NCERT Exemplar - MCQs

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

## (D) none of these

This question is similar to Ex 5.2, 9 - Chapter 5 Class 12 - Continuity and Differentiability

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### Transcript

Question 15 The set of points where the function f given by f (x) = |2xβ1| sin x is differentiable is R (B) R β {1/2} (C) (0, β) (D) none of these f(x) = |2π₯β1| sinβ‘π₯ = {β((2π₯β1) sinβ‘π₯, 2π₯β1β₯[email protected]β(2π₯β1) sinβ‘π₯, 2π₯β1<0)β€ = {β((2π₯β1) sinβ‘π₯, π₯β₯1/[email protected]β(2π₯β1) sinβ‘γπ₯ ,γ π₯<1/2)β€ Now, f(x) is a differentiable at x = 1/2 if LHD = RHD (πππ)β¬(π‘βπ) (π(π) β π(π β π))/π = (πππ)β¬(hβ0) (π(1/2) β π(1/2 β β))/β =(πππ)β¬(hβ0) (|2(1/2)β1| sinβ‘(1/2)β|2(1/2 β β)β 1| sinβ‘(1/2 ββ))/β = (πππ)β¬(hβ0) (0 β|1 β β β 1| γsin γβ‘(1/2 β β))/β = (πππ)β¬(hβ0) (0 β|β β| sinβ‘(1/2 β β))/β = (πππ)β¬(hβ0) (ββ γsin γβ‘(1/2 β β))/β = (πππ)β¬(hβ0) βsinβ‘(1/2 β β) = βsinβ‘(1/2β0) = βπ ππ 1/2 = β π/π (πππ)β¬(π‘βπ) (π(π+π) β π(π ))/π = (πππ)β¬(hβ0) (π(1/2+β) β π(1/2))/β = (πππ)β¬(hβ0) (|(2(1/2+β)β1| π ππβ‘(1/2+β)β|2(1/2)β1| π ππβ‘(1/2))/β = (πππ)β¬(hβ0) (|1+ β β1| γsin γβ‘γ(1/2 + β) γβ 0 )/β = (πππ)β¬(hβ0) (|β| γsin γβ‘(1/2+β))/β = (πππ)β¬(hβ0) (β γsin γβ‘(1/2+β))/β = (πππ)β¬(hβ0) sinβ‘(1/2+β) = γsin γβ‘(1/2+0) = sin 1/2 = π/π Since LHD β  RHD β΄ f(x) is not differentiable at x = 1/2 Hence, we can say that f(x) is differentiable on R β {π/π} So, the correct answer is (B)