Continuity and Differentiability Class 12 Mathematics Extra Question

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Continuity and Differentiability Class 12 Mathematics Extra Question. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in myCBSEguide website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 5 Continuity and Differentiability Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.

Class 12 Chapter 5 Maths Extra Questions

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CBSE Continuity and Differentiability Class 12 Maths

Chapter 5 Continuity and Differentiability


  1. Let f (x + y) = f(x) + f(y) {tex}\forall {/tex} x, y {tex} \in {\mathbf{R}}{/tex}. Suppose that f (6) = 5 and f ‘ (0) = 1, then f ‘ (6) is equal to

    1. 1
    2. 30
    3. None of these
    4. 25
  2. Derivative of log|x| w.r.t. |x| is

    1. None of these
    2. {tex}\frac{1}{x}{/tex}
    3. {tex} \pm \frac{1}{x}{/tex}
    4. {tex}\frac{1}{{\left| x \right|}}{/tex}
  3. The function f (x) = 1 + |sin x| is

    1. differentiable everywhere
    2. continuous everywhere
    3. differentiable nowhere
    4. continuous nowhere
  4. {tex}\mathop {Lt}\limits_{x \to 0} \;\;\frac{{1 – \cos x}}{{x\sin x}}{/tex} is equal to

    1. 1
    2. 2
    3. 0
    4. {tex}\frac{1}{2}{/tex}
  5. {tex}\mathop {Lt}\limits_{x \to \pi /4} \;\;\;\frac{{\cos x – \sin x}}{{x – \frac{\pi }{4}}}{/tex} is equal to

    1. {tex} – \frac{2}{{\sqrt 2 }}{/tex}
    2. -1
    3. {tex} – \frac{1}{{\sqrt 2 }}{/tex}
    4. {tex}\frac{2}{{\sqrt 2 }}{/tex}
  6. The value of c in Mean value theorem for the function f(x) = x(x – 2), x {tex}\in{/tex}[1, 2] is ________.
  7. The set of points where the function f given by f(x) = |2x – 1| sin x is differentiable is ________.
  8. Differential coefficient of sec (tan-1x) w.r.t. x is ________.
  9. Discuss the continuity of the function {tex}f(x) = \sin x.\cos x{/tex}.

  10. Determine the value of ‘k’ for which the following function is continuous at x = 3 : f(x) = {tex}\left\{ \begin{array} { l } { \frac { ( x + 3 ) ^ { 2 } – 36 } { x – 3 } , x \neq 3 } \\ { k \quad , x = 3 } \end{array} \right.{/tex}.

  11. Determine the value of the constant ‘k’ so that the function f(x) = {tex}\left\{ \begin{array} { l l } { \frac { k x } { | x | } , } & { \text { if } x < 0 } \\ { 3 , } & { \text { if } x \geq 0 } \end{array} \right.{/tex} is continuous at x= 0.

  12. Find {tex}\frac{{dy}}{{dx}},{/tex} {tex}y = {\cos ^{ – 1}}\left( {\frac{{1 – {x^2}}}{{1 + {x^2}}}} \right),0 < x < 1{/tex}

  13. Show that the function defined by {tex}f\left( x \right) = \cos \left( {{x^2}} \right){/tex} is a continuous function.

  14. Determine if f defined by {tex}f\left( x \right) = \left\{ \begin{gathered} {x^2}\sin \frac{1}{x},if\,x \ne 0 \hfill \\ 0,\,if\,\,x = 0 \hfill \\ \end{gathered} \right.{/tex} is a continuous function.

  15. Find the value of k so that the following function is continuous at x = 2.
    f(x) = {tex} \left\{ {\begin{array}{*{20}{c}} {\frac{{{x^3} + {x^2} – 16x + 20}}{{{{(x – 2)}^2}}},}&{x \ne 2} \\ {k,}&{x = 2} \end{array}} \right\}{/tex}

  16. If xy + yx = ab, then find {tex}\frac { dy } { d x }{/tex}.

  17. If ey(x + 1) = 1, then show that {tex}\frac { d ^ { 2 } y } { d x ^ { 2 } } = \left( \frac { d y } { d x } \right) ^ { 2 }{/tex}.

  18. Find {tex}\frac{{dy}}{{dx}}{/tex} if {tex}{y^x} + {x^y} + {x^x} = {a^b}{/tex}.

Chapter 5 Continuity and Differentiability


Solution

    1. 1
      Explanation: {tex}\begin{gathered} f'(6) = \mathop {\lim }\limits_{h \to 0} \frac{{f(6 + h) – f(6)}}{h}\end{gathered} {/tex} = {tex}\mathop {\lim }\limits_{h \to 0} \frac{{f(6 + h) – f(6 + 0)}}{h}{/tex}
      {tex} = \mathop {\lim }\limits_{h \to 0} \frac{{f(6) + f(h) – \left\{ {f(6) + f(0)} \right\}}}{h}{/tex}
      {tex} = \mathop {\lim }\limits_{h \to 0} \frac{{f(h) – f(0)}}{h} = f'(0) = 1{/tex}
    1. {tex}\frac{1}{{\left| x \right|}}{/tex}
      Explanation: {tex}\frac{d}{{d\left| x \right|}}\left( {\log \left| x \right|} \right) = \frac{1}{{\left| x \right|}}{/tex}
    1. continuous everywhere
      Explanation: f(x) = 1 + |sinx| is not derivable at those x for which x for which sinx = 0, however, 1 + |sinx| is continuous everywhere (being the sum of two continuous functions)
    1. {tex}\frac{1}{2}{/tex}
      Explanation: {tex}\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{x\sin x}} {/tex}{tex} = \mathop {\lim }\limits_{x \to 0} \frac{{1 – {{\cos }^2}x}}{{x\sin x(1 + \cos x)}}\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}.\frac{1}{{1 + \cos x}} = 1.\frac{1}{{1 + 1}} = \frac{1}{2}{/tex}
    1. {tex} – \frac{2}{{\sqrt 2 }}{/tex}
      Explanation: {tex}\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\cos x – \sin x}}{{x – \frac{\pi }{4}}}{/tex} {tex} = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{ – \sin x – \cos x}}{1}= – \sin \frac{\pi }{4} – \cos \frac{\pi }{4}= – \frac{2}{{\sqrt 2 }} = – \sqrt 2 {/tex}
  1. {tex}\frac{3}{2}{/tex}
  2. R – {tex}\left\{\frac12\right\}{/tex}
  3. {tex}\frac x{\sqrt{1+x^2}}{/tex}
  4. Since sin x and cos x are continuous functions and product of two continuous function is a continuous function, therefore {tex}f(x) = \sin x.\cos x{/tex} is a continuous function.
  5. Given, f(x) = {tex}\left\{ \begin{array} { l l } { \frac { ( x + 3 ) ^ { 2 } – 36 } { x – 3 } , } & { x \neq 3 } \\ { x , } & { x = 3 } \end{array} \right.{/tex}
    We shall use definition of continuity to find the value of k.
    If f(x) is continuous at x = 3,
    Then, we have {tex}\mathop {\lim }\limits_{x \to 3} f ( x ) = f ( 3 ){/tex}
    {tex}\Rightarrow \quad \mathop {\lim }\limits_{x \to 3} \frac { ( x + 3 ) ^ { 2 } – 36 } { x – 3 } = k{/tex}
    {tex}\Rightarrow \quad \mathop {\lim }\limits_{x \to 3} \frac { ( x + 3 ) ^ { 2 } – 6 ^ { 2 } } { x – 3 } = k{/tex}
    {tex}\Rightarrow \mathop {\lim }\limits_{x \to 3} \frac { ( x + 3 – 6 ) ( x + 3 + 6 ) } { x – 3 } = k{/tex} [ {tex}\because{/tex} a2 – b2 = (a – b)(a + b)] {tex}\Rightarrow \quad \mathop {\lim }\limits_{x \to 3} \frac { ( x – 3 ) ( x + 9 ) } { ( x – 3 ) } = k{/tex}
    {tex}\Rightarrow \quad \mathop {\lim }\limits_{x \to 3} ( x + 9 ) = k{/tex}
    {tex}\Rightarrow{/tex} 3 + 9 = k {tex}\Rightarrow{/tex} k = 12
  6. Let f(x) ={tex}\left\{ \begin{array} { l l } { \frac { k x } { | x | } , } & { \text { if } x < 0 } \\ { 3 , } & { \text { if } x \geq 0 } \end{array} \right.{/tex} be continuous at x = 0
    Then,{tex}\mathop {\lim }\limits_{ x \rightarrow 0 ^ { + } } f ( x ) = \mathop {\lim }\limits_{ x \rightarrow 0 ^ { – } } f ( x ) = f ( 0 ){/tex}
    {tex}\Rightarrow \quad \mathop {\lim }\limits_{ h \rightarrow 0 } f ( 0 + h ) = \mathop {\lim }\limits_{ h \rightarrow 0 } f ( 0 – h ) = f ( 0 ){/tex}
    {tex}\Rightarrow \quad 3 = \mathop {\lim }\limits_{ h \rightarrow 0 } \frac { k ( – h ) } { |- h | } = 3{/tex}
    {tex}\Rightarrow \quad \mathop {\lim }\limits_{ h \rightarrow 0 } \left( \frac { – k h } { h } \right) = 3{/tex}
    {tex}\mathop {\lim }\limits_{ h \rightarrow 0 } ( – k ) = 3{/tex}
    {tex}\therefore{/tex} k = – 3
  7. Given: {tex}y = {\cos ^{ – 1}}\left( {\frac{{1 – {x^2}}}{{1 + {x^2}}}} \right),0 < x < 1{/tex}
    Putting {tex}x = \tan \theta{/tex}
    {tex}y = {\cos ^{ – 1}}\left( {\frac{{1 – {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right){/tex}
    {tex}= {\cos ^{ – 1}}\left( {\cos 2\theta } \right) = 2\theta = 2{\tan ^{ – 1}}x{/tex}
    {tex}\therefore \frac{{dy}}{{dx}} = 2.\frac{1}{{1 + {x^2}}} = \frac{2}{{1 + {x^2}}}{/tex}
  8. Let {tex}f\left( x \right) = {x^2}{/tex} and {tex}g\left( x \right) = \cos x{/tex}, then
    {tex}\left( {gof} \right)\left( x \right) = g\left[ {f\left( x \right)} \right] = g\left( {{x^2}} \right) = \cos {x^2}{/tex}
    Now f and g being continuous it follows that their composite (gof) is continuous.
    Hence {tex}\cos {x^2}{/tex} is continuous function.
  9. Here, {tex}\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {x^2}\sin \frac{1}{x} = 0{/tex} x a finite quantity = 0
    {tex}\left[ {\because \sin \frac{1}{x}{\text{lies between – 1 and 1}}} \right]{/tex}
    Also f(0) = 0
    Since, {tex}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right){/tex} therefore, the function f is continuous at x = 0.
    Also,when {tex}x\ne0{/tex} ,then f(x) is the product of two continuous functions and hence Continuous.Hence,f(x) is continuous everywhere.
  10. According to the question, f(x) = {tex} \left\{ {\begin{array}{*{20}{c}} {\frac{{{x^3} + {x^2} – 16x + 20}}{{{{(x – 2)}^2}}},}&{x \ne 2} \\ {k,}&{x = 2} \end{array}} \right\}{/tex} is continuous at {tex}x = 2.{/tex}
    Now, we have f(2) = k
    {tex} \mathop {\lim }\limits_{ x \rightarrow 2 } f ( x ) = \mathop {\lim }\limits_{ x \rightarrow 2 } \frac { x ^ { 3 } + x ^ { 2 } – 16 x + 20 } { ( x – 2 ) ^ { 2 } }{/tex}
    {tex} = \mathop {\lim }\limits_{ x \rightarrow 2 } \frac { ( x – 2 ) \left( x ^ { 2 } + 3 x – 10 \right) } { ( x – 2 ) ^ { 2 } }{/tex}
    {tex} = \mathop {\lim }\limits_{ x \rightarrow 2 } \frac { ( x – 2 ) ( x + 5 ) ( x – 2 ) } { ( x – 2 ) ^ { 2 } }{/tex}
    {tex} = \mathop {\lim }\limits_{ x \rightarrow 2 } ( x + 5 ){/tex} = 2+ 5 = 7
    f(x) is continuous at x = 2.
    {tex} \therefore \mathop {\lim }\limits_{ x \rightarrow 2 } f ( x ){/tex}= f(2) {tex} \Rightarrow{/tex} 7 = k{tex} \Rightarrow{/tex}k = 7
  11. We have, xy + yx = ab………(i)
    Let xy = v and yx = u……(ii)
    Therefore,on putting these values in Eq. (i), we get,
    v + u = ab
    Therefore,on differentiating both sides w.r.t. x, we get,
    {tex}\frac { d v } { d x } + \frac { d u } { d x } = 0{/tex}……..(iii)
    Now consider, xy = v [ from Eq.(ii)] Therefore,on taking log both sides, we get,
    log xy = logv
    {tex}\Rightarrow{/tex} y log x = log v
    Therefore,on differentiating both sides w.r.t. x, we get,
    {tex}y \cdot \frac { 1 } { x } + \log x \cdot \frac { d y } { d x } = \frac { 1 } { v } \frac { d v } { d x }{/tex}
    {tex}\Rightarrow v \left( \frac { y } { x } + \log x \cdot \frac { d y } { d x } \right) = \frac { d v } { d x }{/tex}
    {tex}\Rightarrow \quad \frac { d v } { d x } = x ^ { y } \left( \frac { y } { x } + \log x \frac { d y } { d x } \right){/tex}………(iv) [ From Eq.(ii)] Also, yx = u [From Eq(ii)] Therefore,on taking log both sides, we get,
    log yx = log u {tex}\Rightarrow{/tex} x log y = log u
    Therefore,on differentiating both sides w.r.t. ‘x’, we get,
    {tex}x \cdot \frac { 1 } { y } \frac { d y } { d x } + 1 \cdot \log y = \frac { 1 } { u } \frac { d u } { d x }{/tex}
    {tex}\Rightarrow \quad \frac { x } { y } \frac { d y } { d x } + \log y = \frac { 1 } { u } \frac { d u } { d x }{/tex}
    {tex}\Rightarrow \quad u \left[ \frac { x } { y } \frac { d y } { d x } + \log y \right] = \frac { d y } { d x }{/tex}
    {tex}\Rightarrow \quad y ^ { x } \left[ \frac { x } { y } \frac { d y } { d x } + \log y \right] = \frac { d u } { d x }{/tex}……..(v) [ From Eq(ii)] Therefore,on substituting the values of {tex}\frac { d v } { d x } \text { and } \frac { d u } { d x }{/tex} from Eqs. (iv) and (v) respectively in Eq. (iii), we get
    {tex}x ^ { y } \left( \frac { y } { x } + \log x \cdot \frac { d y } { d x } \right) + y ^ { x } \left( \frac { x } { y } \frac { d y } { d x } + \log y \right) = 0{/tex}
    {tex}\Rightarrow x ^ { y } \frac { y } { x } + x ^ { y } \log x \cdot \frac { d y } { d x } {/tex}{tex}+ y ^ { x } \cdot \frac { x } { y } \frac { d y } { d x } + y ^ { x } \log y = 0{/tex}
    {tex}\Rightarrow x ^ { y } \log x \cdot \frac { d y } { d x } + y ^ { x } \frac { x } { y } \cdot \frac { d y } { d x }{/tex}{tex}= – x ^ { y } \frac { y } { x } – y ^ { x } \log y{/tex}
    {tex}\Rightarrow \quad \frac { d y } { d x } \left[ x ^ { y } \log x + y ^ { x } \cdot \frac { x } { y } \right]{/tex}{tex}= – x ^ { y } \cdot \frac { y } { x } – y ^ { x } \log y{/tex}
    {tex}\therefore \quad \frac { d y } { d x } = \frac { – x ^ { y – 1 } \cdot y – y ^ { x } \log y } { x ^ { y } \log x + y ^ { x – 1 } \cdot x }{/tex}
  12. According to the question, {tex}e^y (x + 1) = 1{/tex}
    Taking log both sides,
    {tex}\Rightarrow log [e^y (x + 1) ]= log 1{/tex}
    {tex}\Rightarrow log e^y + log(x + 1) = log 1{/tex}
    {tex}\Rightarrow{/tex} {tex}y + log(x + 1) = log1{/tex} [{tex}\because{/tex} log ey = y] differentiating both sides w.r.t. x,
    {tex}\Rightarrow\frac { d y } { d x } + \frac { 1 } { x + 1 } = 0{/tex}………(i)
    Differentiating both sides w.r.t. ‘x’,
    {tex}\Rightarrow \frac { d ^ { 2 } y } { d x ^ { 2 } } – \frac { 1 } { ( x + 1 ) ^ { 2 } } = 0{/tex}
    {tex}\Rightarrow \quad \frac { d ^ { 2 } y } { d x ^ { 2 } } – \left( – \frac { d y } { d x } \right) ^ { 2 } = 0{/tex} [ From Equation(i)] {tex}\Rightarrow \quad \frac { d ^ { 2 } y } { d x ^ { 2 } } – \left( \frac { d y } { d x } \right) ^ { 2 } = 0{/tex}
    {tex}\Rightarrow \quad \frac { d ^ { 2 } y } { d x ^ { 2 } } = \left( \frac { d y } { d x } \right) ^ { 2 }{/tex}
  13. Let {tex}u = {y^x},v = {x^y},w = {x^x}{/tex}
    {tex}u + v + w = {a^b}{/tex}
    Therefore {tex}\frac{{du}}{{dx}} + \frac{{dw}}{{dx}} + \frac{{dv}}{{dx}} = 0{/tex} ….(1)
    {tex}u = {y^x}{/tex}
    Taking log both side
    {tex}\log u = \log {y^x}{/tex}
    {tex}\log u = x.\log y{/tex}
    Differentiate both side w.r.t. to x
    {tex}\frac{1}{u}.\frac{{du}}{{dx}} = x.\frac{1}{y}.\frac{{dy}}{{dx}} + \log y.1{/tex}
    {tex}\frac{{du}}{{dx}} = u\left[ {\frac{x}{y}.\frac{{dy}}{{dx}} + \log y} \right]{/tex}
    {tex}\frac{{du}}{{dx}} = {y^x}\left[ {\frac{x}{y}.\frac{{dy}}{{dx}} + \log y} \right]{/tex}…. (2)
    {tex}v = {x^y}{/tex}
    Taking log both side
    {tex}\log v = \log {x^y}{/tex}
    {tex}\log v = y.\log x{/tex}
    {tex}\frac{1}{v}.\frac{{dv}}{{dx}} = y.\frac{1}{x} + \log x.\frac{{dy}}{{dx}}{/tex}
    {tex}\frac{{dv}}{{dx}} = v\left[ {\frac{y}{x} + \log x.\frac{{dy}}{{dx}}} \right]{/tex}
    {tex}\frac{{dv}}{{dx}} = {x^y}\left[ {\frac{y}{x} + \log x.\frac{{dy}}{{dx}}} \right]{/tex}…. (3)
    {tex}w = {x^x}{/tex}
    Taking log both side
    {tex}\log w = \log {x^x}{/tex}
    {tex}\log w = x\log x{/tex}
    {tex}\frac{1}{w}.\frac{{dw}}{{dx}} = x.\frac{1}{x} + \log x.1{/tex}
    {tex}\frac{1}{w}.\frac{{dw}}{{dx}} = 1 + \log x{/tex}
    {tex}\frac{{dw}}{{dx}} = w(1 + \log x){/tex}
    {tex}\frac{{dw}}{{dx}} = {x^x}(1 + \log x){/tex}…. (4)
    {tex}\frac{{dy}}{{dx}} = \frac{{ – {x^x}(1 + \log x) – y.{x^{y – 1}} – {y^x}\log y}}{{x.{y^{x – 1}} + {x^y}\log x.}}{/tex} (by putting 2,3 and 4 in 1)

Chapter Wise Important Questions Class 12 Maths Part I and Part II

  1. Relations and Functions
  2. Inverse Trigonometric Functions
  3. Matrices
  4. Determinants
  5. Continuity and Differentiability
  6. Application of Derivatives
  7. Integrals
  8. Application of Integrals
  9. Differential Equations
  10. Vector Algebra
  11. Three Dimensional Geometry
  12. Linear Programming
  13. Probability

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