Question 1 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at Nov. 17, 2021 by Teachoo

Hello! Teachoo has made this answer with days (even weeks!) worth of effort and love. If Teachoo has been of any help to you in your Board exam preparation, then please support us by clicking on this link to make a donation

The function
f
(x) = {8(sinβ‘x/x " + cos x, if x " β " 0" k", if x " =" 0" )β€ is continuous at x = 0, then the value of k is

Hello! Teachoo has made this answer with days (even weeks!) worth of effort and love. If Teachoo has been of any help to you in your Board exam preparation, then please support us by clicking on this link to make a donation

Question 1
The function f (x) = {β 8(sinβ‘π₯/π₯ " + cos x, if x " β " 0" @π ", if x " =" 0" )β€ is continuous at x = 0, then the value of k is
(A) 3 (B) 2
(C) 1 (D) 1.5
At π = 0
f(x) is continuous at π₯=0
if L.H.L = R.H.L = π(0)
if if limβ¬(xβ0^β ) π(π₯) = limβ¬(xβ0^+ ) π(π₯) = π(0)
LHL at x β 0
(πππ)β¬(π±βπ^β ) f(x) = limβ¬(hβ0) f(0 β h)
= (πππ)β¬(π‘βπ) f(βh)
= limβ¬(hβ0) sinβ‘γ(ββ)γ/((ββ)) " + cos (β h)"
= (πππ)β¬(π‘βπ) γβπππγβ‘π/(βπ) " +" (πππ)β¬(π‘βπ) " cos h"
= limβ¬(hβ0) sinβ‘β/β " + " (πππ)β¬(ββ0) " cos h"
Using limβ¬(xβ0) sinβ‘π₯/π₯=1
= 1 + "cos 0"
= 1 + 1
= 2
RHL at x β 0
(πππ)β¬(πβπ^+ ) f(x) = limβ¬(hβ0) f(0 + h)
= (πππ)β¬(π‘βπ) f(h)
= limβ¬(hβ0) sinβ‘β/β " + cos h"
= (πππ)β¬(π‘βπ) πππβ‘π/π " + " (πππ)β¬(π‘βπ) " cos h"
Using limβ¬(xβ0) sinβ‘π₯/π₯=1
= 1 + "cos 0"
= 1 + 1
= 2
At π=π
π(π₯)=π
π(π)=π
Since f(x) is continuous at x = 0.
L.H.L = R.H.L = π(0)
π(0)=2
π=π
So, the correct answer is (B)

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.