RD Sharma XII Vol 1 2017 Solutions for Class 12 Science Math Chapter 9 Continuity are provided here with simple step-by-step explanations. These solutions for Continuity are extremely popular among class 12 Science students for Math Continuity Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XII Vol 1 2017 Book of class 12 Science Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XII Vol 1 2017 Solutions. All RD Sharma XII Vol 1 2017 Solutions for class 12 Science Math are prepared by experts and are 100% accurate.

Page No 9.16:

Question 1:

Test the continuity of the function on f(x) at the origin:
fx=xx,x0  1   ,x=0

Answer:

Given:
fx=xx, x01,   x=0

We observe
(LHL at = 0) =limx0-fx  =  lim      h0f0-h =    lim      h0f-h
 =​limh0-h-h=limh0-hh =limh0-1 =-1

 (RHL at x = 0)​ =limx0+fx    =   lim      h0f0+h=    lim      h0fh
                           =​limh0hh=limh0hh=limh01=1 


limx0+fx limx0-fx

Hence, fx is discontinuous at the origin.

Page No 9.16:

Question 2:

A function f(x) is defined as,
fx=x2-x-6x-3; ifx3        5         ; ifx=3
Show that f(x) is continuous that x = 3.

Answer:

Given:
fx=x2-x-6x-3, x35,   x=3

We observe
(LHL at = 3) = limx3-fx    =   lim      h0f3-h
                   =​limh03-h2-3-h-63-h-3=limh09+h2-6h-3+h-6-h=limh0h2-5h-h=limh05-h = 5

And, (RHL at = 3)​ = limx3+fx    =   lim      h0f3+h
                           =​limh03+h2-3+h-63+h-3=limh09+h2+6h-3-h-6h=limh0h2+5hh=limh05+h = 5 

Also, f3=5

limx3+fx  =limx3-fx = f3

Hence, fx is continuous at x=3.

Page No 9.16:

Question 3:

A function f(x) is defined as
fx=x2-9x-3; ifx3     6     ; ifx=3
Show that f(x) is continuous at x = 3

Answer:

Given:
fx=x2-9x-3, if x36,           if x=3

We observe
(LHL at x = 3) = limx3-fx=limh0f3-h
                        =   ​limh03-h2-93-h-3=limh032+h2-6h-93-h-3=limh0h2-6h-h==limh0hh-6-h=limh06-h=6

(RHL at x = 3) = limx3+fx=limh0f3+h
                        =   ​limh03+h2-93+h-3=limh032+h2+6h-9h=limh0h2+6hh=limh0 h6+hh=limh06+h=6

Given:
f3=6


limx3-fx=limx3+fx=f3

Hence, fx is continuous at x=3.



Page No 9.17:

Question 4:

If fx=x2-1x-1; forx1     2      ; forx=1
Find whether f(x) is continuous at x = 1.

Answer:

Given:
fx=x2-1x-1, if x12,         if x=1

We observe
(LHL at x = 1) = limx1-fx=limh0f1-h
                        =   ​limh01-h2-11-h-1=limh01+h2-2h-11-h-1=limh0h2-2h-h=limh0hh-2-h=limh02-h=2

(RHL at x = 1) = limx1+fx=limh0f1+h
                        =   ​limh01+h2-11+h-1=limh01+h2+2h-11+h-1=limh0h2+2hh=limh0hh+2h=limh02+h=2

Given:
f1=2


limx1-fx=limx1+fx=f1

Hence, fx is continuous at x=1.

Page No 9.17:

Question 5:

If fx=sin 3xx, whenx0     1      , whenx=0
Find whether f(x) is continuous at x = 0.

Answer:

Given:
fx=sin3xx, when x01,   when x=0

We observe 
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h
                        =   ​limh0sin-3h-h=limh0-sin3h-h=limh03sin3h3h=3limh0sin3h3h=3·1=3

(RHL at x = 0) = limx0+fx=limh0fh
                        =   ​limh0sin3hh=limh03sin3h3h=3limh0sin3h3h=3·1=3

Given:
f0=1

It is known that for a function fx to be continuous at xa, 
limxa-fx=limxa+fx=fa

But here,
limx0-fx=limx0+fxf0

Hence, fx is discontinuous at x=0.

Page No 9.17:

Question 6:

If fx=e1/x, ifx01     , ifx=0 find whether f is continuous at x = 0.

Answer:

Given:
fx=e1x, if x01,   if x=0

We observe
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h
                        =   ​limh0e-1h=limh01e1h=1limh0e1h=0

(RHL at x = 0) = limx0+fx=limh0fh
                        =   ​limh0e1h=

Given:
f0=1


It is known that for a function fx to be continuous at x = a, 
limxa-fx=limxa+fx=fa

But here,
limx0-fxlimx0+fx

Hence, fx is discontinuous at x=0.

Page No 9.17:

Question 7:

Let fx=1-cos xx2, whenx0       1         , whenx=0
Show that f(x) is discontinuous at x = 0.

Answer:

Given:
fx=1-cosxx2, when x01,  when x=0

Consider:
limx0fx=limx01-cosxx2limx0fx=limx02sin2x2x2limx0fx=limx02sin2x24x24limx0fx=limx02sinx224x22limx0fx=24limx0sinx2x22limx0fx=12·12=12

Given:
f0=1

∴ ​limx0fxf0

Thus, f(x) is discontinuous at x = 0
.

Page No 9.17:

Question 8:

Show that fx=x-x2, whenx0     2      , whenx=0
is discontinuous at x = 0.

Answer:

The given function can be rewritten as:
fx=x-x2, when x>0x+x2, when x<02,        when x=0
 fx=0, when x>0x, when x<02, when x=0

We observe
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0-h=0
                        

(RHL at x = 0) = limx0+fx=limh0f0-h=limh0fhlimh00=0
 
And, f0=2
∴ ​limx0-fx =limx0+fx  f0

Thus, f(x) is discontinuous at x = 0
.

Page No 9.17:

Question 9:

Show that fx=x-ax-a, whenxa     1      , whenx=a
is discontinuous at x = a.

Answer:

The given function can be rewritten as:
fx=x-ax-a, when x>aa-xx-a, when x<a1,        when x=a
 fx=1, when x>a-1, when x<a1, when x=a

 fx=1, when xa-1, when x<a

We observe
(LHL at x = a) = limxa-fx=limh0fa-h=limh0-1=-1
                        

(RHL at x = a) = limxa+fx=limh0fa+hlimh01=1
 

∴ ​limxa-fxlimxa+fx

Thus, f(x) is discontinuous at x = a.

Page No 9.17:

Question 10:

Discuss the continuity of the following functions at the indicated point(s):
(i) fx=x cos1x,x0     0             ,x=0at x=0

(ii) fx=x2sin1x,x0      0         ,x=0at x=0

(iii) fx=(x-a)sin1x-a,xa            0                 ,x=aat x=a

(iv) fx=ex-1log(1+2x), ifxa        7           , ifx=0at x=0

(v) fx=1-xn1-x,x1n-1     ,x=1nNat x=1

(vi) fx=x2-1x-1, forx1      2       , forx=1at x=1

(vii) fx=2x+x2x,x0       0        , x=0at x=0

(viii) fx=x-asin1x-a,  for xa0, for x=aat x=a

Answer:

(i) Given:
fx=x cos1x,  x00, x=0
We observe
limx0 fx=limx0x cos1xlimx0 fx=limx0xlimx0cos1xlimx0 fx=0 ×limx0cos1x=0 
limx0fx=f0
Hence, f(x) is continuous at x = 0.

(ii) Given:
fx=x2sin1x,  x00,  x=0
We observe
limx0x2 sin1x=limx0x2limx0sin1x=0 ×limx0sin1x=0
limx0fx=f0
Hence, f(x) is continuous at x = 0.

(iii) Given:
fx=x-a sin1x-a,   xa0, x=a

Putting xa = y, we get
limxax-a sin1x-a=limy0y sin1y=limy0ylimy0sin1y=0 × limy0sin1y=0
limxafx=fa=0
Hence, f(x) is continuous at x = a.

(iv) Given:
fx=ex-1log1+2x, if  x07, if x=0

We observe
limx0fx=limx0ex-1log1+2xlimx0fx=limx0ex-12xlog1+2x2xlimx0fx=12limx0ex-1xlog1+2x2xlimx0fx=12×limx0ex-1xlimx0log1+2x2x=12×11=12
And, f0=7
limx0fxf0

Hence, f(x) is discontinuous at x = 0.

(v) Given:
fx=1-xn1-x, x1n-1, x=1

Here, f1=n-1

limx1fx=limx11-xn1-xlimx1fx=limx11-xn-1+C1n1-xn-2x+C2n1-xn-3x2+...+Cn-1n1-x0xn-1
limx1fx=0+0...+1n-1=1f1

Thus, fx is discontinuous at x=1.

(vi) Given:
fx=x2-1x-1,   x12,  x=1
fx=x+1,   x<-1-x-1,  -1 x<1x+1,   x>12,   x=1
We observe
(LHL at = 1) = limx1-fx=limh0f1-h=limh0-1-h-1=limh0-2+h=-2
And, f1=2

limx1-fxf1

Hence, f(x) is discontinuous at x = 1.

(vii) Given:
fx=2x+x2x,  x00, x=0

fx=2x+x2x,   x>0-2x+x2x,   x<00,   x=0

fx=x+2,   x>0x-2,   x<00,   x=0

We observe

(LHL at x = 0) = limx0-fx=limh0f-h=limh0-h-2=-2
(RHL at x = 0) = limx0+fx=limh0fh=limh02+h=2

limx0-fxlimx0+fx

Hence, f(x) is discontinuous at x = 0.

(viii) Given:

fx=x-asin1x-a,  for xa0, for x=a

fx=x-asin1x-a,   x>0x+asin1x+a,   x<00,   x=a

We observe

(LHL at x = a) = limxa-fx=-a+asin1-a+a=0
(RHL at x = a) = limxa+fx=a-asin1a-a=0

limxa-fx=limxa+fx=fa

Hence, f(x) is continuous at x = a.
 



Page No 9.18:

Question 11:

Show that fx=1+x2, if0x12-x  , ifx>1is discontinuous at x = 1.

Answer:

Given:
fx=1+x2, if 0x12-x, if x>1

We observe
(LHL at x = 1) = limx1-fx=limh0f1-h=limh01+1-h2=limh02+h2-2h=2
                        

(RHL at x = 1) = limx1+fx=limh0f1+h=limh02-1+h=limh01-h=1
 

∴ ​limx1-fxlimx1+fx

Thus, f(x) is discontinuous at x = 1.

Page No 9.18:

Question 12:

Show that fx=sin 3xtan 2x       , if x<0        32         , if x=0log(1+3x)e2x-1  , if x>0is continuous at x=0

Answer:

Given:
fx=sin3xtan2x, if x<032, if x=0log1+3xe2x-1, if x>0

We observe
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h
 =limh0sin3-htan2-h=limh0sin3htan2h=limh03sin3h3h2tan2h2h=limh03sin3h3hlimh02tan2h2h=3limh0sin3h3h2limh0tan2h2h=3×12×1=32
                        

(RHL at x = 1) = limx0+fx=limh0f0+h=limh0fh
=limh0log1+3he2h-1=limh03hlog1+3h3h2he2h-12h=32limh0log1+3h3he2h-12h=32limh0log1+3h3hlimh0e2h-12h=3×12×1=32

And, f0=32
 

∴ ​limx0-fx=limx0+fx=f0

Thus, f(x) is continuous at x = 0.

Page No 9.18:

Question 13:

Find the value of 'a' for which the function f defined by
fx=asinπ2(x+1),x0tanx-sinxx3,x>0is continuous at x = 0.

Answer:

Given: fx=a sin π2x+1, x0tan x-sin xx3, x>0
We have

(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0a sin π2-h+1=a sinπ2=a

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh0tan h-sin hh3

limx0+fx=limh0sin hcos h-sin hh3limx0+fx=limh0sin hcos h1-cos hh3limx0+fx=limh01-cos htan hh3limx0+fx=limh02sin2 h2tan h4h24×hlimx0+fx=24limh0sin2h2tan hh24×hlimx0+fx=12limh0sinh2h22limh0tan hhlimx0+fx=12×1×1limx0+fx=12

 If fx is continuous at x=0, thenlimx0-fx=limx0+fxa=12

Page No 9.18:

Question 14:

Examine the continuity of the function
fx=3x-2,x0x+1  ,x>0at x=0
Also sketch the graph of this function.

Answer:

The given function can be rewritten as:
  fx=3x-2, x<030-2, x=0x+1, x>0
fx=3x-2, x<0-2, x=0x+1, x>0

We observe

(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h = ​limh03-h-2=-2

(RHL at x = 0) =  limx0+fx=limh0f0+h=limh0fh = ​limh0h+1=1



limx0-fxlimx0+fx

Hence, fx is discontinuous at x=0.

Page No 9.18:

Question 15:

Discuss the continuity of the function f(x) at the point x = 0, where
fx=   x,   1,-x,x>0x=0x<0

Answer:

Given:
fx=x, x>01, x=0-x, x<0



(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h = ​limh0--h=0

(RHL at x = 0) =  limx0+fx=limh0f0+h=limh0fh = ​limh0h=0

And, ​f0=1



limx0-fx=limx0+fxf0

Hence, fx is discontinuous at x=0.

Page No 9.18:

Question 16:

Discuss the continuity of the function f(x) at the point x = 1/2, where
fx=x,1/2,1-x,0x<1/2x=1/21/2<x1

Answer:

Given:
fx=x, 0x<1212, x=121-x, 12<x1

We observe

(LHL at x = 12) = limx12-fx=limh0f12-h = ​limh012-h=12

(RHL at x = 12) = limx12+fx=limh0f12+hlimh01-12+h=12

Also, ​f12=12


limx12-fx =limx12+fx = f12

Hence, fx is continuous at x=12.

Page No 9.18:

Question 17:

Discuss the continuity of fx=2x-1, x<02x+1, x0 at x=0

Answer:

fx=2x-1, x<02x+1, x0LHL at x=0=limx0-fx=20-1=-1                      RHL at x=0=limx0+fx=20+1=1limx0-fxlimx0+fx
Hence, f(x) is discontinuous at x = 0.

Page No 9.18:

Question 18:

For what value of k is the following function continuous at x = 1?
fx=x2-1x-1,x1     k     ,x=1

Answer:

Given: fx=x2-1x-1,  x1k,   x=1
If fx is continuous at x = 1, then
limx1fx = f1
limx1x2-1x-1= k
limx1x-1x+1x-1= k
limx1x+1= k
k=2

Page No 9.18:

Question 19:

Determine the value of the constant k so that the function
fx=x2-3x+2x-1, ifx1         k          , ifx=1is continuous at x=1

Answer:

Given:
fx=x2-3x+2x-1, if x1k,  if x=1
If fx is continuous at x = 1, then,
limx1fx = f1
limx1x2-3x+2x-1= k
limx1x-2x-1x-1= k
limx1x-2= k
k=-1

Page No 9.18:

Question 20:

For what value of k is the function
fx=sin 5x3x, ifx0      k     , ifx=0is continuous at x=0?

Answer:

Given:
fx=sin5x3x, if x0k,  if x=0
If fx is continuous at x = 0, then
limx0fx = f0
limx0sin5x3x= k
limx05 sin5x3×5x= k
53limx0sin5x5x= k
53×1= k
k=53

Page No 9.18:

Question 21:

Determine the value of the constant k so that the function
fx=kx2, ifx23    , ifx>2is continuous at x=2.

Answer:

Given:
fx=kx2, if x23,  if x>2
If fx is continuous at x = 2, then
limx2-fx =lim x2+fx= f2             ...(1)

Now,

limx2-fx =limh0f2-h =limh0k2-h2 =4k



And, f2=3

From (1), we have

4k=3k=34



Page No 9.19:

Question 22:

Determine the value of the constant k so that the function
fx=sin 2x5x, ifx0      k     , ifx=0is continuous at x=0.

Answer:

Given:
fx=sin2x5x, if x0k,  if x=0
If fx is continuous at x = 0, then
limx0fx = f0
limx0sin2x5x= k
limx02sin2x5×2x= k
25limx0sin2x2x= k
25×1= k
k=25

Page No 9.19:

Question 23:

Find the values of a so that the function
fx=ax+5, ifx2 x-1 , ifx>2is continuous at x=2.

Answer:

Given:
fx=ax+5, if x2x-1,  if x>2

We observe
(LHL at x = 2) = limx2-fx=limh0f2-h =  ​limh0a2-h+5=2a+5

(RHL at x = 2) = limx2+fx=limh0f2+h =  ​limh02+h-1 = 1

And, f2=a2+5=2a+5

Since fx is continuous at x = 2, we have
limx2-fx=limx2+fx=f2
2a+5=1
2a=-4
a=-2

Page No 9.19:

Question 24:

Prove that the function
fx=xx+2x2,x0      k         ,x=0
remains discontinuous at x = 0, regardless the choice of k.

Answer:

The given function can be rewritten as:
fx=xx+2x2, x>0-xx-2x2, x<0k, x=0
fx=12x+1, x>012x-1, x<0k, x=0

We observe
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h
                        =limh01-2h-1=-1

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh
                        =limh012h+1=1


So, ​limx0-fxlimx0+fx such that limx0-fx &limx0+fx are independent of k.

Thus, f(x) is discontinuous at x = 0, regardless of the choice of k.

Page No 9.19:

Question 25:

Find the value of k if f(x) is continuous at x = π/2, where
fx=k cos xπ-2x,xπ/2       3      ,x=π/2

Answer:

Given:
fx=kcosxπ-2x, xπ23, x=π2

If f(x) is continuous at x = π2, then
limxπ2fx = fπ2

limxπ2kcosxπ-2x=3     ...(1)

Putting π2-x=h, we get
limxπ2k cos xπ-2x=limh0k cos π2-hπ-2π2-h

From (1), we have

limh0k cos π2-hπ-2π2-h=3
limh0k sin h2h=3
limh0k sin hh=6
k limh0sin hh=6
k×1=6
k=6

Hence, for k=6 , f(x) is continuous at x = π2.

Page No 9.19:

Question 26:

Determine the values of a, b, c for which the function
fx=sin (a+1)x+sinxx,for x<0             c                 , for x=0 x+bx2-xbx3/2     ,for x>0is continuous at x = 0.

Answer:


The given function can be rewritten as:
fx=sin a+1 x+sin xx, for x<0c                                 , for x=0x+bx2-xbx32        , for x>0
fx=sin a+1x+sin xx, for x<0c                              , for x=01+bx-1bx       , for x>0

We observe
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h
                      =limh0-sin a+1h-sin -hh=limh0-sin a+1hh-sin hh
                      =-a+1limh0sin a+1ha+1h-limh0sin hh=-a-1
                        

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh 
                       =limh01+bh-1bh=limh0bhbh1+bh+1=limh011+bh+1=12

And, f0=c

If fx is continuous at x = 0, then

 ​limx0-fx=limx0+fx=f0

-a-1 = 12=c
-a-1 = 12 and c=12
a=-32, c=12


Now, 1+bx-1bx exists only if bx0b0.
 
bR-0

Page No 9.19:

Question 27:

If fx=1-cos kxx sin x,x0       12         ,x=0is continuous at x=0, find k.

Answer:

Given: fx=1-coskxxsinx,  x012,  x=0
If fx is continuous at x = 0, then
limx0fx = f0          ...(1)

Consider:
 limx0fx=limx01-cos kxx sin x=limx02 sin2 kx2x sin x
limx0fx=limx02 sin2 kx2x2sin xx
limx0fx=limx02k24sin kx22kx22sin xx
limx0fx=2k24limx0sinkx22kx22sin xx
limx0fx=2k24limx0sin kx22kx22limx0sin xx
limx0fx=2k24×1=k22

 From equation (1), we have
k22=f0
k22=12k=±1

Page No 9.19:

Question 28:

If fx=x-4x-4+a, ifx<4   a+b        , ifx=4 x-4x-4+b, ifx>4is continuous at x = 4, find a, b.

Answer:

Given:
fx=x-4x-4+a, if x<4a+b, if x=4x-4x-4+b, if x>4

We observe
(LHL at x = 4) = limx4-fx=limh0f4-h
                       =limh04-h-44-h-4+a=limh0-h-h+a=a-1
                        

(RHL at x = 4) = limx4+fx=limh0f4+h
                       =limh04+h-44+h-4+b=limh0hh+b=b+1

And f4=a+b

If f(x) is continuous at x = 4, then 

 ​limx4-fx=limx4+fx=f4
a-1=b+1=a+b
a-1=a+b, b+1=a+b
b=-1, a=1

Page No 9.19:

Question 29:

For what value of k is the function
fx=sin 2xx,x0 k          ,x=0
continuous at x = 0?

Answer:

Given:  fx=sin2xx, x0k, x=0

If f(x) is continuous at x = 0, then

limx0fx=f0
limx0sin2xx=k
limx02sin2x2x=k
2limx0sin2x2x=k
2×1=k
k=2

Page No 9.19:

Question 30:

Let fx=log1+xa-log1-xbx, x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

Answer:

Given: fx=log1+xa-log1-xbx, x0

If f(x) is continuous at x = 0, then

limx0fx=f0
limx0log1+xa-log1-xbx=f0
limx0log1+xaaxa-log1-xbbxb=f0
1alimx0log1+xaxa--1blimx0log1-xb-xb=f0
1a×1--1b×1=f0              Using: limx0log1+xx=11a+1b=f0a+bab=f0

Page No 9.19:

Question 31:

If fx=2x+2-164x-16, ifx2         k        , ifx=2is continuous at x = 2, find k.

Answer:

Given:
fx=2x+2-164x-16, if  x2k              , if x=2

If f(x) is continuous at x = 2, then

limx2fx=f2limx22x+2-164x-16=f2limx242x-42x-42x+4=klimx242x+4=k422+4=k48=kk=12



Page No 9.20:

Question 32:

If fx=cos2x-sin2x-1x2+1-1,x0               k               ,x=0 is continuous at x = 0, find k.

Answer:

Given:
fx=cos2x-sin2x-1x2+1-1,  x0k,  x=0

If f(x) is continuous at x = 0, then

limx0fx=f0limx0cos2x-sin2x-1x2+1-1=klimx01-sin2x-sin2x-1x2+1-1=klimx0-2sin2xx2+1-1=klimx0-2sin2xx2+1+1x2+1-1x2+1+1=klimx0-2sin2xx2+1+1x2=k-2limx0sin2xx2+1+1x2=k-2limx0sinxx2limx0x2+1+1=k-2×1×1+1=kk=-4

Page No 9.20:

Question 33:

Extend the definition of the following by continuity
fx=1-cos7 (x-π)5 (x-π)2 at the point x = π.

Answer:

Given:
fx=1-cos7x-π5x-π2,  x=π

If f(x) is continuous at x = π, then


limxπfx=fπlimxπ1-cos7x-π5x-π2=fπ25limxπsin27x-π2x-π2=fπ25×494limxπsin27x-π2494x-π2=fπ25×494limxπsin27x-π272x-π2=fπ25×494limxπsin7x-π272x-π2=fπ25×494×1=fπ15×492×1=fπ4910=fπ


Hence, the given function will be continuous at x=π, if fπ=4910.

Page No 9.20:

Question 34:

If fx=2x+3 sinx3x+2 sinx, x ≠ 0 is continuous at x = 0, then find f (0).

Answer:

Given:
fx=2x+3sinx3x+2sinx,  x0

If f(x) is continuous at x = 0, then

limx0fx=f0limx02x+3sinx3x+2sinx=f0limx0x2+3sinxxx3+2sinxx=f0limx02+3sinxx3+2sinxx=f0limx02+3sinxxlimx03+2sinxx=f02+3limx0sinxx3+2limx0sinxx=f02+3×13+2×1=f055=f0f0=1

Page No 9.20:

Question 35:

Find the value of k for which
fx=1-cos 4x8x2, whenx0             k      , whenx=0is continuous at x = 0;

Answer:

Given: fx=1-cos4x8x2, when  x0k, when  x=0

If f(x) is continuous at x = 0, then
limx0fx=f0limx01-cos4x8x2=f0limx02sin22x8x2=f022limx0sin22x4x2=f022limx0sin2x2x2=f01×1=f0k=1         f0=k

Page No 9.20:

Question 36:

In each of the following, find the value of the constant k so that the given function is continuous at the indicated point;
(i) fx=1-cos 2kxx2, ifx0               8     , ifx=0at x = 0

(ii) fx=(x-1)tanπx2, ifx1                     k  , ifx=1at x = 1

(iii) fx=k(x2-2x), ifx<0       cos x, ifx0at x = 0

(iv) fx=kx+1, ifxπcos x, ifx>πat x = π

(v) fx=kx+1, ifx53x-5, ifx>5at x = 5

(vi) fx=x2-25x-5,x5          k   ,x=5at x = 5

(vii) fx=kx2,x1 4  ,x<1at x = 1

(viii) fx=k(x2+2), ifx03x+1    , ifx>0

(ix) fx=x3+x2-16x+20x-22,   x2k,  x=2

Answer:

(i) Given:
fx=1-cos 2kxx2, if  x08                , if x=0

If f(x) is continuous at x = 0, then
limx0fx=f0limx01-cos2kxx2=8limx02k2sin2kxk2x2=82k2limx0sinkxkx2=82k2×1=8k2=4k=±2

(ii) Given:
fx=x-1tanπx2, if  x1k, if  x=1

If f(x) is continuous at x = 1, then
limx1fx=f1limx1x-1 tanπx2=k

Putting x-1=y, we get 

limy0 y tanπy+12=klimy0  y tanπy2+π2=klimy0  y tanπ2+πy2=k-limy0  y cotπy2=k-2πlimy0  πy2cosπy2sinπy2=k-2π  limy0cosπy2limy0sinπy2πy2=k-2π×11=kk=-2π

(iii) Given:
fx=kx2-2x, if  x<0cosx, if  x0
We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0kh2+2h=0
(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh0cosh=1

limx0-fxlimx0+fx

Thus, no value of k exists for which fx is continuous at x=0. 


(iv) Given:
fx=kx+1, if  xπcosx, if x>π
We have
(LHL at x = π) = limxπ-fx=limh0fπ-h=limh0kπ-h+1=kπ+1
(RHL at x = π) = limxπ+fx=limh0fπ+h=limh0cosπ+h=cosπ=-1

If f(x) is continuous at x = π, then
limxπ-fx=limxπ+fx
kπ+1=-1k=-2π

(v) Given:
fx=kx+1, if x53x-5, if x>5

We have
(LHL at x = 5) = limx5-fx=limh0f5-h=limh0k5-h+1=5k+1
(RHL at x = 5) = limx5+fx=limh0f5+h=limh035+h-5=10

If f(x) is continuous at x = 5, then
limx5-fx=limx5+fx5k+1=10k=95

(vi) Given:
fx=x2-25x-5,   x5k,  x=5
fx=x-5x+5x-5,   x5k,  x=5
fx=x+5,   x5k,  x=5

If f(x) is continuous at x = 5, then
limx5fx=f5limx5x+5=kk=5+5=10

(vii) Given: fx=kx2,  x14, x<1

We have
(LHL at x = 1) = limx1-fx=limh0f1-h=limh04=4
(RHL at x = 1) = limx1+fx=limh0f1+h=limh0k1+h2=k

If f(x) is continuous at x = 1, then
limx1-fx=limx1+fxk=4

(viii) Given:
fx=kx2+2, if x03x+1, if x>0
We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0k-h2+2=2k
(RHL at x = 0) = limx0+fx=limh0f0+h=limh03h+1=1

If f(x) is continuous at x = 0, then
limx0-fx=limx0+fx2k=1k=12

(ix) Given:
fx=x3+x2-16x+20x-22,   x2k,  x=2
fx=x3+x2-16x+20x2-4x+4,   x2k,  x=2
fx=x+5,   x2k,  x=2

If f(x) is continuous at x = 2, then
limx2fx=f2limx2x+5=kk=2+5=7

Page No 9.20:

Question 37:

Find the values of a and b so that the function f given by
fx=        1 ,if x3ax+b ,     if 3<x<5        7 ,if x5is continuous at x = 3 and x = 5.

Answer:

Given: fx=1, if x3ax+b, if 3<x<57, if x5

We have
(LHL at x = 3) = limx3-fx=limh0f3-h=limh01=1                     

(RHL at x = 3) = limx3+fx=limh0f3+h=limh0a3+h+b=3a+b

(LHL at x = 5) = limx5-fx=limh0f5-h=limh0a5-h+b=5a+b                       

(RHL at x = 5) = limx5+fx=limh0f5+h=limh07=7

If f(x) is continuous at x = 3 and 5, then 

∴ ​limx3-fx =lim  x3+fx    and    limx5-fx =limx5+fx
1=3a+b    ...1  and  5a+b=7    ...2

On solving eqs. (1) and (2), we get
a=3 and b=-8



Page No 9.21:

Question 38:

If fx=x22,if 0x12x2-3x+32,if 1<x2. Show that f is continuous at x = 1.

Answer:

Given: fx=x22,  if 0x12x2-3x+32, if 1<x2

We have
(LHL at x = 1) = limx1-fx=limh0f1-h=limh01-h22=12                     

(RHL at x = 1) = limx1+fx=limh0f1+h=limh021+h2-31+h+32=2-3+32=12

Also, f1=122=12

∴ ​limx1-fx =lim  x1+fx =f1  

Hence, the given function is continuous at x=1.

Page No 9.21:

Question 39:

Discuss the continuity of the f(x) at the indicated points:
(i) f(x) = | x | + | x − 1 | at x = 0, 1.
(ii) f(x) = | x − 1 | + | x + 1 | at x = −1, 1.

Answer:

(i) Given: fx=x+x-1

We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh00-h+0-h-1=1     
              
(RHL at x = 0) = limx0+fx=limh0f0+h=limh00+h+0+h-1=1

Also, f0=0+0-1=0+1=1

Now,

(LHL at x = 1) = limx1-fx=limh0f1-h=limh01-h+1-h-1=1+0=1 
                   
(RHL at x =1) = limx1+fx=limh0f1+h=limh01+h+1+h-1=1+0=1

Also, f1=1+1-1=1+0=1

∴ ​limx0-fx =lim  x0+fx  =  f0     and   lim      x1-fx  = lim       x1+fx  =  f1

Hence, fx is continuous at x=0, 1.

(ii) Given: fx=x-1+x+1

We have
(LHL at x = −1) = limx-1-fx=limh0f-1-h=limh0-1-h-1+-1-h+1=2+0=2    
                
(RHL at x = −1) = limx-1+fx=limh0f-1+h=limh0-1+h-1+-1+h+1=2+0=2

Also, f-1=-1-1+-1+1=-2=2

Now,

(LHL at x = 1) = limx1-fx=limh0f1-h=limh01-h-1+1-h+1=0+2=2 
                   
(RHL at x =1) = limx1+fx=limh0f1+h=limh01+h-1+1+h+1=0+2=2

Also, f1=1+1+1-1=2

∴ ​limx-1-fx =lim  x-1+fx  =  f-1     and  lim      x1-fx  = lim       x1+fx  =  f1

Hence, fx is continuous at x=-1, 1.

Page No 9.21:

Question 40:

Prove that fx=x-xx,x0 2          ,x=0is discontinuous at x = 0

Answer:

The given function can be rewritten as

 fx=x-xx, when x>0x+xx, when x<02,        when x=0

 fx=0, when x>02, when x<02, when x=0

We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh02=2                       

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh00=0
 

∴ ​limx0-fxlimx0+fx

Thus,  f(x) is discontinuous at x = 0
.

Page No 9.21:

Question 41:

If fx=2x2+k,if x0-2x2+k,if x<0, then what should be the value of
k so that f(x) is continuous at x = 0.

Answer:

The given function can be rewritten as

 fx=2x2+k, if x0-2x2+k, if x<0

We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0-2-h2+k=k                       

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh02h2+k=k

If fx is continuous at x=0, then

limx0-fx =limx0+fx = f0limx0-fx =limx0+fx = k

k can be any real number.

Page No 9.21:

Question 42:

For what value of λ is the function
fx=λ(x2-2x),if x0     4x+1 ,if x>0
continuous at x = 0? What about continuity at x = ± 1?

Answer:

The given function f is

If f is continuous at x = 0, then

Therefore, there is no value of λ for which f(x) is continuous at x = 0.

At x = 1,

f (1) = 4x + 1 = 4 × 1 + 1 = 5

Therefore, for any values of λ, f is continuous at x = 1
 

At x = -1, we have

f (-1) = λ1+2=3λ

limx-1λ1+2=3λlimx-1fx=f-1

Therefore, for any values of λf is continuous at x = -1

Page No 9.21:

Question 43:

For what value of k is the following function continuous at x = 2?
fx=2x+1 ;if x<2    k     ;x=23x-1 ;x>2

Answer:

Given: fx=2x+1, if x<2k,  x=23x-1, x>2

We have
(LHL at x = 2) = limx2-fx=limh0f2-h=limh022-h+1=5                     

(RHL at x = 2) = limx2+fx=limh0f2+h=limh032+h-1=5

Also, f2=k

If f(x) is continuous at x = 2, then
limx2-fx =lim  x2+fx=f2

 5=5=k

Hence, for k = 5, fx is continuous at x=2.

Page No 9.21:

Question 44:

Let fx=1-sin3 x3 cos2 x  ,if x<π2           a     ,if x=π2b(1-sin x)(π-2x)2,if x>π2.If f(x) is continuous at x = π2, find a and b.

Answer:

Given: fx=1-sin3x3cos2x, if x<π2a,  if x=π2b1-sinxπ-2x2, if x>π2

We have
(LHL at x = π2) = limxπ2-fx=limh0fπ2-h

                                          =limh01-sin3π2-h3cos2π2-h=limh01-cos3h3sin2h=13limh01-cosh1+cos2h+cosh1-cosh1+cosh=13limh01+cos2h+cosh1+cosh=131+1+11+1=12                    


(RHL at x = π2) = limxπ2+fx=limh0fπ2+h
                                          =limh0b1-sinπ2+hπ-2π2+h2=limh0b1-cosh-2h2=limh02bsin2h24h2=limh02bsin2h216h24=b8limh0sinh2h22=b8×1=b8

Also, fπ2=a

If f(x) is continuous at x = π2 , then
limxπ2-fx =lim  xπ2+fx = fπ2

12 =b8 = a

a=12 and b=4

Page No 9.21:

Question 45:

If the functions f(x), defined below is continuous at x = 0, find the value of k.
fx=1-cos 2x2x2,x<0            k      ,x=0          xx    ,x>0

Answer:

Given: fx=1-cos2x2x2,  x<0k,   x=0xx,  x>0


fx=1-cos2x2x2,  x<0k,   x=01,  x>0


We have
(LHL at x = 0) = limx0-fx=limh0f0-h

                                          =limh01-cos2-h2-h2=limh01-cos2h2h2=12limh02sin2hh2=22limh0sin2hh2=22limh0sinhh2=1×1=1                    

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh01=1
Also, f0=k

If f(x) is continuous at x = 0, then
 ​limx0-fx =lim  x0+fx=f0

1=1=k

Hence, the required value of k is 1.

Page No 9.21:

Question 46:

Find the relationship between 'a' and 'b' so that the function 'f' defined by
fx=ax+1,if x3bx+3,if x>3
is continuous at x = 3.

Answer:

Given: fx=ax+1,  if x3bx+3, if  x>3

We have
(LHL at x = 3) = limx3-fx=limh0f3-h=limh0a3-h+1=3a+1

(RHL at x = 3) = limx3+fx=limh0f3+h=limh0b3+h+3=3b+3                                                                  

 If fx is continuous at x=3, thenlimx3-fx=limx3+fx3a+1=3b+33a-3b=2

Hence, the required relationship between a & b is 3a-3b=2.



Page No 9.34:

Question 1:

Prove that the function fx=sin xx,x<0x+1,x0is everywhere continuous.

Answer:

When x < 0, we have 
fx=sinxx

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function sinxx is continuous at each x < 0.

When x > 0, we have fx=x+1, which is a polynomial function.
Therefore, fx is continuous at each x > 0.

Now,
Let us consider the point x = 0.

Given: fx=sinxx, x<0x+1, x0

We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0sin-h-h=limh0sinhh=1

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh0h+1=1

Also,
f0=0+1=1

limx0-fx=limx0+fx=f0

Thus, fx is continuous at x = 0.

Hence, fx is everywhere continuous.

Page No 9.34:

Question 2:

Discuss the continuity of the function fx=xx,x0  0   ,x=0.

Answer:

Given: fx=xx, x00, x=0

x=x, x0-x, x<0fx=1, x>0-1, x<00, x=0

We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0-1=-1

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh01=1

limx0-fxlimx0+fx

Thus, fx is discontinuous at x = 0.

Page No 9.34:

Question 3:

Find the points of discontinuity, if any, of the following functions:
(i) fx=x3-x2+2x-2,if x1             4            ,if x=1

(ii) fx=x4-16x-2,if x2     16      ,if x=2

(iii) fx=sin xx,if x<02x+3,x0

(iv) fx=sin 3xx,if x0     4      ,if x=0

(v) fx=sin xx+cos x,if x0            5           ,if x=0

(vi) fx=x4+x3+2x2tan-1 x,if x0          10          ,if x=0

(vii) fx=ex-1loge(1+2x),if x0          7           ,if x=0

(viii) fx=x-3,if x1x24-3x2+134,if x<1

(ix) fx=x+3   ,if x-3-2x      ,  if-3<x<36x+2   ,if x>3

(x) fx=x10-1,if x1x2       ,if x>1

(xi) fx=2x ,ifx<0 0   ,if   0x14x ,ifx>1

(xii) fx=sin x-cos x ,if x0        -1         ,if x=0

(xiii) fx=-2 ,ifx-12x ,if-1<x<12  ,ifx1

Answer:

(i)

When x 1, then
fx=x3-x2+2x-2

We know that a polynomial function is everywhere continuous.
So, fx=x3-x2+2x-2 is continuous at each x 1.

At x = 1, we have

(LHL at x = 1) = limx1-fx=limh0f1-h=limh01-h3-1-h2+21-h-2=1-1+2-2=0

(RHL at x = 1) = limx1+fx=limh0f1+h=limh01+h3-1+h2+21+h-2=1-1+2-2=0

Also, f1=4
limx1-fx=limx1+fxf1

Thus, fx is discontinuous at x = 1.

Hence, the only point of discontinuity for fx is x = 1.

(ii)
Given: fx=x4-16x-2, if x216,  if x=2 

When x 2, then
fx=x4-16x-2=x4-24x-2=x2+4x-2x+2x-2=x2+4x+2

We know that a polynomial function is everywhere continuous.
Therefore, the functions x2+4 and x+2 are everywhere continuous.
So, the product function x2+4x+2 is everywhere continuous.
Thus, f(x) is continuous at every x 2.

At x = 2, we have

(LHL at x = 2) = limx2-fx=limh0f2-h=limh02-h2+42-h+2=84=32

(RHL at x = 2) = limx2+fx=limh0f2+h=limh02+h2+42+h+2=84=32

Also, f2=16

limx2-fx=limx2+fxf2

Thus, fx is discontinuous at x = 2.

Hence, the only point of discontinuity for fx is x = 2.

(iii)

When x < 0, then
fx=sinxx

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function sinxx is continuous at each x < 0.

When x > 0, then
fx=2x+3, which is a polynomial function.
Therefore, fx is continuous at each x > 0.

Now,
Let us consider the point x = 0.
Given: fx=sinxx, if x<02x+3,  if x0 

We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0sin-h-h=limh0sinhh=1

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh02h+3=3

limx0-fxlimx0+fx

Thus, fx is discontinuous at x = 0.

Hence, the only point of discontinuity for fx is x = 0.

(iv)

When x 0, then
fx=sin3xx

We know that sin 3x as well as the identity function x are everywhere continuous.
So, the quotient function sin3xx is continuous at each x 0.

Let us consider the point x = 0.
Given: fx=sin3xx, if x04,  if x=0 

We have

(LHL at x = 0) = limx0-fx =limh0f0-h =limh0f-h =limh0sin -3h-h =limh03 sin 3h3h = 3

(RHL at x = 0) = limx0+fx =limh0f0+h =limh0fh =limh0sin 3hh =limh03 sin 3h3h = 3
Also, f0=4

limx0-fx=limx0+fxf0
Thus, fx is discontinuous at x = 0.

Hence, the only point of discontinuity for fx is x = 0.

(v)

When x 0, then
fx=sinxx+cosx

We know that sin x as well as the identity function x both are everywhere continuous.
So, the quotient function sinxx is continuous at each x 0.
Also, cos x is everywhere continuous.
Therefore, sin xx+cos x is continuous at each x 0.

Let us consider the point x = 0.
Given: fx=sinxx+cosx, if x05,  if x=0 

We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0sin-h-h+cos-h=limh0sin-h-h+limh0cos-h=1+1=2

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh0sinhh+cosh=limh0sinhh+limh0cosh=1+1=2
Also, f0=5

limx0-fx=limx0+fxf0
Thus, fx is discontinuous at x = 0.

Hence, the only point of discontinuity for  fx is x = 0.

(vi)

When x 0, then
fx=x4+x3+2x2tan-1x

We know that x4+x3+2x2 is a polynomial function which is everywhere continuous.
Also, tan-1x is everywhere continuous.
So, the quotient function x4+x3+2x2tan-1x is continuous at each x 0.

Let us consider the point x = 0.

Given: fx=x4+x3+2x2tan-1x, if x010,  if x=0 

We have
(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0-h4+-h3+2-h2tan-1-h=limh0h3-h2+2h-tan-1hh=0-1=0

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh0h4+h3+2h2tan-1h=limh0h3+h2+2htan-1hh=01=0
Also, f0=10

limx0-fx=limx0+fxf0
Thus, fx is discontinuous at x = 0.

Hence, the only point of discontinuity for  fx is x = 0.

(vii)
Given: fx=ex-1loge1+2x, if x07, if x=0

We have
limx0fx=limx0ex-1loge1+2x=limx0ex-1x2 loge1+2x2x=12×limx0ex-1xlimx0loge1+2x2x=12
It is given that f0=7

limx0fxf0

Hence, the given function is discontinuous at x = 0 and continuous elsewhere.

(viii)

When x > 1, then
fx=x-3

Since modulus function is a continuous function, fx is continuous for each x > 1.

When x < 1, then
fx=x24-3x2+134

Since, x2  &  3x  are continuous being polynomial functions, x24 & 3x2 will also be continuous.
Also, 134 is continuous being a polynomial function.

x24-3x2+134 is continuous for each x<1.

fx is continuous for each x < 1.

At x = 1, we have
(LHL at x=1) = lim x1-fx=lim h0f1-h=lim h01-h24-31-h2+134=14-32+134=2

(RHL at x=1) = lim x1+fx=lim h0f1+h=lim h01+h-3=-2=2
Also, f1=1-3=-2=2

Thus, lim x1-fx=lim x1+fx=f1

Hence, fx is continuous at x= 1.

Thus, the given function is nowhere discontinuous.

(ix)

At x-3, we have
 fx=x+3

Since modulus function and constant function are continuous, fx=x+3 is continuous for each x-3.

At -3<x<3, we have
fx=-2x
Since polynomial function is continuous and constant function is continuous, fx=-2x is continuous for each-3<x<3.

At x>3, we have
    fx=6x+2

Since polynomial function is continuous and constant function is continuous, fx=6x+2 is continuous for each x>3.

Now, we check the continuity of the function at the point x=3.

We have
(LHL at x=3) = limx3-fx=limh0f3-h=limh0-23-h=-6

(RHL at x=3) = limx3+fx=limh0f3+h=limh063+h+2=20
limx3-fxlimx3+fx

Hence, the only point of discontinuity of the given function is x=3

(x)
Given:

The given function f is defined at all the points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < 1

Case II:

If c = 1, then the left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

It is observed that the left and right hand limit of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case III:

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity of f.

(xi) The given function is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < 0

Case II:

The left hand limit of f at x = 0 is,

The right hand limit of f at x = 0 is,

Therefore, f is continuous at x = 0

Case III:

Therefore, f is continuous at all points of the interval (0, 1).

Case IV:

The left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

It is observed that the left and right hand limits of f at x = 1 do not coincide.

Therefore, f is not continuous at x = 1

Case V:

Therefore, f is continuous at all points x, such that x > 1

Hence, f is not continuous only at x = 1


(xii)

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

Therefore, f is continuous at all points x, such that x 0

Case II:

Therefore, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

(xiii)
The given function
f is

The given function is defined at all points of the real line.

Let c be a point on the real line.

Case I:

Therefore, f is continuous at all points x, such that x < −1

Case II:

The left hand limit of f at x = −1 is,

The right hand limit of f at x = −1 is,

Therefore, f is continuous at x = −1

Case III:

Therefore, f is continuous at all points of the interval (−1, 1).

Case IV:

The left hand limit of f at x = 1 is,

The right hand limit of f at x = 1 is,

Therefore, f is continuous at x = 2

Case V:

Therefore, f is continuous at all points x, such that x > 1

Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.



Page No 9.35:

Question 4:

In the following, determine the value(s) of constant(s) involved in the definition so that the given function is continuous:
(i) fx=sin 2x5x,if x0    3k     ,if x=0

(ii) fx=kx+5,if x2x-1,if x>2

(iii) fx=k(x2+3x),if x<0   cos 2x  ,if x0

(iv) fx= 2 ,if x3ax+b,  if 3<x<5  9  ,if x5

(v) fx=4 ,if x-1ax2+b,if -1<x<0cos x,if x0

(vi) fx=1+px-1-pxx,if -1x<0           2x+1x-2            ,if 0x1

(vii) fx=     5    ,ifx2ax+b,if2<x<10   21   ,ifx10

(viii) fx=k cos xπ-2x ,x<π2        3       ,x=π23 tan 2x2x-π,x>π2

Answer:

(i) Given: fx=sin 2x5x, if x03k, if x=0

If fx is continuous at x = 0, then 
limx0fx=f0

limx0sin 2x5x=f0limx02sin 2x2×5x=f025limx0sin 2x2x=f025=3kk=215

(ii) Given: fx=kx+5, if x2x-1, if x>2

If fx is continuous at x = 2, then 
limx2-fx=limx2+fx

limh0f2-h=limh0f2+hlimh0k2-h+5=limh02+h-12k+5=12k=-4k=-2

(iii) Given: fx=kx2+3x, if x<0cos 2x, if x0

If fx is continuous at x = 0, then 
limx0-fx=limx0+fx

limh0f-h=limh0fhlimh0k-h2-3h=limh0cos 2h0=1          It is not possible

Hence, there does not exist any value of k, which can make the given function continuous.


(iv) Given: fx=2, if x3ax+b, if 3<x<59, if x5

If fx is continuous at x = 3 and 5, then 
limx3-fx =limx3+fx   and  limx5-fx =limx5+fx

limh0f3-h=limh0f3+h  and  limh0f5-h=limh0f5+h limh02=limh0a3+h+b  and  limh0a5-h+b=limh092=3a+b  and  5a+b=92=3a+b  and  5a+b=9a=72 and b=-172

(v)

Given: fx=4, if x-1ax2+b, if -1<x<0cos x, if x0

If fx is continuous at x = −1 and 0, then 
limx-1-fx =limx-1+fx   and  limx0-fx =limx0+fx

limh0f-1-h =limh0f-1+h  and  limh0f-h =limh0fh limh04 =limh0a-1+h2+b and  limh0a-h2+b =limh0cos h4=a+b  and  b=1a=3  and  b=1

(vi)

Given: fx=1+px-1-pxx, if -1x<02x+1x-2, if 0x1

If fx is continuous at x = 0, then 
limx0-fx=limx0+fx

 limh0f-h=limh0fh limh01-ph-1+ph-h=limh02h+1h-2limh01-ph-1+ph1-ph+1+ph-h1-ph+1+ph=limh02h+1h-2limh01-ph-1-ph-h1-ph+1+ph=limh02h+1h-2limh0-2ph-h1-ph+1+ph=limh02h+1h-2limh02p1-ph+1+ph=limh02h+1h-22p2=1-2p=-12

(vii)

Given: fx=5, if x2ax+b, if 2<x<1021, if x10

If fx is continuous at x = 2 and 10,  then 
limx2-fx =limx2+fx   and  limx10-fx =limx10+fx

 limh0f2-h =limh0f2+h   and  limh0f10-h =limh0f10+h limh05 =limh0a2+h+b  and  limh0a10-h+b =limh021 5=2a+b   ...1     and      10a+b=21     ...2On solving eqs. 1 and 2, we geta=2 and b=1


(viii)

 Given: fx=k cos xπ-2x,  x<π23    ,  x=π23 tan 2x2x-π,  x>π2

If fx is continuous at x = π2, then 
limxπ2-fx=fπ2

limh0fπ2-h = fπ2limh0fπ2-h = 3limh0k cos π2-hπ-2π2-h = 1limh0k sin hπ-π+2h = 1limh0k sin h2h = 1k2limh0sin hh = 1k2 = 1k = 2



Page No 9.36:

Question 5:

The function fx= x2/a   ,if 0x<1   a    ,if 1x<22b2-4bx2,if 2x<
is continuous on (0, ∞), then find the most suitable values of a and b.

Answer:

Given:  f is continuous on 0,

∴  f is continuous at x = 1 and 2


At x = 1, we have

limx1-fx =limh0f1-h =limh01-h2a = 1a

limx1+fx=limh0f1+h=limh0a=a

Also,

At x = 2, we have

limx2-fx =limh0f2-h =limh0a = a

limx2+fx =limh0f2+h =limh02b2-4b2+h2 = 2b2-4b2 = b2-2b

f is continuous at x = 1 and 2

limx1-fx=limx1+fx   and limx2-fx=limx2+fx

1a=a  and  b2-2b=aa2=1  and  b2-2b=a a=±1  and  b2-2b=a       ...1

If a = 1, then

b2-2b = 1        From eq. (1)b2-2b-1 = 0b = 2±4+42 = 2±222 = 1±2

If a = −1, then

b2-2b = -1              From eq. (1)b2-2b+1 = 0b-12 = 0b = 1

Hence, the most suitable values of a and b are

a = −1, b = 1  or a = 1, b=1±2

Page No 9.36:

Question 6:

Find the values of a and b so that the function f(x) defined by
fx=x+a2sin x        ,if 0x<π/42x cot x+b           ,if π/4x<π/2a cos 2x-b sin x,if π/2xπ
becomes continuous on [0, π].

Answer:

Given: f is continuous on 0, π.

f is continuous at x = π4 and π2


At x = π4, we have

limxπ4-fx =limh0fπ4-h =limh0π4-h+a2sin π4-h = π4+a2 sin π4 = π4+a

limxπ4+fx =limh0fπ4+h =limh02π4+h cot π4+h+b = π2 cot π4+b = π2+b


At x = π2, we have

limxπ2-fx =limh0fπ2-h =limh02π2-h cot π2-h+b = b

limxπ2+fx =limh0fπ2+h =limh0a cos 2π2+h-b sin π2+h = -a-b

Since f is continuous at x = π4 and x = π2, we get

limxπ2-fx =limxπ2+fx   and limxπ4-fx =limxπ4+fx

-b-a = b  and π4+a = π2+bb = -a2        ...1  and -π4 = b-a     ...2 -π4 = -3a2        Substituting the value of b in eq. 2 a = π6 b = -π12                 From eq.1

Page No 9.36:

Question 7:

The function f(x) is defined as follows:
fx=x2+ax+b ,0x<23x+2         ,2x42ax+5b    ,4<x8
If f is continuous on [0, 8], find the values of a and b.

Answer:

Given: f is continuous on 0, 8.

f is continuous at x = 2 and x = 4

At x = 2, we have
limx2-fx =limh0f2-h =limh02-h2+a2-h+b = 4+2a+b

limx2+fx =limh0f2+h =limh032+h+2 = 8

Also,
At x = 4, we have

limx4-fx =limh0f4-h =limh034-h+2 = 14

limx4+fx =limh0f4+h =limh02a4+h+5b = 8a+5b

f is continuous at x = 2 and x = 4

limx2-fx =limx2+fx  and  limx4-fx =limx4+fx

4+2a+b=8  and 8a+5b=142a+b=4    ...1  and  8a+5b=14    ...2

On simplifying eqs. (1) and (2), we get

a=3 and b=-2

Page No 9.36:

Question 8:

If fx=tanπ4-xcot 2x for x ≠ π/4, find the value which can be assigned to f(x) at x = π/4 so that the function f(x) becomes continuous every where in [0, π/2].

Answer:

When xπ4, tan π4-x and cot 2x are continuous in 0, π2.

Thus, the quotient function tan π4-xcot 2x is continuous in 0,π2 for each xπ4.

So, if fx is continuous at x=π4, then it will be everywhere continuous in 0, π2.

Now,
Let us consider the point x = π4.

Given: fx = tan π4-xcot 2x, xπ4

We have
(LHL at x = π4) = limxπ4-fx =limh0fπ4-h =limh0tanπ4-π4+hcotπ2-2h = limh0tan htan 2h =limh0tan hh2 tan 2h2h = 12limh0tan hhlimh0tan 2h2h = 12

(RHL at x = π4) = limxπ4+fx =limh0fπ4+h =limh0tan π4-π4-hcot π2+2h =limh0tan -h-tan 2h =limh0tan htan 2h =limh0tan hh2 tan 2h2h =12limh0tan hhlimh0tan 2h2h = 12

If fx is continuous at x=π4, then

    lim       xπ4-fx  =limxπ4+fx = fπ4

∴  fπ4=12

Hence, for ​fπ4=12, the function fx will be everywhere continuous in ​0, π2.

Page No 9.36:

Question 9:

Discuss the continuity of the function
fx=2x-1 ,if x<23x2    ,if x2

Answer:

When x < 2, we have 
fx = 2x-1

We know that a polynomial function is everywhere continuous.
So, fx is continuous for each x < 2.

When x>2, we have 
fx=3x2

The functions 3x and 2 are continuous being the polynomial and constant function, respectively.
Thus, the quotient function 3x2 is continuous at each x > 2.

Now,
Let us consider the point x = 2.
Given: fx=2x-1, if x<23x2, if x2

We have
(LHL at x = 2) = limx2-fx =limh0f2-h =limh022-h-1 = 4-1 = 3

(RHL at x = 2) = limx2+fx =limh0f2+h =limh03h+22 = 3

Also,
f2=322=3

∴ limx2-fx=limx2+fx=f2

Thus, fx is continuous at x = 2.

Hence, fx is everywhere continuous.

Page No 9.36:

Question 10:

Discuss the continuity of f(x) = sin | x |.

Answer:

This function f is defined for every real number and f can be written as the composition of two functions as,

f = h o g, where

 hogx = hgx = hx = sin x

It has to be proved first that are continuous functions.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

Therefore, g is continuous at all points x < 0

Case II:

Therefore, g is continuous at all points x > 0

Case III:

Therefore, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = sin x

It is evident that h (x) = sin x is defined for every real number.

Let c be a real number.
Put
x = c + k

If x c, then k 0

h (c) = sin c

So, h is a continuous function.

 fx = hogx = hgx = hx = sin x is a continuous function.



Page No 9.37:

Question 11:

Prove that
fx=sin xx ,x<0x+1 ,x0
is everywhere continuous.

Answer:

When x < 0, we have 
fx=sinxx

We know that sin x as well as the identity function x are everywhere continuous.
So, the quotient function sinxx is continuous at each x < 0.

When x > 0, we have
fx=x+1, which is a polynomial function.
Therefore, fx is continuous at each x > 0.

Now,
Let us consider the point x = 0.
Given: fx = sin xx, x<0x+1, x0

We have
(LHL at x = 0) = limx0-fx =limh0f0-h =limh0f-h =limh0sin -h-h =limh0sin hh = 1

(RHL at x = 0) = limx0+fx =limh0f0+h =limh0fh =limh0h+1 = 1
Also,
f0=0+1=1

limx0-fx=limx0+fx=f0

Thus, fx is continuous at x = 0.

Hence, fx is everywhere continuous.

Page No 9.37:

Question 12:

Show that the function g (x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.

Answer:

Given:

It is evident that g is defined at all integral points.

Let nZ.

Then,

The left hand limit of f at x = n is,

The right hand limit of f at x = n is,

It is observed that the left and right hand limits of f at x = n do not coincide.
i.e. limxn-gx limxn+gx

So, f is not continuous at x = n, nZ

Hence, g is discontinuous at all integral points.

Page No 9.37:

Question 13:

Discuss the continuity of the following functions:
(i) f(x) = sin x + cos x
(ii) f(x) = sin x − cos x
(iii) f(x) = sin x cos x

Answer:

It is known that if g and h are two continuous functions, then g+h, g-h and g×h are also continuous.

It has to proved first that g (x) = sin x and h (x) = cos x are continuous functions.

Let g (x) = sin x

It is evident that g (x) = sin x is defined for every real number.

Let c be a real number. Put x = c + h

If x c, then h 0

So, g is a continuous function.

Let h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x c, then h 0

h (c) = cos c

So, h is a continuous function.

Therefore, it can be concluded that

(i) f (x) = g (x) + h (x) = sin x + cos x is a continuous function.

(ii) f (x) = g (x) − h (x) = sin x − cos x is a continuous function.

(iii) f (x) = g (x× h (x) = sin x cos x is a continuous function.

Page No 9.37:

Question 14:

Show that f (x) = cos x2 is a continuous function.

Answer:

Given: f (x) = cos (x2)

This function f is defined for every real number and f can be written as the composition of two functions as

f = g o h, where g (x) = cos x and h (x) = x2

It has to be first proved that g (x) = cos x and h (x) = x2 are continuous functions.

It is evident that g is defined for every real number.

Let c be a real number.

Then, g (c) = cos c

So, g (x) = cos x is a continuous function.

Now,
h
(x) = x2

Clearly, h is defined for every real number.

Let k be a real number, then h (k) = k2

So, h is a continuous function.

It is known that for real valued functions g and h, such that (g o h) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then, (f o g) is continuous at x = c.

Therefore, is a continuous function.

Page No 9.37:

Question 15:

Show that f (x) = | cos x | is a continuous function.

Answer:

The given function is

This function f is defined for every real number and f can be written as the composition of two functions as,

f = g o h, where

It has to be first proved that are continuous functions.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

So, g is continuous at all points x < 0.

Case II:

So, g is continuous at all points x > 0.

Case III:

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Now, h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number.
Put
x = c + h

If x c, then h 0

h (c) = cos c

So, h (x) = cos x is a continuous function.

It is known that for real valued functions g and h,such that (g o h) is defined at x = c, if g is continuous at x = c and if f is continuous at g (c), then (f o g) is continuous at x = c.

Therefore, is a continuous function.

Page No 9.37:

Question 16:

Find all the points of discontinuity of f defined by f (x) = | x | − | x + 1 |.

Answer:

Given:

The two functions, g and h, are defined as

Then, f = g h

The continuity of g and h is examined first.

Clearly, g is defined for all real numbers.

Let c be a real number.

Case I:

So, g is continuous at all points x < 0.

Case II:

So, g is continuous at all points x > 0.

Case III:

So, g is continuous at x = 0

From the above three observations, it can be concluded that g is continuous at all points.

Clearly, h is defined for every real number.

Let c be a real number.

Case I:

So, h is continuous at all points x < −1.

Case II:

So, h is continuous at all points x > −1.

Case III:

So, h is continuous at x = −1

From the above three observations, it can be concluded that h is continuous at all points of the real line.

So, g and h are continuous functions.

Thus,
f = gh is also a continuous function.

Therefore, f has no point of discontinuity.

Page No 9.37:

Question 17:

Determine if fx=x2 sin1x ,x0      0        ,x=0is a continuous function?

Answer:

The given function f is

It is evident that f is defined at all points of the real line.

Let c be a real number.

Case I:

So, f is continuous at all points x 0

Case II:

limx0-fx=limx0-x2 sin 1x=limx0x2 sin 1xIt is known that -1sin 1x1, x0.-x2  x2 sin 1x  x2limx0-x2  limx0x2 sin 1x  limx0x20  limx0x2 sin 1x  0limx0x2 sin 1x = 0limx0-fx = 0Similarly, limx0+fx =limx0+x2 sin 1x =limx0x2 sin 1x = 0

So, f is continuous at x = 0

From the above observations, it can be concluded that f is continuous at every point of the real line.

Thus, f is a continuous function.

Page No 9.37:

Question 18:

Given the function fx=1x+2. Find the points of discontinuity of the function f(f(x)).

Answer:

ffx=11x+2+2=x+22x+5
So, ffx is not defined at x+2=0 and 2x+5=0If x+2=, then x=-2If 2x+5=0, then x=-52
Hence, the function is discontinuous at x=-52 and -2

Page No 9.37:

Question 19:

Find all point of discontinuity of the function ft=1t2+t-2, where t=1x-1

Answer:

ft=1t2+t-2Now, let u=1x-1fu=1u2+2u-u-2=1u2+u-2=1u+2u-1So, fu is not defined at u=-2 and u=1If u=-2, then-2=1x-12x=1x=12If u=1, then1=1x-1x=2
Hence, the function is discontinuous at x=12, 2



Page No 9.41:

Question 1:

Define continuity of a function at a point.

Answer:

Continuity at a point:

A function fx is said to be continuous at a point xa of its domain, iff limxafx=fa.
Thus, fx is continuous at x=a.limxafx = fa limxa-fx =limxa+fx = fa

Page No 9.41:

Question 2:

What happens to a function f (x) at x = a, if limxa f (x) = f (a)?

Answer:

If fx is a function defined in its domain such that limxafx=fa, then fx becomes continuous at x=a.

Page No 9.41:

Question 3:

Find f (0), so that fx=x1-1-x becomes continuous at x = 0.

Answer:

If fxis continuous at x = 0, then limx0fx=f0          ...(1)

Given: fx=x1-1-x

fx = x1+1-x1-1-x1+1-xfx = x1+1-x1-1-xfx = 1+1-x


limx01+1-x=f0        From eq. (1)
f0=2

So, for f0=2, the function f(x) becomes continuous at x = 0.

Page No 9.41:

Question 4:

If fx=xsin 3x,x0      k     ,x=0is continuous at x = 0, then write the value of k.

Answer:

If fx is continuous at x=0, then
limx0fx=f0

limx0xsin 3x = klimx01sin 3xx = klimx013 sin 3x3x = k131limx0sin 3x3x = kk = 13

Page No 9.41:

Question 5:

If the function fx=sin 10xx, x0 is continuous at x = 0, find f (0).

Answer:

 Given: fx=sin 10xx, x0 is continuous at x=0.
limx0fx=f0

limx0sin 10xx=f0

limx010 sin 10x10x=f010limx0sin 10x10x=f0f0=10



Page No 9.42:

Question 6:

If fx=x2-16x-4,if x4           k  ,if x=4 is continuous at x = 4, find k.

Answer:

Given: fx=x2-16x-4, if x4k, if x=4

If fx is continuous at x=4, then
limx4fx=f4

limx4x2-16x-4=k

limx4x+4x-4x-4=klimx4x+4=kk=8

Page No 9.42:

Question 7:

Determine whether fx=sin x2x,x0       0    ,x=0 is continuous at x = 0 or not.

Answer:

Given: fx=sin x2x, x00, x=0

We have
limx0fx=limx0sin x2x             =limx0x sin x2x2            =limx0sin x2x2 limx0x           =1×0           =0           =f0

 limx0fx=f0

Hence, fx is continuous at x=0.

Page No 9.42:

Question 8:

If fx=1-cos xx2,x0           k     ,x=0is continuous at x = 0, find k.

Answer:

Given: fx=1-cos xx2, x0k, x=0
If fx is continuous at x=0, then
limx0fx=f0

limx01-cos xx2=klimx02sin x224x22=k12limx0sinx22x22=k1×12=kk=12

Page No 9.42:

Question 9:

If fx=sin-1 xx,x0       k      ,x=0is continuous at x = 0, write the value of k.

Answer:

Given, fx=sin-1xx, x0k, x=0

If fx is continuous at x=0, then

limx0fx=f0

limx0sin-1xx=f0

limx0sin-1xx=kk=1     limx0sin-1xx =1        

Page No 9.42:

Question 10:

Write the value of b for which fx=      5x-40<x14x2+3bx1<x<2 is continuous at x = 1.

Answer:

Given: fx=5x-4, 0<x14x2+3bx, 1<x<2

If fx is continuous at x=1, then
limx1-fx=limx1+fx=f1                   ...(1)

Now,
limx1-fx=limh0f1-h=limh051-h-4=5-4=1

limx1+fx=limh0f1+h=limh041+h2+3b1+h=4+3b

Also, f1=51-4=1


limx1-fx =limx1+fx = f1           From eq. (1)1=4+3b=1

1=4+3b-3=3bb=-1

Thus, for b=-1, the function fx is continuous at x=1.

Page No 9.42:

Question 1:

The function fx=4-x24x-x3
(a) discontinuous at only one point
(b) discontinuous exactly at two points
(c) discontinuous exactly at three points
(d) none of these

Answer:

(C) discontinuous exactly at three points

Given: fx=4-x24x-x3
fx=4-x2x4-x2
fx=1x, x0 and 4-x20 or x0, ±2

Clearly, fx is defined and continuous for all real numbers except 0, ±2.

Therefore, fx is discontinuous exactly at three points.

Page No 9.42:

Question 2:

If f (x) = | xa | ϕ (x), where ϕ (x) is continuous function, then
(a) f' (a+) = ϕ (a)
(b) f' (a) = −ϕ (a)
(c) f' (a+) = f' (a)
(d) none of these

Answer:

(a)  f'a+=ϕa
(b)  f'a-=-ϕa

Here, fx=x-a ϕx

 f'a+ =limh0fa+h-fah =limh0h+a-aϕa+h-a-aϕah =limh0h ϕa+hh=limh0ϕa+h=ϕa

Also,

f'a- =limh0fa-h-fah=limh0a-h-a ϕa-h-a-a ϕah=limh0-h ϕa-hh=limh0-ϕa-h=-ϕa

Page No 9.42:

Question 3:

If fx=log10 x, then at x = 1
(a) f (x) is continuous and f' (1+) = log10 e
(b) f (x) is continuous and f' (1+) = log10 e
(c) f (x) is continuous and f' (1) = log10 e
(d) f (x) is continuous and f' (1) = −log10 e

Answer:

(a) f (x) is continuous and f' (1+) = log10e
(d) f (x) is continuous and f' (1) = log10e

Given:
fx = log10 x=loge xloge 10=loge x×log10 e=log10 e loge x

f'1+=limh0f1+h-f1h= limh0log10 e loge1+h-log10 e loge 1h=log10 elimh0loge 1+hh=log10e×1=log10 e


Also,

f'1-=limh0f1-h-f1h=limh0log10 e loge 1-h-log10 e loge 1h=-log10 elimh0loge 1-h-h=-log10 e×1=-log10 e



 

Page No 9.42:

Question 4:

If fx=36x-9x-4x+12-1+cos x,x0        k                      ,x=0is continuous at x = 0, then k equals
(a) 162 log 2 log 3
(b) 162 ln 6
(c) 162 ln 2 ln 3
(d) none of these

Answer:


c 162 ln2 ln3

Given: fx=36x-9x-4x+12-1+cosx, x0k, x=0
If fx is continuous at x=0, then 
limx0fx=f0

limx036x-9x-4x+12-1+cos x=klimx09x4x-9x-4x+12-1+cos x=klimx09x4x-1-14x-12-1+cos x=klimx09x-14x-12-1+cos x=klimx09x-14x-12-2cos x2=klimx09x-14x-121-cos x2=klimx09x-14x-122sin2x4=klimx089x-14x-1162x2sin2x4x2=klimx089x-14x-12x2sin2x4x216=k82limx09x-14x-1x2sin2x4x42=k82limx09x-1x limx04x-1xlimx0sin x4x42=k82×ln 9×ln 41=k               limx0ax-1x=a82×2 ln 3×2 ln 21=k         322×ln 3 ln 21=k3222×ln 3 ln 21=kk=162 ln 2 ln 3



Page No 9.43:

Question 5:

If f (x) defined by fx=x2-xx2-x,x0, 1   1         ,x=0  -1      ,x=1 then f (x) is continuous for all
(a) x
(b) x except at x = 0
(c) x except at x = 1
(d) x except at x = 0 and x = 1.

Answer:

(d) x except at x = 0 and x = 1.

Given: fx=x2-xx2-x, x0, 11      , x=0-1    , x=1


fx=x x-1xx-1, x0, 11      , x=0-1     , x=1


fx=1, x>11, x<0-1, 0<x<11, x=0-1, x=1


fx=1, x>11, x0-1,  0<x1

So, 

limx0-fx=limh0f-h=1

Also,

limx0+fx=limh0fh=-1

limx0+fxlimx0-fx

Thus, fx is discontinuous at x=0.

Now,

limx1-fx=limh0f1-h=-1

limx1+fx=limh0f1+h=1

limx1+fxlimx1-fx

So, fx is discontinuous at x=1.

Hence, fx is continuous for all except at x=0 and x = 1.

Page No 9.43:

Question 6:

If fx=1-sin xπ-2x2.log sin xlog1+π2-4πx+4x2,xπ2                  k                                          ,x=π2
is continuous at x = π/2, then k =
(a) -116

(b) -132

(c) -164

(d) -128

Answer:

c -164

If fx is continuous at x=π2, then
limxπ2fx=fπ2If π2-x=t, thenlimt0 fπ2-t=fπ2limt01-sin π2-t4t2×log sin π2-tlog1+π2-4ππ2-t+4π2-t2=klimt01-cos t4t2×log cos tlog1+π2-2π2+4πt+4π24+t2-πt=klimt01-cos t4t2×log cos tlog1-π2+4πt+π2+4t2-4πt=klimt01-cos t4t2×log cos tlog 1+4t2=klimt02 sin2 t216×t24×log cos tlog 1+4t2=k216limt0sin2 t2t24×log cos t4t2 log 1+4t24t2=k18limt0sin2 t2t22×log cos t4t2log 1+4t24t2=k18limt0sin2 t2t22×log 1-sin2 t4t2log1+4t24t2=k18limt0sin2t2t22×log1-sin2t8t2log1+4t24t2=k164limt0sin2t2t22×log1-sin2tt2log1+4t24t2=k164lim t0sint2t22×limt0log1-sin2tt2limt0log1+4t24t2=k1641×limt0-sin2t log 1-sin2tt2-sin2t=k-164limt0sin2t log 1-sin2tt2-sin2t=k-164limt0sintt2limt0log 1-sin2t-sin2t=k-164limt0sintt2limt0log1-sin2t-sin2t=k    k=-164           lim x0log1-xx=1

Page No 9.43:

Question 7:

If f (x) = (x + 1)cot x be continuous at x = 0, then f (0) is equal to
(a) 0
(b) 1/e
(c) e
(d) none of these

Answer:

(c) e

Suppose fx is continuous at x=0.


Given: fx=x+1cotx

log fx=cot x log x+1       Taking log on both sideslimx0log fx=limx0cot x log x+1limx0log fx=limx0log x+1tan xlimx0log fx=limx0log x+1xtan xxlimx0log fx=limx0log x+1xlimx0tan xxlog limx0fx=limx0log x+1xlimx0tan xx           fx is continuous at x=0log limx0fx=1limx0fx=ef0=e           fx is continuous at x=0

Page No 9.43:

Question 8:

If fx=log1+ax-log1-bxx,x0                     k                    ,x=0
and f (x) is continuous at x = 0, then the value of k is
(a) ab
(b) a + b
(c) log a + log b
(d) none of these

Answer:

b a+b

Given: fx=log1+ax-log1-bxx, x0k, x=0

If f(x) is continuous at x = 0, then
limx0fx=f0

limx0log1+ax-log1-bxx=k

limx0alog1+axax-blog1-bxbx=kalimx0log1+axax-blimx0log1-bxbx=kalimx0log1+axax+blimx0log1-bx-bx=ka×1+b×1=k                  limx0log1+xx=1k=a+b

Page No 9.43:

Question 9:

The function fx=e1/x-1e1/x+1,x0       0       ,x=0
(a) is continuous at x = 0
(b) is not continuous at x = 0
(c) is not continuous at x = 0, but can be made continuous at x = 0
(d) none of these

Answer:

(b) is not continuous at x = 0

Given: fx=e1x-1e1x+1, x00, x=0

We have
limx0fx=limx0e1x-1e1x+1

If e1x=t, then
x0,  t

limx0fx=limtt-1t+1=limt1-1t1+1t=1-01+0=1
Also, f0=0

 limx0fxf0

Hence, fx is discontinuous at x=0.

Page No 9.43:

Question 10:

Let fx=x-4x-4+a,x<4   a+b       ,x=4x-4x-4+b,x>4.
Then, f (x) is continuous at x = 4 when
(a) a = 0, b = 0
(b) a = 1, b = 1
(c) a = −1, b = 1
(d) a = 1, b = −1.

Answer:

(d) a = 1, b = −1.

Given: fx=x-4x-4+a, if x<4a+b, if x=4x-4x-4+b, if x>4

We have
(LHL at x = 4) = limx4-fx=limh0f4-h

                       =limh04-h-44-h-4+a=limh0-h-h+a=a-1
                        

(RHL at x = 4) = limx4+fx=limh0f4+h

                       =limh04+h-44+h-4+b=limh0hh+b=b+1

Also,
f4=a+b

If f(x) is continuous at x = 4, then 

limx4-fx=limx4+fx=f4

a-1=b+1=a+b
a-1=a+b and b+1=a+b
b=-1 and a=1

Page No 9.43:

Question 11:

If the function fx=cos x1/x,x0    k           ,x=0 is continuous at x = 0, then the value of k is
(a) 0
(b) 1
(c) −1
(d) e.

Answer:

(b)
Given: fx=cosx1xk, x=0, x0

Iffx is continuous at x=0, then
limx0fx=f0

limx0cos x1x=kIf limxafx=1 and limxagx=0, then limxafxgx=elimxafx-1×gxelimx0cos x-1x= ke0=k           limx0cos x-1x=0k=1

Page No 9.43:

Question 12:

Let f (x) = | x | + | x − 1|, then
(a) f (x) is continuous at x = 0, as well as at x = 1
(b) f (x) is continuous at x = 0, but not at x = 1
(c) f (x) is continuous at x = 1, but not at x = 0
(d) none of these

Answer:

(a) f (x) is continuous at x = 0, as well as at x = 1

Since modulus function is everywhere continuous , x and x-1 are also everywhere continuous.

Also, 
It is known that if f and g are continuous functions, then g will also be continuous.

Thus, ​ ​x+x-1 is everywhere continuous.

Hence, fx is continuous at x=0 and x=1.



Page No 9.44:

Question 13:

Let fx=x4-5x2+4x-1 x-2,x1, 2             6            ,x=1           12           ,x=2. Then, f (x) is continuous on the set
(a) R
(b) R − {1}
(c) R − {2}
(d) R − {1, 2}

Answer:

(d) R − {1, 2}

Given: fx=x4-5x2+4x-1x-2, x1, 26,    x=112,   x=2

Now,x4-5x2+4=x4-x2-4x2+4=x2x2-1-4x2-1=x2-1x2-4=x-1x+1x-2x+2fx=x-1x+1x-2x+2x-2x-1, x1, 26,  x=112,   x=2        


fx=x+1x+2,   x<1-x+1x+2,     1<x<2x+1x+2,  x>26,           x=112,          x=2


So, 

limx1-fx=limh0f1-h=limh01-h+11-h+2=2×3=6

limx1+fx=limh0f1+h=-limh01+h+11+h+2=-2×3=-6

Also,

limx2-fx=limh0f2-h=-limh02-h+12-h+2=-12

limx2+fx=limh0f2+h=limh02+h+12+h+2=12

Thus, limx1+fxlimx1-fx andlimx2+fxlimx2-fx

Therefore, the only points of discontinuities of the function fx are x=1 and x=2 .

Hence, the given function is continuous on the set  R − {1, 2}
.
.

Page No 9.44:

Question 14:

If fx=sin (a+1) x+sin xx ,x<0                 c                 ,x=0      x+bx2-xbxx ,x>0is continuous at x = 0, then
(a) a = -32, b = 0, c = 12
(b) a = -32, b = 1, c = -12
(c) a = -32, bR − {0}, c = 12
(d) none of these

Answer:

(c) a = -32, bR − {0}, c12

The given function can be rewritten as
 

fx=sin a+1 x+x sin xx, for x<0c                              , for x=0x+bx2-xbx32       , for x>0
 

fx=sin a+1x+ sin xx, for x<0c                              , for x=01+bx-1bx       , for x>0

We have
(LHL at x = 0) = limx0-fx =limh0f0-h =limh0f-h

                      =limh0-sin a+1h- sin -hh=limh0-sin a+1hh-sin hh

                      =-a+1limh0sin a+1ha+1h-lim h0sin hh=-a-1
                        

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh 

                       =limh01+bh-1bh=limh0bhbh1+bh+1=limh011+bh+1=12

Also, f0=c

If fx is continuous at x = 0, then
 ​limx0-fx=limx0+fx=f0

-a-1 = 12=c
-a-1 = 12 and c=12
a=-32, c=12

Now, 1+bx-1bx exists only if bx0b0.

Thus, bR-0.

 

 

 

 

 

 

 

 

 

 

 

 

Page No 9.44:

Question 15:

If fx=mx+1   ,xπ2sin x+n,x>π2is continuous at x=π2, then
(a) m = 1, n = 0

(b) m=nπ2+1

(c) n=mπ2

(d) m=n=π2

Answer:

(c) n=mπ2n=mπ2


Here,

fπ2=2+1

We have
(LHL at x=π2) = lim    xπ2- fx =lim   h0fπ2-h=lim   h0mπ2-h+1=2+1

(RHL at x=π2) = lim    xπ2+ fx =lim   h0fπ2+h=lim   h0sinπ2+h+n=n+1

Thus,
If fx is continuous at x=π2, then 
lim    xπ2- fx =lim    xπ2+ fx 


2+1=n+12=n

Page No 9.44:

Question 16:

The value of f (0), so that the function
fx=a2-ax+x2-a2+ax+x2a+x-a-x becomes continuous for all x, given by
(a) a3/2
(b) a1/2
(c) −a1/2
(d) −a3/2

Answer:

(c) -a12

Given: fx=a2-ax+x2-a2+ax+x2a+x-a-x

 fx=a2-ax+x2-a2+ax+x2a2-ax+x2+a2+ax+x2a+x-a-xa2-ax+x2+a2+ax+x2 fx=a2-ax+x2-a2+ax+x2a+x-a-xa2-ax+x2+a2+ax+x2 fx=-2axa+x+a-xa+x-a-xa2-ax+x2+a2+ax+x2a+x+a-x fx=-2axa+x+a-xa+x-a+xa2-ax+x2+a2+ax+x2 fx=-2axa+x+a-x2xa2-ax+x2+a2+ax+x2 fx=-aa+x+a-xa2-ax+x2+a2+ax+x2

If fx is continuous for all x, then it will be continuous at x = 0 as well.

So, if fx is continuous at x = 0, then
limx0fx=f0

 limx0-aa+x+a-xa2-ax+x2+a2+ax+x2=f0 -2aaa2+a2=f0 -2aaa+a=f0 f0=-a

Page No 9.44:

Question 17:

The function
fx=  1   ,x11n2 ,1n<x<1n-1, n=2,3, ...  0   ,x=0
(a) is discontinuous at finitely many points
(b) is continuous everywhere
(c) is discontinuous only at x=±1n, nZ − {0} and x = 0
(d) none of these

Answer:

Given: fx=1,             x11n2, 1n<x<1n-10,            x=0


fx=1,            -1x11n2, 1n<x<1n-10,            x=0


Case 1: x>1 or x<-1 and x>1

Here,
fx=1, which is the constant function
So, fx is continuous for all x1 or x-1 and x1.

Case 2: 1n<x<1n-1, n=2, 3, 4, ...

Here,
fx=1n2, n=2, 3, 4, ... fx=1n2, n=2, 3, 4, ..., which is also a constant function.

So, fx is continuous for all 1n<x<1n-1, n=2, 3, 4, ....

Case 3: Consider the points x = -1 and x = 1.

We have
LHL at x=-1=limx-1-fx=limx-1-1=1RHL at x=-1=limx-1+fx=limx-1+14=14          fx=14 for -1<x<12, when n=2Clearly, limx-1-fxlimx-1+fx at x=-1So, fx is discontinuous at x=-1.

Similarly,  f(x) is discontinuous at x = 1.

Case 4: Consider the point x = 0.

We have
limx0-fx=limh0f1n-h=limh0f1n-h=1n-12

limx0+fx=limh0f1n+h=limh0f1n+h=1n2

limx0+fxlimx0-fx

Thus, fx is discontinuous at x=0.

At = 0, we have
limx0-fx0=f0

So, fx is discontinuous at x=0.

Case 5: Consider the point x=1n, n=2, 3, 4, ...

We have
limx1n-fx=limh0f1n-h=limh0f1n-h=1n-12

limx1n+fx=limh0f1n+h=limh0f1n+h=1n2

limx1n+fxlimx1n-fx

Hence, fx is discontinuous only at  x=±1nnZ-0 and x=0.

Page No 9.44:

Question 18:

The value of f (0), so that the function
fx=27-2x1/3-39-3 243+5x1/5x0 is continuous, is given by
(a) 23
(b) 6
(c) 2
(d) 4

Answer:

(c) 2

For f(x) to be continuous at x = 0, we must have
 limx0fx=f0f0= limx0fx=limx027-2x13-39-3243+5x15f0=limx027-2x13-2713324315-243+5x15=13limx027-2x13-2713x24315-243+5x15x=-13limx027-2x13-2713x243+5x15-24315x=215limx027-2x13-2713-2x243+5x15-243155x=215limx027-2x13-271327-2x-27243+5x15-24315243+5x-243=215×13×27-2315×243-45=215×13×1272315×124345=2
 

Page No 9.44:

Question 19:

The value of f (0) so that the function
fx=2-256-7x1/85x+321/5-2, x ≠ 0 is continuous everywhere, is given by
(a) −1
(b) 1
(c) 26
(d) none of these

Answer:

(d) none of these

Given:  fx=2-256-7x185x+3215-2

For fx to be continuous at x = 0, we must have
limx0fx=f0
f0=limx0fx=limx02-256-7x185x+3215-2f0=limx025618-256-7x185x+3215-3215=-limx0256-7x18-25618x5x+3215-3215x=-75limx0256-7x18-256187x5x+3215-32155x=75limx0256-7x18-25618256-7x-2565x+3215-32155x+32-32=75×18×256-7815×32-45=75×18×2415×27=764



Page No 9.45:

Question 20:

fx=1+px-1-pxx,-1x<02x+1x-2                       ,0x1
is continuous in the interval [−1, 1], then p is equal to
(a) −1
(b) −1/2
(c) 1/2
(d) 1

Answer:

(b) -12
Given: fx=1+px-1-pxx, if -1x<02x+1x-2, if 0x1

If fx is continuous at x = 0,  then 
limx0-fx=limx0+fx

 limh0f-h=limh0fh limh01-ph-1+ph-h=limh02h+1h-2limh01-ph-1+ph1-ph+1+ph-h1-ph+1+ph=limh02h+1h-2limh01-ph-1-ph-h1-ph+1+ph=limh02h+1h-2limh0-2ph-h1-ph+1+ph=limh02h+1h-2limh02p1-ph+1+ph=limh02h+1h-22p2=1-2p=-12

Page No 9.45:

Question 21:

The function fx=x2/a       ,0x<1   a           ,1x<22b2-4bx2,2x<
is continuous for 0 ≤ x < ∞, then the most suitable values of a and b are
(a) a = 1, b = −1
(b) a = −1, b = 1 + 2
(c) a = −1, b = 1
(d) none of these

Answer:

(c) a = -1, b = 1

Given: fx is continuous for 0 ≤ x < ∞.

This means that fx is continuous for x=1, 2
.

Now,


If fx is continuous at x = 1, then
limx1-fx=f1limh0f1-h=a1-h2a=a1a=aa2=1a=±1

If fx is continuous at x = 2, then​

limx2-fx=f2limh0f2-h=2b2-4b2limh0a=b2-2ba=b2-2bb2-2b-a=0

∴ For a = 1, we have

b2-2b-1=0b=2±4-4-12=1±2

Also,
For a = −1, we have

b2-2b+1=0b-12=0b=1

Thus, a=-1 and b=1

Page No 9.45:

Question 22:

If fx=1-sin xπ-2x2, when x ≠ π/2 and f (π/2) = λ, then f (x) will be continuous function at x = π/2, where λ =
(a) 1/8
(b) 1/4
(c) 1/2
(d) none of these

Answer:

(a) 18

If fx is continuous at x=π2, then
limxπ2fx=fπ2

limxπ21-sin xπ-2x2=fπ2     ...(1)

Suppose π2-x=t, then

limt01-sin π2-t2t2=fπ2                 From eq. (1)limt01-cos t4t2=fπ214limt02 sin2 t2t2=fπ214limt024 sin2 t2t24=fπ218limt0sin2 t2t24=fπ218limt0sin t2t22=fπ2fπ2=λ=18

Page No 9.45:

Question 23:

The value of a for which the function
fx=4x-13sinx/a log 1+x2/3,x012log 43                            ,x=0may be continuous at x = 0 is
(a) 1
(b) 2
(c) 3
(d) none of these

Answer:

(d) none of these

For f(x) to be continuous at x=0, we must have
limx0fx=f0

limx04x-13sinxa log1+x23=12log 43


limx04x-13x3sinxalog1+x23x3=12log 43limx0a4x-1x3sinxaxalog1+x23x2=12log 433alimx04x-1x3sinxaxalog1+x23x23=12log 433alimx04x-1x3limx0sinxaxalimx0log1+x23x23=12log 433alog 43=12log 43               limx0ax-1x=log a, limx0log1+xx=1 andlimx0sin xx=1a=4

Page No 9.45:

Question 24:

The function f (x) = tan x is discontinuous on the set
(a) {n π : nZ}
(b) {2n π : nZ}
(c) 2n+1π2: n  Z
(d) nπ2: n  Z

Answer:

c 2n+1π2: nZ

When tan2n+1π2= tannπ+π2=-cotnπ, it is not defined at the integral points. nZ

Hence, fx is discontinuous on the set 2n+1π2: nZ.

Page No 9.45:

Question 25:

The function fx=sin 3xx,x0 k2       ,x=0 is continuous at x = 0, then k =
(a) 3
(b) 6
(c) 9
(d) 12

Answer:

(b) 6

Given:  fx=sin 3xxk2, x=0, x0


If fx is continuous at x=0, then
limx0fx=f0

limx0sin 3xx=f03limx0sin 3x3x=k23×1=k2k2=3k=6

Page No 9.45:

Question 26:

If the function
fx=2x-sin-1 x2x+tan-1 x
is continuous at each point of its domain, then the value of f (0) is
(a) 2

(b) 13

(c) -13

(d) 23

Answer:

(b) 13

Given:  fx=2x-sin-1x2x+tan-1x

If f(x) is continuous at x = 0, then

limx0fx=f0limx02x-sin-1x2x+tan-1x=f0limx0x2-sin-1xxx2+tan-1xx=f0limx02-sin-1xx2+tan-1xx=f02-limx0sin-1xx2+limx0tan-1xx=f02-12+1=f0f0=13

Page No 9.45:

Question 27:

The value of b for which the function
fx=5x-4       ,0<x14x2+3bx ,1<x<2is continuous at every point of its domain, is
(a) −1

(b) 0

(c) 133

(d) 1

Answer:

(a) −1

Given: fx is continuous at every point of its domain. So, it is continuous at x=1.

limx1+fx=f1limh0f1+h=f1limh041+h2+3b1+h=51-44+3b=1-3=3bb=-1

 

Page No 9.45:

Question 28:

If fx=11-x, then the set of points discontinuity of the function f (f(f(x))) is
(a) {1}
(b) {0, 1}
(c) {−1, 1}
(d) none of these

Answer:

(b) {0, 1}

Given: fx=11-x

Clearly, f:R-1R

Now, 
ffx=f11-x=11-11-x=1-x-x=x-1x
fof:R-0, 1R 

Now,

fffx=fx-1x=11-x-1x=x

fofof:R-0, 1R 

Thus, ​fffx is not defined at x=0, 1.

Hence, ​fffx is discontinuous at {0, 1}.

 



Page No 9.46:

Question 29:

Let fx=tanπ4-xcot 2x, xπ4. The value which should be assigned to f (x) at x=π4, so that it is continuous everywhere is
(a) 1
(b) 1/2
(c) 2
(d) none of these

Answer:

(b) 12

If fx is continuous at x=π4, then 
limxπ4fx=fπ4

limxπ4tan π4-xcot 2x=fπ4


If π4-x=y, then xπ4 and y0.

 limy0tan ycot 2π4-y=fπ4limy0tanycotπ2-2y=fπ4limy0tan ytan 2y=fπ4limy0tan yytan 2yy=fπ4limy0tan yy2 tan 2y2y=fπ412limy0tan yytan 2y2y=fπ412limy0tan yylimy0tan 2y2y=fπ41211=fπ4fπ4=12

 

Page No 9.46:

Question 30:

The function fx=x3+x2-16x+20x-2 is not defined for x = 2. In order to make f (x) continuous at x = 2, Here f (2) should be defined as
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

Here,

x3+x2-16x+20=x3-2x2+3x2-6x-10x+20=x2x-2+3xx-2-10x-2=x-2x2+3x-10=x-2x-2x+5=x-22 x+5

So, the given function can be rewritten as

 fx=x-22x+5x-2

fx=x-2x+5

If fx is continuous at x=2, then
limx2fx=f2

limx2x-2x+5=f2f2=0

Hence, in order to make fx continuous at x=2, f2 should be defined as 0.

Page No 9.46:

Question 31:

If fx=a sinπ2x+1,x0tan x-sin xx3,x>0is continuous at x = 0, then a equals
(a) 12

(b) 13

(c) 14

(d) 16

Answer:

(a) 12


Given: fx=a sin π2x+1 , x0tan x-sin xx3 , x>0
We have

(LHL at x = 0) = limx0-fx=limh0f0-h=limh0f-h=limh0a sin π2-h+1=a sin π2=a

(RHL at x = 0) = limx0+fx=limh0f0+h=limh0fh=limh0tan h-sin hh3

                                                                    =limh0sin hcos h-sin hh3=limh0sin hcos h1-cos hh3=limh01-cos h tan hh3=limh02 sin2 h2 tan h4×h24×h=24limh0sin2 h2 tan hh24×h=12limh0sin h2h22×limh0tanhh=12×1×1=12

 If fx is continuous at x=0, thenlimx0-fx=limx0+fxa=12

Page No 9.46:

Question 32:

If fx=ax2+b ,0x<1   4        ,x=1x+3     ,1<x2,
then the value of (a, b) for which f (x) cannot be continuous at x = 1, is
(a) (2, 2)
(b) (3, 1)
(c) (4, 0)
(d) (5, 2)

Answer:


(d) (5, 2)

If f(x) is continuous at x = 1, then
limx1-fx=f1


limh0f1-h=4               f1=4limh0a1-h2+b=4 a+b=4   

Thus, the possible values of (a, b) can be 2, 2, 3, 1, 4, 0. But a, b5, 2.

Hence, for a, b=5, 2fx cannot be continuous at = 1.

Disclaimer: The question in the book has some error. The solution here is created according to the question given in the book.

Page No 9.46:

Question 33:

If the function f (x) defined by
fx=log 1+3x-log 1-2xx,x0                    k                       ,x=0is continuous at x = 0, then k =
(a) 1
(b) 5
(c) −1
(d) none of these

Answer:

b 5

Given: fx=log 1+3x-log 1-2xx, x0k, x=0

If f(x) is continuous at x = 0, then​limx0fx=f0.

limx0log1+3x-log1-2xx=k

limx03 log 1+3x3x-2 log 1-2x2x=k3limx0log 1+3x3x-2limx0log 1-2x2x=k3limx0log 1+3x3x+2limx0log 1-2x-2x=k3×1+2×1=k                            limx0log 1+xx=1k=3+2=5

Page No 9.46:

Question 34:

If fx=1-cos 10xx2         , x<0         a                     ,x=0x625+x-25,x>0
then the value of a so that f (x) may be continuous at x = 0, is
(a) 25
(b) 50
(c) −25
(d) none of these

Answer:

(b) 50

If fx is continuous at x=0, then
limx0-fx=f0

limh0f-h=f0

limh01-cos -10h-h2=f0limh01-cos 10hh2=f0limh02 sin2 5hh2=alimh02×25sin2 5h25h2=a50limh0sin2 5h5h2=a50limh0sin 5h5h2=aa=50

Page No 9.46:

Question 35:

If fx=x sin1x, x0, then the value of the function at x = 0, so that the function is continuous at x = 0, is
(a) 0
(b) −1
(c) 1
(d) indeterminate

Answer:

(a) 0

Given:
fx=x sin1x,  x0

Here,
limx0x sin 1x=limx0x limx0sin1x=0×limx0sin1x=0


If f(x) is continuous at x = 0, thenlimx0fx=f0.

f0=0

Page No 9.46:

Question 36:

The value of k which makes
fx=sin1x,x0  k      ,x=0continuous at x = 0, is
(a) 8
(b) 1
(c) −1
(d) none of these

Answer:

(d) none of these

If fx is continuous at x=0, then 
limx0fx=f0limx0sin1x=kBut limx0sin1x does not exist. Thus, there does not exist any k that makes fx a continuous function.

 

Page No 9.46:

Question 37:

The values of the constants a, b and c for which the function
fx=1+ax1/x      ,x<0  b                    ,x=0x+c1/3-1x+11/2-1,x>0may be continuous at x = 0, are

(a) a=loge23, b=-23, c=1

(b) a=loge23, b=23, c=-1

(c) a=loge23, b=23, c=1

(d) none of these

Answer:

c a=log23, b=23, c=1


Given: fx=1+ax1x, x<0b,  x=0x+c13-1x+112-1, x>0

If fx is continuous at x=0, then
limx0-fx=limx0+fx =f0

limx0-fx=f0limh0f-h=f0limh01-ah-1h=f0limh0alog 1-ah-ah=log ba×1=log b           limx0log 1+xx=1a=log b

Also,
limx0+fx =f0

limx0+fx=f0limh0fh=f0limh0h+c13-1h+112-1=f0limh0h+c13-1h+112-1×h+112+1h+112+1=f0limh0h+c13-1h×h+112+1=blimh0h+c13-1h×limh0h+112+1=blimh0h+c13-1h×2=blimh0h+c13-113h+c-c=b2c13-13=b2        limxaxn-anx-a=nan-1, where c=113=b223=b a=log23



Page No 9.47:

Question 38:

The points of discontinuity of the function
fx=2x  ,0x14-2x ,1<x<522x-7 ,52x4is are

(a) x = 1, x=52

(b) x=52

(c) x=1,52,4

(d) x = 0, 4

Answer:

b x=52=5

If 0x1, then  fx=2x.

Since fx=2x is a polynomial function, it is continuous.
Thus, fx is continuous for every 0x1.

If 1<x<52, then  fx=4-2x. Since 2x is a polynomial function and 4 is a constant function, both of them are continuous. So, their difference will also be continuous.
Thus, fx is continuous for every 1<x<52.

If 52x4, then  fx=2x-7. Since 2x is a polynomial function and 7 is continuous function, their difference will also be continuous.
Thus, fx is continuous for every 52x4.

Now,
Consider the point x=1. Here,

limx1-fx=limh0f1-h=limh021-h=2

limx1+fx=limh0f1+h=limh04-21+h=2

Also, f1=21=2

limx1-fx=limx1+fx=f1

Thus, fx is continuous at x=1.

Now,
Consider the point x=52. Here,

limx52-fx=limh0f52-h=limh04-252-h=-1

limx52+fx=limh0f52+h=limh0252-h-7=-2

limx52+fx  limx52-fx

Thus, fx is discontinuous at x=52.



 

 

 

 

 

 

 

 

Page No 9.47:

Question 39:

If fx=1-sin2 x3 cos2 x   ,x<π2          a          ,x=π2b 1-sin xπ-2x2,x>π2. Then, f (x) is continuous at x=π2, if
(a) a=13, b = 2

(b) a=13, b=83

(c) a=23, b=83

(d) none of these

Answer:

(b) a=13 , b=83


Given:  fx=1-sin2x3cos2x, if x<π2a,  if x=π2b1-sinxπ-2x2, if x>π2

We have
(LHL at x = π2) = limxπ2-fx=limh0fπ2-h

                                          =limh01-sin2 π2-h3 cos2 π2-h=limh01-cos2 h3 sin2 h=13limh0 sin2 h sin2 h=13                    

(RHL at x = π2) = limxπ2+fx=limh0fπ2+h

                                          =limh0b1-sin π2+hπ-2π2+h2=limh0b1-cos h-2h2=limh02b sin2h24h2=limh02b sin2h216h24=b8limh0sinh2h22=b8×1=b8

Also, fπ2=a

If f(x) is continuous at x = π2, then 

 ​limxπ2-fx =lim  xπ2+fx=fπ2

13 =b8 = a

a=13  and b=83

Page No 9.47:

Question 40:

The points of discontinuity of the function
fx=152x2+3 ,x16-5x          ,1<x<3x-3            ,x3is are
(a) x = 1
(b) x = 3
(c) x = 1, 3
(d) none of these

Answer:

(b) x = 3

If x1, then fx=152x2+3.
Since 2x2+3 is a polynomial function and 15 is a constant function, both of them are continuous. So, their product will also be continuous.
Thus, fx is continuous at x1.

If 1<x<3, then fx=6-5x.

Since 5x is a polynomial function and 6 is a constant function, both of them are continuous. So, their difference will also be continuous.
Thus, fx is continuous for every 1<x<3.

If x3, then fx=x-3.
Since x-3 is a polynomial function, it is continuous. So, fx is continuous for every x3.

Now,
Consider the point x=1. Here,

limx1-fx=limh0f1-h=limh01521-h2+3=1

limx1+fx=limh0f1+h=limh06-51+h=1

Also,
f1=15212+3=1

Thus,
limx1-fx=limx1+fx=f1

Hence, fx is continuous at x=1.

Now,
Consider the point x=3. Here,

limx3-fx=limh0f3-h=limh06-53-h=-9

limx3+fx=limh0f3+h=limh03+h-3=0

Also,
f1=15212+3=1

Thus,
limx3-fxlimx3+fx

Hence, fx is discontinuous at x=3.

So, the only point of discontinuity of fx is x=3.

Page No 9.47:

Question 41:

The value of a for which the function
fx=5x-4      ,if 0<x14x2+3ax,if 1<x<2is continuous at every point of its domain, is

(a) 133
(b) 1
(c) 0
(d) −1

Answer:

(d) −1

Given: fx=5x-4, if 0<x14x2+3ax, if 1<x<2

If fx is continuous in its domain, then it will be continuous at x=1.

Now,
limx1-fx=limh0f1-h=limh051-h-4=5-4=1limx1+fx=limh0f1+h=limh041+h2+3a1+h=4+3a

Since f(x) is continuous at x = 1,
limx1-fx=limx1+fx

4+3a=13a=-3a=-1

Page No 9.47:

Question 42:

If fx=sin cos x-cos xπ-2x2,xπ2          k                      ,x=π2is continuous at x = π/2, then k is equal to
(a) 0
(b) 12
(c) 1
(d) −1

Answer:

(a) 0

Given:  fx=sincos x-cos xπ-2x2, xπ2k, x=π2

If f(x) is continuous at x=π2, then
limxπ2fx=fπ2

limxπ2sincos x-cos xπ-2x2=k


Now,
π2-x=y
π-2x=2y

Also, xπ2, y0

limy0sincosπ2-y-cosπ2-y4y2=k

limy0sinsin y-siny4y2=k

limy02 sinsin y-y2 cossin y+y24y2=k           sin C - sin D=2 sinC-D2 cosC+D212limy0sinsin y-y2 ycossin y+y2y=k12limy0sin y-y2 sinsin y-y2 ysin y-y2cossin y+y2y=k12limy0sin y-y2ysinsin y-y2sin y-y2cossin y+y2y=k12limy0sin y-y2ylimy0sinsin y-y2sin y-y2limy0cossin y+y2y=k14limy0sin yy-1limy0sinsin y-y2sin y-y2limy0cossin y+y2y=k14×0×1×limy0cossin y+y2y=k0=k



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