Find ‘a’ and ‘b’ , if the function given by

f(x) = {ax + b, if x<1,

 2x+1,  if x ≥ 1

is differentiable at x = 1

OR

Determine the values of ‘a’ and ‘b’ such that the following function is continuous at x = 0:

f(x) = {x + sin x / sin (a+1)x, if - π < x < 0,

2, if x = 0,

2 e sin⁡bx - 1 /bx, if x > 0

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/


Find ‘a’ and ‘b’ if f(x) = {ax^2 + b | if f(x) = { x + sin x

Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3
Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4
Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 5 Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 6 Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 7 Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 8 Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 9 Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 10 Question 14 - CBSE Class 12 Sample Paper for 2018 Boards - Part 11

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Transcript

Question 14 Find a and b , at if the function given by f(x) = { ( 2+ , <1@2 +1, 1) is differentiable x = 1f(x) is differentiable at x = 1 if it is continuous. So, first we check if it is continuous. Check continuous f is continuous at x = 1 if L.H.L = R.H.L = (1) i.e. if lim (x 1^ ) ( ) = lim (x 1^+ ) ( ) = (1) lim (x 1^ ) ( ) = lim (x 1^ ) ^2+ Putting x = 1 = a. (1)2 + b = a + b lim (x 1^+ ) ( ) = lim (x 1^+ ) 2 +1 Putting x = 1 = 2(1) +1 = 2 + 1 = 3 lim (x 1^+ ) ( ) = lim (x 1^+ ) 2 +1 Putting x = 1 = 2(1) +1 = 2 + 1 = 3 And (1)=2x+1=2(1)+1=3 Now, lim (x 1^ ) ( ) = lim (x 1^+ ) ( ) = (1) a + b = 3 = 3 Thus, a + b = 3 Now, we check differentiability Check differentiability f(x) = { ( 2+ , <1@2 +1, 1) f(x) is a differentiable at x = 1 if LHD = RHD ( ) (h 0) ( ( ) ( ))/ = ( ) (h 0) ( (1) (1 ))/ = ( ) (h 0) ((2(1) + 1) ( (1 )^2+ ))/ = ( ) (h 0) (3 (1 )^2 )/ From (1), a + b = 3 Putting b = 3 a = ( ) (h 0) (3 (1 )^2 (3 ))/ = ( ) (h 0) (3 (1 )^2 3 + )/ = ( ) (h 0) ( (1 )^2 )/ = ( ) (h 0) ( (1^2 + ^2 2 ))/ ( ) (h 0) ( ( + ) ( ))/ = ( ) (h 0) ( (1 + ) (1))/ = ( ) (h 0) (2(1 + ) + 1 (2(1) + 1))/ = ( ) (h 0) (2 + 2 + 1 (2 + 1))/ = ( ) (h 0) (3 + 2 3)/ = ( ) (h 0) 2 / = ( ) (h 0) (2) = 2= ( ) (h 0) ( + ^2 + 2 )/ = ( ) (h 0) ( ^2 + 2 )/ = ( ) (h 0) +2 = a(0) + 2a = 2a = ( ) (h 0) ( + ^2 + 2 )/ = ( ) (h 0) ( ^2 + 2 )/ = ( ) (h 0) +2 = a(0) + 2a = 2a = ( ) (h 0) ( + ^2 + 2 )/ = ( ) (h 0) ( ^2 + 2 )/ = ( ) (h 0) +2 = a(0) + 2a = 2a = ( ) (h 0) ( + ^2 + 2 )/ = ( ) (h 0) ( ^2 + 2 )/ = ( ) (h 0) +2 = a(0) + 2a = 2a Since f(x) is differentiable at x = 1 LHD = RHD 2a = 2 a = 1 Putting a = 1 in (1) a + b = 3 1 + b = 3 b = 3 1 = 2 So, a = 1, b = 2 Question 14 Determine the values of a and b such that the following function is continuous at x = 0: f(x) = { (( + )/( ( +1) ), < <0 @ @2, =0@ @2 ( ^sin 1)/ , >0@ ) f is continuous at x = 0 if L.H.L = R.H.L = (0) i.e. if lim (x 0^ ) ( ) = lim (x 0^+ ) ( ) = (0) lim (x 0^ ) ( ) = lim (x 0^ ) ( + )/( ( + 1) ) Theory If we put 0 It will become ( + )/( ( + 1) ) = (0 + sin 0)/sin ( +1) 0 = (0 + 0)/sin 0 = 0/0 Since it is of 0/0 form, We use an identity We use lim (x 0) ( )/ = 1 lim (x 0^+ ) ( ) = lim (x 0^+ ) 2 ( ^sin 1)/ Theory If we put 0 It will become 2 ( ^sin 1)/ = 2( ^sin 0 1)/( 0) = 2(1 1)/0 = 2 0/0 Since it is of 0/0 form, We use an identity We use lim (x 0) ( )/ = 1 and lim (x 0) ( ^ 1)/ = 1 Taking x common from numerator = lim (x 0^ ) (1 + ( )/ )/( ( + 1) ) Multiplying and dividing (a + 1)x in denominator = lim (x 0^ ) (1 + ( )/ )/( ( + 1) ( + 1) /( + 1) ) = lim (x 0^ ) (1 + ( )/ )/(( + 1) ( ( + 1) )/( + 1) ) = lim (x 0^ ) ((1 + ( )/ ))/(( + 1) ( ( + 1) )/( + 1) )Multiplying and dividing sin bx = lim (x 0^+ ) 2 ( ^sin 1)/ sin /sin = lim (x 0^+ ) 2 ( ^sin 1)/sin sin / Using lim (x 0) ( )/ = 1 and lim (x 0) ( ^ 1)/ = 1 = 2 1 1 = 2 = 1/((a + 1)) lim (x 0^ ) ((1 + ( )/ ))/( ( ( + 1) )/( + 1) ) Using lim (x 0) ( )/ = 1 = 1/((a + 1)) ((1 + 1))/( 1) = 2/((a + 1)) And (0)=2 Now, lim (x 0^ ) ( ) = lim (x 0^+ ) ( ) = (0) 2/((a + 1)) = 2 = 2 Thus, 2/( + 1) = 2 This is only possible when a = 0 And since there is no b in the equation, b can be of any value a = 0 and b can be of any value

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo