If f:A→B is an onto function such that f(x)=|x|−x+1|x|−x. Then A and B are respectively

f(x)=|x|−x+1|x|−xA is domain of f(x)∴f(x) is defined iff |x|−x> 0⇒ |x| > xx∈(−∞, 0)B is Range of f(x)we know that A.M ≥ G.M⇒ |x|−x+1|x|−x 2≥ |x|−x1|x|−x⇒ |x|−x+1|x|−x≥ 21⇒ f(x)≥2B= Range = [2, ∞)

If f:A→B is an onto function such that f(x)=|x|−x+1|x|−x. Then A and B are respectively

f(x)=|x|−x+1|x|−xA is domain of f(x)∴f(x) is defined iff |x|−x> 0⇒ |x| > xx∈(−∞, 0)B is Range of f(x)we know that A.M ≥ G.M⇒ |x|−x+1|x|−x 2≥ |x|−x1|x|−x⇒ |x|−x+1|x|−x≥ 21⇒ f(x)≥2B= Range = [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

$I f f : A \to B i s a n o n t o f u n c t i o n s u c h t h a t f (x) = \sqrt{| x | - x} + \frac{1}{\sqrt{| x | - x}} . T h e n A a n d B a r e r e s p e c t i v e l y$

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

$(- \infty, \infty) [2, \infty)$

$(- \infty, \infty) (0, \infty)$

$(- \infty, 0] (0, \infty)$

$(0, \infty) (2, \infty)$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = \sqrt{| x | - x} + \frac{1}{\sqrt{| x | - x}} \\ A i s d o m a i n o f f (x) \\ ∴ f (x) i s d e f i n e d i f f | x | - x > 0 \\ \Rightarrow | x | > x \\ x \in (- \infty, 0) \\ B i s R a n g e o f f (x) \\ w e k n o w t h a t A . M \geq G . M \\ \Rightarrow \frac{\sqrt{| x | - x} + \frac{1}{\sqrt{| x | - x}}}{2} \geq \sqrt{\sqrt{| x | - x} \frac{1}{\sqrt{| x | - x}}} \\ \Rightarrow \sqrt{| x | - x} + \frac{1}{\sqrt{| x | - x}} \geq 2 \sqrt{1} \\ \Rightarrow f (x) \geq 2 \\ B = R a n g e = [2, \infty) \end{array}$