Let f(x)=(x2−3x+2) (x3−6x2+11x−6)+|sin(x+π4)| The set of points at which the function f(x) is not differentiable in [0,2π]

f(x)=(x−1)2(x−2)2(x−3)+|sin(x+π4)|since (x−1)2 (x−2)2(x−3) isnot differnitiable at x=3 and sin (x+π4)is not differnitiable at x+π4=π,2π⇒x=3π4,7π4

Let f(x)=(x2−3x+2) (x3−6x2+11x−6)+|sin(x+π4)| The set of points at which the function f(x) is not differentiable in [0,2π]

f(x)=(x−1)2(x−2)2(x−3)+|sin(x+π4)|since (x−1)2 (x−2)2(x−3) isnot differnitiable at x=3 and sin (x+π4)is not differnitiable at x+π4=π,2π⇒x=3π4,7π4

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 $f (x) = (x^{2} - 3 x + 2) |(x^{3} - 6 x^{2} + 11 x - 6)| + | \sin (x + \frac{π}{4}) |$ The set of points at which the function $f (x)$ is not differentiable in $[0, 2 π]$

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

${1, 2, 3, \frac{3 π}{4}, \frac{7 π}{4}}$

${1, 2, 3}$

${3, \frac{3 π}{4}, \frac{7 π}{4}}$

${\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = {(x - 1)}^{2} {(x - 2)}^{2} (x - 3) + | \sin (x + \frac{π}{4}) | \\ \sin c e {(x - 1)}^{2} {(x - 2)}^{2} (x - 3) i s \\ n o t d i f f e r n i t i a b l e a t x = 3 a n d \sin (x + \frac{π}{4}) \\ i s n o t d i f f e r n i t i a b l e a t x + \frac{π}{4} = π, 2 π \\ \Rightarrow x = \frac{3 π}{4}, \frac{7 π}{4} \end{array}$