If the inequation x2−18x+72<x−1, then x lies in

x2−18x+72 0,x2−18x+72 0⇒x>1x2−18x+72 71⇒x>716 .Solution set =[12,∞)

If the inequation x2−18x+72<x−1, then x lies in

x2−18x+72 0,x2−18x+72 0⇒x>1x2−18x+72 71⇒x>716 .Solution set =[12,∞)

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 inequation $\sqrt{x^{2} - 18 x + 72} < x - 1$ , then x lies in

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

$\emptyset$

$[1, 2)$

$[12, \infty)$

$(1, 2]$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \sqrt{x^{2} - 18 x + 72} < x - 1 \Rightarrow x^{2} - 18 x + 72 \geq 0, x - 1 > 0, x^{2} - 18 x + 72 < (x - 1)^{2} \\ x^{2} - 18 x + 72 = 0 \Rightarrow x^{2} - 12 x - 6 x + 72 = 0 \Rightarrow x (x - 12) - 6 (x - 12) = 0 \Rightarrow (x - 12) (x - 6) = 0 \\ \Rightarrow x = 6 or 12 . x^{2} - 18 x + 72 \geq 0 \Rightarrow x \leq 6 or x \geq 12 . x - 1 > 0 \Rightarrow x > 1 \\ x^{2} - 18 x + 72 < (x - 1)^{2} \Rightarrow x^{2} - 18 x + 72 < x^{2} - 2 x + 1 \Rightarrow 16 x > 71 \Rightarrow x > \frac{71}{6} . \end{array}$