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

f(x)=x3+2px2+27x+16f1(x)=3x2+4px+27f(x) is stricetly in (vng function sof1(x)>03x2+4px+27>0Δ<0⇒16p2−4(81)<0⇒p2−814<0⇒p∈−92,92

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

f(x)=x3+2px2+27x+16f1(x)=3x2+4px+27f(x) is stricetly in (vng function sof1(x)>03x2+4px+27>0Δ<0⇒16p2−4(81)<0⇒p2−814<0⇒p∈−92,92

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

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

$(- 9, 9)$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = x^{3} + 2 p x^{2} + 27 x + 16 \\ f^{1} (x) = 3 x^{2} + 4 p x + 27 \end{array}$
f(x) is stricetly in (vng function so
$\begin{array}{l} f^{1} (x) > 0 \\ 3 x^{2} + 4 p x + 27 > 0 \\ Δ < 0 \Rightarrow 16 p^{2} - 4 (81) < 0 \\ \Rightarrow p^{2} - \frac{81}{4} < 0 \Rightarrow p \in (\frac{- 9}{2}, \frac{9}{2}) \end{array}$