Ex 5.1, 27 - 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
-
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, 27 Find the values of k so that the function f is continuous at the indicated point π(π₯)={β(ππ₯2 , ππ π₯β€2@3, ππ π₯>2)β€ at x = 2Given that function is continuous at π₯ = 2 π is continuous at π₯ = 2 if L.H.L = R.H.L = π(2) i.e. limβ¬(xβ2^β ) π(π₯)=limβ¬(xβ2^+ ) " " π(π₯)= π(2) LHL at x β 2 limβ¬(xβ2^β ) f(x) = limβ¬(hβ0) f(2 β h) = limβ¬(hβ0) γπ(2ββ)γ^2 = γπ(2β0)γ^2 = γπ(2)γ^2 = ππ RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) 3 = 3 RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) 3 = 3 RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) 3 = 3 RHL at x β 2 limβ¬(xβ2^+ ) f(x) = limβ¬(hβ0) f(2 + h) = limβ¬(hβ0) 3 = 3 Since LHL = RHL 4k = 3 k = π/π
