fx=x-1+x+1fx=1-x-1-x if x 1 fx=-2x ; if x 1limx→-1-fx=limx→-1+fx=fx⇒f is continuous at x=-1 limx→1-fx=limx→1+fx=f1=2⇒f is continuous at x=1 f1-1-=-2, f1-1+=0 f11-=0, f11+=2 ∴f is not differentiable at x=-1, 1

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

Let $⨍ (x)= |x - 1| + |x + 1|$

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

$⨍ (x)$ is differentiable at $x = \pm 1$

$⨍ (x)$ is not differentiable at $x = \pm 1$

$⨍ (x)$ is neither continuous nor differentiable at $x = \pm 1$

$⨍ (x)$ is not continuous at $x = 0$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = |x - 1| + |x + 1|$

$f (x) = \{\begin{cases} 1 - x - 1 - x if x < - 1 \\ 1 - x + 1 + x if - 1 < x < 1 \\ x - 1 + x + 1 if x > 1 \end{cases} f (x) = \{\begin{cases} - 2 x; if x < - 1 \\ 2; if - 1 < x < 1 \\ 2 x; if x > 1 \end{cases}$

$\lim_{x \to - 1^{-}} f (x) = \lim_{x \to - 1^{+}} f (x) = f (x) \Rightarrow f is continuous at x = - 1 \lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x) = f (1) = 2 \Rightarrow f is continuous at x = 1 f^{1} (- 1^{-}) = - 2, f^{1} (- 1^{+}) = 0 f^{1} (1^{-}) = 0, f^{1} (1^{+}) = 2 ∴ f is not differentiable at x = - 1, 1$