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

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Question 14 Find ‘a’ and ‘b’ , at if the function given by f(x) = {█(𝑎𝑥2+𝑏, 𝑖𝑓 𝑥<1@2𝑥+1, 𝑖𝑓 𝑥 ≥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 = 𝑓(1) i.e. if lim┬(x→1^− ) 𝑓(𝑥) = lim┬(x→1^+ ) 𝑓(𝑥) = 𝑓(1) lim┬(x→1^− ) 𝑓(𝑥) = lim┬(x→1^− ) 𝑎𝑥^2+𝑏 Putting x = 1 = a. (1)2 + b = a + b lim┬(x→1^+ ) 𝑓(𝑥) = lim┬(x→1^+ ) 2𝑥+1 Putting x = 1 = 2(1) +1 = 2 + 1 = 3 lim┬(x→1^+ ) 𝑓(𝑥) = lim┬(x→1^+ ) 2𝑥+1 Putting x = 1 = 2(1) +1 = 2 + 1 = 3 And 𝑓(1)=2x+1=2(1)+1=3 Now, lim┬(x→1^− ) 𝑓(𝑥) = lim┬(x→1^+ ) 𝑓(𝑥) = 𝑓(1) a + b = 3 = 3 Thus, a + b = 3 Now, we check differentiability Check differentiability f(x) = {█(𝑎𝑥2+𝑏, 𝑖𝑓 𝑥<1@2𝑥+1, 𝑖𝑓 𝑥 ≥1)┤ f(x) is a differentiable at x = 1 if LHD = RHD (𝑙𝑖𝑚)┬(h→0) (𝑓(𝑥) − 𝑓(𝑥 − ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) (𝑓(1) − 𝑓(1 − ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) ((2(1) + 1) − (𝑎(1−ℎ)^2+𝑏))/ℎ = (𝑙𝑖𝑚)┬(h→0) (3 − 𝑎(1 − ℎ)^2 − 𝑏)/ℎ From (1), a + b = 3 Putting b = 3 – a = (𝑙𝑖𝑚)┬(h→0) (3 − 𝑎(1 − ℎ)^2 − (3 − 𝑎))/ℎ = (𝑙𝑖𝑚)┬(h→0) (3 − 𝑎(1 − ℎ)^2 − 3 + 𝑎)/ℎ = (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎(1 − ℎ)^2 )/ℎ = (𝑙𝑖𝑚)┬(h→0) (𝑎 −𝑎(1^2 + ℎ^2 − 2ℎ))/ℎ (𝑙𝑖𝑚)┬(h→0) (𝑓(𝑥 + ℎ) − 𝑓(𝑥))/ℎ = (𝑙𝑖𝑚)┬(h→0) (𝑓(1 + ℎ) − 𝑓(1))/ℎ = (𝑙𝑖𝑚)┬(h→0) (2(1 + ℎ) + 1 − (2(1) + 1))/ℎ = (𝑙𝑖𝑚)┬(h→0) (2 + 2ℎ + 1 − (2 + 1))/ℎ = (𝑙𝑖𝑚)┬(h→0) (3 + 2ℎ − 3)/ℎ = (𝑙𝑖𝑚)┬(h→0) 2ℎ/ℎ = (𝑙𝑖𝑚)┬(h→0) (2) = 2= (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎 + 𝑎ℎ^2 + 2𝑎ℎ)/ℎ = (𝑙𝑖𝑚)┬(h→0) (𝑎ℎ^2 + 2𝑎ℎ)/ℎ = (𝑙𝑖𝑚)┬(h→0) 𝑎ℎ+2𝑎 = a(0) + 2a = 2a = (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎 + 𝑎ℎ^2 + 2𝑎ℎ)/ℎ = (𝑙𝑖𝑚)┬(h→0) (𝑎ℎ^2 + 2𝑎ℎ)/ℎ = (𝑙𝑖𝑚)┬(h→0) 𝑎ℎ+2𝑎 = a(0) + 2a = 2a = (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎 + 𝑎ℎ^2 + 2𝑎ℎ)/ℎ = (𝑙𝑖𝑚)┬(h→0) (𝑎ℎ^2 + 2𝑎ℎ)/ℎ = (𝑙𝑖𝑚)┬(h→0) 𝑎ℎ+2𝑎 = a(0) + 2a = 2a = (𝑙𝑖𝑚)┬(h→0) (𝑎 − 𝑎 + 𝑎ℎ^2 + 2𝑎ℎ)/ℎ = (𝑙𝑖𝑚)┬(h→0) (𝑎ℎ^2 + 2𝑎ℎ)/ℎ = (𝑙𝑖𝑚)┬(h→0) 𝑎ℎ+2𝑎 = 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) = {█((𝑥 + 𝑠𝑖𝑛 𝑥)/(𝑠𝑖𝑛 (𝑎+1)𝑥), 𝑖𝑓−𝜋<𝑥<0 @ @2, 𝑖𝑓 𝑥=0@ @2 (𝑒^sin⁡𝑏𝑥 − 1)/𝑏𝑥,𝑖𝑓 𝑥>0@ )┤ f is continuous at x = 0 if L.H.L = R.H.L = 𝑓(0) i.e. if lim┬(x→0^− ) 𝑓(𝑥) = lim┬(x→0^+ ) 𝑓(𝑥) = 𝑓(0) lim┬(x→0^− ) 𝑓(𝑥) = lim┬(x→0^− ) (𝑥 + 𝑠𝑖𝑛 𝑥)/(𝑠𝑖𝑛 (𝑎 + 1)𝑥) Theory If we put 0 It will become (𝑥 + 𝑠𝑖𝑛 𝑥)/(𝑠𝑖𝑛 (𝑎 + 1)𝑥) = (0 + sin⁡0)/sin⁡〖 (𝑎+1) ×0〗 = (0 + 0)/sin⁡0 = 0/0 Since it is of 0/0 form, We use an identity We use lim┬(x→0) (𝑠𝑖𝑛 𝑥)/𝑥 = 1 lim┬(x→0^+ ) 𝑓(𝑥) = lim┬(x→0^+ ) 2 (𝑒^sin⁡𝑏𝑥 − 1)/𝑏𝑥 Theory If we put 0 It will become 2 (𝑒^sin⁡𝑏𝑥 − 1)/𝑏𝑥 = 2(𝑒^sin⁡0 − 1)/(𝑏 × 0) = 2(1 − 1)/0 = 2 × 0/0 Since it is of 0/0 form, We use an identity We use lim┬(x→0) (𝑠𝑖𝑛 𝑥)/𝑥 = 1 and lim┬(x→0) (𝑒^𝑥 − 1)/𝑥 = 1 Taking x common from numerator = lim┬(x→0^− ) 𝑥(1 + (𝑠𝑖𝑛 𝑥)/𝑥)/(𝑠𝑖𝑛 (𝑎 + 1)𝑥) Multiplying and dividing (a + 1)x in denominator = lim┬(x→0^− ) 𝑥(1 + (𝑠𝑖𝑛 𝑥)/𝑥)/(𝑠𝑖𝑛 (𝑎 + 1)𝑥 × (𝑎 + 1)𝑥/(𝑎 + 1)𝑥) = lim┬(x→0^− ) 𝑥(1 + (𝑠𝑖𝑛 𝑥)/𝑥)/((𝑎 + 1) 𝑥 × (𝑠𝑖𝑛 (𝑎 + 1)𝑥)/(𝑎 + 1)𝑥) = lim┬(x→0^− ) ((1 + (𝑠𝑖𝑛 𝑥)/𝑥))/((𝑎 + 1) × (𝑠𝑖𝑛 (𝑎 + 1)𝑥)/(𝑎 + 1)𝑥)Multiplying and dividing sin bx = lim┬(x→0^+ ) 2 (𝑒^sin⁡𝑏𝑥 − 1)/𝑏𝑥 × sin⁡𝑏𝑥/sin⁡𝑏𝑥 = lim┬(x→0^+ ) 2 (𝑒^sin⁡𝑏𝑥 − 1)/sin⁡𝑏𝑥 × sin⁡𝑏𝑥/𝑏𝑥 Using lim┬(x→0) (𝑠𝑖𝑛 𝑥)/𝑥 = 1 and lim┬(x→0) (𝑒^𝑥 − 1)/𝑥 = 1 = 2 ×1 ×1 = 2 = 1/((a + 1)) lim┬(x→0^− ) ((1 + (𝑠𝑖𝑛 𝑥)/𝑥))/( (𝑠𝑖𝑛 (𝑎 + 1)𝑥)/(𝑎 + 1)𝑥) Using lim┬(x→0) (𝑠𝑖𝑛 𝑥)/𝑥 = 1 = 1/((a + 1)) ×((1 + 1))/( 1) = 2/((a + 1)) And 𝑓(0)=2 Now, lim┬(x→0^− ) 𝑓(𝑥) = lim┬(x→0^+ ) 𝑓(𝑥) = 𝑓(0) 2/((a + 1)) = 2 = 2 Thus, 2/(𝑎 + 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
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