Maths_Ch-052.pdf - Class-XII-Maths Continuity and Differentiabil ity CBSE NCERT Solutions for Class 12 Maths Chapter 05 Back of Chapter Questions

Maths_Ch-052.pdf - Class-XII-Maths Continuity and...

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Class-XII-Maths C ontinuity and Differentiability 1 Practice more on Continuity and Differentiability CBSE NCERT Solutions for Class 12 Maths Chapter 05 Back of Chapter Questions Exercise ?. ? 1. Prove that the function ?(?) = 5? − 3 is continuous at ? = 0, at ? = −3 and at ? = 5. Solution: Given function is ?(?) = 5? − 3 At ? = 0, ?(0) = 5(0) − 3 = −3 LHL = lim ?→0 ?(?) = lim ?→0 (5? − 3) = −3 RHL = lim ?→0 + ?(?) = lim ?→0 + (5? − 3) = −3 Here, at ? = 0, LHL = RHL = ?(0) = −3 Hence, the function ? is continuous at ? = 0. Now at ? = −3, ?(−3) = 5(−3) − 3 = −18 LHL = lim ?→−3 ?(?) = lim ?→−3 (5? − 3) = −18 RHL = lim ?→−3 + ?(?) = lim ?→−3 + (5? − 3) = −18 Here, at ? = −3, LHL = RHL = ?(−3) = −18 Hence, the function ? is continuous at ? = −3. At ? = 5, ?(5) = 5(5) − 3 = 22 𝐿𝐻𝐿 = lim ?→5 ?(?) = lim ?→5 (5? − 3) = 22 RHL = lim ?→5 + ?(?) = lim ?→5 + (5? − 3) = 22 Here, at ? = 5, LHL = RHL = ?(5) = 22 Hence, the function ? is continuous at ? = 5. 2. Examine the continuity of the function ?(?) = 2? 2 − 1 at ? = 3.
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Class-XII-Maths C ontinuity and Differentiability 2 Practice more on Continuity and Differentiability Solution: Given function is ?(?) = 2? 2 − 1. At ? = 3, ?(3) = 2(3) 2 − 1 = 17 LHL = lim ?→3 ?(?) = lim ?→3 (2? 2 − 1) = 17 RHL = lim ?→3 + ?(?) = lim ?→3 + (2? 2 − 1) = 17 Here, at ? = 3, LHL = RHL = ?(3) = 17 Therefore, the function ? is continuous at ? = 3. 3. Examine the following functions for continuity: (a) ?(?) = ? − 5 (b) ?(?) = 1 ?−5 , ? ≠ 5 (c) ?(?) = ? 2 −25 ?+5 , ? ≠ −5 (d) ?(?) = |? − 5| Solution: (a) Given function ?(?) = ? − 5 Let ? be any real number. At ? = ?, ?(?) = ? − 5 LHL = lim ?→? ?(?) = lim ?→? (? − 5) = ? − 5 RHL = lim ?→? + ?(?) = lim ?→? + (? − 5) = ? − 5 At ? = ?, LHL = RHL = ?(?) = ? − 5 Hence, the function ? is continuous for all real numbers. (b) Given function ?(?) = 1 ?−5 , ? ≠ 5 Let ?(? ≠ 5) be any real number. At ? = ?, ?(?) = 1 ?−5 LHL = lim ?→? ?(?) = lim ?→? ( 1 ?−5 ) = 1 ?−5 RHL = lim ?→? + ?(?) = lim ?→? + ( 1 ?−5 ) = 1 ?−5 At ? = ?, LHL = RHL = ?(?) = 1 ?−5 Hence, the function ? is continuous for all real numbers (except 5 ).
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Class-XII-Maths C ontinuity and Differentiability 3 Practice more on Continuity and Differentiability (c) Given function ?(?) = ? 2 −25 ?+5 , ? ≠ −5 Let ?(? ≠ −5) be any real number. At ? = ?, ?(?) = ? 2 −25 ?+5 = (?+5)(?−5) (?+5) = (? + 5) LHL = lim ?→? ?(?) = lim ?→? ( ? 2 −25 ?+5 ) = lim ?→? ( (?+5)(?−5) (?+5) ) = ? + 5 RHL = lim ?→? + ?(?) = lim ?→? + ( ? 2 −25 ?+5 ) = lim ?→? + ( (?+5)(?−5) (?+5) ) = ? + 5 At ? = ?, LHL = RHL = ?(?) = ? + 5 Hence, the function ? is continuous for all real numbers (except −5 ). (d) Given function is ?(?) = |? − 5| = { 5 − ?, ? < 5 ? − 5, ? ≥ 5 Let ? be any real number. According to question, ? can be ? < 5 or ? = 5 or ? > 5. First case: If ? < 5, ?(?) = 5 − ? and lim ?→? ?(?) = lim ?→? (5 − ?) = 5 − ?. Here, lim ?→? ?(?) = ?(?) Hence, the function ? is continuous for all real numbers less than 5. Second case: If ? = 5, ?(?) = ? − 5 and lim ?→? ?(?) = lim ?→? (? − 5) = ? − 5. Here, lim ?→? ?(?) = ?(?) Hence, the function ? is continuous at ? = 5. Third case: If ? > 5, ?(?) = ? − 5 and lim ?→? ?(?) = lim ?→? (? − 5) = ? − 5. Here, lim ?→? ?(?) = ?(?) Hence, the function ? is continuous for all real numbers greater than 5. Hence, the function ? is continuous for all real numbers. 4. Prove that the function ?(?) = ? ? , is continuous at ? = ?, where ? is a positive integer. Solution: Given function is ?(?) = ? ? . At ? = ?, ?(?) = ? ? lim ?→? ?(?) = lim ?→? (? ? ) = ? ? Here, at ? = ?, lim ?→? ?(?) = ?(?) = ? ? Since lim ?→? ?(?) = ?(?) = ? ?
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Class-XII-Maths C ontinuity and Differentiability 4 Practice more on Continuity and Differentiability Hence, the function ? is continuous at ? = ?, where ? is positive integer. 5. Is the function ? defined by ?(?) = { ?, ? ≤ 1 5, ? > 1 continuous at ? = 0? At ? = 1? At ? = 2?
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