{"id":162701,"date":"2023-10-31T11:56:16","date_gmt":"2023-10-31T06:26:16","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=162701"},"modified":"2023-10-31T12:44:19","modified_gmt":"2023-10-31T07:14:19","slug":"ml-aggarwal-class-12-maths-solutions-section-a-chapter-5-ex-5-1","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-12-maths-solutions-section-a-chapter-5-ex-5-1\/","title":{"rendered":"ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.1"},"content":{"rendered":"

Parents can use ISC Mathematics Class 12 Solutions<\/a> Chapter 5 Continuity and Differentiability Ex 5.1 to provide additional support to their children.<\/p>\n

ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.1<\/h2>\n

Question 1.
\nExamine the following functions for continuity at the inclined points:
\n(i) f(x) = \\(\\left\\{\\begin{array}{ccc}
\nx^3+3 & , & x \\neq 0 \\\\
\n1 & , & x=0
\n\\end{array}\\right.\\) at x = 0 (NCERT)
\n(ii) f(x) = x3<\/sup> + 2x2<\/sup> – 1 at x = 1. (NCERT Exampler)
\nSolution:
\n(i) \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) (x3<\/sup> + 3)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) [h3<\/sup> + 3] = 3
\n\\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) (x3<\/sup> + 3)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) [(- h)3<\/sup> + 3] = 3
\n\u2234 \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) \u2260 f(0)
\n[\u2235 f(0) = 1]
\nThus f is continuous at x = 0.<\/p>\n

(ii) Given f(x) = x3<\/sup> + 2x2<\/sup> – 1
\n\u2234 f(1) = 13<\/sup> + 2 – 12<\/sup> – 1
\n= 3 – 1 = 2
\n\\(\\underset{x \\rightarrow 1}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1}{\\mathrm{Lt}}\\) (x3<\/sup> + 2x2<\/sup> – 1)
\n= 1 + 2 – 1 = 2
\n\u2234 \\(\\underset{x \\rightarrow 1}{\\mathrm{Lt}}\\) f(x) = f(1).
\nThus f(x) is continuous at x = 1.<\/p>\n

\"ML<\/p>\n

Question 2.
\nIf f(x) = \\(\\frac{x^2-1}{x-1}\\) for x \u2260 1 annd f(x) = 2 when x = 1, show that the function is continuous at x = 1.
\nSolution:
\n(i) L.H.L. = \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) f (1 – h)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{(1-h)^2-1}{1-h-1}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h^2-2 h}{-h}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h(h-2)}{-h}\\)
\n= 2<\/p>\n

R.H.L. = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) f (1 + h)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{(1+h)^2-1}{1+h-1}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h^2+2 h}{h}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h(h+2)}{h}\\)
\n= 2
\nalso, f(1) = 2
\n\u2234 \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x) = f(1) = 2.<\/p>\n

Question 2(old).
\nProve that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5. (NCERT)
\nSolution:
\nGiven, f(x) = 5x – 3<\/p>\n

at x = 0 :
\n\\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) 5x – 3
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) 5 (0 – h) – 3
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) – 5h – 3 = – 3
\n\\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) 5x – 3
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) 5 (0 + h) – 3 = – 3
\nAlso f(0) = 5 . 0 – 3
\n= – 3
\nThus \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = f(0)
\n\u2234 f is continuous at x = 0<\/p>\n

at x = – 3
\n\\(\\underset{x \\rightarrow 3^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 3^{-}}{\\mathrm{Lt}}\\) 5x – 3
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) 5 (- 3 – h) – 3
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) – 5h – 18
\n= – 18
\n\\(\\underset{x \\rightarrow 3^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\ {Lt}_{x \\rightarrow-3^{+}}\\) 5x – 3
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) 5 (- 3 + h) – 3
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) (5h – 18) = 18
\nAlso f(- 3) = 5 (- 3) – 3 = – 18
\n\u2234 \\(\\underset{x \\rightarrow 3^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 3^{+}}{\\mathrm{Lt}}\\) f(x) = f(3)
\nHence f(x) is continuous at x = – 3.<\/p>\n

at x = 5
\n\\(\\underset{x \\rightarrow 5^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 5^{-}}{\\mathrm{Lt}}\\) (5x – 3)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) (5 (5 – h) – 3)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) (22 – 5h)
\n= 22
\n\\(\\underset{x \\rightarrow 5^{+}}{\\mathrm{Lt}}\\)] f(x) = \\(\\underset{x \\rightarrow 5^{+}}{\\mathrm{Lt}}\\) (5x – 3)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) 5 (5 + h) – 3
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) (22 + 5h)
\n= 22
\nAlso, f(5) = 5 \u00d7 5 – 3
\n= 22
\n\u2234 \\(\\underset{x \\rightarrow 5^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 5^{+}}{\\mathrm{Lt}}\\) f(x) = f(5)
\nHence f(x) is continuous at x = 5.<\/p>\n

\"ML<\/p>\n

Question 3.
\nA function f is defined as f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{x^2-x-6}{x-3} & , x \\neq 3 \\\\
\n5 & , x=3
\n\\end{array}\\right.\\). show that f is continuous at x = 3.
\nSolution:
\nL.H.L. = \\(\\underset{x \\rightarrow 3^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) f (3 – h)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{(3-h)^2-(3-h)-6}{3-4-3}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h^2-5 h}{-h}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h(h-5)}{-h}\\)
\n= 5
\nR.H.L. = \\(\\underset{x \\rightarrow 3^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) f (3 + h)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{(3+h)^2-(3+h)-6}{3+h-3}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h^2+5 h}{h}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h(h+5)}{h}\\)
\n= + 5
\n\u2234 L.H.L. = R.H.L. and f(3) = 5.
\nThus \\(\\underset{x \\rightarrow 3^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 3^{+}}{\\mathrm{Lt}}\\) f(x) = 5 = f(3)
\n\u2234 f(x) is continuous at x = 3.<\/p>\n

Question 4.
\nIs the function f defined by f(x) = \\(\\left\\{\\begin{array}{lll}
\nx, & \\text { if } & x \\leq 1 \\\\
\n5, & \\text { if } & x>1
\n\\end{array}\\right.\\) continuous at
\n(i) x = 0
\n(ii) x = 1
\n(iii) x = 2 ? (NCERT)
\nSolution:
\nGiven, f(x) = \\(\\left\\{\\begin{array}{lll}
\nx, & \\text { if } & x \\leq 1 \\\\
\n5, & \\text { if } & x>1
\n\\end{array}\\right.\\)
\nat x = 0;
\n\\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) x = 0
\nand \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) x = 0
\n\u2234 \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = f(0)
\n\u2234 f is continuous at x = 0<\/p>\n

at x = 1 ;
\n\\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) 5 = 5
\nand \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) x = 1
\n\u2234 \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x) \u2260 \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x)
\n\u2234 f is not continuous at x = 1.<\/p>\n

at x = 2 ;
\n\\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) 5 = 5
\nand \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) 5 = 5
\n\u2234 \\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) f(x) = 5 = f(2)
\n\u2234 f is continuous at x = 2.<\/p>\n

\"ML<\/p>\n

Question 5.
\nIs the function f defined by f(x) = \\(\\left\\{\\begin{array}{ccc}
\n3 x+5 & , \\text { if } & x \\geq 2 \\\\
\nx^2 & , \\text { if } & x<2
\n\\end{array}\\right.\\) continuous at x = 2 ? (NCERT Exampler)
\nSolution:
\nHere f(2) = 3 \u00d7 2 + 5
\n= 6 + 5
\n= 11
\n\\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) 3x + 5
\n= 6 + 5 = 11
\n\\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) x2<\/sup>
\n= 22<\/sup> = 4
\n\u2234 \\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) f(x) \u2260 \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) f(x)
\nThus f is not continuous at x = 2.<\/p>\n

Question 6.
\nIf the function f defined by f(x) = \\(\\left\\{\\begin{array}{ccc}
\n\\frac{2 x^2-3 x-2}{x-5} & , \\text { if } & x \\neq 2 \\\\
\n5 & , \\text { if } & x=2
\n\\end{array}\\right.\\) continuous at x = 2? (NCERT Exampler)
\nSolution:
\nHere, \\(\\underset{x \\rightarrow 2}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 2}{\\mathrm{Lt}}\\) \\(\\frac{2 x^2-3 x-2}{x-2}\\)
\n= \\(\\underset{x \\rightarrow 2}{\\mathrm{Lt}}\\) \\(\\frac{(x-2)(2 x+1)}{x-2}\\)
\n= \\(\\underset{x \\rightarrow 2}{\\mathrm{Lt}}\\) (2x + 1)
\n= 4 + 1 = 5
\nHere f(2) = 5
\n[since f(x) = 5 at x = 2]
\n\u2234 \\(\\underset{x \\rightarrow 2}{\\mathrm{Lt}}\\) f(x) = f(2)
\nThus, f is continuous at x = 2.<\/p>\n

\"ML<\/p>\n

Question 7.
\nIf f(x) = \\(\\left\\{\\begin{array}{cc}
\nk x^2+5, & x \\leq 1 \\\\
\n2, & x>1
\n\\end{array}\\right.\\), find k so that f may be continuous at x = 1.
\nSolution:
\n\\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) 2 = 2
\n\\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) (kx2<\/sup> + 5)
\n= k \u00d7 12<\/sup> + 5
\n= k + 5
\nNow f is given to be continuous at x = 1.
\n\u2234 \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x) = f(1)
\nk + 5 = 2
\nk = – 3.<\/p>\n

Question 7 (old).
\nIs the function f defined by f(x) = \\(\\left\\{\\begin{array}{ccc}
\n\\frac{x}{\\sin 2 x} & \\text {, when } & x \\neq 0 \\\\
\n2 & \\text {, when } & x=0
\n\\end{array}\\right.\\) continuous at x = 0?
\nSolution:
\n\\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\)
\n= \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{x}{\\sin 2 x}\\)
\n= \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{1}{2}\\left(\\frac{2 x}{\\sin 2 x}\\right)\\)
\n= \\(\\frac{1}{2}\\) \u00d7 1
\n= \\(\\frac{1}{2}\\)
\n[\u2235 \\(\\ {Lt}_{\\theta \\rightarrow 0} \\frac{\\theta}{\\sin \\theta}\\)]
\nalso, f(0) = 2
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) \u2260 f(0)
\nThus, f(x) is discontinuous at x = 0.<\/p>\n

Question 8.
\nIf f(x) = \\(\\left\\{\\begin{array}{ccc}
\n3 x-8 & , & x \\leq 5 \\\\
\n2 k & , & x>5
\n\\end{array}\\right.\\), find k so that may be continuous at x = 5.
\nSolution:
\n\\(\\underset{x \\rightarrow 5^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 5^{+}}{\\mathrm{Lt}}\\) 2k = 2k
\n\\(\\underset{x \\rightarrow 5^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 5^{-}}{\\mathrm{Lt}}\\) (3x – 8)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) 3 (5 – h) – 8
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) (7 – 3h)
\n= 7
\nSince f(x) is continuous at x = 5
\n\u2234 \\(\\underset{x \\rightarrow 5^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 5^{-}}{\\mathrm{Lt}}\\) f(x) = f(5)
\n\u21d2 2k = 7
\n\u21d2 k = \\(\\frac{7}{2}\\)<\/p>\n

Question 8 (old).
\nIf f(x) = \\(\\left\\{\\begin{array}{cc}
\nk x^2 & , x \\leq 2 \\\\
\n3 & , x>2
\n\\end{array}\\right.\\), find k so that f may be continuous atx = 2. (NCERT)
\nSolution:
\nl.H.L. = \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 2}{\\mathrm{Lt}}\\) kx2<\/sup> = 4k
\nand R.H.L. = \\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 2}{\\mathrm{Lt}}\\) 3 = 3
\nand f(2) = 4k
\nSince f(x) is continuous at x = 2
\n\u2234 \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) f(x) = f(2)
\n\u21d2 4k = 3
\n\u21d2 k = \\(\\frac{3}{4}\\).<\/p>\n

\"ML<\/p>\n

Question 9.
\nIf f(x) = \\(\\begin{cases}3 x-4, & 0 \\leq x \\leq 2 \\\\ 2 x+\\lambda, & 2<x \\leq 5\\end{cases}\\)
\n\u2234 L.H.L. = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) – x = 0
\nR.H.L. = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) x = 0
\nf(0) = 0
\n\u2234 \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\)
\nf(x) = f(0)
\nThus, f is continuous at x = 0.<\/p>\n

Question 10.
\n(i) f(x) = \\(\\frac{x^2-9}{x-3}\\) is not defined at x = 3. What value should be assigned to f(3) for continuity of f(x) at x = 3?
\n(ii) If f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{(x+3)^2-36}{x-3}, & x \\neq 3 \\\\
\nk, & x=3
\n\\end{array}\\right.\\), find k so that the function f may be continuous at x = – 1 (ISC 2018)
\n(iii) Determine the value of ‘k’for which the following function is continuous at x = 3 ; f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{(x+3)^2-36}{x-3}, & x \\neq 3 \\\\
\nk, & x=3
\n\\end{array}\\right.\\).
\n(iv) If f(x) = \\(\\left\\{\\begin{array}{ccc}
\n\\frac{x^2-x-6}{x^2-2 x-3} & , & x \\neq 3 \\\\
\nk & , & x=3
\n\\end{array}\\right.\\), find k so that the function f may be continuous at x = 3.
\nSolution:
\nL.H.L. = \\(\\underset{x \\rightarrow 3^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) f (3 – h)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{(3-h)^2-9}{3-h-3}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h^2-6 h+9-9}{-h}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h(h-6)}{-h}\\)
\n= 6
\nR.H.L. = \\(\\underset{x \\rightarrow 3^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) f (3 + h)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{(3+h)^2-9}{3+h-3}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{9+h^2+6 h-9}{h}\\)
\n= \\(\\underset{h \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{h(h+6)}{h}\\)
\n= 0 + 6 = 6
\n\u2234 L.H.L. = R.H.L.
\nThus \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) \\(\\frac{x^2-9}{x-3}\\) = 6
\nNow f(x) is continuous at x = 3 if \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) f(x) = f(3) i.e., if f(3) = 6.<\/p>\n

\"ML<\/p>\n

(ii) Given f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{(x+3)^2-36}{x-3}, & x \\neq 3 \\\\
\nk, & x=3
\n\\end{array}\\right.\\)
\n\\(\\) f(x) = \\(\\underset{x \\rightarrow-1}{\\mathrm{Lt}\\) \\(\\frac{x^2-2 x-3}{x+1}\\)
\n= \\(\\underset{x \\rightarrow-1}{\\mathrm{Lt}}\\) \\(\\frac{(x+1)(x-3)}{x+1}\\)
\n= \\(\\underset{x \\rightarrow-1}{\\mathrm{Lt}}\\) x – 3
\n= – 1 – 3 = – 4
\nalso f (- 1) = k
\nSince f may be continuous at x = – 1
\n\u2234 \\(\\underset{x \\rightarrow-1}{\\mathrm{Lt}}\\) f(x) = f(- 1)
\n\u21d2 – 4 = k<\/p>\n

(iii) \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) \\(\\frac{(x+3)^2-36}{x-3}\\)
\n= \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) \\(\\frac{(x+3)^2-6^2}{x-3}\\)
\n= \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) \\(\\frac{(x+3-6)(x+3+6)}{x-3}\\)
\n= \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) \\(\\frac{(x-3)(x+9)}{x-3}\\)
\n= \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) (x + 9) = 12
\nAlso f(3) = k
\nSince f is continuous at x = 3
\n\u2234 \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) f(x) = f(3)
\n\u21d2 12 = k<\/p>\n

(iv) \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) \\(\\frac{x^2-x-6}{x^2-2 x-3}\\)
\n= \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) \\(\\frac{(x-3)(x+2)}{(x-3)(x+1)}\\)
\n= \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) \\(\\frac{x+2}{x+1}\\)
\n= \\(\\frac{3+2}{3+1}\\)
\n= \\(\\frac{5}{4}\\)
\nHence f(3) = k (given)
\nSince f is continuous at x = 3
\n\u2234 \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) f(x) = f(3)
\n\u21d2 \\(\\frac{5}{4}\\) = k
\nHence the required value of k be \\(\\frac{5}{4}\\).<\/p>\n

\"ML<\/p>\n

Question 11.
\nIf f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{\\tan 3 x}{k x}, & x \\neq 0 \\\\
\n1 & , x=0
\n\\end{array}\\right.\\), find k so that the function f may be continuous at x = 0.
\nSolution:
\nHere, \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{\\tan 3 x}{k x}\\)
\n= \\(\\frac{3}{k}\\) \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{\\tan 3 x}{k x}\\)
\n= \\(\\frac{3}{k} {Lt}_{x \\rightarrow 0} \\frac{\\tan 3 x}{3 x}\\)
\n= \\(\\frac{3}{k} \\times 1\\)
\n= \\(\\frac{3}{k}\\)
\n[\u2235 \\(\\ {Lt}_{\\theta \\rightarrow 0} \\frac{\\tan \\theta}{\\theta}\\) = 1]
\nAlso, given f(0) = 1
\nSince f is continuous at x = 0
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = f(0)
\n\u21d2 \\(\\frac{3}{k}\\) = 1
\n\u21d2 k = 3.<\/p>\n

Question 12.
\nIs the function f defined by f(x) = tan x continuous at x = \\(\\frac{\\pi}{2}\\)?
\nSolution:
\nGiven f(x) = tan x
\nSince \\(\\underset{x \\rightarrow \\frac{\\pi^{-}}{2}}{\\ {Lt}}\\) f(x) = \\(\\underset{x \\rightarrow \\frac{\\pi^{-}}{2}}{\\ {Lt}}\\) tan x \u2192 \u221e
\nand \\(\\underset{x \\rightarrow \\frac{\\pi^{+}}{2}}{\\ {Lt}}\\) f(x) = \\(\\underset{x \\rightarrow \\frac{\\pi^{+}}{2}}{\\ {Lt}}\\) + tan x \u2192 – \u221e
\n\u2234 \\(\\mathrm{Lt}_{x \\rightarrow \\frac{\\pi}{2}}\\) f(x) does not exists.
\nHence f(x) is not continuous at x = \\(\\frac{\\pi}{2}\\).<\/p>\n

\"ML<\/p>\n

Question 13.
\nIs the function f defined by f(x) = |x| continuous at x = 0? (NCERT)
\nSolution:
\nf(x) = |x|
\n= \\(\\left\\{\\begin{aligned}
\nx & ; \\quad x>0 \\\\
\n0 ; & x=0 \\\\
\n-x ; & x<0 \\end{aligned}\\right.\\)
\n\u2234 L.H.L. = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) – x = 0
\nR.H.L. = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) x = 0
\nf(0) = 0
\n\u2234 \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x) = f(0)
\nThus, f is continuous at x = 0.<\/p>\n

Question 14.
\nIs the function f defined by f(x) = |x – 1| continuous at x = 1?
\nSolution:
\nGiven f(x) = |x – 1|
\n= \\(\\left\\{\\begin{array}{rll} x-1 & ; & x>1 \\\\
\n-(x-1) & ; & x<1 \\\\
\n0 & ; & x=1
\n\\end{array}\\right.\\)
\n\u2234 L.H.L. = \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) – (x – 1)
\n= – (1 – 1) = 0
\nR.H.L. = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) x – 1
\n= 1 – 1 = 0
\nand f(1) = 1 – 1 = 0
\n\u2234 L.H.L = R.H.L.
\n= f(1) = 0
\nThus f is continuous at x = 1.<\/p>\n

Question 15.
\nIs the function f defined by f(x) = x – |x| continuous at x = 0?
\nSolution:
\nGiven f(x) = x – |x|
\n= \\(\\left\\{\\begin{array}{rc}
\nx-x ; & x \\geq 0 \\\\
\nx-(-x) & ; \\quad x<0
\n\\end{array}\\right.\\)
\n= \\(\\left\\{\\begin{aligned}
\n0 & ; \\quad x \\geq 0 \\\\
\n2 x & ; \\quad x<0
\n\\end{aligned}\\right.\\)
\n\u2234 L.H.L. = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) 2x
\n= 2 \u00d7 0 = 0
\nR.H.L. = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) 0 = 0
\nHence, f is continuous at x = 0.<\/p>\n

\"ML<\/p>\n

Question 16.
\nExamine the following functions for continuity at the inclined points :
\n(i) f(x) = \\(\\begin{cases}x^2 & , \\quad x \\geq 0 \\\\ -x & , \\quad x<0\\end{cases}\\) at x = 0
\n(ii) f(x) = \\(\\left\\{\\begin{array}{cl}
\n5 x-4 & , \\text { if } x<1 \\\\
\n4 x^2-3 x & , \\text { if } x \\geq 1
\n\\end{array}\\right.\\) at x = 1
\nSolution:
\n(i) L.H.L. = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) – x = 0
\nR.H.L. = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) x2<\/sup> = 0
\nand f(0) = 02<\/sup> = 0
\n\u2234 L.H.L. = R.H.L. = f(0)
\nThus, f is continuous at x = 0.<\/p>\n

(ii) L.H.L. = \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) 5x – 4
\n= 5 \u00d7 1 – 4 = 1
\nand R.H.L. = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) 4x2<\/sup> – 3x
\n= 4 – 3 = 1
\nalso f(1) = 4 \u00d7 12<\/sup> – 3 \u00d7 1 = 1
\n\u2234 L.H.L. = R.H.L. = f(1)
\nThus, f is continuous at x = 1.<\/p>\n

\"ML<\/p>\n

Question 17.
\nIs the function f defined by f(x) = \\(f(x)=\\left\\{\\begin{array}{ccc}
\n2 x^2-3 & , \\text { if } & x \\leq 1 \\\\
\n5 & , \\text { if } & x>1
\n\\end{array}\\right.\\) continuous at x = 0? what about its continuity at x = 1 and x = 2?
\nSolution:
\nL.H.L. = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) (2x2<\/sup> – 3)
\n= 0 – 3
\n[as x \u2192 0–<\/sup> \u21d2 x < 0]
\n= – 3
\nR.H.L. = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) (2x2<\/sup> – 3)
\n= 2 \u00d7 02<\/sup> – 3
\n= – 3
\n[as x \u2192 0+<\/sup>
\n\u21d2 x > 0 but x < 1
\n\u2234 f(x) = 2x2<\/sup> – 3]
\n\u2234 L.H.L. = R.H.L. = f(0)
\nThus, f is continuous at x = 0.<\/p>\n

at x = 1 :
\nL.H.L. = \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) (2x2<\/sup> – 3)
\n= 2 \u00d7 12<\/sup> – 3
\n= 2 – 3 = – 1
\nR.H.L. = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) 5 = 5
\n\u2234 L.H.L. \u2260 R.H.L.
\nThus, f is not continuous at x = 1.<\/p>\n

at x = 2 :
\nWhen x = 2 > 1
\n\u2234 f(x) = 5
\ni.e. when x \u2192 2+<\/sup> or x \u2192 2–<\/sup>
\n\u2234 f(x) = 5
\n\u2234 L.H.L. = R.H.L. = 5 = f(2)
\nThus f is continuous at x = 2.<\/p>\n

Question 18.
\nExamine the following functions for continuity :
\n(i) f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{x}{|x|}, & x \\neq 0 \\\\
\n0, & x=0
\n\\end{array}\\right.\\) at x = 0
\n(ii) f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{x}{|x|}, & x \\neq 0 \\\\
\n0, & x=0
\n\\end{array}\\right.\\) at x = 4
\nSolution:
\n(i) L.H.L. = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) \\(\\frac{x}{|x|}\\)
\n= \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) \\(\\frac{x}{-x}\\)
\n= – 1
\n[\u2235 x \u2192 0–<\/sup>
\n\u21d2 x < 0
\n\u21d2 |x| = – x]
\nR.H.L. = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) \\(\\frac{x}{|x|}\\)
\n= \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) \\(\\frac{x}{x}\\)
\n= 1
\n[\u2235 x \u2192 0+<\/sup>
\n\u21d2 x > 0
\n\u21d2 |x| = x]
\n\u2234 L.H.L. \u2260 R.H.L.
\nThus, f is not continuous at x = 0.<\/p>\n

(ii) L.H.L. = \\(\\underset{x \\rightarrow 4^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 4^{-}}{\\mathrm{Lt}}\\) \\(\\frac{x-4}{2|x-4|}\\)
\n= \\(\\underset{x \\rightarrow 4^{-}}{\\mathrm{Lt}}\\) \\(\\frac{x-4}{-2(x-4)}=-\\frac{1}{2}\\)
\n[\u2235 x \u2192 4–<\/sup>
\n\u21d2 x < 4
\n\u21d2 x – 4 < 0
\n\u2234 |x – 4| = – (x – 4)]
\nand R.H.L. = \\(\\underset{x \\rightarrow 4^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 4^{+}}{\\mathrm{Lt}}\\) \\(\\frac{x-4}{2|x-4|}\\)
\n= \\(\\underset{x \\rightarrow 4^{+}}{\\mathrm{Lt}}\\) \\(\\frac{x-4}{2(x-4)}=\\frac{1}{2}\\)
\n[\u2235 x \u2192 4+<\/sup>
\n\u21d2 x > 4
\n\u21d2 x – 4 > 0
\n\u2234 |x – 4| = -x – 4]
\n\u2234 \\(\\underset{x \\rightarrow 4^{-}}{\\mathrm{Lt}}\\) f(x) \u2260 \\(\\underset{x \\rightarrow 4^{+}}{\\mathrm{Lt}}\\) f(x)
\nHence f is not continuous at x = 4.<\/p>\n

\"ML<\/p>\n

Question 19.
\nExamine the following functions for continuity :
\n(i) f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{x}{\\sin 3 x} & , \\text { when } x \\neq 0 \\\\
\n3 & \\text {, when } x=0
\n\\end{array}\\right.\\) at x = 0
\n(ii) f(x) = \\(\\left\\{\\begin{array}{c}
\n\\frac{\\sin 2 x}{\\sin 3 x}, \\text { when } x \\neq 0 \\\\
\n2, \\text { when } x=0
\n\\end{array}\\right.\\) at x = 1
\nSolution:
\n(i) \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{x}{\\sin 3 x}\\)
\n= \\(\\frac{1}{3}\\) \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{3 x}{\\sin 3 x}\\)
\n= \\(\\frac{1}{3}\\) \u00d7 1
\n= \\(\\frac{1}{3}\\)
\n[\u2235 \\(\\underset{\\theta \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{\\theta}{\\sin \\theta}\\) = 1]
\nalso f(0) = 3
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) \u2260 f(0)
\nThus, f is discontinuous at x = 0<\/p>\n

(ii) \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{\\sin 2 x}{\\sin 3 x}\\)
\n= \\(\\frac{\\operatorname{Lt}_{x \\rightarrow 0} \\frac{\\sin 2 x}{2 x} \\times 2 x}{\\operatorname{Lt}_{x \\rightarrow 0} \\frac{\\sin 3 x}{3 x} \\times 3 x}\\)
\n= \\(\\frac{2}{3} \\times \\frac{1}{1}=\\frac{2}{3}\\)
\n[\u2235 \\(\\underset{\\theta \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{\\theta}{\\sin \\theta}\\) = 1 and
\n\\(\\operatorname{Lt}_{x \\rightarrow a}\\left(\\frac{f(x)}{g(x)}\\right)=\\frac{\\operatorname{Lt}_{x \\rightarrow a} f(x)}{\\operatorname{Lt}_{x \\rightarrow a} g(x)}\\)]
\nalso, f(0) = 2
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) \u2260 f(0)
\nThus, f is discontinuous at x = 0.<\/p>\n

Question 20.
\n(i) If f(x) = \\(\\left\\{\\begin{array}{lll}
\nk x+1 & \\text {, if } & x \\leq 5 \\\\
\n3 x-5 & \\text { if } & x>5
\n\\end{array}\\right.\\) is continuous at x = 5, find the value of k.
\n(ii) Find the value of k so that the function f(x) = \\(\\left\\{\\begin{array}{lll}
\nk x+1 & \\text {, if } & x \\leq 5 \\\\
\n3 x-5 & \\text {, if } & x>5
\n\\end{array}\\right.\\) is continuous at x = 5. (NCERT)
\n(iii) For what value of k is the following function continuous at x = 2? f(x) = \\(\\left\\{\\begin{array}{cc}
\n2 x+1, & x<2 \\\\ k, & x=2 \\\\ 3 x-1, & x>2
\n\\end{array}\\right.\\)
\n(iv) Find the value of k so that the function f defined by f(x) = \\(\\begin{cases}k x+1, & \\text { if } x \\leq \\pi \\\\ \\cos x & , \\text { if } x>\\pi\\end{cases}\\) is continuous at x = \u03c0.
\n(v) For what value of k is the function f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{\\tan 5 x}{\\sin 2 x}, & x \\neq 0 \\\\
\nk & , x=0
\n\\end{array}\\right.\\) continuous at x = 0?
\n(vi) Find the value of the constant k so that the function f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{1-\\cos 4 x}{8 x^2} & , x \\neq 0 \\\\
\nk & , x=0
\n\\end{array}\\right.\\) is continuous at x = 0. (NCERT Exampler)
\n(vii) Determine the value of constant ‘k’ so that the function f(x) = \\(\\left\\{\\begin{array}{cc}
\n\\frac{k x}{|x|}, & \\text { if } x<0 \\\\
\n3, & \\text { if } x \\geq 0
\n\\end{array}\\right.\\) is continuous at x = 0.
\nSolution:
\n(i) Since f(x) is continuous at x = 5.
\n\u2234 \\(\\underset{x \\rightarrow 5}{\\mathrm{Lt}}\\) f(x) = f(5)
\n\u21d2 \\(\\underset{x \\rightarrow 5}{\\mathrm{Lt}}\\) \\(\\frac{x^2-25}{x-5}\\) = k
\n\u21d2 \\(\\underset{x \\rightarrow 5}{\\mathrm{Lt}}\\) \\(\\frac{(x-5)(x+5)}{x-5}\\) = k
\n\u21d2 \\(\\underset{x \\rightarrow 5}{\\mathrm{Lt}}\\) (x + 5) = kx
\n\u21d2 5 + 5 = k
\n\u21d2 k = 10
\nThus f is continuous at x = 5 if k = 10.<\/p>\n

(ii) L.H.L. = \\(\\underset{x \\rightarrow 5^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 5}{\\mathrm{Lt}}\\) kx + 1
\n= 5k + 1
\nR.H.L. = \\(\\underset{x \\rightarrow 5^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 5}{\\mathrm{Lt}}\\) 3x – 5
\n= 15 – 5
\n= 10
\n[as x \u2192 5
\n\u21d2 x > 5
\n\u2234 f(x) = 3x – 5]
\nalso f(5) = 5k + 1
\nSince f(x) is continuous at x = 5.
\n\u2234 L.H.L. = R.H.L. = f(5)
\n\u21d2 5k + 1 = 10
\n\u21d2 5k = 9
\n\u21d2 k = \\(\\frac{9}{5}\\).<\/p>\n

(iii) \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) 2x + 1
\n= 2 \u00d7 2 + 1 = 5
\n\\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) (3x – 1)
\n= 6 – 1 = 5
\nalso f(2) = 5
\nSince f is continuous at x = 2.
\n\u2234 \\(\\underset{x \\rightarrow 2^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 2^{+}}{\\mathrm{Lt}}\\) f(x) = f(2)
\nThus, 5 = k.<\/p>\n

\"ML<\/p>\n

(iv) L.H.L = \\(\\underset{x \\rightarrow \\pi^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow \\pi}{\\mathrm{Lt}}\\) kx + 1
\n= k\u03c0 + 1
\nR.H.L. = \\(\\underset{x \\rightarrow \\pi^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow \\pi}{\\mathrm{Lt}}\\) cos x
\n= cos \u03c0
\n= – 1
\nand f(\u03c0) = k\u03c0 + 1
\nSince f(x) is continuous at x = \u03c0
\n\u2234 L.H.L. = R.H.L. = f(\u03c0)
\n\u21d2 k\u03c0 + 1 = – 1
\n\u21d2 k = – \\(\\frac{2}{\\pi}\\).
\nThus f(x) is continuous at x = \u03c0 if k = – \\(\\frac{2}{\\pi}\\)<\/p>\n

(v) \"ML<\/p>\n

Also, f(0) = k
\nSince f is continuous at x = 0.
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = f(0)
\n\u21d2 \\(\\frac{5}{2}\\) = k<\/p>\n

(vi) Since f(x) is continuous at x = 0.
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = f(0)
\n\u21d2 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{1-\\cos 4 x}{8 x^2}\\) = k
\n\u21d2 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{2 \\sin ^2 2 x}{8 x^2}\\) = k
\n\u21d2 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{\\sin ^2 2 x}{(2 x)^2}\\) = k
\n\u21d2 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\left(\\frac{\\sin 2 x}{2 x}\\right)^2\\) = k
\n[\u2235 \\(\\underset{\\theta \\rightarrow 0}{\\mathrm{Lt}}\\) = \\(\\frac{\\sin \\theta}{\\theta}\\) = 1]
\nThus f(x) is continuous at x = 0 if k = 1.<\/p>\n

(vii) \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) \\(\\frac{k x}{|x|}\\)
\n= \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) \\(\\frac{k x}{-x}\\) = – k
\n[as x \u2192 0–<\/sup>
\n\u21d2 x < 0 \u21d2 |x| = – x] \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) 3 = 3
\nand f(0) = 3
\nSince f is continuous at x = 0
\n\u2234 \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\)
\nf(x) = f(0) – k = 3
\n\u21d2 k = – 3.<\/p>\n

\"ML<\/p>\n

Question 21.
\nFind the relation between a and b so that the function f defined by f(x) = \\(\\left\\{\\begin{array}{l} a x+1, \\text { if } x \\leq 3 \\\\ b x+3, \\text { if } x>3
\n\\end{array}\\right.\\) is continuous at x = 3.
\nSolution:
\nL.H.L. = \\(\\underset{x \\rightarrow 3^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) ax + 1
\n= 3a + 1
\nand R.H.L. = \\(\\underset{x \\rightarrow 3^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 3}{\\mathrm{Lt}}\\) bx + 3
\n= 3b + 3
\nalso f(3) = 3a + 1
\nSince f(x) is given to be continuous at x = 3.
\n\u2234 L.H.L. = R.H.L. = f(3)
\n\u21d2 3a + 1 = 3b + 3
\n\u21d2 3a – 3b = 2<\/p>\n

Question 22.
\nIf the function f is defined by f(x) = \\(\\left\\{\\begin{array}{ccc}
\n3 a x+b & \\text {, if } & x>1 \\\\
\n11 & \\text {, if } & x=1 \\\\
\n5 a x-2 b & \\text {, if } & x<1 \\end{array}\\right.\\) is continuous at x = 1, find the values of a and b. Solution: Given f(x) = \\(\\left\\{\\begin{array}{ccc} 3 a x+b & \\text {, if } & x>1 \\\\
\n11 & \\text {, if } & x=1 \\\\
\n5 a x-2 b & \\text {, if } & x<1
\n\\end{array}\\right.\\)
\n\u2234 \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) (3ax + b)
\nPut x = 1 + h
\nas x \u2192 1+<\/sup>
\n\u21d2 h \u2192 0+<\/sup>
\n= \\(\\underset{h \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) 3a (1 + h) + b
\n= 3a + b
\n\\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) (5ax – 2b)
\nPut x = 1 – h
\nas x \u2192 1–<\/sup>
\n\u21d2 h \u2192 0+<\/sup>
\n= \\(\\underset{h \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) 5a (1 – h) – 2b
\n= 5a – 2b
\nalso f(1) = 11
\nSince f(x) is continuous at x = 1
\n\u2234 \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x) = f(1)
\n\u2234 3a + b = 5a – 2b = 11
\ni.e. 3a+ b = 11 ………..(1)
\n5a – 2b = 11 …………..(2)
\nOn solving (1) and (2) ;
\n11a = 33
\n\u21d2 a = 3 ; b = 2.<\/p>\n

\"ML<\/p>\n

Question 23.
\nConsider the function defined as follows, for x \u2260 0. In each case, what choice (if any) of f(0) will make the function continuous at x = 0?
\n(i) f(x) = \\(\\frac{\\sqrt[3]{1+x}-1}{x}\\)
\n(ii) f(x) = \\(\\frac{5|x|-3 \\tan x+\\sin 2 x}{x}\\).
\nSolution:
\n\\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) \\(\\frac{\\sqrt[3]{1+x}-1}{x}\\)<\/p>\n

\"ML<\/p>\n

For f(x) is to be continuous at x = 0
\nwe have \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = f(0)
\ni.e. f(0) = \\(\\frac{1}{3}\\)<\/p>\n

(ii) Now, \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x)<\/p>\n

\"ML<\/p>\n

Thus, f is discontinuous at x = 0, irrespective of the choice of f(0).
\nHence, there is no choice of f(0) that makes f(x) is continuous at x = 0.<\/p>\n

\"ML<\/p>\n

Question 24.
\nExamine the function f(x) = \\(\\left\\{\\begin{array}{cc}
\nx \\sin \\frac{1}{x}, & x \\neq 0 \\\\
\n0 & , x=0
\n\\end{array}\\right.\\), for continuity at x = 0. (NCERT Exampler)
\nSolution:
\nLet g(x) = x
\nand h(x) = sin \\(\\frac{1}{x}\\)
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) g(x) = \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) x = 0
\nand |h(x)| = |sin \\(\\frac{1}{x}\\)| \u2264 1 \u2200 x \u2208 R, x \u2260 0
\n[\u2235 |sin t| \u2264 1 \u2200 x \u2208 R]
\ni.e. h (x) is bounded in the deleted neighbourhood of 0.
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) g(x) h(x) = 0
\n\u21d2 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) x sin \\(\\frac{1}{x}\\) = 0
\n\u21d2 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = 0, also f(0) = 0
\n[\u2235 f(x) = 0 at x = 0
\n\u2234 f(0) = 0]
\n\u2234 \\(\\underset{x \\rightarrow 0}{\\mathrm{Lt}}\\) f(x) = f(0)
\nThus, f is continuous at x = 0.<\/p>\n

Question 25.
\nShow that the functionf(x) = |x| + |x – 1|, x \u2208 R, is continuous both at x = 0 and x = 1. (NCERT Exampler)
\nSolution:
\nGiven f(x) = |x| + |x – 1|
\n\u2234 f(x) = \\(\\left\\{\\begin{array}{ccc}
\n-x-(x-1) & \\text { if } & x<0 \\\\
\nx-(x-1) & \\text { if } & 0 \\leq x<1 \\\\
\nx+x-1 & \\text { if } & x \\geq 1
\n\\end{array}\\right.\\)
\ni.e. f(x) = \\(\\left\\{\\begin{array}{ccc}
\n-2 x+1 & ; & x<0 \\\\
\n1 & ; & 0 \\leq x<1 \\\\
\n2 x-1 & ; & x \\geq 1
\n\\end{array}\\right.\\)<\/p>\n

Continuity at x = 0:
\nL.H.L. = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) (- 2x + 1)
\n= 0 + 1 = 1
\nand R.H.L.
\n= \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x)
\n= \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) 1 = 1
\nalso f(0) = 1
\n\u2234 \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x)
\n= f(0) = 1
\nThus, f(x) is continuous at x = 0.<\/p>\n

Continuity at x = 1 :
\n\\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) 1 = 1
\n\\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) 2x – 1
\n= 2 \u00d7 1 – 1 = 1
\nalso f(1) = 2 \u00d7 1 – 1 = 1
\n\u2234 \\(\\underset{x \\rightarrow 1^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 1^{+}}{\\mathrm{Lt}}\\) f(x)
\n= f(1) = 1
\nThus, f(x) is continuous at x = 1.<\/p>\n

\"ML<\/p>\n

Question 26.
\nExamine the function f(x) = \\(\\left\\{\\begin{array}{cc}
\ne^{1 \/ x}, & x \\neq 0 \\\\
\n1, & x=0
\n\\end{array}\\right.\\) for continuity at x = 0.
\nSolution:
\n\\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) e1\/x<\/sup>
\nPut x = 0 – h
\nas x \u2192 0–<\/sup>
\n\u21d2 h \u2192 0+<\/sup>
\n= \\(\\underset{h \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) e– 1\/h<\/sup>
\n= e-\u221e<\/sup>
\n= \\(\\frac{1}{\\infty}\\) = 0
\n\\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x) = \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) e1\/\ud835\udf0f<\/sup>
\n= e+ \u221e<\/sup> \u2192 + \u221e
\nThus \\(\\underset{x \\rightarrow 0^{-}}{\\mathrm{Lt}}\\) f(x) \u2260 \\(\\underset{x \\rightarrow 0^{+}}{\\mathrm{Lt}}\\) f(x)
\nHence function f(x) is discontinuous at x = 0.<\/p>\n","protected":false},"excerpt":{"rendered":"

Parents can use ISC Mathematics Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.1 to provide additional support to their children. ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.1 Question 1. Examine the following functions for continuity at the inclined points: (i) f(x) = at x = …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162701"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=162701"}],"version-history":[{"count":12,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162701\/revisions"}],"predecessor-version":[{"id":162716,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162701\/revisions\/162716"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=162701"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=162701"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=162701"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}