Solutions Of Questions Of Continuity & Differentiability From NCERT Class XII

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1 Solutions Of Questions Of Continuity & Differentiability From NCERT Class XII Compiled By OP Gupta [ ] Q0. 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. [NCERT Ex-5. Q9 Sol. The given function g (x) is defined at all integral points. Let n be an integer. Then g (n) = n [n] = n n = 0. LHL (at x = n): lim g ( x ) lim( x [ x ]) n ( n ). xn xn RHL (at x = n): lim g ( x ) lim( x [ x ]) n ( n ) 0. xn xn Since LHL (at x = n) RHL (at x = n). Therefore g is not continuous at x = n i.e., g (x) is discontinuous at all integral points. Q0. Is the function defined by f (x) = x sin x + 5 continuous at x = π? [NCERT Ex-5. Q0 Sol. We have f (x) = x sin x + 5 f (π) (π) sin π 5 5 π (i) LHL (at x = π ): lim f ( x) lim ( x sin x 5) [Put x π h so that as x π, h 0 RHL (at x = π ): xπ0 xπ0 lim((π h) sin(π h) 5) h0 h h h0 lim f ( x) lim ( x sin x 5) xπ0 xπ0 lim((π h) sin(π h) 5) h0 h h h0 lim((π ) sin 5) ((π 0) sin 0 5) π 5 (ii) [Put x π h so that as x π, h 0 lim((π ) sin 5) ((π 0) sin 0 5) π 5 (iii) By (i), (ii) & (iii), it is clearly evident that LHL (at x = π ) = RHL (at x = π ) = f ( π ). So, f is continuous at x = π. Q03. 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 [NCERT Ex-5. Q Sol. Since we know that if g and h are two continuous functions then, g + h, g h, g.h are also continuous. (A) So, in order to prove all (a), (b) and (c), it will be sufficient to prove that sin x and cos x are continuous functions. Let g (x) = sin x. Let c be any real number. Then g (c) = sin c (i) [As sin x is defined for every real number] LHL (at x = c): lim g( x) lim sin x [Put x c h so that as x c, h 0 xc h0 xc limsin( c h) sin( c 0) sin c (ii) RHL (at x = c): lim g( x) lim sin x [Put x c h so that as x c, h 0 xc h0 xc limsin( c h) sin( c 0) sin c (iii) By (i), (ii) & (iii), it is clearly evident that LHL (at x = c) = RHL (at x = c) = g (c). So, sin x is continuous at all real values of x. Also, let h (x) = cos x. Let c be any real number. Then h (c) = cos c (iv) [As cos x is defined for every real number] LHL (at x = c): lim h( x) lim cos x [Put x c h so that as x c, h 0 xc h0 xc lim cos( c h) cos( c 0) cosc (v) RHL (at x = c): lim h( x) lim cos x [Put x c h so that as x c, h 0 xc xc Please visit for various stuffs on Mathematics :

2 lim cos( c h) cos( c 0) cosc (vi) h0 By (iv), (v) & (vi), it is clearly evident that LHL (at x = c) = RHL (at x = c) = h (c). So, cos x is continuous at all real values of x. Hence by using the statement (A) mentioned above, we can say: (a) f (x) = sin x + cos x is continuous for all real values of x. (b) f (x) = sin x cos x is continuous for all real values of x. (c) f (x) = sin x. cos x is continuous for all real values of x. Q04. Discuss continuity of cosine, cosecant, secant and cotangent functions. [NCERT Ex-5. Q Sol. Note that if f and g are continuous functions, then f ( x) (i), g ( x ) 0 is continuous. g( x) (ii), g ( x ) 0 is continuous. g( x) (iii), f ( x ) 0 is continuous. f ( x) Now, for the discussion over continuity of cosine function, please see the previous question. Also, if cos x is continuous (see the solution of previous question) then sec x is continuous as π well for all x R (n ), n Z as sec x cos x, cos x 0 (since cos x = 0 at all π x (n ), n Z ). Again, if sin x is continuous (see the solution of previous question) then cosec x is continuous as well for all x R nπ, n Z as cosec x sin x, sin x 0 (since sin x = 0 at all x nπ, n Z). cos x Finally, cot x is continuous for all x R nπ, n Z as cot x, sin x 0 (since sin x = sin x 0 at all x nπ, n Z). Q05. x sin, if x 0 Determine if f defined by f ( x) x is a continuous function? 0, if x 0 [NCERT Ex-5. Q Sol. LHL (at x = 0): lim f ( x) lim f (0 h) lim f ( h) [Put x 0 h so that as x 0, h 0 x0 h 0 h 0 lim( h) sin h0 h lim[ h sin ] h0 h = 0. [ sin x RHL (at x = 0): lim f ( x) lim f (0 h) lim f ( h) [Put x 0 h so that as x 0, h 0 x0 h 0 h 0 lim( h) sin h0 h lim h sin h0 h = 0. [ sin x And f (0) = 0. Since LHL (at x = 0) = RHL (at x = 0) = f (0). Therefore f is continuous at x = 0. Please visit for various stuffs on Mathematics :

3 Q06. Show that the function f (x) = cos (x ) is a continuous function. [NCERT Ex-5. Q3 Sol. The function f is defined for all real number and it can be expressed as the composition of two functions f = g o h, where g (x) = cos x and h (x) = x. [ (g o h) (x) = g (h (x)) = g (x ) = cos x = f (x)] So, we need to prove that g (x) and h (x) are continuous functions. We have g (x) = cos x. Let c be any real number. Then g (c) = cos c (i) [As cos x is defined for every real number] LHL (at x = c): lim g( x) lim cos x [Put x c h so that as x c, h 0 xc h0 xc lim cos( c h) cos( c 0) cosc (ii) RHL (at x = c): lim g( x) lim cos x [Put x c h so that as x c, h 0 xc h0 xc lim cos( c h) cos( c 0) cosc (iii) By (i), (ii) & (iii), it is clearly evident that LHL (at x = c) = RHL (at x = c) = f (c). So, g (x) is continuous at all real values of x. [Continuity of cos x has been discussed in Q03 as well.] Also, we have h (x) = x. Clearly, the function h is defined for every real number. Let k be a real number, then h (k) = k. And lim h( x) lim x k. xk xk lim h( x) h( k) h( x) is a continuous function. xk Since it is known that for real valued functions g and h, such that (g o h) is defined at c, if h is continuous at c and if g is continuous at h (c), then (g o h) is continuous at c. Therefore f (x) = (g o h) (x) = cos (x ) is continuous function. Q07. Show that the function f (x) = cos x is a continuous function. [NCERT Ex-5. Q3 Sol. The function f is defined for all real number and it can be expressed as the composition of two functions f = g o h, where g (x) = x and h (x) = cos x. [ (g o h) (x) = g (h (x)) = g (cos x) = cos x = f (x)] So, we need to prove that g (x) and h (x) are continuous functions. x, if x 0 We have g( x) x x, if x 0 Case I: If c < 0, then g (c) = c and lim g( x) lim( x) c xc xc lim g ( x ) g ( c ) xc g( x) is continuous at all points x, such that x < 0. Case II: If c > 0, then g (c) = c and lim g( x) lim( x) c lim g( x) g( c) xc xc xc g( x) is continuous at all points x, such that x > 0. Case III: If c = 0, then g (0) = 0. LHL (at x = 0): lim g( x) lim( x) 0 0. RHL (at x = 0): x0 x0 lim g( x) lim( x) 0. x0 x0 Clearly LHL (at x = 0) = RHL (at x = 0) = g (0). g( x) is continuous at x = 0. Also h (x) = cos x. Let c be any real number. Then h (c) = cos c (i) [As cos x is defined for every real number] LHL (at x = c): lim h( x) lim cos x [Put x c h so that as x c, h 0 xc xc Please visit for various stuffs on Mathematics : 3

4 lim cos( c h) cos( c 0) cosc (ii) h0 RHL (at x = c): lim h( x) lim cos x [Put x c h so that as x c, h 0 xc h0 xc lim cos( c h) cos( c 0) cosc (iii) By (i), (ii) & (iii), it is clearly evident that LHL (at x = c) = RHL (at x = c) = h (c). So, h (x) is continuous at all real values of x. Since it is known that for real valued functions g and h, such that (g o h) is defined at c, if h is continuous at c and if g is continuous at h (c), then (g o h) is continuous at c. Therefore f (x) = (g o h) (x) = g (h (x)) = g (cos x) = cos x is continuous function. Q08. Examine that sin x is a continuous function. [NCERT Ex-5. Q33 Sol. The function f is defined for all real number and it can be expressed as the composition of two functions f = g o h, where g (x) = sin x and h (x) = x. [ (g o h) (x) = g (h (x)) = g ( x ) = sin x = f (x)] So, we need to prove that g (x) and h (x) are continuous functions. We have g (x) = sin x. Let c be any real number. Then g (c) = sin c (i) [As sin x is defined for every real number. LHL (at x = c): lim g( x) lim sin x [Put x c h so that as x c, h 0 xc h0 xc limsin( c h) sin( c 0) sin c (ii) RHL (at x = c): lim g( x) lim sin x [Put x c h so that as x c, h 0 xc h0 xc limsin( c h) sin( c 0) sin c (iii) By (i), (ii) & (iii), it is clearly evident that LHL (at x = c) = RHL (at x = c) = g (c). So, sin x is continuous at all real values of x. x, if x 0 Also we have h( x) x x, if x 0 Case I: If c < 0, then h (c) = c and lim h( x) lim( x) c xc xc lim h ( x ) h ( c ) xc h( x) is continuous at all points x, such that x < 0. Case II: If c > 0, then h (c) = c and lim h( x) lim( x) c lim h( x) h( c) xc xc xc h( x) is continuous at all points x, such that x > 0. Case III: If c = 0, then h (0) = 0. LHL (at x = 0): lim h( x) lim( x) 0 0. RHL (at x = 0): x0 x0 lim h( x) lim( x) 0. x0 x0 Clearly LHL (at x = 0) = RHL (at x = 0) = h (0). h( x) is continuous at x = 0. Since it is known that for real valued functions g and h, such that (g o h) is defined at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous at c. Therefore f (x) = (g o h) (x) = g (h (x)) = g ( x ) = sin x is continuous function. Q09. Find all the points of discontinuity of f defined by f (x) = x x +. [NCERT Ex-5. Q3 Sol. We have f (x) = x x + i.e., f (x)= g (x) + h (x) where, g (x) = x and h (x) = x +. We shall firstly, examine the continuity of g (x) and h (x). Please visit for various stuffs on Mathematics : 4

5 x, if x 0 We have g( x) x x, if x 0 Case I: If c < 0, then g (c) = c and lim g( x) lim( x) c lim g( x) g( c) xc xc xc g( x) is continuous at all points x, such that x < 0. Case II: If c > 0, then g (c) = c and lim g( x) lim( x) c lim g( x) g( c) xc xc xc g( x) is continuous at all points x, such that x > 0. Case III: If c = 0, then g (0) = 0. LHL (at x = 0): lim g( x) lim( x) 0 0. RHL (at x = 0): x0 x0 lim g( x) lim( x) 0. x0 x0 Clearly LHL (at x = 0) = RHL (at x = 0) = g (0). g( x) is continuous at x = 0. x, if x Now h (x) = x + ( x ), if x Case I: If c <, then h (c) = (c + ) and lim h( x) lim[ ( x )] ( c ) lim h( x) h( c) xc xc xc h( x) is continuous at all points x, such that x <. Case II: If c >, then h (c) = c + and lim h( x) lim( x ) c lim h( x) h( c) xc xc xc h( x) is continuous at all points x, such that x >. Case III: If c =, then h ( ) = + = 0. LHL (at x = ): lim h( x) lim ( x ) ( ) 0. RHL (at x = ): x x lim h( x) lim( x ) 0. x x0 Clearly LHL (at x = ) = RHL (at x = ) = h ( ). h( x) is continuous at x =. Since g and h are continuous functions, therefore f (x) = g(x) + h (x) is also continuous. Hence, there is no point of discontinuity for the function f (x). Q0. Show that the function f defined by f (x) = x + x, where x is any real number, is a continuous function. [NCERT Example 0 Sol. Let g (x) = x + x and h (x) = x for all real x. Then, (h o g) (x) = h (g (x)) = h ( x + x ) = x + x = f (x). We know that x is a continuous function. Hence h (x) is continuous function. So, g (x) being the sum of a polynomial function and the modulus function is also continuous. Finally the function f being a composite of two continuous functions is continuous as well. Q. Find all the points of discontinuity of the greatest integer function defined by f (x) = [x], where [x] denotes the greatest integer less than or equal to x. [NCERT Example 5 Sol. We have f (x) = [x]. Case I: Let c be a real number which is not equal to any integer. It is evident that for all real numbers close to c the value of the function is equal to [c]; i.e., lim f ( x) lim[ x] [ c]. xc xc Also f (c) = [c] and hence the greatest integer function is continuous at all non-integral real numbers (since lim f ( x ) f ( c ) [ c ] ). xc 5 Please visit for various stuffs on Mathematics :

6 Case II: Let c be an integer. Then lim[ x] c and lim[ x] c. xc xc Since lim f ( x) lim f ( x), therefore the greatest integer function is discontinuous at all xc xc integral points. Q. Prove that f (x) = x, x R is not differentiable at x =. [NCERT Ex-5. Q09 Sol. x, if x We have f (x) = x. ( x ), if x Then f () = = 0. We know that a function is differentiable at a point x = m in its domain if LHD and RHD at x = m are both finite and equal to each other. Differentiability at x = : f ( x) f () LHD (at x = ): lim x x [ ( x )] (0) ( x ) lim lim lim( ). x x x x x f ( x) f () RHD (at x = ): lim x x ( x ) (0) ( x ) lim lim lim(). x x x x x Since LHD (at x = ) RHD (at x = ) so, f is not differentiable at x =. Q3. Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at x = and x =. [NCERT Ex-5. Q0 Sol. We have f (x) = [x], 0 < x < 3. As a function is differentiable at a point x = m in its domain if LHD and RHD at x = m are both finite and equal to each other. Differentiability at x = : f ( x) f () f ( h) f () LHD (at x = ): lim lim [Put x h so that as x, h 0 x x h0 ( h ) [ h] [] 0 lim lim lim. h0 h h0 h h0 h 0 f ( x) f () f ( h) f () RHD (at x = ): lim lim [Put x h so that as x, h 0 x x h0 ( h ) [ h] [] 0 lim lim lim lim(0) 0. h0 h h0 h h0 h h0 Since LHD (at x = ) RHD (at x = ) so, f is not differentiable at x =. [NOTE that here we didn t need to evaluate RHD as LHD is already not defined (i.e., ), which is an enough reason for f to not be differentiable at x =.] Differentiability at x = : f ( x) f () LHD (at x = ): lim x x [ x] [] lim lim lim. x x x x x x 0 Since LHD (at x = ) doesn t exist so, f is not differentiable at x =. Hii. Here is a short message I have to convey. I ve devoted myself for the service of Mathematics.. to help the students in need in all possible ways. It will be a thing of pleasure for me if my work/collection serves any purpose in your life. Wish You All The Very Best! Lots of love and blessings! OP Gupta [ , , theopgupta@gmail.com] 6 Please visit for various stuffs on Mathematics :