Tiêu đề 5. CONTINUITY & DIFFERENTIABILITY.pdf ✅

TOPIC-1 Continuity Concepts Covered:  Left hand Limit  Right Hand Limit Revision Notes FORMULAE FOR LIMITS: (a) lim cos x x → = 0 1 (b) lim sin x x → x = 0 1 (c) lim tan x x → x = 0 1 (d) lim sin x x → x − = 0 1 1 (e) lim tan x x → x − = 0 1 1 (f) lim log , x x e a x a a → − = > 0 1 0 (g) lim x x e → x − = 0 1 1 (h) lim log ( ) x e x → x + = 0 1 1 (i) lim x a n n x a n x a na → − − − = 1 z For a function f(x), lim x m → f(x) exists if lim x m → − f(x) = lim x m → + f(x). z A function f(x) is continuous at a point x = m if, lim ( ) lim ( ) ( ) xm xm fx fx f m → − → + = = , where lim ( ) x m f x → − is Left Hand Limit of f(x) at x = m and lim ( ) x m f x → + is Right Hand Limit of f(x) at x = m. Also f(m) is the value of function f(x) at x = m. [Board 2020] z A function f(x) is continuous at x = m (say) if, f(m) = lim x m → f(x), i.e., a function is continuous at a point in its domain if the limit value of the function at that point equals the value of the function at the same point. z For a continuous function f(x) atx = m, lim x m → f(x) can be directly obtained by evaluating f(m). z Indeterminate forms or meaningless forms: 0 0 0 1 00 0 , , , , , , . ∞ ∞ × ∞ ∞ − ∞ ∞ ∞ Example-1 Find the value of k for which the function f(x) = sin cos , , x x x x k x − − ≠ =       4 4 4 π π π is continuous at x = x 4 . Sol. lim sin cos x x x x f → − − =       π π π 4 4 4 lim sin cos x x x x k → −       − = π π 4 2 1 2 1 2 4 lim sin cos cos .sin x x x x k → −       − = π π π π 4 2 4 4 4 lim sin x x x k → −       −       = π π π 4 2 4 4 4 2 4 = k k = 1 2 2 UNIT – III : CALCULUS 5 CONTINUITY & DIFFERENTIABILITY CHAPTER Learning Objectives After going through this Chapter, the student would be able to:  Learn the continuity and differentiability of the function.  Relate continuity and differentiability.  Solve problems related to differentiability and continuity. LIST OF TOPICS Topic-1: Continuity Topic-2: Differentiability
Continuity & Differentiability SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) 1. Determine the value of the constant ‘k’ so that the function f(x) = kx x x x | | , , if if is < ≥      0 3 0 is continuous at x = 0. R&U [SQP 2023-24] [Delhi, 2017] Concept Applied A function f(x) is continuous at a point x = m, if limx→m+ f(x) = limx→m– f(x) = f(m) Sol. Since, f is continuous at x = 0, lim x→ − 0 f(x)= lim x→ + 0 f(x) = f(0) 1⁄2 Here, f(0) = 3, LHL = lim x→ − 0 f(x) = lim x | | kx → x − 0 = lim x kx → x − 0 − = – k \ – k = 3 or k = –3. 1⁄2 [Marking Scheme SQP, 2023-24] 2. If the function f defined as f x x x x k x ( ) , , = − − ≠ =    2 9 3 3 3 is continuous at x = 3, find the value of k. R&U [Delhi Set-1, 2, 3, 2020] OR Determine the value of ‘k’ for which the following function is continuous at x = 3: f x x x x k x ( ) ( ) , , = + − − ≠ =      3 36 3 3 3 2 R&U [Outside Delhi, 2017] Sol. lim , x x x k → − − = ∴ = 3 2 9 3 6 6 1⁄2 + 1⁄2 [CBSE Marking Scheme, OD 2020] Commonly Made Error Some students do not know how to evaluate limits of the form 0 0 . From this part questions are mainly asked from indeterminate form of limits. Learn all the different forms of limits. Answering Tip 3. If the following function f(x) is continuous at x = 0, then write the value of k. f(x) = sin , , 3 2 0 0 x x x k x ≠ =       R&U [Outside Delhi Comptt., 2017] Sol. lim sin x x →0 x 3 2 = lim . sin x x →0 x 3 2 3 2 3 2 1⁄2 or k = 3 2 1⁄2 [Marking Scheme OD Comptt., 2017] Concept Applied In this case make the denominator same to the angle of sine, to obtain limx → 0 sinx x form Short Answer Type Questions-I (2 marks each) 1. Find the value(s) of ‘l’, if the function f(x) =  λ  ≠    = 2 2 sin , if 0 1 , if 0 x x x x is continuous at x = 0. R&U [Delhi Set-1, 2023] Sol. f(x) = sin , 2 2 0 1 0 λx x x x if , if ≠ =      is continuous at x = 0. f(0) = 1 f x x ( ) → −0 = lim sin ( ) h ( ) h → h − − 0 − 2 2 0 0 λ = lim sin h h →0 h 2 2 λ = lim sin h h → h       0 2 2 λ λ λ = λ2 × 1 = λ2 1 f(0) = lim ( ) x f x → −0 1 = λ2 ∴ λ = ± 1 1 [Marking Scheme, Delhi, 2023] 2. Find the value of k for which the function f given as f(x) = 1 2 0 0 2 − ≠ =      cos , , x x x k x if if is continuous at x = 0 R&U [Delhi Set-3, 2023] This Question is for practice and its solution is given at the end of the chapter.
MATHEMATICS, Class-XII Sol. f(x) = 1 2 0 0 2 − ≠ =      cos , , x x x k x if if is continuous at x = 0 LHL –0 L.H.L.= lim ( ) x f x → = lim cos( ) h ( ) h → h − − 0 − 2 1 0 2 0 = lim cos h h → h − 0 2 1 2 = lim sin ( / ) h h →0 h 2 2 2 2 2 = lim sin / h / h → h       = 0 2 1 4 2 2 2 1 4 f(x) is continuous at x = 0 \ k = 0 1 (0) lim ( ) 4 x f fx → −   ∴ =     0 1 (0) lim ( ) 4 x f fx → −   ∴ =     [Marking Scheme, Delhi, 2023] 3. Find the value(s) of k so that the following function is continuous at x = 0 f x kx x x x x ( ) cos sin if = − ≠ =       1 0 1 2 if 0 R&U [SQP 2020-21] Concept Applied A function f(x) is continuous at a point x = m, if limx→m+ f(x) = limx→m– f(x) = f(m) Sol. lim cos x sin kx → x x − 0 1 = lim sin x sin kx → x x       0 2 2 2 = lim sin x sin kx x x x x →       0 2 2 2 2 2 = lim sin lim sin x x kx kx k x x → →             ×       0 2 2 2 0 2 2 2 2 = 2 1 4 1 2 × × k 11⁄2  f(x) is continuous at x = 0 \ lim ( ) x f x →0 = f(0) Þ k2 2 = 1 2 Þ k2 = 1 Þ k = ±1 1⁄2 [Marking Scheme SQP, 2020-21] 4. Find the value of k for which the function: f(x) = x x x x k x 2 +3 10 2 , 2 , = 2 − − ≠      is continuous at x = 2. R&U [Delhi Comptt., 2017] 5. Find the value of p for which the function f(x) = 1 cos4 , 0 , =0 2 − ≠      x x x p x is continuous at x = 0. R&U [Delhi Comptt., 2017] Sol. lim x→0 f(x) = f(0) lim sin x x → x × 0 2 2 4 2 2 4 = p 1 8 lim sin x x → x       0 2 2 2 = p 1⁄2 p = 8 1⁄2 [Marking Scheme Delhi Comptt., 2017] Short Answer Type Questions-II (3 marks each) 1. Determine the values of 'a' such that the following function is continuous at x = 0: f(x) = x x a x x x e bx x bx + + − < < = − >        sin sin( ) , , , sin 1 0 2 0 2 1 0 if if if π  R&U [SQP, 2017-18] Sol. lim x→ − 0 f(x) = lim sin x sin( ) x x → a x − + 0 + 1 = lim sin sin( ) ( ) ( ) x x x a x a x a → − + + + + 0 1 1 1 1 = 2 a + 1 1⁄2 lim x→ + 0 f(x) = lim sin x bx e → bx + − 0 2 1 1⁄2 = lim sin sin sin x bx e bx bx → bx + − × 0 2 1 1⁄2 f(0) = 2. 1⁄2 For the function to be continuous at 0, we must have lim x→ + 0 f(x) = lim x→ − 0 f(x) = f(0) i.e., we must have 2 a + 1 = 2 or a = 0; b may be any real number other than 0. 1 [Marking Scheme SQP, 2017-18] (Modified) This Question is for practice and its solution is given at the end of the chapter.
Continuity & Differentiability 2. Find the values of p and q, for which: f x x x x p x q x x x ( )= 1 sin 3cos , if < 2 , if = 2 (1 sin ) ( 2 ) , if > 2 3 2 2 - p p - p - p          , is continuous at = 2 x p . A [Delhi Set-1, 2, 3, 2016] Sol. LHL at x=       π 2 = lim ( ) x f x → −π 2 = lim ( sin ) cos x x x → − − π 2 3 2 1 3 = lim x x x → − − π − 2 3 3 2 2 1 3 1 ( sin ) ( sin ) 1⁄2 = lim x x x x x x → − − + + π − + 2 2 1 1 3 1 1 ( sin )( sin sin ) ( sin )( sin ) = lim sin sin sin x x x x → − + + π ( ) + 2 2 1 3 1 1⁄2 = 111 3 1 1 1 2 + + + = ( ) 1⁄2 RHL ( ) at x=π 2 = lim . ( ) ( ) x f x → +π 2 = lim x q x x → + − π π − 2 2 1 2 ( sin ) ( ) Put x = π π 2 2 + → h x . , As h → 0 1⁄2 = q h h lim sin h 0 → − +       − +             1 2 2 2 2 π π π = q h q h h h lim lim h 0 h → → − = 1 4 2 2 4 2 0 2 2 cos sin = q h h q 2 h 2 2 1 0 4 8 2 lim →             × = sin 1⁄2 Also, f π 2       = p Since, f(x) is continuous at x = p 2 \ LHL = RHL = f π 2       \ 1 2 = q p 8 = or q = 4 and p = 1 2 1⁄2 Commonly Made Error Many students commit errors in finding the Left hand limit and Right hand limit. Sufficient time needs to be spent on this topic. Answering Tip 3. For what value of k is the following function continuous at x = − π 6 ? f(x) = 3 6 6 6 sin cos , , x x x x k x + + ≠ − = −        π π π A [SQP 2016-17] Sol. lim ( ) x f x →− π 6 = lim sin cos x x x →− x + + π π 6 3 6 lim ( ) x f x →− π 6 = lim sin cos · x x x →− x +         + π π 6 2 3 2 1 2 6 lim ( ) x f x →− π 6 = lim sin x x →− x +       + π π π 6 2 6 6 1 = 2 1 f −      π 6 = k 1⁄2 For the continuity of f(x) at x = − π 6 , f −      π 6 = lim ( ) x f x →− π 6 or k = 2 1⁄2 [Marking Scheme SQP, 2016-17] (Modified) 4. Find k, if f(x) = k x x x x x x sin ( ), tan sin , π 2 1 0 0 3 + ≤ − >       is continuous at x = 0. R&U [Outside Delhi 2016] This Question is for practice and its solution is given at the end of the chapter.