{"id":3790,"date":"2026-06-06T22:39:26","date_gmt":"2026-06-06T17:09:26","guid":{"rendered":"https:\/\/mymockmate.com\/notes\/?p=3790"},"modified":"2026-09-15T19:41:14","modified_gmt":"2026-09-15T14:11:14","slug":"ncert-class-12-maths-exercise-5-1-solutions-continuity","status":"publish","type":"post","link":"https:\/\/mymockmate.com\/notes\/ncert-class-12-maths-exercise-5-1-solutions-continuity\/","title":{"rendered":"NCERT Class 12 Maths Exercise 5.1 Solutions | Continuity"},"content":{"rendered":"
\"\"<\/a><\/div><\/div>\n

Short Introduction<\/h2>\n\n\n\n

Continuity is one of the most important concepts of Calculus. It helps us understand whether a function behaves smoothly around a point without any sudden breaks or jumps. In CBSE Class 12 Mathematics, Chapter 5 “Continuity and Differentiability” introduces students to the formal definition of continuity and its applications.<\/p>\n\n\n\n

Exercise 5.1 focuses on checking the continuity of polynomial, rational, modulus, piecewise and trigonometric functions at specific points.<\/p>\n\n\n\n

This complete solution guide by www.mymockmate.com<\/a><\/strong> provides easy-to-understand step-by-step explanations that are highly useful for CBSE Board Exams and competitive examinations.<\/p>\n\n\n\n

\n\n\n\n

Quick Information Box<\/h1>\n\n\n\n
Particular<\/th>Details<\/th><\/tr><\/thead>
Chapter<\/td>5<\/td><\/tr>
Chapter Name<\/td>Continuity and Differentiability<\/td><\/tr>
Exercise<\/td>5.1<\/td><\/tr>
Board<\/td>CBSE<\/td><\/tr>
Class<\/td>12<\/td><\/tr>
Difficulty Level<\/td>Easy to Moderate<\/td><\/tr>
Important For<\/td>Board Exams, JEE, CUET<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\n

Concepts Used (Topics Covered)<\/h1>\n\n\n\n

✔ Definition of Continuity<\/p>\n\n\n\n

✔ Left Hand Limit (LHL)<\/p>\n\n\n\n

✔ Right Hand Limit (RHL)<\/p>\n\n\n\n

Related Topic to Read more<\/span>Relations and Functions Exercise 1.1 Solutions Class 12<\/span><\/a><\/div>

✔ Existence of Function Value<\/p>\n\n\n\n

✔ Continuous Functions<\/p>\n\n\n\n

✔ Polynomial Functions<\/p>\n\n\n\n

✔ Rational Functions<\/p>\n\n\n\n

✔ Modulus Functions<\/p>\n\n\n\n

✔ Piecewise Functions<\/p>\n\n\n\n

Related Topic to Read more<\/span>Relations and Functions Exercise 1.2 Solutions Class 12<\/span><\/a><\/div>

✔ Trigonometric Functions<\/p>\n\n\n\n

✔ Greatest Integer Function<\/p>\n\n\n\n

✔ Composite Functions<\/p>

\"\"<\/a><\/div><\/div>\n\n\n\n
\n\n\n\n

Important Formulas<\/h1>\n\n\n\n

Definition of Continuity<\/h3>\n\n\n\n

A function f(x) is continuous at x = a if<\/p>\n\n\n\n

lim x→a f(x) = f(a)<\/p>\n\n\n\n

That is,<\/p>\n\n\n\n

Related Topic to Read more<\/span>Miscellaneous Exercise Solutions Class 12 Relations Functions<\/span><\/a><\/div>

LHL = RHL = f(a)<\/p>\n\n\n\n

\n\n\n\n

Polynomial Functions<\/h3>\n\n\n\n

Every polynomial function is continuous for all real numbers.<\/p>\n\n\n\n

\n\n\n\n

Rational Functions<\/h3>\n\n\n\n

A rational function is continuous wherever its denominator is not zero.<\/p>\n\n\n\n

\n\n\n\n

Modulus Function<\/h3>\n\n\n\n

f(x)=|x|<\/p>\n\n\n\n

is continuous for every real number.<\/p>\n\n\n\n

\n\n\n\n

Sum, Difference and Product Rule<\/h3>\n\n\n\n

If f(x) and g(x) are continuous at x=a, then<\/p>\n\n\n\n

    \n
  • f(x)+g(x)<\/li>\n\n\n\n
  • f(x)-g(x)<\/li>\n\n\n\n
  • f(x)×g(x)<\/li>\n<\/ul>\n\n\n\n

    are also continuous.<\/p>\n\n\n\n

    \n\n\n\n

    Question 1<\/h2>\n\n\n\n

    1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.<\/strong><\/p>\n\n\n\n

    Solution<\/h3>\n\n\n\n

    Since f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>5<\/mn>x<\/mi>−<\/mo>3<\/mn><\/mrow>f(x)=5x-3<\/annotation><\/semantics><\/math> is a polynomial function, it is continuous for every real number.<\/p>\n\n\n\n

    Still, we verify the three points.<\/p>\n\n\n\n

    At x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>5<\/mn>(<\/mo>0<\/mn>)<\/mo>−<\/mo>3<\/mn>=<\/mo>−<\/mo>3<\/mn><\/mrow>f(0)=5(0)-3=-3<\/annotation><\/semantics><\/math>lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>(<\/mo>5<\/mn>x<\/mi>−<\/mo>3<\/mn>)<\/mo>=<\/mo>−<\/mo>3<\/mn><\/mrow>\\lim_{x\\to0}f(x) =\\lim_{x\\to0}(5x-3) =-3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>f<\/mi>(<\/mo>0<\/mn>)<\/mo><\/mrow>\\lim_{x\\to0}f(x)=f(0)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hence, ff is continuous at x=0x=0.<\/strong><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>−<\/mo>3<\/mn><\/mrow>x=-3<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>−<\/mo>3<\/mn>)<\/mo>=<\/mo>5<\/mn>(<\/mo>−<\/mo>3<\/mn>)<\/mo>−<\/mo>3<\/mn>=<\/mo>−<\/mo>18<\/mn><\/mrow>f(-3)=5(-3)-3=-18<\/annotation><\/semantics><\/math>lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>−<\/mo>3<\/mn><\/mrow><\/munder>(<\/mo>5<\/mn>x<\/mi>−<\/mo>3<\/mn>)<\/mo>=<\/mo>−<\/mo>18<\/mn><\/mrow>\\lim_{x\\to-3}(5x-3)=-18<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Thus,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>−<\/mo>3<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>f<\/mi>(<\/mo>−<\/mo>3<\/mn>)<\/mo><\/mrow>\\lim_{x\\to-3}f(x)=f(-3)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hence, ff is continuous at x=−3x=-3.<\/strong><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>5<\/mn><\/mrow>x=5<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>5<\/mn>)<\/mo>=<\/mo>25<\/mn>−<\/mo>3<\/mn>=<\/mo>22<\/mn><\/mrow>f(5)=25-3=22<\/annotation><\/semantics><\/math>lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>5<\/mn><\/mrow><\/munder>(<\/mo>5<\/mn>x<\/mi>−<\/mo>3<\/mn>)<\/mo>=<\/mo>22<\/mn><\/mrow>\\lim_{x\\to5}(5x-3)=22<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore, ff is continuous at x=5x=5.<\/strong><\/p>\n\n\n\n

    Answer<\/h3>\n\n\n\n

    f(x)=5x−3 is continuous at 0,−3, and 5.​<\/p>\n\n\n\n

    \"image\"<\/figure>\n\n\n\n
    \n\n\n\n

    Question 2<\/h2>\n\n\n\n

    Examine the continuity of the function f(x) = 2×2 – 1 at x = 3.<\/strong><\/p>\n\n\n\n

    Solution<\/h3>\n\n\n\n

    First calculate f<\/mi>(<\/mo>3<\/mn>)<\/mo><\/mrow>f(3)<\/annotation><\/semantics><\/math>:f<\/mi>(<\/mo>3<\/mn>)<\/mo>=<\/mo>2<\/mn>(<\/mo>3<\/mn>)<\/mo>2<\/mn><\/msup>−<\/mo>1<\/mn>=<\/mo>18<\/mn>−<\/mo>1<\/mn>=<\/mo>17<\/mn><\/mrow>f(3)=2(3)^2-1=18-1=17<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Now,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>3<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>3<\/mn><\/mrow><\/munder>(<\/mo>2<\/mn>x<\/mi>2<\/mn><\/msup>−<\/mo>1<\/mn>)<\/mo>=<\/mo>2<\/mn>(<\/mo>3<\/mn>)<\/mo>2<\/mn><\/msup>−<\/mo>1<\/mn>=<\/mo>17<\/mn><\/mrow>\\lim_{x\\to3}f(x) =\\lim_{x\\to3}(2x^2-1) =2(3)^2-1 =17<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>3<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>f<\/mi>(<\/mo>3<\/mn>)<\/mo>=<\/mo>17<\/mn><\/mrow>\\lim_{x\\to3}f(x)=f(3)=17<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hence, f(x) is continuous at x=3.​<\/p>\n\n\n\n

    \"image\"<\/figure>\n\n\n\n

    Question 3<\/h2>\n\n\n\n

    Examine the following functions for continuity.<\/strong><\/p>\n\n\n\n

    (a) f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>−<\/mo>5<\/mn><\/mrow>f(x)=x-5<\/annotation><\/semantics><\/math><\/h2>\n\n\n\n

    This is a polynomial function.<\/p>\n\n\n\n

    For any a<\/mi>∈<\/mo>R<\/mi><\/mrow>a\\in\\mathbb R<\/annotation><\/semantics><\/math>,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>a<\/mi><\/mrow><\/munder>(<\/mo>x<\/mi>−<\/mo>5<\/mn>)<\/mo>=<\/mo>a<\/mi>−<\/mo>5<\/mn>=<\/mo>f<\/mi>(<\/mo>a<\/mi>)<\/mo><\/mrow>\\lim_{x\\to a}(x-5)=a-5=f(a)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>−<\/mo>5<\/mn> is continuous for all <\/mtext>x<\/mi>∈<\/mo>R<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f(x)=x-5\\text{ is continuous for all }x\\in\\mathbb R.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
    \n\n\n\n

    (b)<\/h2>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>1<\/mn>x<\/mi>−<\/mo>5<\/mn><\/mrow><\/mfrac>,<\/mo><\/mspace>x<\/mi>≠<\/mo>5<\/mn><\/mrow>f(x)=\\frac1{x-5},\\qquad x\\ne5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    The denominator is zero at x<\/mi>=<\/mo>5<\/mn><\/mrow>x=5<\/annotation><\/semantics><\/math>, so the function is not defined there.<\/p>\n\n\n\n

    For every a<\/mi>≠<\/mo>5<\/mn><\/mrow>a\\ne5<\/annotation><\/semantics><\/math>,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>a<\/mi><\/mrow><\/munder>1<\/mn>x<\/mi>−<\/mo>5<\/mn><\/mrow><\/mfrac>=<\/mo>1<\/mn>a<\/mi>−<\/mo>5<\/mn><\/mrow><\/mfrac>=<\/mo>f<\/mi>(<\/mo>a<\/mi>)<\/mo><\/mrow>\\lim_{x\\to a}\\frac1{x-5} =\\frac1{a-5} =f(a)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hence,f<\/mi> is continuous on <\/mtext>R<\/mi>−<\/mo>{<\/mo>5<\/mn>}<\/mo>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is continuous on }\\mathbb R-\\{5\\}.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    It is not defined<\/strong>, and hence discontinuous as an extended real-line function, at x<\/mi>=<\/mo>5<\/mn><\/mrow>x=5<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
    \n\n\n\n

    (c)<\/h2>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>2<\/mn><\/msup>−<\/mo>25<\/mn><\/mrow>x<\/mi>+<\/mo>5<\/mn><\/mrow><\/mfrac>,<\/mo><\/mspace>x<\/mi>≠<\/mo>−<\/mo>5<\/mn><\/mrow>f(x)=\\frac{x^2-25}{x+5},\\qquad x\\ne-5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Factor the numerator:x<\/mi>2<\/mn><\/msup>−<\/mo>25<\/mn>=<\/mo>(<\/mo>x<\/mi>−<\/mo>5<\/mn>)<\/mo>(<\/mo>x<\/mi>+<\/mo>5<\/mn>)<\/mo><\/mrow>x^2-25=(x-5)(x+5)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>(<\/mo>x<\/mi>−<\/mo>5<\/mn>)<\/mo>(<\/mo>x<\/mi>+<\/mo>5<\/mn>)<\/mo><\/mrow>x<\/mi>+<\/mo>5<\/mn><\/mrow><\/mfrac>=<\/mo>x<\/mi>−<\/mo>5<\/mn>,<\/mo><\/mspace>x<\/mi>≠<\/mo>−<\/mo>5<\/mn><\/mrow>f(x)=\\frac{(x-5)(x+5)}{x+5}=x-5,\\qquad x\\ne-5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Thus, for every x<\/mi>≠<\/mo>−<\/mo>5<\/mn><\/mrow>x\\ne-5<\/annotation><\/semantics><\/math>, f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>−<\/mo>5<\/mn><\/mrow>f(x)=x-5<\/annotation><\/semantics><\/math>, which is continuous.<\/p>\n\n\n\n

    Hence,f<\/mi> is continuous for every <\/mtext>x<\/mi>≠<\/mo>−<\/mo>5.<\/mn><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is continuous for every }x\\ne-5.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>−<\/mo>5<\/mn><\/mrow>x=-5<\/annotation><\/semantics><\/math>, the original function is not defined.<\/p>\n\n\n\n
    \n\n\n\n

    (d)<\/h2>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>∣<\/mi>x<\/mi>−<\/mo>5<\/mn>∣<\/mi><\/mrow>f(x)=|x-5|<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    The modulus function is continuous, and x<\/mi>−<\/mo>5<\/mn><\/mrow>x-5<\/annotation><\/semantics><\/math> is continuous.<\/p>\n\n\n\n

    Therefore, ∣x−5∣ is continuous for all real x.​<\/p>\n\n\n\n

    \"image\"<\/figure>\n\n\n\n
    \n\n\n\n

    Question 4<\/h2>\n\n\n\n

    Prove that the function f(x) = xn<\/sup> is continuous at x = n, where n is a positive integer.<\/strong><\/p>\n\n\n\n

    Solution<\/h3>\n\n\n\n

    We havef<\/mi>(<\/mo>n<\/mi>)<\/mo>=<\/mo>n<\/mi>n<\/mi><\/msup><\/mrow>f(n)=n^n<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Also,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>n<\/mi><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>n<\/mi><\/mrow><\/munder>x<\/mi>n<\/mi><\/msup>=<\/mo>n<\/mi>n<\/mi><\/msup><\/mrow>\\lim_{x\\to n}f(x) =\\lim_{x\\to n}x^n =n^n<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Thus,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>n<\/mi><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>f<\/mi>(<\/mo>n<\/mi>)<\/mo><\/mrow>\\lim_{x\\to n}f(x)=f(n)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>n<\/mi><\/msup> is continuous at <\/mtext>x<\/mi>=<\/mo>n<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f(x)=x^n\\text{ is continuous at }x=n.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    In fact, x<\/mi>n<\/mi><\/msup><\/mrow>x^n<\/annotation><\/semantics><\/math> is continuous for every real x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
    \"image\"<\/figure>\n\n\n\n
    \n\n\n\n
    \"image\"<\/figure>\n\n\n\n

    At x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>0<\/mn><\/mrow>f(0)=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Since the first rule applies near 0<\/mn><\/mrow>0<\/annotation><\/semantics><\/math>,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>x<\/mi>=<\/mo>0<\/mn><\/mrow>\\lim_{x\\to0}f(x)=\\lim_{x\\to0}x=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Thus,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>f<\/mi>(<\/mo>0<\/mn>)<\/mo><\/mrow>\\lim_{x\\to0}f(x)=f(0)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    So, continuous at x=0x=0.<\/strong><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>1<\/mn>)<\/mo>=<\/mo>1<\/mn><\/mrow>f(1)=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Left-hand limit:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>−<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>−<\/mo><\/msup><\/mrow><\/munder>x<\/mi>=<\/mo>1<\/mn><\/mrow>\\lim_{x\\to1^-}f(x)=\\lim_{x\\to1^-}x=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Right-hand limit:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>+<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>5<\/mn><\/mrow>\\lim_{x\\to1^+}f(x)=5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Since1<\/mn>≠<\/mo>5<\/mn><\/mrow>1\\ne5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    the function is not continuous at x=1x=1.<\/strong><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    For x<\/mi>><\/mo>1<\/mn><\/mrow>x>1<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>5<\/mn><\/mrow>f(x)=5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>2<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>5<\/mn>=<\/mo>f<\/mi>(<\/mo>2<\/mn>)<\/mo><\/mrow>\\lim_{x\\to2}f(x)=5=f(2)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hence, continuous at x=2x=2.<\/strong><\/p>\n\n\n\n

    Answer<\/h3>\n\n\n\n

    Continuous at x=0,2;discontinuous at x=1.​<\/p>\n\n\n\n

    \"image\"<\/figure>\n\n\n\n

    Find all points of discontinuity of f, where f is defined by<\/strong><\/p>\n\n\n\n

    Question 6<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>2<\/mn>x<\/mi>+<\/mo>3<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≤<\/mo>2<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>2<\/mn>x<\/mi>−<\/mo>3<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>><\/mo>2<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} 2x+3,&x\\le2\\\\ 2x-3,&x>2 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    The possible point of discontinuity is x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math>. f<\/mi>(<\/mo>2<\/mn>)<\/mo>=<\/mo>2<\/mn>(<\/mo>2<\/mn>)<\/mo>+<\/mo>3<\/mn>=<\/mo>7<\/mn><\/mrow>f(2)=2(2)+3=7<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>2<\/mn>−<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>(<\/mo>2<\/mn>)<\/mo>+<\/mo>3<\/mn>=<\/mo>7<\/mn><\/mrow>\\lim_{x\\to2^-}f(x)=2(2)+3=7<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>2<\/mn>+<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>2<\/mn>(<\/mo>2<\/mn>)<\/mo>−<\/mo>3<\/mn>=<\/mo>1<\/mn><\/mrow>\\lim_{x\\to2^+}f(x)=2(2)-3=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Since7<\/mn>≠<\/mo>1<\/mn><\/mrow>7\\ne1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    the function is discontinuous at x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Answer<\/h3>\n\n\n\n

    x<\/mi>=<\/mo>2<\/mn> is the only point of discontinuity.<\/mtext><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{x=2\\text{ is the only point of discontinuity.}}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
    \n\n\n\n

    Question 7<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>∣<\/mi>x<\/mi>∣<\/mi>+<\/mo>3<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≤<\/mo>−<\/mo>3<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>−<\/mo>2<\/mn>x<\/mi>,<\/mo><\/mrow><\/mstyle><\/mtd>−<\/mo>3<\/mn><<\/mo>x<\/mi><<\/mo>3<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>6<\/mn>x<\/mi>+<\/mo>2<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≥<\/mo>3<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} |x|+3,&x\\le-3\\\\ -2x,&-3<x<3\\\\ 6x+2,&x\\ge3 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Possible discontinuities occur at x<\/mi>=<\/mo>−<\/mo>3<\/mn><\/mrow>x=-3<\/annotation><\/semantics><\/math> and x<\/mi>=<\/mo>3<\/mn><\/mrow>x=3<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    At x<\/mi>=<\/mo>−<\/mo>3<\/mn><\/mrow>x=-3<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>−<\/mo>3<\/mn>)<\/mo>=<\/mo>∣<\/mi>−<\/mo>3<\/mn>∣<\/mi>+<\/mo>3<\/mn>=<\/mo>6<\/mn><\/mrow>f(-3)=|-3|+3=6<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>−<\/mo>3<\/mn>−<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>∣<\/mi>−<\/mo>3<\/mn>∣<\/mi>+<\/mo>3<\/mn>=<\/mo>6<\/mn><\/mrow>\\lim_{x\\to-3^-}f(x)=|-3|+3=6<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>−<\/mo>3<\/mn>+<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>−<\/mo>2<\/mn>(<\/mo>−<\/mo>3<\/mn>)<\/mo>=<\/mo>6<\/mn><\/mrow>\\lim_{x\\to-3^+}f(x)=-2(-3)=6<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Thus,LHL<\/mtext>=<\/mo>RHL<\/mtext>=<\/mo>f<\/mi>(<\/mo>−<\/mo>3<\/mn>)<\/mo>=<\/mo>6<\/mn><\/mrow>\\text{LHL}=\\text{RHL}=f(-3)=6<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    So f<\/mi><\/mrow>f<\/annotation><\/semantics><\/math> is continuous at −<\/mo>3<\/mn><\/mrow>-3<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    At x<\/mi>=<\/mo>3<\/mn><\/mrow>x=3<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>3<\/mn>)<\/mo>=<\/mo>6<\/mn>(<\/mo>3<\/mn>)<\/mo>+<\/mo>2<\/mn>=<\/mo>20<\/mn><\/mrow>f(3)=6(3)+2=20<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>3<\/mn>−<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>−<\/mo>2<\/mn>(<\/mo>3<\/mn>)<\/mo>=<\/mo>−<\/mo>6<\/mn><\/mrow>\\lim_{x\\to3^-}f(x)=-2(3)=-6<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>3<\/mn>+<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>6<\/mn>(<\/mo>3<\/mn>)<\/mo>+<\/mo>2<\/mn>=<\/mo>20<\/mn><\/mrow>\\lim_{x\\to3^+}f(x)=6(3)+2=20<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Since−<\/mo>6<\/mn>≠<\/mo>20<\/mn><\/mrow>-6\\ne20<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    f<\/mi><\/mrow>f<\/annotation><\/semantics><\/math> is discontinuous at x<\/mi>=<\/mo>3<\/mn><\/mrow>x=3<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Answer<\/h3>\n\n\n\n

    x<\/mi>=<\/mo>3<\/mn> is the only point of discontinuity.<\/mtext><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{x=3\\text{ is the only point of discontinuity.}}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
    \n\n\n\n

    Question 8<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>∣<\/mi>x<\/mi>∣<\/mi><\/mrow>x<\/mi><\/mfrac><\/mstyle>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≠<\/mo>0<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>0<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>=<\/mo>0<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} \\dfrac{|x|}{x},&x\\ne0\\\\ 0,&x=0 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math>:<\/p>\n\n\n\n

    For x<\/mi><<\/mo>0<\/mn><\/mrow>x<0<\/annotation><\/semantics><\/math>,∣<\/mi>x<\/mi>∣<\/mi><\/mrow>x<\/mi><\/mfrac>=<\/mo>−<\/mo>x<\/mi><\/mrow>x<\/mi><\/mfrac>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\frac{|x|}{x}=\\frac{-x}{x}=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>−<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0^-}f(x)=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    For x<\/mi>><\/mo>0<\/mn><\/mrow>x>0<\/annotation><\/semantics><\/math>,∣<\/mi>x<\/mi>∣<\/mi><\/mrow>x<\/mi><\/mfrac>=<\/mo>1<\/mn><\/mrow>\\frac{|x|}{x}=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>+<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0^+}f(x)=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Since−<\/mo>1<\/mn>≠<\/mo>1<\/mn><\/mrow>-1\\ne1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    the limit does not exist.<\/p>\n\n\n\n

    Hence,f<\/mi> is discontinuous at <\/mtext>x<\/mi>=<\/mo>0.<\/mn><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is discontinuous at }x=0.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    It is continuous for x<\/mi>≠<\/mo>0<\/mn><\/mrow>x\\ne0<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
    \n\n\n\n

    Question 9<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>x<\/mi>∣<\/mi>x<\/mi>∣<\/mi><\/mrow><\/mfrac><\/mstyle>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi><<\/mo>0<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>−<\/mo>1<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≥<\/mo>0<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} \\dfrac{x}{|x|},&x<0\\\\ -1,&x\\ge0 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>−<\/mo>1<\/mn><\/mrow>f(0)=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>−<\/mo><\/msup><\/mrow><\/munder>x<\/mi>∣<\/mi>x<\/mi>∣<\/mi><\/mrow><\/mfrac>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0^-}\\frac{x}{|x|} =-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>+<\/mo><\/msup><\/mrow><\/munder>(<\/mo>−<\/mo>1<\/mn>)<\/mo>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0^+}(-1)=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Thus,LHL<\/mtext>=<\/mo>RHL<\/mtext>=<\/mo>f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\text{LHL}=\\text{RHL}=f(0)=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,f<\/mi> is continuous at <\/mtext>x<\/mi>=<\/mo>0.<\/mn><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is continuous at }x=0.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Each part is continuous on its respective open interval, sof<\/mi> is continuous for all real <\/mtext>x<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is continuous for all real }x.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
    \n\n\n\n

    Question 10<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>x<\/mi>+<\/mo>1<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≥<\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi><<\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} x+1,&x\\ge1\\\\ x^2+1,&x<1 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>1<\/mn>)<\/mo>=<\/mo>1<\/mn>+<\/mo>1<\/mn>=<\/mo>2<\/mn><\/mrow>f(1)=1+1=2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>−<\/mo><\/msup><\/mrow><\/munder>(<\/mo>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn>)<\/mo>=<\/mo>2<\/mn><\/mrow>\\lim_{x\\to1^-}(x^2+1)=2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>+<\/mo><\/msup><\/mrow><\/munder>(<\/mo>x<\/mi>+<\/mo>1<\/mn>)<\/mo>=<\/mo>2<\/mn><\/mrow>\\lim_{x\\to1^+}(x+1)=2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,LHL<\/mtext>=<\/mo>RHL<\/mtext>=<\/mo>f<\/mi>(<\/mo>1<\/mn>)<\/mo>=<\/mo>2<\/mn><\/mrow>\\text{LHL}=\\text{RHL}=f(1)=2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hence,f<\/mi> is continuous for all real <\/mtext>x<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is continuous for all real }x.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
    \n\n\n\n

    Question 11<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>x<\/mi>3<\/mn><\/msup>−<\/mo>3<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≤<\/mo>2<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>><\/mo>2<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} x^3-3,&x\\le2\\\\ x^2+1,&x>2 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>2<\/mn>)<\/mo>=<\/mo>2<\/mn>3<\/mn><\/msup>−<\/mo>3<\/mn>=<\/mo>5<\/mn><\/mrow>f(2)=2^3-3=5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>2<\/mn>−<\/mo><\/msup><\/mrow><\/munder>(<\/mo>x<\/mi>3<\/mn><\/msup>−<\/mo>3<\/mn>)<\/mo>=<\/mo>5<\/mn><\/mrow>\\lim_{x\\to2^-}(x^3-3)=5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>2<\/mn>+<\/mo><\/msup><\/mrow><\/munder>(<\/mo>x<\/mi>2<\/mn><\/msup>+<\/mo>1<\/mn>)<\/mo>=<\/mo>5<\/mn><\/mrow>\\lim_{x\\to2^+}(x^2+1)=5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Thus,LHL<\/mtext>=<\/mo>RHL<\/mtext>=<\/mo>f<\/mi>(<\/mo>2<\/mn>)<\/mo>=<\/mo>5<\/mn><\/mrow>\\text{LHL}=\\text{RHL}=f(2)=5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,f<\/mi> is continuous for all real <\/mtext>x<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is continuous for all real }x.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
    \n\n\n\n

    Question 12<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>x<\/mi>10<\/mn><\/msup>−<\/mo>1<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≤<\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>x<\/mi>2<\/mn><\/msup>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>><\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} x^{10}-1,&x\\le1\\\\ x^2,&x>1 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>1<\/mn>)<\/mo>=<\/mo>1<\/mn>10<\/mn><\/msup>−<\/mo>1<\/mn>=<\/mo>0<\/mn><\/mrow>f(1)=1^{10}-1=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>−<\/mo><\/msup><\/mrow><\/munder>(<\/mo>x<\/mi>10<\/mn><\/msup>−<\/mo>1<\/mn>)<\/mo>=<\/mo>0<\/mn><\/mrow>\\lim_{x\\to1^-}(x^{10}-1)=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>+<\/mo><\/msup><\/mrow><\/munder>x<\/mi>2<\/mn><\/msup>=<\/mo>1<\/mn><\/mrow>\\lim_{x\\to1^+}x^2=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Since0<\/mn>≠<\/mo>1<\/mn><\/mrow>0\\ne1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    the function is discontinuous at x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Answer<\/h3>\n\n\n\n

    x=1 is the only point of discontinuity.​<\/p>\n\n\n\n

    \"image\"<\/figure>\n\n\n\n
    \"image\"<\/figure>\n\n\n\n

    At x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>1<\/mn>)<\/mo>=<\/mo>1<\/mn>+<\/mo>5<\/mn>=<\/mo>6<\/mn><\/mrow>f(1)=1+5=6<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>−<\/mo><\/msup><\/mrow><\/munder>(<\/mo>x<\/mi>+<\/mo>5<\/mn>)<\/mo>=<\/mo>6<\/mn><\/mrow>\\lim_{x\\to1^-}(x+5)=6<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>+<\/mo><\/msup><\/mrow><\/munder>(<\/mo>x<\/mi>−<\/mo>5<\/mn>)<\/mo>=<\/mo>−<\/mo>4<\/mn><\/mrow>\\lim_{x\\to1^+}(x-5)=-4<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Since6<\/mn>≠<\/mo>−<\/mo>4<\/mn><\/mrow>6\\ne-4<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    the function is discontinuous at x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Answer<\/h3>\n\n\n\n

    The function is not continuous.<\/mtext><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\text{The function is not continuous.}}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    The exercise asks this continuity check explicitly.<\/p>\n\n\n\n

    \"image\"<\/figure>\n\n\n\n

    Discuss the continuity of the function f, where f is defined by<\/strong><\/p>\n\n\n\n

    Question 14<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>3<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>0<\/mn>≤<\/mo>x<\/mi>≤<\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>4<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>1<\/mn><<\/mo>x<\/mi><<\/mo>3<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>5<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>3<\/mn>≤<\/mo>x<\/mi>≤<\/mo>10<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} 3,&0\\le x\\le1\\\\ 4,&1<x<3\\\\ 5,&3\\le x\\le10 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Possible points are x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math> and x<\/mi>=<\/mo>3<\/mn><\/mrow>x=3<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    At x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>1<\/mn>)<\/mo>=<\/mo>3<\/mn><\/mrow>f(1)=3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL =<\/mo>3<\/mn><\/mrow>=3<\/annotation><\/semantics><\/math>, while RHL =<\/mo>4<\/mn><\/mrow>=4<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Therefore, discontinuous at x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    At x<\/mi>=<\/mo>3<\/mn><\/mrow>x=3<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>3<\/mn>)<\/mo>=<\/mo>5<\/mn><\/mrow>f(3)=5<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL =<\/mo>4<\/mn><\/mrow>=4<\/annotation><\/semantics><\/math>, while RHL =<\/mo>5<\/mn><\/mrow>=5<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Therefore, discontinuous at x<\/mi>=<\/mo>3<\/mn><\/mrow>x=3<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Answer<\/h3>\n\n\n\n

    Points of discontinuity: <\/mtext>x<\/mi>=<\/mo>1<\/mn>,<\/mo>3.<\/mn><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\text{Points of discontinuity: }x=1,3.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
    \n\n\n\n

    Question 15<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>2<\/mn>x<\/mi>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi><<\/mo>0<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>0<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>0<\/mn>≤<\/mo>x<\/mi>≤<\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>4<\/mn>x<\/mi>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>><\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} 2x,&x<0\\\\ 0,&0\\le x\\le1\\\\ 4x,&x>1 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Possible points: 0<\/mn>,<\/mo>1<\/mn><\/mrow>0,1<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    At x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>0<\/mn><\/mrow>f(0)=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>−<\/mo><\/msup><\/mrow><\/munder>2<\/mn>x<\/mi>=<\/mo>0<\/mn><\/mrow>\\lim_{x\\to0^-}2x=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>+<\/mo><\/msup><\/mrow><\/munder>0<\/mn>=<\/mo>0<\/mn><\/mrow>\\lim_{x\\to0^+}0=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Thus, continuous at 0<\/mn><\/mrow>0<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    At x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>1<\/mn>)<\/mo>=<\/mo>0<\/mn><\/mrow>f(1)=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>−<\/mo><\/msup><\/mrow><\/munder>0<\/mn>=<\/mo>0<\/mn><\/mrow>\\lim_{x\\to1^-}0=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>1<\/mn>+<\/mo><\/msup><\/mrow><\/munder>4<\/mn>x<\/mi>=<\/mo>4<\/mn><\/mrow>\\lim_{x\\to1^+}4x=4<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Since 0<\/mn>≠<\/mo>4<\/mn><\/mrow>0\\ne4<\/annotation><\/semantics><\/math>,x<\/mi>=<\/mo>1<\/mn> is a point of discontinuity.<\/mtext><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{x=1\\text{ is a point of discontinuity.}}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
    \n\n\n\n

    Question 16<\/h1>\n\n\n\n

    f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>−<\/mo>2<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≤<\/mo>−<\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>2<\/mn>x<\/mi>,<\/mo><\/mrow><\/mstyle><\/mtd>−<\/mo>1<\/mn><<\/mo>x<\/mi>≤<\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>2<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>><\/mo>1<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} -2,&x\\le-1\\\\ 2x,&-1<x\\le1\\\\ 2,&x>1 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>−<\/mo>1<\/mn><\/mrow>x=-1<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>−<\/mo>1<\/mn>)<\/mo>=<\/mo>−<\/mo>2<\/mn><\/mrow>f(-1)=-2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:−<\/mo>2<\/mn><\/mrow>-2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:2<\/mn>(<\/mo>−<\/mo>1<\/mn>)<\/mo>=<\/mo>−<\/mo>2<\/mn><\/mrow>2(-1)=-2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hence continuous at −<\/mo>1<\/mn><\/mrow>-1<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    At x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

    f<\/mi>(<\/mo>1<\/mn>)<\/mo>=<\/mo>2<\/mn><\/mrow>f(1)=2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:2<\/mn>(<\/mo>1<\/mn>)<\/mo>=<\/mo>2<\/mn><\/mrow>2(1)=2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:2<\/mn><\/mrow>2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hence continuous at 1<\/mn><\/mrow>1<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Therefore, f is continuous for all real x.​<\/p>\n\n\n\n

    \"image\"<\/figure>\n\n\n\n
    \"image\"<\/figure>\n\n\n\n

    For continuity at x<\/mi>=<\/mo>3<\/mn><\/mrow>x=3<\/annotation><\/semantics><\/math>,LHL<\/mtext>=<\/mo>RHL<\/mtext>=<\/mo>f<\/mi>(<\/mo>3<\/mn>)<\/mo><\/mrow>\\text{LHL}=\\text{RHL}=f(3)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:3<\/mn>a<\/mi>+<\/mo>1<\/mn><\/mrow>3a+1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:3<\/mn>b<\/mi>+<\/mo>3<\/mn><\/mrow>3b+3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Therefore,3<\/mn>a<\/mi>+<\/mo>1<\/mn>=<\/mo>3<\/mn>b<\/mi>+<\/mo>3<\/mn><\/mrow>3a+1=3b+3<\/annotation><\/semantics><\/math>3<\/mn>a<\/mi>−<\/mo>3<\/mn>b<\/mi>=<\/mo>2<\/mn><\/mrow>3a-3b=2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Hence,a<\/mi>−<\/mo>b<\/mi>=<\/mo>2<\/mn>3<\/mn><\/mfrac><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{a-b=\\frac23}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    or a=b+2\/3​​<\/p>\n\n\n\n

    \"image\"<\/figure>\n\n\n\n
    \"image\"<\/figure>\n\n\n\n

    At x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math><\/h2>\n\n\n\n

    f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>λ<\/mi>(<\/mo>0<\/mn>−<\/mo>0<\/mn>)<\/mo>=<\/mo>0<\/mn><\/mrow>f(0)=\\lambda(0-0)=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>−<\/mo><\/msup><\/mrow><\/munder>λ<\/mi>(<\/mo>x<\/mi>2<\/mn><\/msup>−<\/mo>2<\/mn>x<\/mi>)<\/mo>=<\/mo>0<\/mn><\/mrow>\\lim_{x\\to0^-}\\lambda(x^2-2x)=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>+<\/mo><\/msup><\/mrow><\/munder>(<\/mo>4<\/mn>x<\/mi>+<\/mo>λ<\/mi>)<\/mo>=<\/mo>λ<\/mi><\/mrow>\\lim_{x\\to0^+}(4x+\\lambda)=\\lambda<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    For continuity,0<\/mn>=<\/mo>λ<\/mi><\/mrow>0=\\lambda<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Thus,λ<\/mi>=<\/mo>0<\/mn><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\lambda=0}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    At x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math><\/h2>\n\n\n\n

    Since 1<\/mn>><\/mo>0<\/mn><\/mrow>1>0<\/annotation><\/semantics><\/math>, near x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>4<\/mn>x<\/mi>+<\/mo>λ<\/mi><\/mrow>f(x)=4x+\\lambda<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    which is a linear function and therefore continuous.<\/p>\n\n\n\n

    Hence,f<\/mi> is continuous at <\/mtext>x<\/mi>=<\/mo>1<\/mn> for every <\/mtext>λ<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is continuous at }x=1\\text{ for every }\\lambda.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    For the value required for continuity at x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math>, λ<\/mi>=<\/mo>0<\/mn><\/mrow>\\lambda=0<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n
    \"image\"<\/figure>\n\n\n\n
      \n
    1. Show that the function defined by g(x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.<\/strong><\/li>\n<\/ol>\n\n\n\n

      Let n<\/mi><\/mrow>n<\/annotation><\/semantics><\/math> be any integer.<\/p>\n\n\n\n

      At x<\/mi>=<\/mo>n<\/mi><\/mrow>x=n<\/annotation><\/semantics><\/math>,[<\/mo>n<\/mi>]<\/mo>=<\/mo>n<\/mi><\/mrow>[n]=n<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

      Therefore,g<\/mi>(<\/mo>n<\/mi>)<\/mo>=<\/mo>n<\/mi>−<\/mo>n<\/mi>=<\/mo>0<\/mn><\/mrow>g(n)=n-n=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

      For x<\/mi>→<\/mo>n<\/mi>−<\/mo><\/msup><\/mrow>x\\to n^-<\/annotation><\/semantics><\/math>,[<\/mo>x<\/mi>]<\/mo>=<\/mo>n<\/mi>−<\/mo>1<\/mn><\/mrow>[x]=n-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

      Hence,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>n<\/mi>−<\/mo><\/msup><\/mrow><\/munder>g<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>n<\/mi>−<\/mo><\/msup><\/mrow><\/munder>[<\/mo>x<\/mi>−<\/mo>(<\/mo>n<\/mi>−<\/mo>1<\/mn>)<\/mo>]<\/mo>=<\/mo>1<\/mn><\/mrow>\\lim_{x\\to n^-}g(x) =\\lim_{x\\to n^-}[x-(n-1)] =1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

      For x<\/mi>→<\/mo>n<\/mi>+<\/mo><\/msup><\/mrow>x\\to n^+<\/annotation><\/semantics><\/math>,[<\/mo>x<\/mi>]<\/mo>=<\/mo>n<\/mi><\/mrow>[x]=n<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

      Therefore,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>n<\/mi>+<\/mo><\/msup><\/mrow><\/munder>g<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>0<\/mn><\/mrow>\\lim_{x\\to n^+}g(x)=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

      Thus,LHL<\/mtext>=<\/mo>1<\/mn>,<\/mo><\/mspace>RHL<\/mtext>=<\/mo>0<\/mn><\/mrow>\\text{LHL}=1,\\qquad \\text{RHL}=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

      Since LHL ≠<\/mo><\/mrow>\\ne<\/annotation><\/semantics><\/math> RHL, g(x) is discontinuous at every integral point.<\/p>\n\n\n\n
      \"image\"<\/figure>\n\n\n\n

      20. Is the function defined by f(x) = x2<\/sup> – sin x + 5 continuous at x = π?<\/strong><\/p>\n\n\n\n

      The functions x<\/mi>2<\/mn><\/msup><\/mrow>x^2<\/annotation><\/semantics><\/math>, sin<\/mi>⁡<\/mo>x<\/mi><\/mrow>\\sin x<\/annotation><\/semantics><\/math>, and the constant function 5<\/mn><\/mrow>5<\/annotation><\/semantics><\/math> are continuous everywhere.<\/p>\n\n\n\n

      Therefore, their sum\/difference is also continuous.<\/p>\n\n\n\n

      Hence,f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>2<\/mn><\/msup>−<\/mo>sin<\/mi>⁡<\/mo>x<\/mi>+<\/mo>5<\/mn> is continuous at <\/mtext>x<\/mi>=<\/mo>π<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f(x)=x^2-\\sin x+5\\text{ is continuous at }x=\\pi.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
      \"image\"<\/figure>\n\n\n\n
        \n
      1. Discuss the continuity of the following functions:
        (a) f(x) = sin x + cos x
        (b) f(x) = sin x – cos x
        (c) f(x) = sin x . cos x<\/strong><\/li>\n<\/ol>\n\n\n\n

        (a) f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>sin<\/mi>⁡<\/mo>x<\/mi>+<\/mo>cos<\/mi>⁡<\/mo>x<\/mi><\/mrow>f(x)=\\sin x+\\cos x<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

        Both sin<\/mi>⁡<\/mo>x<\/mi><\/mrow>\\sin x<\/annotation><\/semantics><\/math> and cos<\/mi>⁡<\/mo>x<\/mi><\/mrow>\\cos x<\/annotation><\/semantics><\/math> are continuous for all real x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        Therefore,sin<\/mi>⁡<\/mo>x<\/mi>+<\/mo>cos<\/mi>⁡<\/mo>x<\/mi> is continuous on <\/mtext>R<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\sin x+\\cos x\\text{ is continuous on }\\mathbb R.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        (b) f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>sin<\/mi>⁡<\/mo>x<\/mi>−<\/mo>cos<\/mi>⁡<\/mo>x<\/mi><\/mrow>f(x)=\\sin x-\\cos x<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

        Difference of two continuous functions is continuous.sin<\/mi>⁡<\/mo>x<\/mi>−<\/mo>cos<\/mi>⁡<\/mo>x<\/mi> is continuous on <\/mtext>R<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\sin x-\\cos x\\text{ is continuous on }\\mathbb R.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        (c) f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>sin<\/mi>⁡<\/mo>x<\/mi>⋅<\/mo>cos<\/mi>⁡<\/mo>x<\/mi><\/mrow>f(x)=\\sin x\\cdot\\cos x<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

        Product of continuous functions is continuous.sin<\/mi>⁡<\/mo>x<\/mi>cos<\/mi>⁡<\/mo>x<\/mi> is continuous on <\/mtext>R<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\sin x\\cos x\\text{ is continuous on }\\mathbb R.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        The three functions are specified in the exercise statement.<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n

        22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.<\/strong><\/p>\n\n\n\n

        Discuss the continuity of cosine, cosecant, secant and cotangent functions.<\/p>\n\n\n\n

        1. Cosine function<\/h3>\n\n\n\n

        f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>cos<\/mi>⁡<\/mo>x<\/mi><\/mrow>f(x)=\\cos x<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Cosine is continuous for every real number.cos<\/mi>⁡<\/mo>x<\/mi> is continuous on <\/mtext>R<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\cos x\\text{ is continuous on }\\mathbb R}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        2. Cosecant function<\/h3>\n\n\n\n

        csc<\/mi>⁡<\/mo>x<\/mi>=<\/mo>1<\/mn>sin<\/mi>⁡<\/mo>x<\/mi><\/mrow><\/mfrac><\/mrow>\\csc x=\\frac1{\\sin x}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        It is continuous wherever sin<\/mi>⁡<\/mo>x<\/mi>≠<\/mo>0<\/mn><\/mrow>\\sin x\\ne0<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        Sincesin<\/mi>⁡<\/mo>x<\/mi>=<\/mo>0<\/mn><\/mspace>at <\/mtext>x<\/mi>=<\/mo>n<\/mi>π<\/mi><\/mrow>\\sin x=0\\quad\\text{at }x=n\\pi<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        we havecsc<\/mi>⁡<\/mo>x<\/mi> is continuous for <\/mtext>x<\/mi>≠<\/mo>n<\/mi>π<\/mi>,<\/mo> <\/mtext>n<\/mi>∈<\/mo>Z<\/mi>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\csc x\\text{ is continuous for }x\\ne n\\pi,\\ n\\in\\mathbb Z.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        3. Secant function<\/h3>\n\n\n\n

        sec<\/mi>⁡<\/mo>x<\/mi>=<\/mo>1<\/mn>cos<\/mi>⁡<\/mo>x<\/mi><\/mrow><\/mfrac><\/mrow>\\sec x=\\frac1{\\cos x}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        It is continuous wherever cos<\/mi>⁡<\/mo>x<\/mi>≠<\/mo>0<\/mn><\/mrow>\\cos x\\ne0<\/annotation><\/semantics><\/math>.cos<\/mi>⁡<\/mo>x<\/mi>=<\/mo>0<\/mn><\/mrow>\\cos x=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        atx<\/mi>=<\/mo>(<\/mo>2<\/mn>n<\/mi>+<\/mo>1<\/mn>)<\/mo>π<\/mi><\/mrow>2<\/mn><\/mfrac><\/mrow>x=\\frac{(2n+1)\\pi}{2}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Therefore,sec<\/mi>⁡<\/mo>x<\/mi> is continuous for <\/mtext>x<\/mi>≠<\/mo>(<\/mo>2<\/mn>n<\/mi>+<\/mo>1<\/mn>)<\/mo>π<\/mi><\/mrow>2<\/mn><\/mfrac>.<\/mi><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{\\sec x\\text{ is continuous for }x\\ne\\frac{(2n+1)\\pi}{2}.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        4. Cotangent function<\/h3>\n\n\n\n

        cot<\/mi>⁡<\/mo>x<\/mi>=<\/mo>cos<\/mi>⁡<\/mo>x<\/mi><\/mrow>sin<\/mi>⁡<\/mo>x<\/mi><\/mrow><\/mfrac><\/mrow>\\cot x=\\frac{\\cos x}{\\sin x}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        It is continuous wherever sin<\/mi>⁡<\/mo>x<\/mi>≠<\/mo>0<\/mn><\/mrow>\\sin x\\ne0<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        Therefore, cotx is continuous for x Not = nπ.<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n
        \"image\"<\/figure>\n\n\n\n

        LHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>−<\/mo><\/msup><\/mrow><\/munder>sin<\/mi>⁡<\/mo>x<\/mi><\/mrow>x<\/mi><\/mfrac>=<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0^-}\\frac{\\sin x}{x}=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        RHL:lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>+<\/mo><\/msup><\/mrow><\/munder>(<\/mo>x<\/mi>+<\/mo>1<\/mn>)<\/mo>=<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0^+}(x+1)=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Also,f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>0<\/mn>+<\/mo>1<\/mn>=<\/mo>1<\/mn><\/mrow>f(0)=0+1=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Therefore,LHL<\/mtext>=<\/mo>RHL<\/mtext>=<\/mo>f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>1<\/mn><\/mrow>\\text{LHL}=\\text{RHL}=f(0)=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Hence, the function is continuous at 0<\/mn><\/mrow>0<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        Answer<\/h3>\n\n\n\n

        There are no points of discontinuity.​<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n
        \"image\"<\/figure>\n\n\n\n

        At x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math>,−<\/mo>1<\/mn>≤<\/mo>sin<\/mi>⁡<\/mo>1<\/mn>x<\/mi><\/mfrac>≤<\/mo>1<\/mn><\/mrow>-1\\le\\sin\\frac1x\\le1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Multiplying by x<\/mi>2<\/mn><\/msup>≥<\/mo>0<\/mn><\/mrow>x^2\\ge0<\/annotation><\/semantics><\/math>,−<\/mo>x<\/mi>2<\/mn><\/msup>≤<\/mo>x<\/mi>2<\/mn><\/msup>sin<\/mi>⁡<\/mo>1<\/mn>x<\/mi><\/mfrac>≤<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow>-x^2\\le x^2\\sin\\frac1x\\le x^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        As x<\/mi>→<\/mo>0<\/mn><\/mrow>x\\to0<\/annotation><\/semantics><\/math>,−<\/mo>x<\/mi>2<\/mn><\/msup>→<\/mo>0<\/mn>,<\/mo><\/mspace>x<\/mi>2<\/mn><\/msup>→<\/mo>0<\/mn><\/mrow>-x^2\\to0,\\qquad x^2\\to0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Hence, by the squeeze theorem,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>x<\/mi>2<\/mn><\/msup>sin<\/mi>⁡<\/mo>1<\/mn>x<\/mi><\/mfrac>=<\/mo>0<\/mn><\/mrow>\\lim_{x\\to0}x^2\\sin\\frac1x=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Sincef<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>0<\/mn><\/mrow>f(0)=0<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        we getlim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>f<\/mi>(<\/mo>0<\/mn>)<\/mo><\/mrow>\\lim_{x\\to0}f(x)=f(0)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Therefore,f<\/mi> is continuous at <\/mtext>x<\/mi>=<\/mo>0.<\/mn><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is continuous at }x=0.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        It is also continuous for x<\/mi>≠<\/mo>0<\/mn><\/mrow>x\\ne0<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        Thus, f is continuous on R.​<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n
        \"image\"<\/figure>\n\n\n\n

        At x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>−<\/mo>1<\/mn><\/mrow>f(0)=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Also,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>(<\/mo>sin<\/mi>⁡<\/mo>x<\/mi>−<\/mo>cos<\/mi>⁡<\/mo>x<\/mi>)<\/mo>=<\/mo>sin<\/mi>⁡<\/mo>0<\/mn>−<\/mo>cos<\/mi>⁡<\/mo>0<\/mn>=<\/mo>0<\/mn>−<\/mo>1<\/mn>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0}(\\sin x-\\cos x) =\\sin0-\\cos0 =0-1=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Therefore,lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0}f(x)=f(0)=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Hence,f<\/mi> is continuous at <\/mtext>x<\/mi>=<\/mo>0.<\/mn><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{f\\text{ is continuous at }x=0.}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Since sin<\/mi>⁡<\/mo>x<\/mi>−<\/mo>cos<\/mi>⁡<\/mo>x<\/mi><\/mrow>\\sin x-\\cos x<\/annotation><\/semantics><\/math> is continuous elsewhere, f is continuous for all real x.<\/p>\n\n\n\n
        \"image\"<\/figure>\n\n\n\n

        Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.<\/strong><\/p>\n\n\n\n

        Question 26<\/h1>\n\n\n\n

        f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>k<\/mi>cos<\/mi>⁡<\/mo>x<\/mi><\/mrow>π<\/mi>−<\/mo>2<\/mn>x<\/mi><\/mrow><\/mfrac><\/mstyle>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≠<\/mo>π<\/mi>2<\/mn><\/mfrac><\/mrow><\/mstyle><\/mtd><\/mtr>3<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>=<\/mo>π<\/mi>2<\/mn><\/mfrac><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} \\dfrac{k\\cos x}{\\pi-2x},&x\\ne\\frac\\pi2\\\\ 3,&x=\\frac\\pi2 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Find k<\/mi><\/mrow>k<\/annotation><\/semantics><\/math> for continuity at x<\/mi>=<\/mo>π<\/mi>2<\/mn><\/mfrac><\/mrow>x=\\frac\\pi2<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        We requirelim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>π<\/mi>\/<\/mi>2<\/mn><\/mrow><\/munder>k<\/mi>cos<\/mi>⁡<\/mo>x<\/mi><\/mrow>π<\/mi>−<\/mo>2<\/mn>x<\/mi><\/mrow><\/mfrac>=<\/mo>3<\/mn><\/mrow>\\lim_{x\\to\\pi\/2}\\frac{k\\cos x}{\\pi-2x}=3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Put x<\/mi>=<\/mo>π<\/mi>2<\/mn><\/mfrac>+<\/mo>h<\/mi><\/mrow>x=\\frac\\pi2+h<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        Thencos<\/mi>⁡<\/mo>(<\/mo>π<\/mi>2<\/mn><\/mfrac>+<\/mo>h<\/mi>)<\/mo><\/mrow>=<\/mo>−<\/mo>sin<\/mi>⁡<\/mo>h<\/mi><\/mrow>\\cos\\left(\\frac\\pi2+h\\right)=-\\sin h<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        andπ<\/mi>−<\/mo>2<\/mn>(<\/mo>π<\/mi>2<\/mn><\/mfrac>+<\/mo>h<\/mi>)<\/mo><\/mrow>=<\/mo>−<\/mo>2<\/mn>h<\/mi><\/mrow>\\pi-2\\left(\\frac\\pi2+h\\right)=-2h<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Therefore,lim<\/mi>⁡<\/mo><\/mrow>h<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>k<\/mi>(<\/mo>−<\/mo>sin<\/mi>⁡<\/mo>h<\/mi>)<\/mo><\/mrow>−<\/mo>2<\/mn>h<\/mi><\/mrow><\/mfrac>=<\/mo>k<\/mi>2<\/mn><\/mfrac>lim<\/mi>⁡<\/mo><\/mrow>h<\/mi>→<\/mo>0<\/mn><\/mrow><\/munder>sin<\/mi>⁡<\/mo>h<\/mi><\/mrow>h<\/mi><\/mfrac>=<\/mo>k<\/mi>2<\/mn><\/mfrac><\/mrow>\\lim_{h\\to0}\\frac{k(-\\sin h)}{-2h} = \\frac{k}{2}\\lim_{h\\to0}\\frac{\\sin h}{h} =\\frac{k}{2}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        For continuity,k<\/mi>2<\/mn><\/mfrac>=<\/mo>3<\/mn><\/mrow>\\frac{k}{2}=3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Hence,k<\/mi>=<\/mo>6<\/mn><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{k=6}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
        \n\n\n\n

        Question 27<\/h1>\n\n\n\n

        f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>k<\/mi>x<\/mi>2<\/mn><\/msup>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≤<\/mo>2<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>3<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>><\/mo>2<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} kx^2,&x\\le2\\\\ 3,&x>2 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        For continuity at x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>2<\/mn>)<\/mo>=<\/mo>4<\/mn>k<\/mi><\/mrow>f(2)=4k<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        LHL:4<\/mn>k<\/mi><\/mrow>4k<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        RHL:3<\/mn><\/mrow>3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Thus,4<\/mn>k<\/mi>=<\/mo>3<\/mn><\/mrow>4k=3<\/annotation><\/semantics><\/math>k<\/mi>=<\/mo>3<\/mn>4<\/mn><\/mfrac><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{k=\\frac34}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
        \n\n\n\n

        Question 28<\/h1>\n\n\n\n

        f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>k<\/mi>x<\/mi>+<\/mo>1<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≤<\/mo>π<\/mi><\/mrow><\/mstyle><\/mtd><\/mtr>cos<\/mi>⁡<\/mo>x<\/mi>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>><\/mo>π<\/mi><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} kx+1,&x\\le\\pi\\\\ \\cos x,&x>\\pi \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        At x<\/mi>=<\/mo>π<\/mi><\/mrow>x=\\pi<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>π<\/mi>)<\/mo>=<\/mo>k<\/mi>π<\/mi>+<\/mo>1<\/mn><\/mrow>f(\\pi)=k\\pi+1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        LHL:k<\/mi>π<\/mi>+<\/mo>1<\/mn><\/mrow>k\\pi+1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        RHL:cos<\/mi>⁡<\/mo>π<\/mi>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\cos\\pi=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        For continuity,k<\/mi>π<\/mi>+<\/mo>1<\/mn>=<\/mo>−<\/mo>1<\/mn><\/mrow>k\\pi+1=-1<\/annotation><\/semantics><\/math>k<\/mi>π<\/mi>=<\/mo>−<\/mo>2<\/mn><\/mrow>k\\pi=-2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Hence,k<\/mi>=<\/mo>−<\/mo>2<\/mn>π<\/mi><\/mfrac><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{k=-\\frac2\\pi}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n
        \n\n\n\n

        Question 29<\/h1>\n\n\n\n

        f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>{<\/mo>k<\/mi>x<\/mi>+<\/mo>1<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>≤<\/mo>5<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr>3<\/mn>x<\/mi>−<\/mo>5<\/mn>,<\/mo><\/mrow><\/mstyle><\/mtd>x<\/mi>><\/mo>5<\/mn><\/mrow><\/mstyle><\/mtd><\/mtr><\/mtable><\/mrow><\/mrow>f(x)= \\begin{cases} kx+1,&x\\le5\\\\ 3x-5,&x>5 \\end{cases}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        At x<\/mi>=<\/mo>5<\/mn><\/mrow>x=5<\/annotation><\/semantics><\/math>,f<\/mi>(<\/mo>5<\/mn>)<\/mo>=<\/mo>5<\/mn>k<\/mi>+<\/mo>1<\/mn><\/mrow>f(5)=5k+1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        RHL:3<\/mn>(<\/mo>5<\/mn>)<\/mo>−<\/mo>5<\/mn>=<\/mo>10<\/mn><\/mrow>3(5)-5=10<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        For continuity,5<\/mn>k<\/mi>+<\/mo>1<\/mn>=<\/mo>10<\/mn><\/mrow>5k+1=10<\/annotation><\/semantics><\/math>5<\/mn>k<\/mi>=<\/mo>9<\/mn><\/mrow>5k=9<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Therefore, k=9\/5​​<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n
        \"image\"<\/figure>\n\n\n\n

        For continuity at x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math>,<\/mtd>2<\/mn>a<\/mi>+<\/mo>b<\/mi>=<\/mo>5<\/mn><\/mrow><\/mtd><\/mtd>(1)<\/mtext><\/mtd><\/mtr><\/mtable>2a+b=5 \\tag{1}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        For continuity at x<\/mi>=<\/mo>10<\/mn><\/mrow>x=10<\/annotation><\/semantics><\/math>,<\/mtd>10<\/mn>a<\/mi>+<\/mo>b<\/mi>=<\/mo>21<\/mn><\/mrow><\/mtd><\/mtd>(2)<\/mtext><\/mtd><\/mtr><\/mtable>10a+b=21 \\tag{2}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Subtract (1) from (2):8<\/mn>a<\/mi>=<\/mo>16<\/mn><\/mrow>8a=16<\/annotation><\/semantics><\/math>a<\/mi>=<\/mo>2<\/mn><\/mrow>a=2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Substitute into (1):2<\/mn>(<\/mo>2<\/mn>)<\/mo>+<\/mo>b<\/mi>=<\/mo>5<\/mn><\/mrow>2(2)+b=5<\/annotation><\/semantics><\/math>b<\/mi>=<\/mo>1<\/mn><\/mrow>b=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Hence,a<\/mi>=<\/mo>2<\/mn>,<\/mo><\/mspace>b<\/mi>=<\/mo>1<\/mn><\/mrow><\/mstyle><\/mstyle><\/mstyle><\/menclose><\/mrow>\\boxed{a=2,\\qquad b=1}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        The original exercise asks for these values so that the entire piecewise function is continuous.<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n

        31. Show that the function defined by f(x) = cos (x2<\/sup>) is a continuous function.<\/strong><\/p>\n\n\n\n

        We know:x<\/mi>2<\/mn><\/msup><\/mrow>x^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        is continuous for every real x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math>, andcos<\/mi>⁡<\/mo>x<\/mi><\/mrow>\\cos x<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        is continuous for every real x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        The composition of continuous functions is continuous.<\/p>\n\n\n\n

        Therefore, cos(x2) is continuous for all x∈R.​<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n

        32. Show that the function defined by f(x) = |cos x| is a continuous function.<\/strong><\/p>\n\n\n\n

        We know thatcos<\/mi>⁡<\/mo>x<\/mi><\/mrow>\\cos x<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        is continuous for every real x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        Also, the modulus function∣<\/mi>x<\/mi>∣<\/mi><\/mrow>|x|<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        is continuous.<\/p>\n\n\n\n

        Therefore, the composition∣<\/mi>cos<\/mi>⁡<\/mo>x<\/mi>∣<\/mi><\/mrow>|\\cos x|<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        is continuous.<\/p>\n\n\n\n

        Hence, ∣cosx∣ is continuous for all real x.​<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n

        33. Examine that sin |x| is a continuous function.<\/strong><\/p>\n\n\n\n

        The function∣<\/mi>x<\/mi>∣<\/mi><\/mrow>|x|<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        is continuous for every real x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        The functionsin<\/mi>⁡<\/mo>x<\/mi><\/mrow>\\sin x<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        is also continuous for every real x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        Therefore, their compositionsin<\/mi>⁡<\/mo>∣<\/mi>x<\/mi>∣<\/mi><\/mrow>\\sin|x|<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        is continuous.<\/p>\n\n\n\n

        Hence, sin∣x∣ is continuous for all x∈R.​<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n

        34. Find all the points of discontinuity of f defined by f(x) = |x| – |x + 1|.<\/strong><\/p>\n\n\n\n

        The possible points are where the expressions inside the modulus become zero:x<\/mi>=<\/mo>0<\/mn>,<\/mo><\/mspace>x<\/mi>=<\/mo>−<\/mo>1<\/mn><\/mrow>x=0,\\qquad x=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        We examine these points.<\/p>\n\n\n\n

        For x<\/mi><<\/mo>−<\/mo>1<\/mn><\/mrow>x<-1<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

        ∣<\/mi>x<\/mi>∣<\/mi>=<\/mo>−<\/mo>x<\/mi>,<\/mo><\/mspace>∣<\/mi>x<\/mi>+<\/mo>1<\/mn>∣<\/mi>=<\/mo>−<\/mo>(<\/mo>x<\/mi>+<\/mo>1<\/mn>)<\/mo><\/mrow>|x|=-x,\\qquad |x+1|=-(x+1)<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Thus,f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>−<\/mo>x<\/mi>+<\/mo>(<\/mo>x<\/mi>+<\/mo>1<\/mn>)<\/mo>=<\/mo>1<\/mn><\/mrow>f(x)=-x+(x+1)=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        For −<\/mo>1<\/mn>≤<\/mo>x<\/mi><<\/mo>0<\/mn><\/mrow>-1\\le x<0<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

        ∣<\/mi>x<\/mi>∣<\/mi>=<\/mo>−<\/mo>x<\/mi>,<\/mo><\/mspace>∣<\/mi>x<\/mi>+<\/mo>1<\/mn>∣<\/mi>=<\/mo>x<\/mi>+<\/mo>1<\/mn><\/mrow>|x|=-x,\\qquad |x+1|=x+1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Therefore,f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>−<\/mo>x<\/mi>−<\/mo>(<\/mo>x<\/mi>+<\/mo>1<\/mn>)<\/mo>=<\/mo>−<\/mo>2<\/mn>x<\/mi>−<\/mo>1<\/mn><\/mrow>f(x)=-x-(x+1)=-2x-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        For x<\/mi>≥<\/mo>0<\/mn><\/mrow>x\\ge0<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

        ∣<\/mi>x<\/mi>∣<\/mi>=<\/mo>x<\/mi>,<\/mo><\/mspace>∣<\/mi>x<\/mi>+<\/mo>1<\/mn>∣<\/mi>=<\/mo>x<\/mi>+<\/mo>1<\/mn><\/mrow>|x|=x,\\qquad |x+1|=x+1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Hence,f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>−<\/mo>(<\/mo>x<\/mi>+<\/mo>1<\/mn>)<\/mo>=<\/mo>−<\/mo>1<\/mn><\/mrow>f(x)=x-(x+1)=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Now check the critical points.<\/p>\n\n\n\n

        At x<\/mi>=<\/mo>−<\/mo>1<\/mn><\/mrow>x=-1<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

        lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>−<\/mo>1<\/mn>−<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>1<\/mn><\/mrow>\\lim_{x\\to-1^-}f(x)=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        andlim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>−<\/mo>1<\/mn>+<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>−<\/mo>2<\/mn>(<\/mo>−<\/mo>1<\/mn>)<\/mo>−<\/mo>1<\/mn>=<\/mo>1<\/mn><\/mrow>\\lim_{x\\to-1^+}f(x) =-2(-1)-1=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Also,f<\/mi>(<\/mo>−<\/mo>1<\/mn>)<\/mo>=<\/mo>1<\/mn><\/mrow>f(-1)=1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Therefore, f<\/mi><\/mrow>f<\/annotation><\/semantics><\/math> is continuous at −<\/mo>1<\/mn><\/mrow>-1<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        At x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math><\/h3>\n\n\n\n

        lim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>−<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>−<\/mo>2<\/mn>(<\/mo>0<\/mn>)<\/mo>−<\/mo>1<\/mn>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0^-}f(x)=-2(0)-1=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        andlim<\/mi>⁡<\/mo><\/mrow>x<\/mi>→<\/mo>0<\/mn>+<\/mo><\/msup><\/mrow><\/munder>f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>−<\/mo>1<\/mn><\/mrow>\\lim_{x\\to0^+}f(x)=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Also,f<\/mi>(<\/mo>0<\/mn>)<\/mo>=<\/mo>−<\/mo>1<\/mn><\/mrow>f(0)=-1<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

        Therefore, f<\/mi><\/mrow>f<\/annotation><\/semantics><\/math> is continuous at 0<\/mn><\/mrow>0<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

        Final Answer<\/h3>\n\n\n\n

        There are no points of discontinuity<\/strong>. f(x)=∣x∣−∣x+1∣ is continuous for every real x.​<\/p>\n\n\n\n

        \"image\"<\/figure>\n\n\n\n

        Quick Answer Key<\/h1>\n\n\n\n
        Q.<\/th>Answer<\/th><\/tr><\/thead>
        1<\/td>Continuous at 0<\/mn>,<\/mo>−<\/mo>3<\/mn>,<\/mo>5<\/mn><\/mrow>0,-3,5<\/annotation><\/semantics><\/math><\/td><\/tr>
        2<\/td>Continuous at x<\/mi>=<\/mo>3<\/mn><\/mrow>x=3<\/annotation><\/semantics><\/math><\/td><\/tr>
        3(a)<\/td>Continuous on R<\/mi><\/mrow>\\mathbb R<\/annotation><\/semantics><\/math><\/td><\/tr>
        3(b)<\/td>Continuous on R<\/mi>−<\/mo>{<\/mo>5<\/mn>}<\/mo><\/mrow>\\mathbb R-\\{5\\}<\/annotation><\/semantics><\/math><\/td><\/tr>
        3(c)<\/td>Continuous on R<\/mi>−<\/mo>{<\/mo>−<\/mo>5<\/mn>}<\/mo><\/mrow>\\mathbb R-\\{-5\\}<\/annotation><\/semantics><\/math><\/td><\/tr>
        3(d)<\/td>Continuous on R<\/mi><\/mrow>\\mathbb R<\/annotation><\/semantics><\/math><\/td><\/tr>
        4<\/td>Continuous at x<\/mi>=<\/mo>n<\/mi><\/mrow>x=n<\/annotation><\/semantics><\/math><\/td><\/tr>
        5<\/td>Discontinuous at x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math><\/td><\/tr>
        6<\/td>Discontinuous at x<\/mi>=<\/mo>2<\/mn><\/mrow>x=2<\/annotation><\/semantics><\/math><\/td><\/tr>
        7<\/td>Discontinuous at x<\/mi>=<\/mo>3<\/mn><\/mrow>x=3<\/annotation><\/semantics><\/math><\/td><\/tr>
        8<\/td>Discontinuous at x<\/mi>=<\/mo>0<\/mn><\/mrow>x=0<\/annotation><\/semantics><\/math><\/td><\/tr>
        9<\/td>Continuous everywhere<\/td><\/tr>
        10<\/td>Continuous everywhere<\/td><\/tr>
        11<\/td>Continuous everywhere<\/td><\/tr>
        12<\/td>Discontinuous at x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math><\/td><\/tr>
        13<\/td>Discontinuous at x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math><\/td><\/tr>
        14<\/td>Discontinuous at x<\/mi>=<\/mo>1<\/mn>,<\/mo>3<\/mn><\/mrow>x=1,3<\/annotation><\/semantics><\/math><\/td><\/tr>
        15<\/td>Discontinuous at x<\/mi>=<\/mo>1<\/mn><\/mrow>x=1<\/annotation><\/semantics><\/math><\/td><\/tr>
        16<\/td>Continuous everywhere<\/td><\/tr>
        17<\/td>a<\/mi>−<\/mo>b<\/mi>=<\/mo>2<\/mn>3<\/mn><\/mfrac><\/mrow>a-b=\\frac23<\/annotation><\/semantics><\/math><\/td><\/tr>
        18<\/td>λ<\/mi>=<\/mo>0<\/mn><\/mrow>\\lambda=0<\/annotation><\/semantics><\/math> for continuity at 0<\/mn><\/mrow>0<\/annotation><\/semantics><\/math>; continuous at 1<\/mn><\/mrow>1<\/annotation><\/semantics><\/math> for every λ<\/mi><\/mrow>\\lambda<\/annotation><\/semantics><\/math><\/td><\/tr>
        19<\/td>Discontinuous at every integer<\/td><\/tr>
        20<\/td>Continuous at x<\/mi>=<\/mo>π<\/mi><\/mrow>x=\\pi<\/annotation><\/semantics><\/math><\/td><\/tr>
        21<\/td>All three continuous on R<\/mi><\/mrow>\\mathbb R<\/annotation><\/semantics><\/math><\/td><\/tr>
        22<\/td>Continuous on their respective domains<\/td><\/tr>
        23<\/td>No point of discontinuity<\/td><\/tr>
        24<\/td>Continuous everywhere<\/td><\/tr>
        25<\/td>Continuous everywhere<\/td><\/tr>
        26<\/td>k<\/mi>=<\/mo>6<\/mn><\/mrow>k=6<\/annotation><\/semantics><\/math><\/td><\/tr>
        27<\/td>k<\/mi>=<\/mo>3<\/mn>4<\/mn><\/mfrac><\/mrow>k=\\frac34<\/annotation><\/semantics><\/math><\/td><\/tr>
        28<\/td>k<\/mi>=<\/mo>−<\/mo>2<\/mn>π<\/mi><\/mfrac><\/mrow>k=-\\frac2\\pi<\/annotation><\/semantics><\/math><\/td><\/tr>
        29<\/td>k<\/mi>=<\/mo>9<\/mn>5<\/mn><\/mfrac><\/mrow>k=\\frac95<\/annotation><\/semantics><\/math><\/td><\/tr>
        30<\/td>a<\/mi>=<\/mo>2<\/mn>,<\/mo> <\/mtext>b<\/mi>=<\/mo>1<\/mn><\/mrow>a=2,\\ b=1<\/annotation><\/semantics><\/math><\/td><\/tr>
        31<\/td>Continuous everywhere<\/td><\/tr>
        32<\/td>Continuous everywhere<\/td><\/tr>
        33<\/td>Continuous everywhere<\/td><\/tr>
        34<\/td>No point of discontinuity<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

        Common Mistakes<\/h1>\n\n\n\n

        ❌ Forgetting to check whether the function is defined.<\/p>\n\n\n\n

        ❌ Assuming every rational function is continuous everywhere.<\/p>\n\n\n\n

        ❌ Not comparing LHL and RHL in piecewise functions.<\/p>\n\n\n\n

        ❌ Ignoring denominator equal to zero.<\/p>\n\n\n\n

        ❌ Missing modulus function properties.<\/p>\n\n\n\n

        \n\n\n\n

        Exam Tips<\/h1>\n\n\n\n

        ✅ Always write:<\/p>\n\n\n\n

        Function Value<\/p>\n\n\n\n

        ↓<\/p>\n\n\n\n

        LHL<\/p>\n\n\n\n

        ↓<\/p>\n\n\n\n

        RHL<\/p>\n\n\n\n

        ↓<\/p>\n\n\n\n

        Conclusion<\/p>\n\n\n\n

        ✅ Remember:<\/p>\n\n\n\n

        Polynomial → Continuous Everywhere<\/p>\n\n\n\n

        Rational → Continuous except denominator = 0<\/p>\n\n\n\n

        Modulus → Continuous Everywhere<\/p>\n\n\n\n

        Greatest Integer → Discontinuous at Integers<\/p>\n\n\n\n

        \n\n\n\n

        Practice MCQs<\/h1>\n\n\n\n

        1.<\/h3>\n\n\n\n

        Which function is continuous everywhere?<\/p>\n\n\n\n

        A. 1\/x<\/p>\n\n\n\n

        B. |x|<\/p>\n\n\n\n

        C. 1\/(x−2)<\/p>\n\n\n\n

        D. [x]<\/p>\n\n\n\n

        Answer: B<\/p>\n\n\n\n

        \n\n\n\n

        2.<\/h3>\n\n\n\n

        The function 1\/(x−3) is discontinuous at<\/p>\n\n\n\n

        A. 1<\/p>\n\n\n\n

        B. 2<\/p>\n\n\n\n

        C. 3<\/p>\n\n\n\n

        D. 4<\/p>\n\n\n\n

        Answer: C<\/p>\n\n\n\n

        \n\n\n\n

        3.<\/h3>\n\n\n\n

        Every polynomial function is<\/p>\n\n\n\n

        A. Discontinuous<\/p>\n\n\n\n

        B. Continuous<\/p>\n\n\n\n

        C. Undefined<\/p>\n\n\n\n

        D. None<\/p>\n\n\n\n

        Answer: B<\/p>\n\n\n\n

        \n\n\n\n

        4.<\/h3>\n\n\n\n

        The greatest integer function is discontinuous at<\/p>\n\n\n\n

        A. Rational numbers<\/p>\n\n\n\n

        B. Irrational numbers<\/p>\n\n\n\n

        C. Integers<\/p>\n\n\n\n

        D. Positive numbers<\/p>\n\n\n\n

        Answer: C<\/p>\n\n\n\n

        \n\n\n\n

        FAQ Section<\/h1>\n\n\n\n

        Q1. What is continuity?<\/h3>\n\n\n\n

        A function is continuous if<\/p>\n\n\n\n

        LHL = RHL = Function Value.<\/p>\n\n\n\n

        \n\n\n\n

        Q2. Are polynomial functions always continuous?<\/h3>\n\n\n\n

        Yes. Every polynomial function is continuous for all real numbers.<\/p>\n\n\n\n

        \n\n\n\n

        Q3. Where are rational functions discontinuous?<\/h3>\n\n\n\n

        At points where the denominator becomes zero.<\/p>\n\n\n\n

        \n\n\n\n

        Q4. Is modulus function continuous?<\/h3>\n\n\n\n

        Yes, |x| is continuous everywhere.<\/p>\n\n\n\n

        \n\n\n\n

        Q5. Which questions are most important for CBSE Boards?<\/h3>\n\n\n\n

        Piecewise functions, rational functions and greatest integer functions.<\/p>\n\n\n\n

        \n\n\n\n

        Prepare smarter with detailed NCERT solutions, chapter-wise mock tests, previous year questions and All India Rank based practice tests only at:<\/p>\n\n\n\n

        www.mymockmate.com<\/a><\/h2>\n\n\n\n

        Mock Test Makes Preparation Easy!<\/strong><\/p>\n

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