Ex 5.1, 29 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at August 11, 2026 by Teachoo
Checking continuity using LHL and RHL
Example 12
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Chapter 5 Class 12 Continuity and Differentiability
Concept wise
- Checking continuity at a given point
- Checking continuity at any point
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Checking continuity using LHL and RHL
- Algebra of continous functions
- Continuity of composite functions
- Checking if funciton is differentiable
- Finding derivative of a function by chain rule
- Finding derivative of Implicit functions
- Finding derivative of Inverse trigonometric functions
- Finding derivative of Exponential & logarithm functions
- Logarithmic Differentiation - Type 1
- Logarithmic Differentiation - Type 2
- Derivatives in parametric form
- Finding second order derivatives - Normal form
- Finding second order derivatives- Implicit form
- Proofs
- Verify Rolles theorem
- Verify Mean Value Theorem
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Transcript
Ex 5.1, 29 Find the values of k so that the function f is continuous at the indicated point π(π₯)={β(ππ₯+1, ππ π₯β€5@3π₯β5, ππ π₯>5)β€ at x = 5 Given that function is continuous at π₯ =5 π is continuous at π₯ =5 if L.H.L = R.H.L = π(5) i.e. limβ¬(xβ5^β ) π(π₯)=limβ¬(xβ5^+ ) " " π(π₯)= π(5) LHL at x β 5 (πππ)β¬(π₯β5^β ) f(x) = (πππ)β¬(ββ0) f(5 β h) = limβ¬(hβ0) k(5 β h) + 1 = k(5 β 0) + 1 = 5k + 1 RHL at x β 5 (πππ)β¬(π₯β5^+ ) f(x) = (πππ)β¬(ββ0) f(5 + h) = limβ¬(hβ0) 3(5 + h) β 5 = 3(5 + 0) β 5 = 3 Γ 5 β 5 = 15 β 5 = 10 Since L.H.L = R.H.L 5k + 1 = 10 5k = 10 β 1 5k = 9 π= π/π
