Chapter 5 Class 12 Continuity and Differentiability

Class 12
Important Questions for exams Class 12

Transcript

Question 5 Verify Mean Value Theorem, if š (š„) = š„3 ā 5š„2 ā 3š„ in the interval [a, b], where a = 1 and b = 3. Find all š ā (1, 3) for which š ā²(š) = 0.š (š„) = š„3 ā 5š„2 ā 3š„ in [a, b], where a = 1 and b = 3 Condition 1 š (š„) = š„3 ā 5š„2 ā 3š„ š(š„) is a polynomial & every polynomial function is continuous ā“ š(š„) is continuous at š„ā[1, 3] Conditions of Mean value theorem š(š„) is continuous at (š, š) š(š„) is derivable at (š, š) If both conditions satisfied, then there exist some c in (š, š) such that šā²(š) = (š(š) ā š(š))/(š ā š) Condition 2 š(š„) = š„3 ā 5š„2 ā 3š„ š(š„) is a polynomial & every polynomial function is differentiable ā“ š(š„) is differentiable at š„ā[1, 3] Now, š(š„)" = " š„3 ā 5š„2 ā 3š„ š^ā² (š„)" = 3" š„2 ā10š„ ā 3 š„ā[1, 3] So, šā²(š) = " 3" š^2ā10šā3 Also, š(š„)" = " š„3 ā 5š„2 ā 3š„ š(š)" = " š(1) = (1)^3ā5(1)^2ā3(1) = 1ā5ā3 = ā7 š(š)" = " š(3) = (3)^3ā5(3)^2ā3(3) = 27ā45ā9 = ā27 By Mean Value Theorem š^ā² (š) = (š(š) ā š(š))/(š ā š) "3" š^2ā10šā3 = (ā27 ā (ā7))/(3 ā 1) "3" š^2ā10šā3 = (ā27 + 7)/2 "3" š^2ā10šā3 = (ā20)/2 "3" š^2ā10šā3 = ā10 "3" š^2ā10šā3+10 = 0 "3" š^2ā10š+7 = 0 "3" š^2ā3šā7š+7 = 0 "3" š(šā1)ā7(šā1) = 0 (3šā7)(šā1) = 0 So, c = 7/3 & c = 1 Since c = 7/3 lies between 1 & 3 c = š/š ā[1, 3] Thus, Mean Value Theorem is verified. From our question Find all š ā (1, 3) for which š ā²(š) = 0. We need to find cā[1, 3] For which š^ā² (š) = 0 š^ā² (š) = 0 "3" š^2ā10šā3 = 0 The above equation is of the form š“š„^2+šµš„+š¶ x = (āšµ Ā± ā(šµ^2 ā4š“š¶) )/2š“ c = (ā(ā10) Ā± ā((ā10)^2 ā 4(ā3)(3) ) )/2š“ c = (10 Ā± ā(100 + 36) )/2(ā3) c = (10 Ā± ā136 )/6 c = (10 Ā± ā(2 Ć 2 Ć 34))/6 c = (10 Ā± 2ā34)/6 c = 2(5 Ā± ā34 )/6 c = (5 Ā± ā34)/3 So, c = (5 + ā34)/3 c = (5 + 5.83)/3 c = 10.83/3 c = 3.61 c = (5 ā ā34)/3 c = (5 ā 5.83)/3 c = (ā0.83)/3 c = ā0.28 Thus, c = 3.61 & c = ā0.28 But both values do not lie between [1, 3] Hence, there exists no value of šā[1, 3] for which š^ā² (š) = 0