NCERT Exemplar - MCQs

Chapter 5 Class 12 Continuity and Differentiability
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## (D) None of these

This question is similar to Ex 5.1, 32 - Chapter 5 Class 12 and Ex 5.2, 9 - Chapter 5 Class 12 Continuity and Differentiability

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Question 5 Let f (x) = |cos x|. Then, (A) f is everywhere differentiable. (B) f is everywhere continuous but not differentiable at n = nπ, n β Z (C) f is everywhere continuous but not differentiable at x = (2n + 1)π/2, n β Z (D) None of these f(π₯) = |cos π₯| We need to check continuity and differentiability of f(π₯) Continuity of f(π) Let π(π₯)=|π₯| & β(π₯)=cosβ‘π₯ Then, πππ(π)=π(β(π₯)) =π(cosβ‘π₯ ) =|cosβ‘π₯ | =π(π) β΄ π(π₯)=ππβ(π₯) We know that, π(π)=πππβ‘π is continuous as cos is continuous π(π)=|π| is continuous as it is a modulus function Hence, g(π₯) & h(π₯) both are continuous And If two functions g(π₯) & h(π₯) are continuous then their composition ππβ(π₯) is also continuous β΄ π(π) is continuous Differentiability of π(π) π(π₯)=|cosβ‘π₯ | Since, it is a modulus function so we check differentiability when cosβ‘π₯=0 i.e., π=((ππ +π)π)/π, πβπ π(π₯) is differentiable at π₯=((2π +1)π)/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(((ππ +π)π)/π) β π(((ππ +π)π)/π β β))/β = (πππ)β¬(hβ0) (|cosβ‘γ((ππ +π)π)/πγ | β|cosβ‘γ((ππ +π)π/π β β)γ |)/β = (πππ)β¬(hβ0) (|0| β |cosβ‘γ((ππ +π)π/π β β)γ |)/β Using cos (A β B) = cos Acos B + sin Asin B = (πππ)β¬(hβ0) (0 β |cosβ‘γ((ππ +π)π)/π cosβ‘γβ+ sinβ‘γ((ππ +π)π)/π sinβ‘β γ γ γ |)/β = (πππ)β¬(hβ0) ( β |0 β γsin γβ‘γ((ππ +π)π)/π sinβ‘β γ |)/β = (πππ)β¬(hβ0) ( βsinβ‘γ((ππ +π)π)/π sinβ‘β γ)/β = βπ ππ ((2π +1)π)/2Γ(πππ)β¬(hβ0) sinβ‘β/β Using (πππ)β¬(π₯β0) π ππβ‘π₯/π₯=1 =βπ ππ ((2π +1)π)/2Γ1 =βπππ ((ππ +π)π)/π (πππ)β¬(π‘βπ) (π(π + π) β π(π ))/π = (πππ)β¬(hβ0) (π(((ππ +π)π)/π + β) β π(((ππ +π)π)/π))/β = (πππ)β¬(hβ0) (|cosβ‘γ((2π + 1)π/2 + β )γ |β|cosβ‘γ((2π +1)π)/2γ |)/β = (πππ)β¬(hβ0) (|πππ β‘(((2π + 1)π)/2 +β) |β|0|)/( β) Using cos (A + B) = cos Acos B -- sin Asin B = (πππ)β¬(hβ0) (|γcos γβ‘γ((2π +1)π)/2 cosβ‘γβ β sinβ‘γ((2π +1)π)/2 sinβ‘β γ γ γ | β 0)/β = (πππ)β¬(hβ0) |0 β sinβ‘γ((2π + 1)π)/2 sinβ‘β γ |/β = (πππ)β¬(hβ0) γsin γβ‘γ((2π +1)π)/2 sinβ‘β γ/β =π in ((2π +1)π)/2Γ(πππ)β¬(hβ0) sinβ‘β/β Using (πππ)β¬(π₯β0) π ππβ‘π₯/π₯=1 = π in ((2π +1)π)/2Γ1 = sin ((ππ +π)π)/π Since, LHD β  RHD β΄ π(π₯) is not differentiable at π=((ππ +π)π)/π Hence, π(π₯) is continuous everywhere but not differentiable at π₯β((2π +1)π)/2, πβπ So, the correct answer is (C)