The least natural number ‘a’ for which x+ax−2>2 ∀ x∈(0,∞) is

f(x)=x+ax−2−2 f1(x)=1−2ax−3=0 ⇒x=(2a)1/3 f11(x)=6ax−4 f1((2a)1/3)>0 This x=(2a)1/3 is the point of minimaFor x+ax−2−2>0 ∀ x we must have f((2a)1/3)>0 ⇒(2a)1/3+a(2a)−2/3−2>0 ⇒3a>2(2a)2/3 ⇒27a3>32a2 ⇒a>3227 Hence least value of ‘a’ is 2

The least natural number ‘a’ for which x+ax−2>2 ∀ x∈(0,∞) is

f(x)=x+ax−2−2 f1(x)=1−2ax−3=0 ⇒x=(2a)1/3 f11(x)=6ax−4 f1((2a)1/3)>0 This x=(2a)1/3 is the point of minimaFor x+ax−2−2>0 ∀ x we must have f((2a)1/3)>0 ⇒(2a)1/3+a(2a)−2/3−2>0 ⇒3a>2(2a)2/3 ⇒27a3>32a2 ⇒a>3227 Hence least value of ‘a’ is 2

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

The least natural number ‘a’ for which $x + a x^{- 2} > 2 \forall x \in (0, \infty)$ 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

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = x + a x^{- 2} - 2 f^{1} (x) = 1 - 2 a x^{- 3} = 0 \Rightarrow x = {(2 a)}^{1 / 3} f^{11} (x) = 6 a x^{- 4} f^{1} ({(2 a)}^{1 / 3}) > 0$
This $x = {(2 a)}^{1 / 3}$ is the point of minima
For $x + a x^{- 2} - 2 > 0 \forall x$ we must have
$f ({(2 a)}^{1 / 3}) > 0 \Rightarrow {(2 a)}^{1 / 3} + a {(2 a)}^{- 2 / 3} - 2 > 0 \Rightarrow 3 a > 2 {(2 a)}^{2 / 3} \Rightarrow 27 a^{3} > 32 a^{2} \Rightarrow a > \frac{32}{27}$
Hence least value of ‘a’ is 2