Let S = { ( λ , μ ) ∈ R × R : f ( t ) = ( ∣ λ ∣ e t − μ ) ⋅ sin ( 2∣ t ∣ ) , t ∈ R , is a differentiable function } . Then S is a subest of?

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S = { ( λ , μ ) ∈ R × R : f ( t ) = ( ∣ λ ∣ e ∣ t − μ ) sin ( 2 t ∣ ) , t ∈ R f ( t ) = ( ∣ λ ∣ e ∣ t ∣ − μ ) sin ( 2∣ t ∣ ) { ( ∣ λ ∣ e t − μ ) sin 2 t t > 0 ( ∣ λ ∣ e − t − μ ) ( − sin 2 t ) t < 0 f ′ ( t ) { ( ∣ λ ∣ e t ) sin 2 t + ( ∣ λ ∣ e t − μ ) ( 2 cos 2 t ) t > 0 ∣ λ ∣ e − t sin 2 t + ( ∣ λ ∣ e − t − μ ) ( − 2 cos 2 t ) < t < 0 As, f ( t ) is differentiable ∴ LHD = RHD at t = 0 ∣ λ ∣ ⋅ sin 2 ( 0 ) + ( ∣ λ ∣ e 0 − μ ) 2 cos ( 0 ) = ∣ λ ∣ e − 0 ⋅ sin 2 ( 0 ) − 2 cos ( 0 ) ( ∣ λ ∣ e − 0 − μ ) ⇒ 0 + ( ∣ λ ∣ − μ ) 2 = 0 − 2 ( ∣ λ ∣ − μ ) 4 ( ∣ λ ∣ − μ ) = 0 ∣ λ ∣ = μ So, S ≡ ( λ , μ ) = { λ ∈ R & μ ∈ [ 0 , ∞ )} Therefore set S is subset of R × [ 0 , ∞ )

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