sol-m1204

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HINTS/SOLUTIONS for M1204(Continuity & Differentiability)

Classroom Discussion Exercise

1. (b) cosx in continuous for all x R tanx is continuous

for all x R

2

1n2

1x

xsinlim0x

and f(0) = 1

f(x) =

0x1

0xx

xsinis continuous

x < 0 f(x) = 1x

|x|

x > 0 f(x) = 1 f(x) = 1 x R is also continuous.

2. (c) f is continuous at all points.

3. (b) f(2 ) =

2xlim

x2xcot

=

0xlim x2

xtan =21

4. (c) f(1) =1x

lim

(1-x) tan2x =

0ylim

y cot2y

=0y

lim

2ytan

x

=2

5. (b)2x

lim

k2x

)8x( 3

= 12k = f(2)

12k = 3 k =41

6. (c) The function discontinuous at -1 and 1.

7. (b) k =0x

lim f(x) = 0 =0x

lim f(x).

8. (b) f(0)=0x

lim x

1ex x

)x1log(lim0x

= 1.

9. (a) f is discontinuous at 0 f1(0) does not exist.

10. (d) 32

112

3

xcos1xsin

dxdy

11. (d) y = ex 1dxdy = ex

1y1e

dydx x

12. (c) f’(x) = 2x1

1

.

x21 =

x)x1(21

13. (c) The function is21 sin 2x2; derivative is

2x cos 2x2 = 2x (cos2x2 – sin2x2).

14. (c) y2 =x1x1 = 1 + x1

2

2ydxdy = 2)x1(

2

15. (a) tany2x3y2x3

=7

1tany2x3y2x3

(7)

xy =k (a constant) y = kx

dxdy =k =

xy .

16. (c) x = . )ysin(ysin

dydx

)y(sinysin)ycos(ycos)ysin(

2

= ysinsin.

2

17. (c) y = xsinx

logy = sinx.logx

xcos.xlogx

xsindxdy.

y1

xlog.xcos

xxsiny

dxdy

18. (d)2ttan

)tcos1(2tsin2

dtdx

dtdy

dxdy

.

19. (b) 1y1

x1

y =1x

x

21x1

dxdy

20. (c) f(x) =2 - cos-1 xtan2

2x1x1 1

2

2

)x('f = 2x12

21. (c)

xtan.321

xtan32

tany 1 =

x

32tantantan 11

y = tan1 x32

1

dxdy



22. (b) sin-1x =2 cos-1 x the derivative is derivative

of

t2

with respect to t = 1.

23. (b) Let u = sec2x and v = tanx u = 1 + v2

xtan2v2dvdu

24. (a) y =x1

x = -1 +

x11

dxdy

2)x1(1

32

2

)x1(2

dxyd

25. (a) y = log 2y2 x1xex1x

ey . y1 =2

y

2

2

2 x1

e

1x

x1x

1x

x1

y1 = 1yx1x1

1 21

22

again differentiate with respect to x(1 + x2) . 2y1y2 + (y1)2 . 2x = 0Divided by 2y1 (1 + x2) y2 + xy1 = 0

Regular Homework Exercise

1. (b)0x

lim f(x) = 2 = f(0) =0x

lim f(x) , in (b).

2. (b) f(o) =0x

lim

x13

x16

x12

x16

xx

xx

=3log6log2log6log

=2log3log = log23

3. (b) f(x) is not continuous at x = 1 and 2Since )x(flim)x(flim

1x1x

)x(flim)x(flim2x2x

4. (a) f(x) = x |x|

f(x) =

0x,00x,x2

is continuous for all x.

5. (b) The function is discontinuous at 0

6. (b)0x

lim f(x) = 2 =0x

lim f(x) = f(0)

It is continuous at x = 0.

7. (d) f(0) = 1 =0x

lim f(x) = a;

f(1) = 0 = ba)x(flim1x

b = 1.

8. (b) f(x) = 2

2

)1x()3x2()1x(

-5 as x (1)

9. (c) 3,225 and f(x) = [x 1] = 1

)3,2(x0)x('f

10. (c) The derivative isx

)xcos(log

11. (b) y = e-xdxdy = e-x = y

y1

dydx

12. (b) x2 y2 = x2y2 y2 =1x

x2

2

y =1x

x2

1x1x

1dxdy

22

322

1xdydx

13. (c) log 3ey5x2y5x2

key5x2y5x2 3e

xy =k (a constant) y = kx

xyk

dxdy

14. (d) The derivative is 2x . sec2x2 2xtan2

extan2

1

15. (d) y = yxsinyyxsin 2

y2 + y = sinx

(2y + 1) xcosdxdy

1y2xcos

dxdy

16. (a))cossinsin(a)sincos(cosa

dxdy

= tan

17. (c) y = tan-12x1

x2

+ cot-1 2x1x2

=2

dxdy = 0.

18. (a) derivative of x with respect to ex



=x.t.r.weofderivative

1x = x

x ee1

19. (a) y’ =x

)xsin(logx

)xcos(log

xy' = cos(logx) sin(logx)

xy” + y’ . 1 = x

xlogcosx

xlogsin

x2y” + xy’ = y

20. (a) 2

1

1x1

xsin2y

(1 x2)y12 = 4(sin1x)2 = 4y

(1 x2) 2y1y2 + y12.(2x) = 2y1

Divided by 2y1 (1 x2)y2 xy1 = 2

Assignment Exercise

1. (b) sinlim0x x

1

x

1sinlim0x

x

1sinlim0x

does not exist

2. (b) 1xx

xsinlim 2

2

0x = 1x

1x

xsinlim2

0x

= 1

= f(0) it is continuous at x = 0

1xxxsinlim 2

2

1x does not exist

it is not continuous at x = 1

3. (b) The only discontinuity is at x = -1.

4. (a) f(0) =0x

lim

x

)x21log(x

)x21log(

= 2 + (2) = 0.

5. (c) k = f(0) =x

x1x1lim0x

.1x1x1

2lim0x

6. (d) The domain is R – {0}. And the function isdiscontinuous at all integral points; so Z–{0}.

7. (d)x

0x1sinx

lim2

0x

exists (=0).

8. (c) f1(0) = 1; f1 (0+) = 1 f1(0) does notexist.

9. (a) |x| > 1 ( 1) (1, ) In (, 1)

y = |x|log = xlog

y2 = log(x)

2y.x11.

x1

dxdy

yx21

dxdy

In (1, ), y2 = logx

yx21

dxdy

yx21

dxdy , in |x| > 1

10. (c) x2 (1 + y) = y2 (1 + x) (x y)(x + y + xy) = 0( x y)

x + y + xy = 0 y =x1

x

2)x1(1

dxdy

11. (d) y2 = xy represent two straight lines y = x and y = 0

dxdy = 1 or 0

12. (a) 2

2a

adxdy

13. (b) y = tan1

tan1tan1

= 4

y = xtan4

1

2x11

dxdy

14. (c) x2 + y2 = r2

x + yy’ = 0 y’ = yx

y” = 2y

'y.x1y

y2y” = y + x.yx

y3y” = (x2 + y2) = r2

15. (c) The function defined by x is not differentiable at

0 (1, 1). Hence Rolle’s Theorem is notapplicable.



Additional Practice Exercise

1. (d)ax

lim f(x), ax

lim f(x) do not exist for any a.

2. (b) f is continuous at all points in R – {0};

Now0x

lim

x2 sinx1 = 0 f is discontinuous

at 0. f(0) = 1

3. (d) |x|.

4. (d) x

x31logx51loglim0x

= 5 x3

x31loglim3x5

x51loglim0x30x5

= 5 1 3 1 = 8

5. (b) a. 0 + b.0 + 3 = -a + b b – a = 32a (-1) – b(1) = -2a. 0 + b.0

2a + b = 0a = ─1, b = 2.

6. (c) f is continuous at all points, except at 0.

7. (c) k = f(0) =

x

13x

14limxx

0x xx2sin

=2

3log4log =34log

21

= log32

8. (b) k = f(0) =0x

lim )x21log(

x.xcos1

xxtanx

1e 2

3

xtanx3

= 1. 121.

211

9. (d) Obviously, choice is (d).

10. (d) a differentiable function is necessarilycontinuous.

11. (a) f(-x) = f(x) f’(x) = f’(-x)

f’(1) = f’(1) = 21

12. (d) y =elog10xlog10

elogxlogxlog 10

1010

e10

y =elogxlog = logex

x1

dxdy

13. (d) obviously, the result is not validwhere g1(x) = 0.

14. (d) y = 1e xlog 1

= loge x

x1

dxdy .

15. (a) y = log e x =x21

dxdyxlog

21

16. (d)dxd 10x = 10x log 10.

17. (b)elogxxlogxxlog x

ex = log e x dxdy =

x1

18. (a) y – x = x log x dxdy = 1 + 1+ log x

19. (b) It is the derivative of2 -2 tan-1x with respect to

2 ─2 tan-1x is the required derivative is 1

20. (d) It is the derivative of21 cos-12x

=2x41

2.21

.

21. (c) y = sin─1

21x1xcos

1x1x

2

21

2

2

0dxdy .

22. (a) tan2x = sec2x-1 thederivative of y 1 with respect to y is 1

23. (d) log tanx = log cot x

derivative of (– log y) with respect to y = y1

= tan x

24. (a) f’(x) = derivative of 3 tan-1x with respect to x.

= 2x13

25. (c) f’(x) = derivative of xtan22

1

= 2x12

26. (d) If 0 < x < 1,putting x = tan, we get y = 2tan1x

For x > 1, puttingx1 = tan, we get y = 2tan1

x1

Hence y =

1x,x1tan2

1x0,xtan21

1

Although y is continuous at x = 1, it is not differentiablethere

27. (a) f(x) =

2xcos

2xsin21

2xcos

2xsin21

tan1



=

2

2

1

2xsin

2xcos

2xsin

2xcos

tan

=

2xsin

2xcos

2xsin

2xcos

tan 1 ,

since sin2x > cos

2x in

2 < x <

f(x) = tan1

2x

4tan

2x

4

f’(x) =21

28. (d)dxdy

21

2tsec

2ttan

1tsin8dtdx 2

= 8tsin

tcos8tsin

1tsin2

dtdy = 8cos t

dtdxdtdy

dxdy = tan t

2

2

dxyd = ttan

dxd

dxdy

dxd

= sec2t .dxdt = sec2t

tcos8tsin2

=tcos8

tsin4

at t =6 ,

2

2

dxyd =

6cos86sin

4

=

916

21

81

=91

29. (b) Since f(x) is differentiable at x = 1, it is continuous m + a + 1 = 1 + 1 = 2 m + a = 1 (1)f’(1) = f’(1+) m = 2 (2)From (1) and (2), m = 2, a = 1

30. (d)

1x3

5xdxd

1x35x'f

dxdy

= 21x3

141x3

5x'f

at x = 1,27).2('f

dxdy

=2x4 e)x('fcesine.

27

sol-m1204

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