Question 2 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at May 29, 2023 by Teachoo

The value of k (k < 0) for which the function
f
defined as
f (x)={((1 -coskx)/(x sinx ),x≠0 1/2,x=0)┤ is continuous at x = 0 is :

(a) ± 1 (b) −1 (c) ± 1/2 (d) 1/2

This question is
inspired from -
Question 21
- CBSE Class 12 Sample Paper for 2021 Boards

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Question 2 The value of k (k < 0) for which the function f defined as π (π₯)={β((1 βcosβ‘ππ₯)/(π₯ sinβ‘π₯ ),π₯β 0@1/2, π₯=0)β€ is continuous at x = 0 is : (a) Β± 1 (b) β1 (c) Β± 1/2 (d) 1/2
Given that function is continuous at x = 0
π(π₯) is continuous at x = 0
i.e. (π₯π’π¦)β¬(π±βπ) π(π)=π(π)
Limit at x β 0
(πππ)β¬(π₯β0) f(x) = (πππ)β¬(ββ0) f(h)
= limβ¬(hβ0) (1 β cosβ‘πβ)/(β (sinβ‘β) )
= limβ¬(hβ0) (π γπ¬π’π§γ^πβ‘γππ/πγ)/(β (sinβ‘β))
= limβ¬(hβ0) (2 sin^2β‘γπβ/2γ)/1 Γ1/(β (sinβ‘β))
= (πππ)β¬(π‘βπ) (π γπππγ^πβ‘γππ/πγ)/(ππ/π)^π Γ (ππ/π)^π/(π (πππβ‘π))
= limβ¬(hβ0) (2 sin^2β‘γπβ/2γ)/(πβ/2)^2 Γ (π^2 β^2)/(4β (sinβ‘β))
= limβ¬(hβ0) sin^2β‘γπβ/2γ/(πβ/2)^2 Γ (π^π π)/(π(πππβ‘π))
= π^π/π limβ¬(hβ0) sin^2β‘γπβ/2γ/(πβ/2)^2 Γ β/sinβ‘β
= π^2/2 limβ¬(hβ0) sin^2β‘γπβ/2γ/(πβ/2)^2 Γlimβ¬(hβ0) β/sinβ‘β
= π^2/2 Γ 1 Γ 1
= π^π/π
Now,
(π₯π’π¦)β¬(π±βπ) π(π)=π(π)
π^2/2 = 1/2
π^2 =1
π =Β±π
But, given that k < 0
Thus, only value is k = β1
So, the correct answer is (b)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!