒p∈(−92,92)

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If the function $f (x) = x^{3} + 2 p x^{2} + 27 x + 16$ is strictly increasing for all $x \in R,$ then the range of $P$ is

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$(- \infty, \frac{- 9}{2}) \cup (\frac{9}{2}, \infty)$

$(\frac{- 9}{2}, \frac{9}{2})$

$(- \infty, - 9) \cup (9, \infty)$

$(- 9, 9)$

answer is B.

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Detailed Solution

$f' (x) > 0 \Rightarrow 3 x^{2} + 4 p x + 27 > 0$

$Δ < 0 \Rightarrow p^{2} - \frac{81}{4} < 0 \Rightarrow p \in (\frac{- 9}{2}, \frac{9}{2})$

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If the function f(x)=x3+2px2+27x+16 is strictly increasing for all x∈R, then the range of P is

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If the function $f (x) = x^{3} + 2 p x^{2} + 27 x + 16$ is strictly increasing for all $x \in R,$ then the range of $P$ is