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

Transcript

Question 18 Let f (x) = |sin x|. Then (A) f is everywhere differentiable (B) f is everywhere continuous but not differentiable at x = nπ, nβ Z. (C) f is everywhere continuous but not differentiable at x = (2n + 1) π/2, nβ Z. (d) None of these f(π₯) = |sin π₯| We need to check continuity and differentiability of f(π₯) Continuity of f(π) Let π(π₯)=|π₯| & β(π₯)=sinβ‘π₯ Then, πππ(π)=π(β(π₯)) =π(sinβ‘π₯ ) =|sinβ‘π₯ | =π(π) β΄ π(π₯)=ππβ(π₯) We know that, π(π)=π¬π’π§β‘π is continuous as sin 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₯)=|sinβ‘π₯ | Since, it is a modulus function so we check differentiability when sinβ‘π₯=0 i.e., π=ππ, πβπ π(π₯) is differentiable at π₯=ππ, 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) β π(ππβ β))/β = (πππ)β¬(hβ0) (|sinβ‘ππ | β|sinβ‘γ(ππ β β)γ |)/β = (πππ)β¬(hβ0) (|0| β |sinβ‘γ(ππ β β)γ |)/β Using sin (A β B) = sin Acos B β cos Asin B = (πππ)β¬(hβ0) (0 β |sinβ‘γππ cosβ‘γβ β cosβ‘γππ sinβ‘β γ γ γ |)/β = (πππ)β¬(hβ0) ( β |0 β cosβ‘γππ sinβ‘β γ |)/β = (πππ)β¬(hβ0) ( βcosβ‘γππ sinβ‘β γ)/β = βcos ππ Γ(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cos ππ Γ1 =β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+ β) β π(ππ))/β = (πππ)β¬(hβ0) (|sinβ‘γ(ππ+β)γ |β|sinβ‘ππ |)/β = (πππ)β¬(hβ0) (|sinβ‘γ(ππ+β)γ |β|0|)/( β) Using sin (A + B) = sin Acos B + cos Asin B = (πππ)β¬(hβ0) (|sinβ‘γππ cosβ‘γβ + cosβ‘γππ sinβ‘β γ γ γ | β 0)/β = (πππ)β¬(hβ0) |0 + cosβ‘γππ sinβ‘β γ |/β = (πππ)β¬(hβ0) cosβ‘γππ sinβ‘β γ/β = cos ππ Γ(πππ)β¬(hβ0) sinβ‘β/β Using (πππ)β¬(π₯β0) π ππβ‘π₯/π₯=1 = cos ππ Γ1 = ππ¨π¬ ππ Since, LHD β  RHD β΄ π(π₯) is not differentiable at π=ππ Hence, π(π₯) is continuous everywhere but not differentiable at πβππ, πβπ So, the correct answer is (B)