Question 15 - NCERT Exemplar - MCQs - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

Last updated at Nov. 17, 2021 by Teachoo

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The set of points where the function f given by f (x) = |2xβ1| sin x is differentiable is

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Question 15
The set of points where the function f given by f (x) = |2xβ1| sin x is differentiable is
R (B) R β {1/2} (C) (0, β) (D) none of these
f(x) = |2π₯β1| sinβ‘π₯
= {β((2π₯β1) sinβ‘π₯, 2π₯β1β₯0@β(2π₯β1) sinβ‘π₯, 2π₯β1<0)β€
= {β((2π₯β1) sinβ‘π₯, π₯β₯1/2@β(2π₯β1) sinβ‘γπ₯ ,γ π₯<1/2)β€
Now,
f(x) is a differentiable at x = 1/2 if
LHD = RHD
(πππ)β¬(π‘βπ) (π(π) β π(π β π))/π
= (πππ)β¬(hβ0) (π(1/2) β π(1/2 β β))/β
=(πππ)β¬(hβ0) (|2(1/2)β1| sinβ‘(1/2)β|2(1/2 β β)β 1| sinβ‘(1/2 ββ))/β
= (πππ)β¬(hβ0) (0 β|1 β β β 1| γsin γβ‘(1/2 β β))/β
= (πππ)β¬(hβ0) (0 β|β β| sinβ‘(1/2 β β))/β
= (πππ)β¬(hβ0) (ββ γsin γβ‘(1/2 β β))/β
= (πππ)β¬(hβ0) βsinβ‘(1/2 β β)
= βsinβ‘(1/2β0)
= βπ ππ 1/2
= β π /π
(πππ)β¬(π‘βπ) (π(π+π) β π(π ))/π
= (πππ)β¬(hβ0) (π(1/2+β) β π(1/2))/β
= (πππ)β¬(hβ0) (|(2(1/2+β)β1| π ππβ‘(1/2+β)β|2(1/2)β1| π ππβ‘(1/2))/β
= (πππ)β¬(hβ0) (|1+ β β1| γsin γβ‘γ(1/2 + β) γβ 0 )/β
= (πππ)β¬(hβ0) (|β| γsin γβ‘(1/2+β))/β
= (πππ)β¬(hβ0) (β γsin γβ‘(1/2+β))/β
= (πππ)β¬(hβ0) sinβ‘(1/2+β)
= γsin γβ‘(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)

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