If the function ⨍ (x)=2+cosx-1(π-x)2,x≠π k,x=π iscontinuous at x=π, then k equals

limx→π 2+cos⁡x−1(π−x)2Rationalising =limx→π 1+cos⁡x(π−x)2(2+cos⁡x+1) =limx→π 1−cos⁡(π−x)(π−x)2(2+cos⁡x+1)=limx→π 2sin2⁡(π−x)2(π−x)2(2+cos⁡x+1)limx→π2sin(π−x)2π−x22×1412+cos⁡x+1=14

If the function ⨍ (x)=2+cosx-1(π-x)2,x≠π k,x=π iscontinuous at x=π, then k equals

limx→π 2+cos⁡x−1(π−x)2Rationalising =limx→π 1+cos⁡x(π−x)2(2+cos⁡x+1) =limx→π 1−cos⁡(π−x)(π−x)2(2+cos⁡x+1)=limx→π 2sin2⁡(π−x)2(π−x)2(2+cos⁡x+1)limx→π2sin(π−x)2π−x22×1412+cos⁡x+1=14

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 function $⨍ (x)= \{\begin{cases} \frac{\sqrt{2 + cosx} -1}{{(π - x)}^{2}}, & x \neq π \\ k, & x = π \end{cases}$ is
continuous at $x = π,$ then $k$ equals

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

1/4

1/2

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$lim_{x \to π} \frac{\sqrt{2 + \cos x} - 1}{(π - x)^{2}} Rationalising = lim_{x \to π} \frac{1 + \cos x}{(π - x)^{2} (\sqrt{2 + \cos x} + 1)} \begin{aligned} = lim_{x \to π} \frac{1 - \cos (π - x)}{(π - x)^{2} (\sqrt{2 + \cos x} + 1)} \\ = lim_{x \to π} \frac{2 \sin^{2} \frac{(π - x)}{2}}{(π - x)^{2} (\sqrt{2 + \cos x} + 1)} \\ lim_{x \to π} 2 {(\frac{\sin \frac{(π - x)}{2}}{\frac{π - x}{2}})}^{2} \times \frac{1}{4} (\frac{1}{\sqrt{2 + \cos x} + 1}) = \frac{1}{4} \end{aligned}$