Limits Continuity And Differentiability Question 12
Question: The number of points in (1, 3), where $ f(x)={a^{[x^{2}]}},a>1 $ , is not differentiable, where [x] denotes the integral part of x.
Options:
A) 5
B) 7
C) 9
D) 11
Show Answer
Answer:
Correct Answer: B
Solution:
Here $ 1<x<3 $ and in this interval $ x^{2} $ is an increasing functions, thus, $ 1<x^{2}<9 $
$ [x^{2}]=1,1\le x<\sqrt{2}=2,\sqrt{2}\le x<\sqrt{3} $
$ =3,\sqrt{3}\le x<2=4,2\le x<\sqrt{5} $
$ =5,\sqrt{5}\le x<\sqrt{6}=6,\sqrt{6}\le x<\sqrt{7} $
$ =7\sqrt{7}\le x<\sqrt{8}=8,\sqrt{8}\le x<3 $
Clearly, $ [x^{2}] $ and also $ {a^{[x^{2}]}} $ is discontinuous and not differentiable at only 7 points, $ x=\sqrt{2},\sqrt{3},2,\sqrt{5},\sqrt{6}\sqrt{7},\sqrt{8} $
BETA
