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Page No 9.10:

Question 1:

Show that f(x) = |x − 2| is continuous but not differentiable at x = 2.

Answer:

Given: f(x) = |x-2| = x-2,     x2-x+2,   x<2
 
Continuity at x=2: We have,

(LHL at x = 2)
 =limx2- f(x) = limh0 f(2-h) = lim h0  (-2+h)+2= 0.

(RHL at x = 2)
  =lim     x2+f(x) = limh0 f(2+h) = limh0  2+h-2 = 0.

and f(2) = 0

Thus, limx2- f(x) = limx2+ f(x) = f(2)f(2).
Hence, f(x) is continuous at x=2.

 Differentiability at x = 2: We have,

(LHD at x = 2)
 =limx2- f(x) - f(2)x-2 = limx2 (-x+2) - 0x-2 = limx2 -(x-2)x-2 = limx2 (-1) =-1

(RHD at x=2)
=limx2+ f(x) - f(2)x-2 = limx2 (x-2) - 0x-2 = limx2 1 = 1

Thus, limx2- f(x) ≠ limx2+ f(x).

Hence, f(x) is not differentiable at x=2 .

Page No 9.10:

Question 2:

Show that f(x) = x1/3 is not differentiable at x = 0.

Answer:

Disclaimer: It might be a wrong question because f(x) is differentiable at x=0

Given: f(x) = x13.
We have,
(LHD at x = 0)
 limx0- f(x) - f(0)x-0=limh0 f(0-h) - f(0)0-h-0=limh0 0-h13-013-h=limh0 -h13-h=limh0  -h-23= 0
           
(RHD at x = 0)
 limx0+ f(x) - f(0)x-0=limh0 f(0+h) - f(0)0+h-0=limh0 0+h13-013-h=limh0 h13h=limh0  h-23= 0                      

LHD at (x = 0)= RHD at (x = 0)

Hence,  f(x) = x13 is differentiable at x = 0

Page No 9.10:

Question 3:

Show that fx=12 x-13, if x32x2+5, if x>3 is differentiable at x = 3. Also, find f'(3).

Answer:

Given: f(x) =12x-13,  x3.2x2 + 5,   x>3.

We have to show that the given function is differentiable at x = 3.

We have,

(LHD at x=3) = lim x3- f(x) - f(3)x-3
 
                      = lim x3 12x-13 - 23x-3= limx3  12x-36x-3= limx3  12 (x-3)x-3= limx3 12 = 12

(RHD at x = 3) = limx3+ f(x) - f(3)x-3

                        = limx3 2x2+5 - 23x-3= limx3 2x2 -18x-3= limx3 2 (x2 -9)x-3=limx3 2(x+3) = 2×6 = 12

Thus, (LHD at x=3) = (RHD at x=3) = 12.

So, f(x) is differentiable at x=3 and f'(3) = 12.

Page No 9.10:

Question 4:

Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat: fx=3x-2,0<x12x2-x,1<x25x-4,x>2

Answer:

Given: 
         f(x) =  3x-2,    0<x12x2-x,  1<x25x-4,     x>2

First , we will show that f(x) is continuos at x=2.

We have,

(LHL at x=2)

=lim      x2- f(x) = lim h0 f(2-h)  = limh0 2(2-h)2 - (2-h) = limh0 (8 + 2h2 - 8h - 2 + h) = 6

(RHL at x = 2) 

=limx2+ f(x) = limh0  f(2+h) = limh0 5(2+h) - 4 = limh0 (10 + 5h -4) = 6

and f(2) = 2×4 - 2 = 6.           

Thus,  limx2- f(x) = limx2+ f(x) = f(2).

Hence the function is continuous at x=2.


Now, we will check whether the given function is differentiable at x = 2.

We have,

(LHD at x = 2)

 limx2- f(x) - f(2)x-2 = limh0 f(2-h) - f(2)-h = limh0 2h2 -7h + 6 - 6-h =limh0 -2h + 7 = 7

(RHD at x = 2)

 limx2+ f(x) - f(2)x-2 = limh0 f(2+h) - f(2)h = limh0 10 + 5h - 4 -6h= 5

Thus, LHD at x=2 ≠ RHD at x = 2.

Hence, function is not differentiable at x = 2.

Page No 9.10:

Question 5:

Discuss the continuity and differentiability of the fx=x+x-1 in the interval -1,2

Answer:

Given: fx=x+x-1x=-x for x<0x=x for x>0x-1=-x-1=-x+1 for x-1<0 or x<1x-1=x-1 for x-1>0 or x>1Now,fx=-x-x+1=-2x+1     x-1,0orfx=x-x+1=1     x0,1orfx=x+x-1=2x-1     x1,2
Now,LHL=limx0- fx=limx0- -2x+1=0+1=1RHL=limx0+ fx=limx0+ 1=1Hence, at x=0, LHL=RHLAgain,LHL=limx1- fx=limx1-1=1RHL=limx1+ fx=limx1+ 2x-1=2-1=1Hence, at x=1, LHL=RHL
Now,
fx=-x-x+1=-2x+1     x-1,0f'x=-2     x-1,0orfx=x-x+1=1     x0,1f'x=0     x0,1orfx=x+x-1=2x-1     x1,2f'x=2     x1,2
Now,LHL=limx0- f'x=limx0- -2=-2RHL=limx0+ f'x=limx0+ 0=0Since, at x=0, LHLRHLHence, fx is not differentiable at x=0Again,LHL=limx1- f'x=limx1- 0=0RHL=limx1+ f'x=limx1+ 2=2Since, at x=1, LHLRHLHence, fx is not differentiable at x=1

Page No 9.10:

Question 6:

Find whether the function is differentiable at x = 1 and x = 2
fx=xx12-x-2+3x-x21x2x>2

Answer:

fx=xx12-x-2+3x-x21x2x>2f'x=1x1-13-2x1x2x>2
Now,LHL=limx1- f'x=limx1- 1=1RHL=limx1+ f'x=limx1+ -1=-1Since, at x=1, LHLRHLHence, fx is not differentiable at x=1Again,LHL=limx2- f'x=limx2- -1=-1RHL=limx2+ f'x=limx2+ 3-2x=3-4=-1Since, at x=2, LHL=RHLHence, fx is differentiable at x=2

Page No 9.10:

Question 7:

Show that the function fx=xm sin1x , x00, x=0

(i) differentiable at x = 0, if m > 1
(ii) continuous but not differentiable at x = 0, if 0 < m < 1
(iii) neither continuous nor differentiable, if m ≤ 0

Answer:

Given:
        f(x) = xm sin1x0      x≠0 , x=0

(i) Let m=2, then the function becomes f(x) = x2 sin1x0 ,  x≠0, x=0

Differentiability at x=0:
limx0 f(x) - f(0)x-0 = limx0 f(x)x = limx0  x sin1x =0.     
 [ ∵ limx0 x sin1x = 0 , as x sin1x - 0 = x sin1x = x sin1x x   (∵sinθ1  for all θ) and hence x sin1x<0  when x-0<εx-0<ε ]
So, f'(0) = 0, which means f is differentiable at x=0.
Hence the given function is differentiable at x=0.

(ii) Let m=12, 0<m<1. Then the function becomes
     f(x) = x120sin1x  ,     x≠0 , x=0

Continuity at x=0:
(LHL at x=0) = limx0- f(x) = limh0 f(0-h) = limh0 (-h)12 sin10-h = limh0 h12 sin1h = limh0 h32  = 0.
(RHL at x=0) = limx0+ f(x) = limh0 f(0+h) = limh0 h12 sin1h = limh0 h32 = 0.
and f(0) = 0
LHL at x=0 = RHL at x=0 = limx0 f(x),
Hence continuous.
Now Differentiabilty at x=0 when 0<m<1.
(LHD at x=0) = limx0- f(x) - f(0)x-0 = limh0 f(0-h) - f(0)0-h-0 =limh0 (-h)12 sin1-h-h

 

Page No 9.10:

Question 8:

Find the values of a and b so that the function fxx2+3x+a,if x1bx+2       ,if x>1 is differentiable at each xR.

Answer:

Given:
        f(x) = x2+3x+a,     x1bx+2,      x>1 

It is given that the function is differentiable at each xR and every differentiable function is continuous.
So, f(x) is continuous at x=1.

Therefore,

   limx1- f(x)=limx1+ f(x) = f(1)

limx1 x2+3x+a = limx1 bx+2 = a+4                   Using def. of f(x)  a+4 = b+2 = a+4             ...(i)                                 


Since, f(x) is differentiable at x=1. So,

(LHD at x = 1) = (RHD at x = 1)

    limx1- f(x) - f(1)x-1 = limx1+ f(x) - f(1)x-1limx1 x2+3x+a-a-4x-1 = limx1 bx+2 -4-ax-1                        Using def. of f(x) limx1  (x+4) (x-1)x-1= limx1 bx-2-ax-1 limx1  (x+4) (x-1)x-1= limx1 bx-bx-1                        Using (i)             limx1  (x+4) (x-1)x-1 = limx1 b(x-1)x-1 5 = b

From (i), we have

    a+4 = b+2a+4 = 5+2a = 7-4 a= 3
Hence, a=3 , b=5.

Page No 9.10:

Question 9:

Show that the function fx=2x-3 x,x 1sin π x2,x<1 is continuous but not differentiable at x = 1.

Answer:

Given: f(x) = 2x-3 [x] ,  x1sinπx2,    x<1

Continuity at x = 1:
(LHL at x = 1) = limx1-f(x)=limh0f(1-h)=limh0 sinπ (1-h)2=sinπ2=1

(RHL at x = 1) = limx1+ f(x) = limh0 f(1+h) = limh0 2(1+h)-31+h=limh0 2(1+h)-3=1

Hence, (LHL at x = 1) = (RHL at x = 1)

Differentiability at x = 1:

LHD at x=1=limx1-fx-f1x-1LHD at x=1=limh0f1-h-f11-h-1LHD at x=1=limh0f1-h-f1-hLHD at x=1=limh0sinπ1-h2-1-hLHD at x=1=limh0cosπh2-1-hLHD at x=1=-π2limh0cosπh2-1π2h=0RHD at x=1=limx1+fx-f1x-1RHD at x=1=limh0f1+h-f11+h-1RHD at x=1=limh0f1+h-f1hRHD at x=1=limh0-21+h-3-1hRHD at x=1=limh0-2hh=-2

LHD ≠ RHD

Hence, the function is continuous but not differentiable at x = 1.



Page No 9.11:

Question 10:

If fx=ax2-b,if x<11x     ,if x1 is differentiable at x = 1, find a, b.

Answer:

Given: f(x) =ax2+b,     x<11x,           x1
 f(x) = -1x ,                          x<-1ax2-b,                     -1<x<11x,                               x1  

It is given that the given function is differentiable at x = 1.

We know every differentiable function is continuous. Therefore it is continuous at x=1. Then,

   lim  x1- f(x) = limx1+ f(x) limx1 ax2-b = limx1 1x a-b = 1                                       ...(i)


It is also differentiable at x=1. Therefore,

 (LHD at x = 1) = (RHD at x = 1)

limx1-f(x) - f(1)x-1 = limx1+f(x) - f(1)x-1 limx1 ax2-b - 1x-1= limx1 1x - 1x-1  limx1 ax2+1-a-1x-1=limx1 -(x-1)x-1                 Using (i)  limx1 a (x+1)= limx1 -1   2a=-1  a =-12
From (i), we have:
          a-b = 1-12 - b = 1 b =- 32

Hence, when a=-12 and b=-32 the function is differentiable at x = 1. 

Page No 9.11:

Question 11:

Find the values of a and b, if the function f defined by fx=x2+3x+a,x1bx+2,x>1 is differentiable at x = 1.

Answer:

Given that f(x) is differentiable at x = 1. Therefore,  f(x) is continuous at x = 1.

limx1-fx=limx1+fx=f1limx1x2+3x+a=limx1bx+2=1+3+a1+3+a=b+2a-b+2=0                              .....1

Again, f(x) is differentiable at x = 1. So,

(LHD at x = 1) = (RHD at x = 1)
limx1-fx-f1x-1=limx1+fx-f1x-1
limx1x2+3x+a-4+ax-1=limx1bx+2-4+ax-1limx1x2+3x-4x-1=limx1bx-2-ax-1limx1x+4x-1x-1=limx1bx-bx-1limx1x+4=limx1bx-1x-15=b

Putting b = 5 in (1), we get

a = 3

Hence, a = 3 and b = 5.



Page No 9.16:

Question 1:

If f is defined by f (x) = x2, find f'(2).

Answer:

Given: f(x) = x2.

We know a  polynomial function is everywhere differentiable. Therefore f(x) is differentiable at x=2.

 f'(2) = limh0 f(2+h) - f(2)h f'(2) = limh0 (2+h)2 - 22h f'(2) = limh0 (4+h2+4h) - 4h f'(2) = limh0 h (h+4)h f'(2) = 4

Page No 9.16:

Question 2:

If f is defined by fx=x2-4x+7, show that f'5=2f'72

Answer:

Given: f(x) = x2-4x+7

Clearly, f(x) being a polynomial function, is everywhere differentiable. The derivative of f at x is given by:

 f'(x) = limh0 f(x+h) - f(x)h f'(x) = limh0 x+h2 -4(x+h) +7 - (x2 -4x+7)h f'(x) = limh0 x2+h2+2xh -4x-4h+7 -x2+4x-7h f'(x) = limh0 h2 +2xh -4hh f'(x) = limh0 h(h+2x-4)h f'(x) = 2x-4

Now, 

f'(5) = 2×5 - 4= 6f'72 = 2×72 - 4 = 3
Therefore,   f'(5) = 2×3 = 2f'72
 Hence proved.

Page No 9.16:

Question 3:

Show that the derivative of the function f given by
fx=2x3-9x2+12x+9, at x = 1 and x = 2 are equal.

Answer:

Given: f(x) = 2x3-9x2+12x+9

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of f at x is given by:

   f'(x) = limh0 f(x+h - f(x)h f'(x) = limh0  2(x+h)3-9(x+h)2+12(x+h) + 9 - 2x3+9x2-12x-9h f'(x) = limh0 2x3 + 2h3+6x2h +6xh2 -9x2-9h2-18xh+12x+12h+9 -2x3+9x2-12x-9h f'(x) = limh0 2h3 +6x2h +6xh2 -9h2 -18xh+12hh f'(x) = limh0  h(h2 +6x2+6xh -9h-18x+12)h f'(x) = 6x2-18x+12

So,

f'(1) = 6x2-3x+2          = 6×(1-3+2)          = 0f'(2) = 6x2-3x+2          = 6×(4-6+2)          = 0

Hence the derivative at x=1 and x=2 are equal.

Page No 9.16:

Question 4:

If for the function Φ x=λx2+7x-4, Φ'5=97, find λ.

Answer:

Given: ϕ(x) = λx2+7x-4

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of ϕ at x is given by:

ϕ'(x) = limh0 ϕ(x+h) - ϕ(x)h ϕ'(x) =limh0 λ(x+h)2 +7(x+h) -4 - λx2-7x+4h ϕ'(x) = limh0 λx2 +λh2+2λxh+7x+7h-4-λx2-7x+4h ϕ'(x) = limh0 λh2 +2λxh+7hh ϕ'(x) = limh0 h(λh +2λx+7)h ϕ'(x) = 2λx+7
It is given ϕ'(5) = 97
Thus,
      ϕ'(5) = 10λ + 7 = 97 10λ+7 =97 10λ = 90 λ = 9

Page No 9.16:

Question 5:

If fx=x3+7x2+8x-9, find f'(4).

Answer:

Given: f(x) = x3+7x2+8x-9

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of f at x is given by:

 f'(x) = limh0f(x+h) - f(x)h f'(x) = limh0 (x+h)3+7(x+h)2+8(x+h)-9 - x3 -7x2-8x+9h f'(x) = limh0 x3+h3+3x2h + 3xh2+7x2+7h2+14xh+8x+8h-9-x3-7x2-8x+9h f'(x) = limh0 h3+3x2h+3xh2+7h2+14xh+8hh f'(x) = limh0 h(h2+3x2+3xh+7h+14x+8)h f'(x) = limh0 h2+3x2+3xh+7h+14x+8 f'(x) = 3x2+14x+8
Thus,
      f'(4) = 3×42+14×4+8          = 48+56+8         =112

Page No 9.16:

Question 6:

Find the derivative of the function f defined by f (x) = mx + c at x = 0.

Answer:

Given: f(x) = mx+c

Clearly, being a polynomial function, is differentiable everywhere. Therefore the derivative of f at x is given by:

f'(x) = limh0 f(x+h) - f(x)h  f'(x) = limh0 m(x+h) +c -mx-ch f'(x) = limh0 mx+mh+c-mx-ch f'(x) = limh0 mhh  f'(x) = m

Thus, f'(0) = m

Page No 9.16:

Question 7:

Examine the differentialibilty of the function f defined by
fx=2x+3if-3 x-2x+1x+2if -2x<0if 0x1

Answer:

fx=2x+3if-3 x-2x+1x+2if -2x<0if 0x1f'x=2if-3 x-211if -2x<0if 0x1
Now,LHL=limx-2- f'x=limx-2- 2=2RHL=limx-2+ f'x=limx-2+ 1=1Since, at x=-2, LHLRHLHence, fx is not differentiable at x=-2Again,LHL=limx0- f'x=limx0- 1=1RHL=limx0+ f'x=limx0+ 1=1Since, at x=0, LHL=RHLHence, fx is differentiable at x=0

Page No 9.16:

Question 8:

Write an example of a function which is everywhere continuous but fails to differentiable exactly at five points.

Answer:

fx=x+x+1+x+2+x+3+x+4
The above function is continuous everywhere but not differentiable at x = 0, 1, 2, 3 and 4

Page No 9.16:

Question 9:

Discuss the continuity and differentiability of f (x) = |log |x||.

Answer:

We have,
f (x) = |log |x||

x=-x        -<x<-1-x        -1<x<0  x        0<x<1  x        1<x<log x=log -x        -<x<-1log -x        -1<x<0  log x        0<x<1 log x        1<x<log x=log -x        -<x<-1-log -x        -1<x<0 -log x        0<x<1 log x        1<x<
LHD at x=-1=limx-1-fx-f-1x+1                             =limx-1-log -x-0x+1                             =limh0log 1+h-1-h+1                             =-limh0log 1+hh=-1

RHD at x=-1=limx-1+fx-f-1x+1                             =limx-1+-log -x-0x+1                             =limh0-log 1-h-1+h+1                             =limh0-log 1-hh=1
Here, LHD ≠ RHD
So, function is not differentiable at x = − 1

At 0 function is not defined.
LHD at x=1=limx1-fx-f1x-1                        =limx1--log x-0x-1                       =limh0-log 1-h1-h-1                       =-limh0log 1-hh=-1
RHD at x=1=limx1+fx-f1x-1                        =limx1+log x-0x-1                       =limh0log 1+h1+h-1                       =limh0log 1+hh=1
Here, LHD ≠ RHD
So, function is not differentiable at x = 1
Hence, function is not differentiable at x = 0 and ± 1
At 0 function is not defined.
So, at 0 function is not continuous.
LHL at x=-1=limx-1-fx                             =limx-1-log -x                             =log 1=0RHL at x=-1=limx-1+fx                             =limx-1+-log -x                             =-log 1=0f-1=0Therefore, fx=log x is continuous at x=-1
LHL at x=1=limx1-fx                        =limx1--log x                        =-log 1=0RHL at x=1=limx1+fx                        =limx1+log x                        =log 1=0f1=0Therefore, at x=1, fx=log x is continuous.

Hence, function f (x) = |log |x|| is not continuous at x = 0

Page No 9.16:

Question 10:

Discuss the continuity and differentiability of f (x) = e|x| .

Answer:

Given: f(x) = ex
       f(x) = ex,           x0e-x,         x<0

Continuity: 

(LHL at x = 0) 

limx0- f(x) = limh0 f(0-h) = limh0 e-(0-h)  = limh0 eh = 1

(RHL at x = 0) 

limx0+ f(x) = limh0 f(0+h) = limh0 e(0+h) = 1

and f(0) = e0 = 1

Thus,  limx0- f(x) = limx0+ f(x) = f(0)

Hence,function is continuous at x = 0 .

Differentiability at x = 0.

(LHD at x = 0)

 limx0-f(x) - f(0)x-0= limh0f(0-h) - f(0)0-h-0= limh0 e-(0-h) - 1-h= limh0 eh-1-h  = -1                    limh0 eh-1h =1 
(RHD at x = 0)

 limx0+f(x) - f(0)x-0= limh0f(0+h) - f(0)0+h-0= limh0 e(0+h) - 1h= limh0 eh-1h  = 1                    limh0 eh-1h =1 

LHD at (x = 0)RHD at (x = 0)

Hence the function is not differentiable at x = 0.

Page No 9.16:

Question 11:

Discuss the continuity and differentiability of

fx=x-c cos 1x-c,xc0                                ,x=c

Answer:

Given: fx=x-c cos 1x-c,xc0                                ,x=c

Continuity:

(LHL at x = c) 

limxc- f(x) = limh0 f(c-h) = limh0 (c-h-c)  cos1c-h-c = limh0 -h cos1h Since , cos 1h  is a bounded function and 0 ×bounded function is 0=0

(RHL at x = c

limxc+ f(x) =limh0 f(c+h) = limh0 (c+h-c) cos1c+h-c = limh0 h cos1h Since, cos1h  is a bounded function and 0× bounded function is 0=0

and 
Differentiability at x = c

(LHD at x = c)
limxc- f(x) - f(c)x-c = limh0f(c-h) - f(c)c-h-c = limh0 -h cos1-h -  0-h           0.cos 1c-c=0, as cos function is bounded function.=lim h0cos1h=A number which oscillates between -1 and 1LHD(x=c) does not exist . Similarly , we can show that RHD(x=c) does not exist . Hence , f(x) is not differentiable at x=c
 

Page No 9.16:

Question 12:

Is |sin x| differentiable? What about cos |x|?

Answer:

Let, f(x) = |sin x|

sin x=0,  for x=nπ,sin x=-sin x      2m-1π< x<2mπ , where mZ sin x       2mπ< x<2m+1π , where mZ- sin x       2m+1π< x<2m+1π , where mZ

LHD at x=2mπ=limx2mπ-fx-f2mπx-2mπ                             =limx2mπ--sinx-0x-2mπ                             =limh0-sin2mπ-h2mπ-h-2mπ                             =limh0sinh-h=-1
RHD at x=2mπ=limx2mπ+fx-f2mπx-2mπ                             =limx2mπ+sinx-0x-2mπ                             =limh0sin2mπ+h2mπ+h-2mπ                             =limh0sinhh=1


Here, LHD  RHD So, function is not differentiable at x=2mπ, where, mZ     .....1                                                                                          
LHD at x=2m+1π=limx2m+1π-fx-f2m+1πx-2m+1π                                   =limx2m+1π-sin x-0x-2m+1π                                   =limh0sin 2m+1π-h2m+1π-h-2m+1π                                   =limh0sin h-h=-1
RHD at x=2m+1π=limx2m+1π+fx-f2m+1πx-2m+1π                                   =limx2m+1π+-sin x-0x-2m+1π                                   =limh0-sin 2m+1π+h2m+1π+h-2m+1π                                   =limh0sin hh=1
Here, LHD  RHD. So, function is not differentiable at x=2m+1π, where, mZ     .....2                                                                                         
From, 1 and 2, we getfx=sin x is not differentiable at x=nπ

We know that,cos x=cos x               For all xRAlso we know that cos x is differentiable at all real points.Therefore, cos x is differentiable everywhere.



Page No 9.17:

Question 1:

Let f (x) = |x| and g (x) = |x3|, then
(a) f (x) and g (x) both are continuous at x = 0
(b) f (x) and g (x) both are differentiable at x = 0
(c) f (x) is differentiable but g (x) is not differentiable at x = 0
(d) f (x) and g (x) both are not differentiable at x = 0

Answer:

Option (a) f (x) and g (x) both are continuous at x = 0

Given: fx =  x ,  gx =  x3 

We know  x is continuous at x=0 but not differentiable at x = 0 as (LHD at x = 0) ≠ (RHD at x = 0).
Now, for the function gx =  x3  = x3,             x0-x3,          x<0
Continuity at x = 0:

(LHL at x = 0) = limx0-gx = limh0 g0-h = limh0--h3 = limh0 h3 = 0.

(RHL at x = 0) = limx0+fx = limh0f0+h = limh0 h3 = 0.

and g0 = 0.
Thus,   limx0-gx = limx0+gx = g0.
Hence, g(x) is continuous at x = 0.

Differentiability at x = 0:

(LHD at x = 0) = limx0-fx - f0x-0 = limh0f0-h - f00-h-0 = limh0h3 -0-h = 0.

(RHD at x = 0) = limxc+fx - f0x-0 = limh0f0+h - f00+h-0 = limh0h3 -0h = limh0h3h =0
Thus, (LHD at x = 0) = (RHD at x = 0). 
Hence, the function  gx is differentiable at x = 0.

Page No 9.17:

Question 2:

The function f (x) = sin−1 (cos x) is
(a) discontinuous at x = 0
(b) continuous at x = 0
(c) differentiable at x = 0
(d) none of these

Answer:

(b) continuous at x = 0

Given: f(x) = sin-1cos x.

Continuity at x = 0:

We have,
(LHL at x = 0) 

limx0- f(x) =limh0 sin-1cos0-h = limh0 sin-1cos h= sin-11= π2

(RHL at x = 0)

  limx0+ fx= limh0 sin-1cos0+h= limh0 sin-1cos h = sin-11 = π2

f(0) = sin-1cos 0        = sin-11       = π2

Differentiability at x = 0:
(LHD at x = 0) 

limx0- fx - f0x-0 = limh0 sin-1cos0-h - π2-h = limh0 sin-1cos-h -π2-h = limh0 sin-1cosh -π2-h= limh0 sin-1sin π2-h -π2-h= limh0-h-h=1

RHD at x = 0

limx0+ fx - f0x-0 = limh0 sin-1cos0+h-π2h = limh0 sin-1cosh-π2h = limh0 sin-1sin π2-h-π2-h= limh0-hh=-1

 LHDRHD

Hence, the function is not differentiable at x = 0 but is continuous at x = 0.

Page No 9.17:

Question 3:

The set of points where the function f (x) = x |x| is differentiable is
(a) -, 
(b) -, 00, 
(c) 0, 
(d) 0, 

Answer:

(a) -, 

We have,fx=xxfx=-x2,        x<0   0 ,        x=0   x2,         x>0When, x<0, we have         fx=-x2 which being a polynomial function is continuous and differentiable in -, 0When, x>0, we have         fx=x2 which being a polynomial function is continuous and differentiable in 0, Thus possible point of non-differentiability of fx is x=0Now, LHD at x=0 =limx0- fx- f0x-0                              =limx0--x2- 0x                              =limh0--h2-h                              =limh0h                              =0And  RHD at x=0 =limx0+ fx- f0x-0                              =limx0+x2- 0x                              =limh0h2h                              =lim h0h                              =0LHD at x=0=RHD at x=0So, fx is also differentiable at x=0i.e. fx is differentiable in -, 

Page No 9.17:

Question 4:

If fx=x+2tan-1 x+2, x-22, x=-2, then f (x) is

(a) continuous at x = − 2
(b) not continuous at x = − 2
(c) differentiable at x = − 2
(d) continuous but not derivable at x = − 2

Answer:

(b) not continuous at x = − 2

Given:
 f(x) = x+2tan-1(x+2) ,    x-2                  2  ,               x=-2
 f(x) = -(x+2)tan-1(x+2),         x<-2 (x+2)tan-1(x+2),      x>-22  ,                      x=-2
Continuity at x = − 2.
(LHL at x= − 2) = limx-2-f(x)=limh0f(-2-h)=limh0-(-2-h+2)tan-1(-2-h+2)=limh0htan-1(-h) =-1.

(RHL at x = −2) = limx-2+f(x=limh0f(-2+h=limh0(-2+h+2)tan-1(-2+h+2)=limh0htan-1(h) =1.

Also f(-2) = 2
Thus,  limx-2-f(x)limx-2+ f(x) ≠ f(-2).
Therefore, given function is not continuous at x = − 2

Page No 9.17:

Question 5:

Let fx=x+x x. Then, for all x
(a) f is continuous
(b) f is differentiable for some x
(c) f' is continuous
(d) f'' is continuous

Answer:

(a) f is continuous
(c) f' is continuous

We have,fx=x+x x       =xx+x2       =xx+x2fx=2x2          x00             x<0To check continuity of fx at x=0LHL at x=0=limx0-fx                       =limx0-0                       =0RHL at x=0=limx0+fx                       =limx0+2x2                       =0And f0=0Here, LHL=RHL=f0Therefore, fx is continuous at x=0Hence, fx is continuous everywhere.

To check the differentiability of fx at x=0LHD at x=0=limx0-fx-f0x-0                        =limx0-0-0x=0RHD at x=0=limx0+fx-f0x-0                        =limx0-2x2-0x                        =limx0-2x2-0x                        =limx0-2x=0LHD=RHDTherefore, fx is derivative at x=0Hence, fx is differentiable everywhere.

f'x=4x         x00             x<0To check continuity of f'x at x=0LHL at x=0=limx0-f'x                       =limx0-0                       =0RHL at x=0=limx0+f'x                       =limx0+4x                       =0And f'0=0Here, LHL=RHL=f'0Therefore, f'x is continuous at x=0Hence, f'x is continuous everywhere.
f''x=4         x00             x<0To check continuity of f''x at x=0LHL at x=0=limx0-f''x                       =limx0-0                       =0RHL at x=0=limx0+f''x                       =limx0+4                       =4Therefore, LHLRHL Therefore, f''x is not continuous at x=0Hence, f''x is not continuous everywhere.

Page No 9.17:

Question 6:

The function f (x) = e|x| is
(a) continuous everywhere but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) none of these

Answer:

(a) continuous everywhere but not differentiable at x = 0



Given: f(x) = e-x = ex,                  x 0e-x,                x<0Continuity :limx0- f(x) = limh0 f(0-h) = limh0 e-(0-h)  = limh0 eh = 1  

RHL at x = 0

limx0+ f(x) = limh0 f(0+h) = limh0 e(0+h) = 1

and f(0) = f(0) = e0 = 1
Thus, limx0- f(x) = limx0+ f(x) = f

Hence, function is continuous at x = 0

Differentiability at x = 0

(LHD at x = 0)

limx0-f(x) - f(0)x-0= limh0f(0-h) - f(0)0-h-0= limh0 e-(0-h) - 1-h= limh0 ehh = 

Therefore, left hand derivative does not exist.
Hence, the function is not differentiable at x = 0.

Page No 9.17:

Question 7:

The function f (x) = |cos x| is
(a) everywhere continuous and differentiable
(b) everywhere continuous but not differentiable at (2n + 1) π/2, nZ
(c) neither continuous nor differentiable at (2n + 1) π/2, nZ
(d) none of these

Answer:

(b) everywhere continuous but not differentiable at (2n + 1) π/2, nZ

We have,fx=cos xfx=cos x,      2nπx<4n+1π20,     x=4n+1π2    -cos x,    4n+1π2<x<4n+3π2             0 ,        x=4n+3π2        cos x,        4n+3π2< x2n+2πWhen, x is in first quadrant, i.e. 2nπx<4n+1π2 , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 2nπ, 4n+1π2When, x is in second quadrant or in third quadrant, i.e., 4n+1π2<x<4n+3π2 , we have         fx=-cos x which being a trigonometrical function is continuous and differentiable in 4n+1π2, 4n+3π2When, x is in fourth quadrant, i.e., 4n+3π2< x2n+2π , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 4n+3π2, 2n+2πThus possible point of non-differentiability of fx are x=4n+1π2, 4n+3π2Now, LHD at x=4n+1π2 =limx4n+1π2- fx- f4n+1π2x-4n+1π2                                               =limx4n+1π2-  cos x- 0x-4n+1π2                                              =limx4n+1π2-  -sin x1-0      By L'Hospital rule                                              =-1And  RHD at x=4n+1π2 =limx4n+1π2+ fx- f4n+1π2x-4n+1π2                                              =limx4n+1π2+  -cos x- 0x-4n+1π2                                              =limx4n+1π2+  sin x1-0      By L'Hospital rule                                              =1lim x4n+1π2-fx limx4n+1π2+fxSo fx is not differentiable at x=4n+1π2Now, LHD at x=4n+3π2 =limx4n+1π2- fx- f4n+3π2x-4n+3π2                                              =limx4n+3π2-  -cos x- 0x-4n+3π2                                              =limx4n+3π2-  sin x1-0      By L'Hospital rule                                              =1And  RHD at x=4n+3π2 =limx4n+3π2+ fx- f4n+3π2x-4n+3π2                                              =limx4n+3π2+  cos x- 0x-4n+3π2                                              =limx4n+3π2+ -sin x1-0      By L'Hospital rule                                              =-1lim x4n+3π2-fx limx4n+3π2+fxSo fx is not differentiable at x=4n+3π2Therefore, fx is neither differentiable at 4n+1π2 nor at 4n+3π2i.e. fx is not differentiable at odd multiples of π2i.e. fx is not differentiable at x=2n+1π2Therefore, f(x) is everywhere continuous but not differentiable at 2n+1π2 .

Page No 9.17:

Question 8:

If fx=1-1-x2, then f x is

(a) continuous on [−1, 1] and differentiable on (−1, 1)
(b) continuous on [−1, 1] and differentiable on -1, 00, 1
(c) continuous and differentiable on [−1, 1]
(d) none of these

Answer:

b continuous on -1,1 and differentiable on -1,00,1

We have,fx=1-1-x2Here, function will be defined for those values of x for which1-x201x2x21x1-1x1Therefore, function is continuous in -1, 1
Now, we need to check the differentiability of fx=1-1-x2 in the interval -1, 1.Now, we will check the differentiability at x=0LHD at x=0=limx0-fx-f0x-0                        =limx0-1-1-x2-0x                        =limx0-1-1-x2x                        =limh01-1-0-h20-h                        =limh01-1-h2-h=-RHD at x=0=limx0+fx-f0x-0                        =limx0+1-1-x2-0x                        =limx0+1-1-x2x                        =limh01-1-0+h20+h                        =limh01-1-h2h= So, the function is not differentiable at x=0.                

Page No 9.17:

Question 9:

If fx=asin x+bex+cx3 and if f (x) is differentiable at x = 0, then
(a) a=b=c=0
(b) a=0, b=0; cR
(c) b=c=0, aR
(d) c=0, a=0, bR

Answer:

(b) a=0, b=0; cR

We have,fx=a sin x+be x+c x3       =a sin x+bex+cx3           0<x<π2-a sin x+be-x-cx3     -π2<x<0Here, fx is differentiable at x=0Therefore, LHD at x=0=RHD at x=0limx0-fx-f0x-0=limx0+fx-f0x-0limx0--a sinx+be-x-cx3-bx=limx0+a sin x+bex+cx3-bxlimh0-a sin0-h+be-0-h-c0-h3-b0-h=limh0a sin 0+h+be0+h+c0+h3-b0+hlimh0a sin h+be h+ch3-b-h=limh0a sin h+beh+ch3-bhlimh0a cos h+beh+3ch2-1=limh0a cos h+beh+3ch21         By L'Hospital rule-a+b=a+b -2a+b=0a+b=0This is true for all value of ccRIn the given options, option b satisfies a+b=0 and cR



Page No 9.18:

Question 10:

If fx=x2+x21+x2+x21+x2+...+x21+x2+....,

then at x = 0, f (x)
(a) has no limit
(b) is discontinuous
(c) is continuous but not differentiable
(d) is differentiable

Answer:

(b) is discontinuous

We have,fx=x2+x21+x2+x21+x2+...+x21+x2+....,When x=0 then x2=0 and x21+x2=0f0=0+0+0+0.......f0=0When, x0Then, x2>0and 1+x2>x20<x21+x2<1limx0 fx=lim x0x2+x21+x2+x21+x2+...+x21+x2+....,              =limx0x21+11+x2+11+x2+...+11+x2+....,              =limx0x211-11+x2             Sum of infinite series where, r=11+x2              =limx0x21+x2x2              =limx01+x2              =1limx0 fxf0fx is discontinuous at x=0

Page No 9.18:

Question 11:

If fx=logex, then
(a) f' 1+=1
(b) f' 1=-1
(c) f' 1=1
(d) f' 1=-1

Answer:

(a) f' 1+=1 and (b) f' 1=-1

fx=logex, = - loge x , for 0 < x < 1loge x , for x 1Differentiablity at x = 1,we have ,   (LHD at x = 1 ) =lim x 1- f(x) - f(1)x -1                           = lim x 1- -log x - log 1x -1                             = -lim x 1- log x x -1                                                    = -lim h 0 log (1 - h) 1 -h - 1                            = -lim h 0 log (1 - h)  -h   = -1                                      

    (RHD at x = 1 )  = lim x 1+ f(x) - f(1)x -1                                                                 = lim x 1+ log x - log (1)x -1                                                               = lim h  0  log (1 + h)x -1 =  lim h  0  log (1 + h)h = 1

Page No 9.18:

Question 12:

If fx=loge |x|, then
(a) f (x) is continuous and differentiable for all x in its domain
(b) f (x) is continuous for all for all × in its domain but not differentiable at x = ± 1
(c) f (x) is neither continuous nor differentiable at x = ± 1
(d) none of these

Answer:

(b) f (x) is continuous for all x in its domain but not differentiable at x = ± 1

We have,fx=loge |x|We know that log function is defined for positive value.Here, x is positive for all non zero x.Therefore, domain of function is R-0
And we know that logarithmic function is continuous in its domain.
Therefore, logex is continuous in its domain.
We will check the differentiability at its critical points. logex=loge-x                     -<x<-1-loge-x                    -1<x<0-logex                       0<x<1logex                    1<x<LHD at x=-1=limx-1-fx-f-1x--1                         =limx-1-loge-x-0x+1                         =limh0loge--1-h-1-h+1                         =limh0loge1+h-h                         =-1RHD at x=-1=limx-1+fx-f-1x--1                         =limx-1+-loge-x-0x+1                          =limh0-loge--1+h-1+h+1                          =limh0-loge1-hh                            =-limh0loge1-hh                            =-1×-1=1Here, LHDRHDTherefore, the given function is not differentiable at x=-1.

LHD at x=1=limx1-fx-f1x-1                      =limx1--logex-0x-1                      =limh0-loge1-h1-h-1                      =limh0loge1-hh                      =-1RHD at x=1=limx1+fx-f1x-1                      =limx1+logex-0x-1                      =limh0loge1+h1+h-1                      =limh0loge1+hh                      =1Here, LHDRHDTherefore,the given function is not differentiable at x=1.
Therefore, given function is continuous for all x in its domain but not differentiable at x = ± 1

Page No 9.18:

Question 13:

Let fx=1xfor x1ax2+bfor x<1

If f (x) is continuous and differentiable at any point, then
(a) a=12, b=-32
(b) a=-12, b=32
(c) a = 1, b = − 1
(d) none of these

Answer:

(b) a=-12, b=32

We have,fx=-1x,       x-1ax2+b,    -1<x<11x,          x1Given: fx is differentiable and continuous at every point.Consider a point x=1limx1-fx=limx1+fxlimx1-ax2+b=limx1+1xa+b=1          ...iIt is also differentiable at x=1limx1-fx-f1x-1=limx1+fx-f1x-1limx1-ax2+b-1x-1=limx1+1x-1x-1limx1-ax2-ax-1=limx1+1-xx-1x       Using ilimx1-ax+1=limx1+-x2a=-1a=-12Plugging a=-12 in i we get,b=32a=-12, b=32

Page No 9.18:

Question 14:

The function f (x) = x − [x], where [⋅] denotes the greatest integer function is
(a) continuous everywhere
(b) continuous at integer points only
(c) continuous at non-integer points only
(d) differentiable everywhere

Answer:

(c) continuous at non-integer points only

We have,fx=x-xConsider n be an integer.fx=x-x=x-n-1       n-1x<n0                   x=nx-n             nx<n+1Now,LHL at x=n=limxn-fx=x-n-1=x-n+1RHL at x=n=limxn+fx=x-n=x-nAs, LHLRHL at x=ni.e., given function is not continuous at n.Now, n is any integer.Therefore, given function is not continuous at integers.

Therefore, given points are continuous at non-integer points only.

Page No 9.18:

Question 15:

Let fxax2+1,x>1x+1/2,x1. Then, f (x) is derivable at x = 1, if

(a) a = 2
(b) a = 1
(c) a = 0
(d) a = 1/2

Answer:

(d) a = 1/2


Given: f(x) = ax2+1,   x>1x+12,     x1
The function is derivable at x = 1, iff left hand derivative and right hand derivative of the function are equal at x = 1.

LHD at x=1=limx1-fx-f1x-1LHD at x=1=limh0f1-h-f11-h-1LHD at x=1=limh0f1-h-f1-hLHD at x=1=limh01-h+12-32-h=1RHD at x=1=limx1+fx-f1x-1RHD at x=1=limh0f1+h-f11+h-1RHD at x=1=limh0f1+h-f1hRHD at x=1=limh0a1+h2+1-32hRHD at x=1=limh0a1+h2+2h-12h LHD=RHDa-12=0a=12

Page No 9.18:

Question 16:

Let f (x) = |sin x|. Then,
(a) f (x) is everywhere differentiable.
(b) f (x) is everywhere continuous but not differentiable at x = n π, nZ
(c) f (x) is everywhere continuous but not differentiable at x=2n+1π2, nZ.
(d) none of these

Answer:

(b) f (x) is everywhere continuous but not differentiable at x = n π, nZ

We have,fx=sin xfx=0 ,           x=2nπ sin x,      2nπ<x<2n+1π0,            x=2n+1π-sin x,   2n+1π<x<2n+2πWhen, x is in first or second quadrant, i.e., 2nπ<x<2n+1π , we have         fx=sin x which being a trigonometrical function is continuous and differentiable in 2nπ, 2n+1πWhen, x is in third or fourth quadrant, i.e., 2n+1π<x<2n+2π , we have         fx=-sin x which being a trigonometrical function is continuous and differentiable in 2n+1π, 2n+2πThus possible point of non-differentiability of fx are x=2nπ and  2n+1πNow, LHD at x=2nπ =limx2nπ- fx- f2nπx-2nπ                                     =limx2nπ-  -sin x- 0x-2nπ                                     =limx2nπ-  -cos x1-0      By L'Hospital rule                                     =-1And  RHD at x=2nπ =limx2nπ+ fx- f2nπx-2nπ                                   =limx2nπ+  sin x- 0x-2nπ                                   =limx2nπ+  cos x1-0      By L'Hospital rule                                   =1lim x2nπ-fx limx2nπ+fxSo fx is not differentiable at x=2nπNow, LHD at x=2n+1π =limx2n+1π- fx- f2n+1πx-2n+1π                                            =limx2n+1π-  sin x- 0x-2n+1π                                            =limx2n+1π-  cos x1-0      By L'Hospital rule                                           =-1And  RHD at x=2n+1π =limx2n+1π+ fx- f2n+1πx-2n+1π                                           =limx2n+1π+  -sin x- 0x-2n+1π                                           =limx2n+1π+ -cos x1-0      By L'Hospital rule                                            =1lim x2n+1π-fx limx2n+1π+fxSo fx is not differentiable at x=2n+1πTherefore, fx is neither differentiable at 2nπ nor at 2n+1πi.e.fx is neither differentiable at even multiple of π nor at odd multiple of πi.e. fx is not differentiable at x=nπTherefore, f(x) is everywhere continuous but not differentiable at nπ.

Page No 9.18:

Question 17:

Let f (x) = |cos x|. Then,
(a) f (x) is everywhere differentiable
(b) f (x) is everywhere continuous but not differentiable at x = n π, nZ
(c) f (x) is everywhere continuous but not differentiable at x=2n+1π2, nZ.
(d) none of these

Answer:

(c) f (x) is everywhere continuous but not differentiable at x=2n+1π2, nZ.

We have,fx=cos xfx=cos x,      2nπx<4n+1π20,     x=4n+1π2    -cos x,    4n+1π2<x<4n+3π2             0 ,        x=4n+3π2        cos x,        4n+3π2< x2n+2πWhen, x is in first quadrant, i.e. 2nπx<4n+1π2 , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 2nπ, 4n+1π2When, x is in second quadrant or in third quadrant, i.e., 4n+1π2<x<4n+3π2 , we have         fx=-cos x which being a trigonometrical function is continuous and differentiable in 4n+1π2, 4n+3π2When, x is in fourth quadrant, i.e., 4n+3π2< x2n+2π , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 4n+3π2, 2n+2πThus possible point of non-differentiability of fx are x=4n+1π2, 4n+3π2Now, LHD at x=4n+1π2 =limx4n+1π2- fx- f4n+1π2x-4n+1π2                                               =limx4n+1π2-  cos x- 0x-4n+1π2                                              =limx4n+1π2-  -sin x1-0      By L'Hospital rule                                              =-1And  RHD at x=4n+1π2 =limx4n+1π2+ fx- f4n+1π2x-4n+1π2                                              =limx4n+1π2+  -cos x- 0x-4n+1π2                                              =limx4n+1π2+  sin x1-0      By L'Hospital rule                                              =1lim x4n+1π2-fx limx4n+1π2+fxSo fx is not differentiable at x=4n+1π2Now, LHD at x=4n+3π2 =limx4n+1π2- fx- f4n+3π2x-4n+3π2                                              =limx4n+3π2-  -cos x- 0x-4n+3π2                                              =limx4n+3π2-  sin x1-0      By L'Hospital rule                                              =1And  RHD at x=4n+3π2 =limx4n+3π2+ fx- f4n+3π2x-4n+3π2                                              =limx4n+3π2+  cos x- 0x-4n+3π2                                              =limx4n+3π2+ -sin x1-0      By L'Hospital rule                                              =-1lim x4n+3π2-fx limx4n+3π2+fxSo fx is not differentiable at x=4n+3π2Therefore, fx is neither differentiable at 4n+1π2 nor at 4n+3π2i.e. fx is not differentiable at odd multiples of π2i.e. fx is not differentiable at x=2n+1π2Therefore, f(x) is everywhere continuous but not differentiable at 2n+1π2 .
 

Page No 9.18:

Question 18:

The function f (x) = 1 + |cos x| is
(a) continuous no where
(b) continuous everywhere
(c) not differentiable at x = 0
(d) not differentiable at x = n π, nZ

Answer:

(b) continuous everywhere

Graph of the function f (x) = 1 + |cos x| is as shown below:



From the graph, we can see that f (x) is everywhere continuous but not differentiable at x=2n+1π2, nZ

Page No 9.18:

Question 19:

The function f (x) =  |cos x| is
(a) differentiable at x = (2n + 1) π/2, nZ
(b) continuous but not differentiable at x = (2n + 1) π/2, nZ
(c) neither differentiable nor continuous at x = nZ
(d) none of these

Answer:

(b) continuous but not differentiable at x = (2n + 1) π/2, nZ

We have,fx=cos xfx=cos x,      2nπx<4n+1π20,     x=4n+1π2    -cos x,    4n+1π2<x<4n+3π2             0 ,        x=4n+3π2        cos x,        4n+3π2< x2n+2πWhen, x is in first quadrant, i.e. 2nπx<4n+1π2 , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 2nπ, 4n+1π2When, x is in second quadrant or in third quadrant, i.e., 4n+1π2<x<4n+3π2 , we have         fx=-cos x which being a trigonometrical function is continuous and differentiable in 4n+1π2, 4n+3π2When, x is in fourth quadrant, i.e., 4n+3π2< x2n+2π , we have         fx=cos x which being a trigonometrical function is continuous and differentiable in 4n+3π2, 2n+2πThus possible point of non-differentiability of fx are x=4n+1π2, 4n+3π2Now, LHD at x=4n+1π2 =limx4n+1π2- fx- f4n+1π2x-4n+1π2                                               =limx4n+1π2-  cos x- 0x-4n+1π2                                              =limx4n+1π2-  -sin x1-0      By L'Hospital rule                                              =-1And  RHD at x=4n+1π2 =limx4n+1π2+ fx- f4n+1π2x-4n+1π2                                              =limx4n+1π2+  -cos x- 0x-4n+1π2                                              =limx4n+1π2+  sin x1-0      By L'Hospital rule                                              =1lim x4n+1π2-fx limx4n+1π2+fxSo fx is not differentiable at x=4n+1π2Now, LHD at x=4n+3π2 =limx4n+1π2- fx- f4n+3π2x-4n+3π2                                              =limx4n+3π2-  -cos x- 0x-4n+3π2                                              =limx4n+3π2-  sin x1-0      By L'Hospital rule                                              =1And  RHD at x=4n+3π2 =limx4n+3π2+ fx- f4n+3π2x-4n+3π2                                              =limx4n+3π2+  cos x- 0x-4n+3π2                                              =limx4n+3π2+ -sin x1-0      By L'Hospital rule                                              =-1lim x4n+3π2-fx limx4n+3π2+fxSo fx is not differentiable at x=4n+3π2Therefore, fx is neither differentiable at 4n+1π2 nor at 4n+3π2i.e. fx is not differentiable at odd multiples of π2i.e. fx is not differentiable at x=2n+1π2Therefore, f(x) is everywhere continuous but not differentiable at 2n+1π2 .

Page No 9.18:

Question 20:

The function fx=sin πx-π4+x2, where [⋅] denotes the greatest integer function, is
(a) continuous as well as differentiable for all x ∈ R
(b) continuous for all x but not differentiable at some x
(c) differentiable for all x but not continuous at some x.
(d) none of these

Answer:


(a) continuous as well as differentiable for all x ∈ R

Here, fx=sin πx-π4+x2
 
Since, we know that πx-π=nπ and sinnπ=0.
    
4+x20

f(x) = 0 for all x

Thus, f(x) is a constant function and it is continuous and differentiable everywhere.



Page No 9.19:

Question 21:

Let f (x) = a + b |x| + c |x|4, where a, b, and c are real constants. Then, f (x) is differentiable at x = 0, if
(a) a = 0
(b) b = 0
(c) c = 0
(d) none of these

Answer:

(b) b = 0

We have,fx=a+bx+cx4fx=a+bx+cx4        x0a-bx+cx4        x<0Here, fx is differentiable at x=0 LHD at x=0 = RHD at x=0limx0-fx-f0x-0=limx0+fx-f0x-0limx0-a-bx+cx4-ax=limx0+a+bx+cx4-axlimh0a-b0-h+c0-h4-a0-h=limh0a+b0+h+c0+h4-a0+hlimh0a+bh+ch4-a-h=limh0a+bh+ch4-ahlimh0bh+ch4-h=limh0bh+ch4hlimh0-b-ch3=limh0b+ch3-b=b2b=0b=0

Page No 9.19:

Question 22:

If f (x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f (x) is
(a) continuous and differentiable at x = 3
(b) continuous but not differentiable at x = 3
(c) differentiable nut not continuous at x = 3
(d) neither differentiable nor continuous at x = 3

Answer:

(d) neither differentiable nor continuous at x = 3

We have,fx=3-x+3+x, where x denotes the least integer greater than or equal to x.fx=3-x+3+3,                  2<x<3-3+x+3+4,                  3<x<4 fx=-x+9                2<x<3x+4                    3<x<4Here,LHL at x=3=limx3-fx=limx3--x+9=-3+9=6RHL at x=3=limx3+fx=limx3-x+4=3+4=7Since, LHL at x=3RHL at x=3Hence, given function is not continuous at x= 3Therefore, the function will also not be differentiable at x=3

Page No 9.19:

Question 23:

If fx=11+e1/x,     x00,     x=0 then f (x) is

(a) continuous as well as differentiable at x = 0
(b) continuous but not differentiable at x = 0
(c) differentiable but not continuous at x = 0
(d) none of these

Answer:

(d) none of these

we have,

(LHL at x = 0 )=  limx 0- f(x) = limh 0 f(0 - h) = limh 0 f(- h)= limh 0 11 + e1/-h= limh 0 11 + 1e1/h            [limh01e1/h = 0]  =   11 + 0= 1(RHL at x = 0) =  limx 0+ f(x) = limh 0 f(0 + h)=  limh 0 11 + e1/h=  11 + e1/0  = 11 + e = 11 +      

So, f(x) is not continuous at x = 0

Differentiability at x = 0

(LHD at x = 0 ) = limx 0- f(x) - f(0)x - 0= limh0 f(0 -h) - f(0)0 -h - 0= limh0 f(-h) - 0 -h= limh0  11 + e1/ - h -h= limh0  11 + 1e1/ h -h         = limh0  11 + 0 -h  = limh0  1 -h = -(RHD at x = 0) = limx 0+ f(x) - f(0)x - 0= limh 0 f(0 + h) - f(0)0 +h - 0= limh 0 f(h) -0 h = limh0  11 + e1/  hh = So, f(x) is also not differentiable at x = 0.




 

Page No 9.19:

Question 24:

If fx=1-cos xx sin x,x012             ,x=0

then at x = 0, f (x) is
(a) continuous and differentiable
(b) differentiable but not continuous
(c) continuous but not differentiable
(d) neither continuous nor differentiable

Answer:

(a) continuous and differentiable
 

we have,

fx=1-cos xx sin x,x012             ,x=0

fx=1-cos xx sin x,x012             ,x=0Continuity at x = 0(LHL at x = 0) = limx 0- f(x)                                               = limh 0 f(0 - h)                               = limh 0 f(- h)                     =  limh 0 1-cos (-h)(-h) sin (-h)                            = limh 0 1-cos hh sin h                       = limh 0 1-cosh   limh 0 1h sin h                       = 1 -cos(0) .  10 sin 0                         = 0                     


(RHL at x = 0) = limx 0+ f(x)                                               = limh 0 f(0 + h)                               = limh 0 f( h)                     =  limh 0 1-cos (h)(h) sin (h)                            = limh 0 1-cos hh sin h                       = limh 0 1-cosh   limh 0 1h sin h                       = 1 - cos 0. 10 sin 0                         = 0


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


For differentiability at x = 0

(LHD at x = 0 ) =limx 0- f(x) - f(0)x -0                                        =  limh0 f(0 - h) - f(0)0 - h -0                           = limh0 f(- h) -12- h                          = limh0 1 - cos(-h)- h sin(-h) -12- h                             =  1h limh0 1 - cosh h sin h  -limh0 12                             =   12 - 0=  12                                       

RHD at x = 0 ) =limx 0+ f(x) - f(0)x -0                                        =  limh0 f(0 + h) - f(0)0 - h -0                           = limh0 f( h) -12- h                          = limh0 1 - cos(h)- h sin(h) -12- h                          =  -1hlimh0 1 - cosh h sin h  -limh0 12                         =  12 - 0 =  12

 

Page No 9.19:

Question 25:

The set of points where the function f (x) given by f (x) = |x − 3| cos x is differentiable, is
(a) R
(b) R − {3}
(c) (0, ∞)
(d) none of these

Answer:

(b) R-3

LHD at x=3=limx3-fx-f3x-3LHD at x=3=limh0f3-h-f33-h-3LHD at x=3=limh0f3-h-f3-hLHD at x=3=limh03-h-3cos3-h-f3-hLHD at x=3=limh0hcos3-h-0-h=-cos3RHD at x=3=limx3+fx-f3x-3RHD at x=3=limh0f3+h-f33+h-3RHD at x=3=limh0f3+h-f3hRHD at x=3=limh03+h-3cos3+h-f3hRHD at x=3=limh0hcos3+h-0h=cos3

So, f(x) is not differentiable at x = 3.

Also, f(x) is differentiable at all other points because both modulus and cosine functions are differentiable and the product of two differentiable function is differentiable.

Page No 9.19:

Question 26:

Let fx=1  ,x-1x,-1<x<10  ,x1 Then, f is

(a) continuous at x = − 1
(b) differentiable at x = − 1
(c) everywhere continuous
(d) everywhere differentiable

Answer:

(b) differentiable at x = − 1

fx=1  ,x-1x,-1<x<10  ,x1

Differentiabilty at x = − 1
(LHD x = − 1)
                                lim x - 1- f(x) - f(-1) x + 1=lim x - 1 f(x) - f(-1) x + 1= lim x - 1 1 - 1 -1 + 1= 0

(RHD x = − 1)
                              = limx -1+f (x) - f(-1)x + 1 = limx -1f (x) - f(-1)x + 1 = limx -1f (x) - f(-1)x + 1= limx -1|x| - |-1|x + 1= 1 - 1|-1 + 1= 0 

Page No 9.19:

Question 27:

The function f(x) = e|x| is
(a) continuous every where but not differentiable at x = 0
(b) continuous and differentiable everywhere
(c) not continuous at x = 0
(d) none of these

Answer:


The given function is f(x) = e|x|.

We know

If f is continuous on its domain D, then f is also continuous on D.

Now, the identity function p(x) = x is continuous everywhere.

So, g(x) = px=x is also continuous everywhere.

Also, the exponential function ax, a > 0 is continuous everywhere.

So, h(x) = ex is continuous everywhere.

The composition of two continuous functions is continuous everywhere.

fx=hogx=ex is continuous everywhere.

Now,

gx=x=x,x0-x,x<0

Lg'0=limh0g0-h-g0-h

Lg'0=limh0--h-0-h

Lg'0=limh0h-h

Lg'0=-1

And

Rg'0=limh0g0+h-g0h

Rg'0=limh0h-0h

Rg'0=limh0hh

Rg'0=1

Lg'0Rg'0

So, gx=x is not differentiable at x = 0.

We know

The exponential function ax, a > 0 is differentiable everywhere.

So, h(x) = ex is differentiable everywhere.

We know that, the composition of differentiable functions is differentiable.
 
Now, ex is differentiable everywhere, but x is not differentiable at x = 0.

fx=hogx=ex is differentiable everywhere except at x = 0.   

Thus, the function f(x) = e|x| is continuous every where but not differentiable at x = 0.

Hence, the correct answer is option (a).

Page No 9.19:

Question 28:

The set of points where the function f(x) = |2x – 1| sin x is differentiable, is

(a) R

(b) R-12

(c) (0, ∞)

(d) none of these

Answer:


Let gx=2x-1 and hx=sinx.

We know that, the trigonometric functions are differentiable in their respective domain.

So, hx=sinx is differentiable for all x ∈ R.

Now,

gx=2x-1=2x-1,x12-2x-1,x<12

(2x − 1) and −(2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x<12 and for all x>12.

So, we need to check the differentiability of g(x) at x=12.

We have

Lg'12=limh0g12-h-g12-h

Lg'12=limh0-212-h-1-2×12-1-h

Lg'12=limh02h-h

Lg'12=-2

And

Rg'12=limh0g12+h-g12h

Rg'12=limh0212+h-1-2×12-1h

Rg'12=limh02hh

Rg'12=2

Lg'12Rg'12

So, gx=2x-1 is not differentiable at x=12.

The function gx=2x-1 is differentiable for all xR-12.

We know that, the product of two differentiable functions is differentiable.

fx=gx×hx=2x-1sinx is differentiable for all xR-12.

Thus, the set of points where the function fx=2x-1sinx is differentiable is R-12.

Hence, the correct answer is option (b).



Page No 9.20:

Question 1:

The function f(x) = |x + 1| is not differentiable at x = ____________.

Answer:


The given function is fx=x+1.

fx=x+1=x+1,x-1-x+1,x<-1

Now, (x + 1) and −(x + 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x>-1 and for all x<-1.

So, we need to check the differentiability of f(x) at x=-1.

We have,

Lf'-1=limh0f-1-h-f-1-h

Lf'-1=limh0--1-h+1--1+1-h

Lf'-1=limh0h-h

Lf'-1=-1

And

Rf'-1=limh0f-1+h-f-1h

Rf'-1=limh0-1+h+1--1+1h

Rf'-1=limh0hh

Rf'-1=1

Lf'-1Rf'-1

So, f(x) is not differentiable at x=-1.

Thus, the function f(x) = |x + 1| is not differentiable at x = −1.


The function f(x) = |x + 1| is not differentiable at x = ___−1___.

Page No 9.20:

Question 2:

The function g(x) = |x – 1| + |x + 1| is not differentiable at x = ____________.

Answer:


x-1=x-1,x1-x-1,x<1

x+1=x+1,x-1-x+1,x<-1

gx=x-1+x+1=-x-1-x+1,x<-1-x-1+x+1,-1x<1x-1+x+1,x1

gx=x-1+x+1=-2x,x<-12,-1x<12x,x1

When x < −1, g(x) = −2x which being a polynomial function is continuous and differentiable.

When −1 ≤ x < 1, g(x) = 2 which being a constant function is continuous and differentiable.

When x ≥ 1, g(x) = 2x which being a polynomial function is continuous and differentiable.

Thus, the possible points of non-differentiability of g(x) are x = −1 and x = 1.

Now,

 Lg'-1=limh0g-1-h-g-1-h

Lg'-1=limh0-2-1-h-2-h

Lg'-1=limh02h-h

Lg'-1=-2

And

Rg'-1=limh0g-1+h-g-1h

Rg'-1=limh02-2-h

Rg'-1=limh00-h

Rg'-1=0

Lg'-1Rg'-1

So, g(x) is not differentiable at x = −1.

Also,

 Lg'1=limh0g1-h-g1-h

Lg'1=limh02-2×1-h

Lg'1=limh00-h

Lg'1=0

And

Rg'1=limh0g1+h-g1h

Rg'1=limh021+h-2×1h

Rg'1=limh02hh

Rg'1=2

Lg'1Rg'1

So, g(x) is not differentiable at x = 1.

Thus, the function gx=x-1+x+1 is not differentiable at x = −1 and x = 1.


The function g(x) = |x – 1| + |x + 1| is not differentiable at x = ___±1___.

Page No 9.20:

Question 3:

The set of points where f(x) = x – [x] not differentiable is ____________.

Answer:


Let g(x) = x and h(x) = [x].

Every polynomial function is differentiable for all x ∈ R. So, g(x) = x is differentiable for all x ∈ R.

Also, the function h(x) = [x] is discontinuous at all integral values of x i.e. x ∈ Z. So, h(x) = [x] is not differentiable at all integral values of x i.e. x ∈ Z.

Now, f(x) = g(x) − h(x) = x − [x]

So, the function f(x) = x − [x] is differentiable for all x ∈ R except at all integral values of x i.e. x ∈ Z. The function f(x) = x − [x] is not differentiable for all x ∈ R − Z. 

Thus, the set of points where f(x) = x – [x] not differentiable is R − Z.


The set of points where f(x) = x – [x] not differentiable is ___R − Z___.

Page No 9.20:

Question 4:

The number of points in [–π, π] where f(x) = sin–1 (sin x) is not differentiable is. ____________.

Answer:



fx=sin-1sinx=-x-π,-πx-π2x,-π2xπ2π-x,π2xπ

Let us check the differentiability of the function at x=-π2 and x=π2.

At x=-π2,

Lf'-π2=limx-π2-fx-f-π2x--π2Lf'-π2=limx-π2-x-π--π2x+π2Lf'-π2=limx-π2-x+π2x+π2Lf'-π2=-1

Rf'-π2=limx-π2+fx-f-π2x--π2Rf'-π2=limx-π2x--π2x+π2Rf'-π2=limx-π2x+π2x+π2Rf'-π2=1

Lf'-π2Rf'-π2

So, the function f(x) is not differentiable at x=-π2.

At x=π2,

Lf'π2=limxπ2-fx-fπ2x-π2Lf'π2=limxπ2x-π2x-π2Lf'π2=1

Rf'π2=limxπ2+fx-fπ2x-π2Rf'π2=limxπ2π-x-π2x-π2Rf'π2=limxπ2-x-π2x-π2Rf'π2=-1

Lf'π2Rf'π2

So, the function f(x) is not differentiable at x=π2.

Thus, the function f(x) = sin–1(sinx), x ∈ [–π, π] is not differentiable at x=-π2 and x=π2.


 The number of points in [–π, π] where f(x) = sin–1 (sin x) is not differentiable is       2      .

Page No 9.20:

Question 5:

The function f(x) = cos–1(cos x), x ∈ (–2π, 2π) is not differentiable at x = ____________.

Answer:


fx=cos-1cosx=x+2π-2πx-π-x,-πx0x,0xπ2π-x,πx2π

Let us check the differentiability of the function at x=-π, x = 0 and x=π.

At x=-π,

Lf'-π=limx-π-fx-f-πx--πLf'-π=limx-πx+2π-πx+πLf'-π=limx-πx+πx+πLf'-π=1

Rf'-π=limx-π+fx-f-πx--πRf'-π=limx-π-x-πx+πRf'-π=limx-π-x+πx+πRf'-π=-1

Lf'-πRf'-π

So, the function f(x) is not differentiable at x=-π.

At x=0,

Lf'0=limx0-fx-f0x-0Lf'0=limx0-x-0xLf'0=limx0-xxLf'0=-1

Rf'0=limx0+fx-f0x-0Rf'0=limx0x-0x-0Rf'0=limx0xxRf'0=1

Lf'0Rf'0

So, the function f(x) is not differentiable at x=0.

At x=π,

Lf'π=limxπ-fx-fπx-πLf'π=limx-πx-πx-πLf'π=1

Rf'π=limxπ+fx-fπx-πRf'π=limxπ2π-x-πx-πRf'π=limxπ-x-πx-πRf'π=-1

Lf'πRf'π

So, the function f(x) is not differentiable at x=π.

Thus, the function f(x) = cos–1(cos x), x ∈ (–2π, 2π) is not differentiable at x=-π, x = 0 and x=π.


The function f(x) = cos–1(cos x), x ∈ (–2π, 2π) is not differentiable at x =       -π,0,π      .

Page No 9.20:

Question 6:

The function fx=sin x, -π2, π2 is not differentiable at x = ____________.

Answer:


We know that, x is not differentiable at x = 0.

Therefore, sinx is not differentiable when sinx=0.

sinx=0

x=nπ, n ∈ Z

x= ...,-2π,-π,0,π,2π,...

Now, the only value of x lying in given interval -π2, π2 at which the function fx=sinx is not differentiable is 0.

Thus, the function fx=sinx, x-π2, π2 is not differentiable at x = 0.


The function fx=sin x, -π2, π2 is not differentiable at x = ___0___.

Page No 9.20:

Question 7:

Let fx=ax2+3,x>1x+52,x1 . If f(x) is differentiable at x = 1, then a = ____________.

Answer:


The given function fx=ax2+3,x>1x+52,x1 is differentiable at x = 1.

Lf'1=Rf'1

limh0f1-h-f1-h=limh0f1+h-f1h

limh01-h+52-1+52-h=limh0a1+h2+3-1+52h

limh0-h-h=limh0a1+h2-12h

1=limh0a-12+a2h+h2h

1=limh0a-12+ah2+hh

1=limh0a-12h+limh0a2+h

Here, LHS is finite.

So, for RHS to be finite, we must have

a-12=0

a=12

Thus, the value of a is 12.


Let fx=ax2+3,x>1x+52,x1 . If f(x) is differentiable at x = 1, then a =      12     .

Page No 9.20:

Question 8:

If f(x) = x |x|, then f' (–1) = ____________.

Answer:


x=x,x0-x,x<0

fx=xx=x2,x0-x2,x<0

Now,

Lf'-1=limh0f-1-h-f-1-h

Lf'-1=limh0--1-h2---12-h

Lf'-1=limh0-1+2h+h2+1-h

Lf'-1=limh0-2h+h2-h

Lf'-1=limh02+h

Lf'-1=2+0=2

Also,

Rf'-1=limh0f-1+h-f-1h

Rf'-1=limh0--1+h2---12h

Rf'-1=limh0-1-2h+h2+1h

Rf'-1=limh0--2h+h2h

Rf'-1=limh02-h

Rf'-1=2-0=2

So, Lf'-1=Rf'-1=2

f'-1=2


If f(x) = x|x|, then f'(–1) = ___2___.

Page No 9.20:

Question 9:

If f(x) = x |x|, then f' (2) =____________.

Answer:


x=x,x0-x,x<0

fx=xx=x2,x0-x2,x<0

Now,

Lf'2=limh0f2-h-f2-h

Lf'2=limh02-h2-22-h

Lf'2=limh04-4h+h2-4-h

Lf'2=limh0-4+hh-h

Lf'2=limh04-h

Lf'2=4-0=4

Also,

Rf'2=limh0f2+h-f2h

Rf'2=limh02+h2-22h

Rf'2=limh04+4h+h2-4h

Rf'2=limh04+hhh

Rf'2=limh04+h

Rf'2=4+0=4

So, Lf'2=Rf'2=4.

f'2=4


If f(x) = x|x|, then f'(2) = ___4____.

Page No 9.20:

Question 10:

The set of point where the function f(x) = |2x – 1| is differentiable, is ____________.

Answer:


The given function is fx=2x-1.

fx=2x-1=2x-1,x12-2x-1,x<12

Now, (2x − 1) and −(2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x>12 and for all x<12.

So, we need to check the differentiability of f(x) at x=12.

We have,
Lf'12=limh0f12-h-f12-h

Lf'12=limh01-212-h-2×12-1-h

Lf'12=limh02h-h

Lf'12=-2

And
Rf'12=limh0f12+h-f12h

Rf'12=limh0212+h-1-2×12-1h

Rf'12=limh02hh

Rf'12=2

Lf'12Rf'12

So, f(x) is not differentiable at x=12.

Thus, the set of points where the function f(x) = |2x – 1| is differentiable is R-12.


The set of point where the function f(x) = |2x – 1| is differentiable, is      R-12     .

Page No 9.20:

Question 11:

The set of points where the function f(x)=x+1,x<22x-1,x2is not differentiable, is ____________.

Answer:


The given function is fx=x+1,x<22x-1,x2.

(x + 1) and (2x − 1) are polynomial functions which are differentiable at each x ∈ R. So, f(x) is differentiable for all x < 2 and for all x > 2.

So, we need to check the differentiability of f(x) at x = 2.

We have

Lf'2=limh0f2-h-f2-h

Lf'2=limh02-h+1-2×2-1-h

Lf'2=limh03-h-3-h

Lf'2=limh0-h-h

Lf'2=1

And

Rf'2=limh0f2+h-f2h

Rf'2=limh022+h-1-2×2-1h

Rf'2=limh03+2h-3h

Rf'2=limh02hh

Rf'2=2

Lf'2Rf'2

So, f(x) is not differentiable at x = 2.

Thus, the set of points where the function f(x) is not differentiable is {2}.


The set of points where the function fx=x+1,x<22x-1,x2is not differentiable, is _____{2}______.

Page No 9.20:

Question 12:

An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is ____________.

Answer:


Consider the function gx=x-1+x+1.

x-1=x-1,x1-x-1,x<1

x+1=x+1,x-1-x+1,x<-1

gx=x-1+x+1=-x-1-x+1,x<-1-x-1+x+1,-1x<1x-1+x+1,x1

gx=x-1+x+1=-2x,x<-12,-1x<12x,x1

When x < −1, g(x) = −2x which being a polynomial function is continuous and differentiable.

When −1 ≤ x < 1, g(x) = 2 which being a constant function is continuous and differentiable.

When x ≥ 1, g(x) = 2x which being a polynomial function is continuous and differentiable.

Let us check the continuity and differentiability of g(x) at x = −1 and x = 1.

At x = −1,

LHL = limx-1-gx=limx-1-2x=-2×-1=2

RHL = limx-1+gx=limx-12=2

g-1=2

Since limx-1-gx=limx-1+gx=g-1, so the function g(x) is continuous at x = −1.

At x = 1,

LHL = limx1-gx=limx12=2

RHL = limx1+gx=limx12x=2×1=2

g1=2×1=2

Since limx1-gx=limx1+gx=g1, so the function g(x) is continuous at x = 1.

Thus, the function g(x) is continuous everywhere i.e. for all x ∈ R.

Now,

 Lg'-1=limh0g-1-h-g-1-h

Lg'-1=limh0-2-1-h-2-h

Lg'-1=limh02h-h

Lg'-1=-2

And

Rg'-1=limh0g-1+h-g-1h

Rg'-1=limh02-2-h

Rg'-1=limh00-h

Rg'-1=0

Lg'-1Rg'-1

So, g(x) is not differentiable at x = −1.

Also,

 Lg'1=limh0g1-h-g1-h

Lg'1=limh02-2×1-h

Lg'1=limh00-h

Lg'1=0

And

Rg'1=limh0g1+h-g1h

Rg'1=limh021+h-2×1h

Rg'1=limh02hh

Rg'1=2

Lg'1Rg'1

So, g(x) is not differentiable at x = 1.

Thus, the function g(x) is differentiable everywhere except at x = −1 and x = 1.


An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is      gx=x-1+x+1     .

Page No 9.20:

Question 13:

The set of points where f(x) = cos |x| is differentiable, is ____________.

Answer:


We know

x=x,x0-x,x<0

fx=cosx=cosx,x0cos-x,x<0

fx=cosx=cosx,x0cosx,x<0                  cos-θ=cosθ

We know that, cosine function is differentiable in its domain. So, f(x) is differentiable for all x < 0 and x > 0.

Let us check the differentiability of fx=cosx at x = 0.

Now,

Lf'0=limh0f0-h-f0-h

Lf'0=limh0cos-h-cos0-h

Lf'0=limh0cosh-1-h

Lf'0=limh0-2sin2h2-h

Lf'0=limh0sinh2h2×limh0sinh2

Lf'0=1×0

Lf'0=0

And

Rf'0=limh0f0+h-f0h

Rf'0=limh0cosh-cos0h

Rf'0=limh0cosh-1h

Rf'0=limh0-2sin2h2h

Rf'0=-limh0sinh2h2×limh0sinh2

Rf'0=-1×0

Rf'0=0

Lf'0=Rf'0

So, f(x) is differentiable at x = 0. Thus, the function f(x) is differentiable everywhere.

Hence, the set of points where fx=cosx is differentiable is R (set of real real numbers).


The set of points where f(x) = cos |x| is differentiable, is _____R_____.

Page No 9.20:

Question 14:

The set of points where f(x) = |sin x| is not differentiable, is ____________.

Answer:


Let gx=x=x,x0-x,x<0

Now,

Lg'0=limh0g0-h-g0-h

Lg'0=limh0--h-0-h

Lg'0=limh0h-h

Lg'0=-1

And

Rg'0=limh0g0+h-g0h

Rg'0=limh0h-0h

Rg'0=limh0hh

Rg'0=1

Lg'0Rg'0

So, x is not differentiable at x = 0.

Therefore, fx=sinx is not differentiable when sinx=0.

sinx=0

x=nπ, nZ

Thus, the set of points where fx=sinx is not differentiable is x=nπ:nZ.


The set of points where f(x) = |sin x| is not differentiable, is      x=nπ:nZ     .

Page No 9.20:

Question 15:

The set of points at which the function fx=1logxis not differentiable, is ____________.

Answer:


The given function is fx=1logx.

For f(x) to be defined,

x0 and logx0

x0 and x1

x0 and x±1

Thus, the function f(x) is not defined when x = −1, x = 0 and x = 1.

We know that, the logarithmic function is differentiable at each point in its domain. Every constant function is differentiable at each x ∈ R. Also, the quotient of two differentiable functions is differentiable.

So, the function fx=1logx is not differentiable at x = −1, x = 0 and x = 1.

Thus, the set of points at which the function fx=1logxis not differentiable is {−1, 0, 1}.


The set of points at which the function fx=1logxis not differentiable, is ___{−1, 0, 1}___.  

Page No 9.20:

Question 16:

The greatest integer function f(x)=x, 0 < x < 2 is not differentiable at x = ________

Answer:

Given: f(x)=x, 0 < x < 2


Thus, the function y=x is not differentiable at x=1 in 0<x<2
Hence, x=1 is a point of non-differentiability.



Page No 9.21:

Question 11:

If limxcfx-fcx-c exists finitely, write the value of limxcfx.

Answer:

Given:  limxcf(x) - f(c)x-c exists finitely. Then,    
 
  limxc f(x) - f(c)x-c = f'(c).

Now,    
limxc f(x) = lim xc f(x) - f(c)x-c (x-c) + f(c)                = limxc f(x) - f(c)x-c (x-c) + f(c)                =  limxc f(x) - f(c)x-c limxc (x-c) + f(c)                = f'(c)×0 + f(c)                 = f(c) 

Page No 9.21:

Question 12:

Write the value of the derivative of f (x) = |x − 1| + |x − 3| at x = 2.

Answer:

Given: f(x) = x-1 + x-3

 f(x) = -(x-1) -(x-3),      x<1x-1 - (x-3),         1x<3          (x-1) + (x-3),        x3

 f(x) =  -2x+4,     x<12,                1x<32x-4,        x3

We check differentiability at x = 2

(LHD at x = 2)

  limx2- f(x) - f(2)x-2 = limh0 f(2-h) - f(2)2-h-2 = limh0 2-2-h =0

Page No 9.21:

Question 13:

If f x=x2+9, write the value of limx4 fx-f4x-4.

Answer:

Given: f(x) = x2 + 9
Now,
       f(4) = 16+9        = 25        =5 

So, f(x) - f(4)x-4 = x2 + 9 - 5x-4

On rationalising the numerator, we get

f(x) - f(4)x-4 = x2+9 -5x-4×x2+9 + 5x2+9 + 5                       = x2+9 - 25(x-4) x2+9 + 5                       = x2 - 16(x-4) x2+9 + 5                      = (x+4)x2+9 + 5

Taking limit x4, we have

limx4 f(x) - f(4)x-4 = limx4 (x+4)x2+9 + 5                                = 810                                = 45



Page No 10.17:

Question 1:

Differentiate the following functions from first principles:

ex

Answer:

Let  fx=e-xfx+h=e-x+h ddxfx=limh0fx+h-fxh                  =limh0e-x+h-e-xh                  =limh0e-x×e-h-e-xh                  =limh0e-xe-h-1-h×-1                  =-e-x limh0e-h-1-h                                                         =-e-x                    limh0e-h-1-h=1So, ddxe-x=-e-x

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Question 2:

Differentiate the following functions from first principles:

e3x

Answer:

Let  fx=e3xfx+h=e3x+hddxfx=limh0fx+h-fxh                =limh0e3x+h-e3xh                =limh0e3xe3h-e3xh                =limh0e3xe3h-13h×3                =3e3xlimh0e3h-13h                =3e3xHence, ddxe3x=3e3x

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Question 3:

Differentiate the following functions from first principles:

eax+b

Answer:

Let  fx=eax+b       fx+h=eax+h+b        ddxfx=limh0fx+h-fxh                             =limh0eax+h+b-eax+bh                             =limh0eax+beah-eax+bh                             =limh0eax+beah-1ah×a                             =aeax+blimh0eah-1ah                             =aeax+bSo, ddxeax+b=aeax+b

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Question 4:

Differentiate the following functions from first principles:

ecos x

Answer:

Let  fx=ecosx       fx+h=ecosx+h        ddxfx=limh0fx+h-fxh                             =limh0ecosx+h-ecosxh                             =limh0 ecosxecosx+h-cosx-1h                             =limh0 ecosxecosx+h-cosx-1cosx+h cosx×cosx+h-cosxh                             =ecosxlim             h0  cosx+h-cosxh ×limh0ecosx+h-cosx-1cosx+h-cosx                             =ecosxlimh0  cosx+h-cosxh                            limh0ex-1x=1                             =ecosxlimh0 -2sinx+h+x2sinx+h-x2h          cosA-cosB=-2sin A+B2sinA-B2                            =ecosxlimh0-sin2x+h21×sinh2h2                            =ecosxlimh0-sin2x+h21× lim       h0sinh2h2                           =ecosxlimh0-sin2x+h2                                         sinxx=1                          =ecosx-sinx                           =-sinxecosxHence, ddxecosx=-sinxecosx

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Question 5:

Differentiate the following functions from first principles:

e2x

Answer:

Let  fx=e2x  fx+h=e2x+h        ddxfx=limh0fx+h-fxh                             =limh0e2x+h-e2xh                             =limh0 e2xe2x+h-2x-1h                             =e2xlimh0 e2x+h-2x-12x+h-2x×limh02x+h-2xh                             =e2xlimh02x+h-2xh                                 limh0eh-1h=1                             =e2xlimh0 2x+h-2xh × 2x+h+2x2x+h+2x       Rationalising the numerator                            =e2xlimh02x+h-2xh2x+h+2x                           =e2xlimh02x+2h-2xh2x+h+2x                                                                =e2xlimh02hh2x+h+2x                           =e2xlimh022x+h+2x                          =e2x2xHence, ddxe2x=e2x2x

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Question 6:

Differentiate the following functions from first principles:

log cos x

Answer:

Let fx=log cosxfx+h=log cosx+hddxfx=limh0fx+h=fxh                     =limh0log cosx+h-log cosxh                     =limh0logcosx+hcosxh                        logA-logB=logAB                     =limh0log1+cosx+hcosx-1h                     =limh0log1+cosx+h-cosxcosxcosx+h-cosxcosx×limh0cosx+h-cosxcosx                     =1×limh0cosx+h-cosxcosx × h                    limx0log1+xx=1                    =limh0-2sinx+h+x2sinx+h-x2cosx × h                     =-2limh0sin2x+h2×sinh22cosx ×h2                     =-2sinx2cosx                            limx0sinxx=1                      =-tanxSo, ddxlog cosx=-tanx

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Question 7:

Differentiate the following function from first principles:

ecot x
 

Answer:

  Let  fx=ecotxfx+h=ecotx+h  ddxfx=limh0fx+h-fxh                      =limh0ecotx+h-ecotxh                      =limh0ecotxecotx+h-cotx-1h                      =ecotxlimh0ecotx+h-cotx-1cotx+h-cotx×cotx+h-cotxh                      =ecotxlimh0cotx+h-cotxh×cotx+h+cotxcotx+h+cotx                            limx0ex-1x=1 and rationalizing the numerator                      =ecotxlimh0cotx+h-cotxhcotx+h+cotx                      =ecotxlimh0cotx+hcotx+1cotx-x-hhcotx+h+cotx                                                                     cotA-B=cotAcotB+1cotB-cotA                      =ecotxlimh0cotx+hcotx+1cot-h×hcotx+h+cotx                      =-ecotxlimh0cotx+hcotx+1htanhcotx+h+cotx                      =ecotx×cot2x+12cotx                                       limx0tanxx=1                      =-ecotx×cosec2x2cotx                                          1+cot2x=cosec2xddxecotx=-ecotx×cosec2x2cotx  

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Question 8:

Differentiate the following functions from first principles:

x
2ex
 

Answer:

 Let  fx=x2exfx+h=x+h2ex+h                 =limh0fx+h-fxh                 =limh0x+h2ex+h-x2exh                 =limh0x2ex+h-x2exh+2xhex+hh+h2ex+hh                 =limh0x2exex+h-x-1h+2xex+h+hex+h                 =limh0x2exeh-1h+2xex+h+hex+h                 =x2ex+2xex+0xex                        limx0ex-1x=1ddxx2ex=exx2+2x

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Question 9:

Differentiate the following functions from first principles:

 log cosec x

Answer:

 Let  fx=log cosecx         fx+h=log cosecx+hddxfx=limh0fx+h-fxh                     =limh0log cosecx+h-log cosecxh                     =limh0logcosecx+hcosecxh                     =limh0log1+sinxsinx+h-1h                     =limh0log1+sinx-sinx+hsinx+hsinx-sinx+hsinx+hsinx-sinx+hsinx+hh                     =limh02cosx+x+h2sinx-x-h2sinx+hh                             limx0log1+xx=1 and sinA-sinB=2cosA+B2sinA-B2                     =limh02cos2x+h2sinx+h -2sin-h2-h2                                 limx0sinxx=1                     =-cotxddxlog cosecx=-cotx

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Question 10:

Differentiate the following functions from first principles:

sin−1 (2x + 3)

Answer:

 Let fx=sin-12x+3        fx+h=sin-12x+h+3        fx+h=sin-12x+2h+3 ddxfx=limh0fx+h-fxh                      =limh0sin-12x+2h+3-sin-12x+3h                      =limh0sin-12x+2h+31-2x+32-2x+31-2x+2h+32h             sin-1x-sin-1y=sin-1x1-y2-y1-x2                      =limh0sin-1zz×zhwhere, z=2x+2h+31-2x+32-2x+31-2x+2h+32 and limh0sin-1hh=1                      =limh0zh                      =limh0 2x+2h+31-2x+32-2x+31-2x+2h+32 h                      =limh02x+2h+321-2x+32-2x+321-2x+2h+32h2x+2h+31-2x+32+2x+31-2x+2h+32           Rationalizing numerator                      =limh02x+32+4h2+4h2x+31-2x+32-2x+321-2x+32-4h2-4h2x+3h2x+2h+31-2x+32+2x+31-2x+2h+32                      =limh02x+32+4h2+4h2x+3-2x+34-4h22x+32-4h2x+33-2x+32+2x+34+4h22x+32+4h2x+33h2x+2h+31-2x+32+2x+31-2x+2h+32                      =limh04hh+2x+3h2x+2h+31-2x+32+2x+31-2x+2h+32                      =42x+32x+31-2x+32+2x+31-2x+32                      =42x+322x+31-2x+32                      =21-2x+32 ddxsin-12x+3=21-2x+32             



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