Solutions Hwt Dc 2

Vidyamandir Classes

Solutions to Home Work Test-4/Mathematics Differential Calculus - 2 3.(D)

y2 

x3 10  x



2 3 dy 3x 10  x   x  1 30 x 2  2 x 3   dx 10  x 2 10  x 2

2y





15 x 2  x3 dy  dx 10  x 2 y

or

HWT - 1



slope of tangent = 2



equation of tangent is  y  5   2  x  5  or



y 2   2 x  5   4 x 2  25  20 x

y  2x  5

2

For co-ordinates of Q

4 x 2  25  20 x 

x3 10  x



5 x3  60 x 2  225 x  250  0



x3  12 x 2  45 x  50  0

 4  2 1  x  0 y    4 1  x2   0



5.(BC)

40 x 2  250  200 x  4 x3  25 x  20 x 2  x3



x2

y  1

and

   4 tan 1 x x  1      0  1 x0 1  x 1  x  y  cos ec 1    sec 1    2x   1  x 2   4 tan 1 x 0 x 1        x 1  2

4.(C)



2

x  1 1  x  0

 y   0 for x   1, 0   1,  

0 x 1 x 1

For f (x) to be defined

2  x2  0 16

2 16

or

x2 

or

 2 tan   x2  16 

or

    x ,   4 4

or

f  x   A

Now, according to the question

2    x 2  0,  16  4 max  f  x    1 9.(A)



f  x   x 4  42 x 2  80 x  32



maximum value of  f  x  





1

is not defined.



3

  4 x  84 x  80 4  x  21x  20   0

f   x   3 x 4  42 x 2  80 x  32 f  x  0

  0, 1  

2

3

3

 4  x  1 x  4  x  5   0

x   4 ,  1   5,  

So f (x) is monotonically increasing in x   4 ,  1   5,  

VMC/Differential Calculus - 2

33

Solutions to HWT-4/Mathematics

Vidyamandir Classes 10.(ABD)

The graph of f (x) is

Differential Calculus - 2 1.(BC)

Let f  x   ax  b



f 1  x  

xb a



f 1   x  



Slope of y  f  x  is a and that of y  f 1   x  is

HWT - 2 x  b a

1 a

f (x) and f 1   x  are orthogonal.

Similarly f   x  and f 1  x  are orthogonal 5.(B)

As x = a is a point of inflection, the first non-zero derivative at x = a should be of odd order.  m should be odd. Also, n should be even, otherwise x = b will be another point of inflection.

7.(A)

14  10x  x 2  x3 is a decreasing function as its derivative is always negative 

it will have minimum value at x = 1 f 1  14  10  1  1  4

Now for this value to be minimum





log10 p 2  4  1 





(As log10 p 2  4 is an increasing function)

p 2  4  10 or p 2  14 . For domain p 2  4  0  p 2  4

Differential Calculus - 2 1.(C) 2.(A)

dy  3x 2  4 x  1  3 dx



x  2,

HWT - 3 2 3

Let line be y = mx. dy  2x  3  m dx



y   2 x  3 x



y  2 x 2  3x

Solve two equations to get :

2 x 2  3x  x 2  3x  4 x2  4



VMC/Differential Calculus - 2

x  2

34

Solutions to HWT-4/Mathematics

Vidyamandir Classes 3.(A)

7.(A)

f   x   sin  cos x  sin x



   In  , 0  f   x  is negative  2 



  In  ,   f   x  is negative. 2 

a  b  88

 

  In  0 ,  f   x  is positive.  2 (A) option is not correct.

Area   88  b  b



Will be maximum when 88  b  b

8.(C)



b  44  a  44



A   44   121 16  1936

f  x 

2

x2  x  1 x2  1

x2  x  1  0

x2  1  0  x  R

and

Differential Calculus - 2 1.(D) 2.(D)

HWT - 4

For f (x) to have local maxima or local minima, we have to check by nth derivative test.





f   x   6 x 2  x  2  6  x  2   x  1 f (x) has local maximum at x  1 And f (x) is increasing from 2 to 1 and 2 to 4 

5.(C)

f  x  

6.(B)

f (x) will have maximum at either 4 or 1 . Check f (4) and f  1 to get values.

ad  bc

 c sin x  d cos x 2 f  x  0

for

bc  ad

f  x   2 n  x  2   x 2  4 x  1 Domain is : x > 2 (as log takes positive inputs)

f  x 

 x  3 x  1 2  2 x  4  2 x2  x  2

For f (x) to increase f   x   0 

2  x  3 x  1

 x  2

But x > 2  

9.(D)

0



 x  3 x  1  0  x  2

x  2  0 and x  1  0

 x  3  0

 x3

Hence x   2 , 3

1  2 x 1  x  0 (decreasing)  f  x   1 0  x  1 (constant)  2 x  1 1  x  2 (increasing)

VMC/Differential Calculus - 2

35

Solutions to HWT-4/Mathematics

Vidyamandir Classes

Differential Calculus - 2 1.(A) 2.(C)

7.(B)

HWT - 5

In Module

dy 1  cos x  0 dx 2 x  sin x 

cos x  1



y  x0  y  x 2

f  x  f  x 



sin x  0



parabola

 2 x  3 x  3   x 2  3x  2  1  x  32 x2  6 x  7

for x > 3

 x  32

Has local minima at x  3  2 f  x 

 8.(B)

1  2   2  2   2  2  3 2

2

2

 3 2 2

f   x   2 x nx  x  x  2nx  1  0 f   x   0 at x  0 , x 

1 but both are < 1 e

For x  1, e  f (x) is increasing  9.(B)

fmin at x  1 and fmax at x = e



fmin = 0

f max  e 2 1  e 2

See from graph :

cos 1 x    



 sin 1 x    

x  sin1, 1

[As sin 1 > cos 1]

Differential Calculus - 2 3.(A)

dy 1 dy 1 1 1 1   1  3  3  3 dx x 1 1 1 2 1  x3 dx 1  x

4.(CD)

x  0 is not in domain .

5.(B)

f   x   5 x 4  20 x3  15 x 2



HWT - 6







 5 x 2 x 2  4 x  3  5 x 2  x  3 x  1  Maximum at 1 and minimum at 3.  p  1, q  3

VMC/Differential Calculus - 2

36

Solutions to HWT-4/Mathematics

Vidyamandir Classes 6.(C)

8.(D)

By graph

lim f  x   1

x  1

lim f  x   1 

discontinuous

x  1

3 x  a  1  cos x 2 And derivative also equal. 

10.(C)

7.(C)

3   sin x  x  2 3



 1 a 2 2 3



a

1   2 2 3

f  x   x 2 e 2x f   x   2 xe 2 x  2 x 2 e 2 x  2e 2 x  x  x 2   2e 2 x x 1  x     

f   x   0 at x  0 , x  1



f (x) is increasing in 0  x  1 and decreasing in x    , 0   1,  

Maximum at x = 1

1 2 2 1 f max  1 e    2 e

Differential Calculus - 2 1.(C)

HWT - 7

g  x  f  4  x  f  2  x g x  0  f  4  x  f  2  x

2.(D)



4 x  2 x

[As f   x  is increasing  x  R as f   x   0  x  R ]



x  1



x   1,  

Graph is



6 solutions (Discontinuity at only x  1 ).

3.(B)

Do by graph.

4.(C)

f (x) is continuous but not diff. as

f   x  f   x

x  0

5.(A)

x  0

4 x  y  xy   2 yy   0

y  x  2 y  4x  y



y 

4 1   2  4x  y 6   x  1, y  2     2 x  2y 1   2 2 1  4

VMC/Differential Calculus - 2

37

Solutions to HWT-4/Mathematics

Vidyamandir Classes 6.(D)

Make it   1 



form

a 1 x

 x  2 Now check it lim    x  x 1

1     lim 1  x  x  1 7.(AC)

  x 1 

x   x 1

x

1  x 1  lim e    x e



None of these

Let P  x1 , y1  be the required point. The given curve is y  x3  2 x 2  x



. . . .(i)

dy  dy   3x 2  4 x  1     3 x12  4 x1  1 dx dx   x1 , y1 

Since the tangent at  x1 , y1  is parallel to the line y  3 x  2 .



Slope of the tangent at  x1 , y1  = slope of the line y  3 x  2



 dy   3  3 x12  4 x1  1  3 .  dx    x1 , y1 



3 x12  4 x1  4  0   x1  2  3 x1  2   0  x1  2 ,

2 3

Since  x1 , y1  lies on (i), therefore y1  x13  2 x12  x1 . When

x1  2 ; y1  23  2  2   2  2

When

x1 

2

3

2

2 2 14  2   2 , y1     2      3 3 27  3   3

 2 14  Thus, the required points are  2 ,  2  and  ,  3 27 

8.(B)

Let  x1 , y1  be the point of intersection of the curves. Then

ax12  by12  1

a x12  b y12  1

. . . .(iii)

. . . .(iv)

Differentiating (i) w.r.t. x, we get :

2ax  2by

dy dy ax 0   dx dx by ax  dy   m1     1 dx by   x1 , y1  1

. . . .(v)

Differentiating (ii) w.r.t. x, we obtain

2a x  2b y

dy dy a x 0   dx dx b y a x  dy   m2     1 b y1  dx  x1 , y1 

. . . .(vi)

The two curves will intersect orthogonally, if

m1m2  1  

ax1 a x   1  1 by1 b y1

 aa x12  bb y12

. . . .(vii)

Subtracting (iv) from (iii), we obtain

 a  a   x12    b  b   y 12 Dividing (viii) by (vii), we get :

VMC/Differential Calculus - 2

. . . .(viii) a  a  b  b 1 1 1 1      aa  bb  a b a  b

38

Solutions to HWT-4/Mathematics

Vidyamandir Classes 9.(D)

4 x 2  9 y 2  36  8 x  18 y 

dy dy 4 y 0   dx dx 9 y

Slope of the tangent 

4x 9y

For this tangent to be perpendicular to the straight line 5 x  2 y  10  0 , we must have

4x 5 10 x    1  y  9y 2 9 Putting this value of y in 4 x 2  9 y 2  36 , we get 64 x 2  324 , which does not have real roots. Hence, at no points on the given curve can the tangent be perpendicular to the given line. 10.(C)

Solving the two equations, we get :

x 2 y  xy  xy  x  1  0  x  0, y  0, x  1 . Since y = 0 does not satisfy the two equations. So, we neglect it. Putting x = 0 in the either equation, we get y = 1. Now, putting x = 1 in one of the two equations we obtain y 

1 . 2

 1 Thus, the two curves intersect at (0, 1) and 1,   2

Now,

x2 y  1  y



x2

dy dy dy 2 xy  2 xy     dx dx dx x2  1



1  dy   dy   0 and     dx    dx 1, 1 2 2   0 , 1



y = 1 and x  2 y  2  0

The equations of the required tangents are 1 1   x  1 2 2 These two tangents intersect at (0, 1). y  1  0  x  0  and y 

Differential Calculus - 2 1.(C)

x1 3  y1 3  a1 3 



HWT - 8

1 2 3 1  2 3 dy x  y 0 3 2 dx

dy x 2 3 y2 3  dy    2 3   2 3     1  dx   a 8 , a 8 dx y x

a a The equation of the tangent at  ,  is given by 8 8 y

a a a   1 x   or x  y   0 4 8 8  

The x and y intercepts of this line on the coordinate axes are each equal to 2

a . So, we have 4

2

a a 4  4  2  a  4    

2.(A)

We have x3  3 xy 2  2  0

. . . .(i)

and

3x 2 y  y3  2  0

. . . .(ii)

Differentiating (i) and (ii) with respect to x, we obtain

2 xy x2  y2  dy   dy   dx   2 xy and  dx   2  c2 x  y 2  c1  dy   dy   dx    dx   1  c1  c2

VMC/Differential Calculus - 2

Hence, the two curves cut at right angles.

39

Solutions to HWT-4/Mathematics

Vidyamandir Classes 3.(C)

We have, dy  0 at (0, 1) and  1, 0  dx



c  0 and 4a  3b  c  0



a



3b and c  0 4

4a  3b  0 and c  0

. . . .(i)

Also, the curve passes through (0, 1) and  1, 0 



d  1 and 0  a  b  c  d



a  b  c 1  0

. . . .(ii)

From (i) and (ii), we get : a  3, b  4 , c  0 and d = 1

4.(A)

dy  12 x3  12 x 2 dx



y  3 x 4  4 x3  1 

Now,

dy  0  12 x3  12 x 2  0  x  1  0  x  1 dx

Let P  x1 , y1  be any point on the curve x n y  a n . Then,

x1n y1  a n

Now,

xn y  an

. . . .(i) dy 0 dx



n x n 1 y  x n



dy y  n dx x



an  dy   n  dx    x1 , y1  x1n 1

 ny1  dy    dx  x1   x1 , y1 



Then equation of the tangent at P  x1 , y1  is y  y1  

n an x1n 1

[Using (i)]

 x  x1 

 x n 1 y   n an 1 This meets the coordinate axes at A  1  x1 , 0  and B  0 , y1  n   na  x1n   



   

1  OA  OB  2

Area of AOB  

n 1  1  x1 y1 na n     x1   y1  n   2  n a x1n  



1  x1   a n na n   x1   n  n 2 n x1   x1



1  n  1 a n x11n 2 n

   

[Using (i)]

2

For the area to be a constant, we must have 1  n  0 i.e. n = 1. 5.(B)

We have,

x  t cos t and y  t sin t 

dx  cos t  t sin t and dt

dy  sin t  t cos t dt

At the origin, we have x  0 , y  0 

t cos t  0 and t sin t  0  t  0

VMC/Differential Calculus - 2

40

Solutions to HWT-4/Mathematics

Vidyamandir Classes The slope of the tangent at t = 0 is  dy  dy  dt   sin t  t cos t    0   dx  dx   cos t  t sin t t  0    dt t  0

So, the equation of the tangent at the origin is y  0  0  x  0   y  0 6.(A)

Let f  x   e x 1  x  2 . Then,

f 1  e0  1  2  0 

x  1 is a real root of the equation f  x   0

Let x   be a real root of f  x   0 such that   1 . Now,

  1    1 or   1

Let us assume that   1 Consider the interval 1,   Clearly, f 1  f  x   0 . So, by Rolle’s theorem f   x   0 has a root in 1,   . But f   x   e x 1  1  1 for all x.



f   x   0 for any x  1,  

This is a contradiction. Hence, f (x) = 0 has no real root other than 1. 7.(BC)

 1 Any point on xy  1 is  t,  , t  0  t dy dy y y0   dx dx x

Now,

xy  1  x



1  dy   dx  1    2  Slope of the normal at   t ,  t 

t

 1 2  t, t  is t .  

Since ax  by  c  0 is normal to xy = 1, therefore slope of the normal is  Thus,

t2  

b a

b  b and a are of opposite signs. a

 either b > 0 and a < 0 or b < 0 and a > 0. 8.(A)

Since (2, 3) lies on y 2  px3  q , therefore 9  8p  q



. . . .(i)

y 2  px3  q  2 y

dy dy 3 px 2  3 px 2   dx dx 2y

12 p  dy       2p 6  dx  2 , 3

Since y  4 x  5 is tangent to y 2  px3  q at (2, 3), therefore  dy  = slope of the line y  4 x  5  dx    2 , 3 



2p  4  p  2

Putting p = 2 in (i), we get : q  7 9.(D)

Differentiating y  e xy  x  0 w.r.t. x, we obtain dy  dy  e xy  x  dx  dx

VMC/Differential Calculus - 2

 y 1  0 



dy ye xy  1 dx 1  xe xy    xy dx 1  xe dy ye xy  1

41

Solutions to HWT-4/Mathematics

Vidyamandir Classes dx 1  0  1  xe xy  0  e xy  dy x

If the curve has a vertical tangent, then

1 x By the algebraic meaning of Rolle’s theorem between any two roots of a polynomial there is always a root of its derivative.

Clearly y = 0, x = 1 satisfies the equations y  e xy  x  0 and e xy  10.(B)

Differential Calculus - 2 1.(C)

We have f  x  



2.(B)

x x and g  x   sin x tan x

sin x  x cos x 2

sin x

and g   x  

tan x  x sec 2 x tan 2 x

Let

  x   sin x  x cos x and   x   tan x  x sec 2 x

Then,

f  x 

  x 2

sin x

and g   x  

  x tan 2 x

   x   sec 2 x  sec 2 x  2 x sec 2 x tan x  2 x sec 2 x tan x

Now,

   x   cos x  cos x  x sin x  x sin x

For

0  x  1 , we have x > 0, sin x  0 , tan x  0 sec x  0



  x   x sin x  0 and    x   0 and 0  x  1



  x  is increasing on (0, 1] and   x  is decreasing on (0, 1]



  x     0  and   x     0 



  x   0 and   x   0



f  x 



f (x) is increasing on (0, 1] and g (x) is decreasing on (0, 1].

  x 2

sin x

 0 and g   x  

and

  x tan 2 x

0

Let f  x   log 1  x   x . Clearly, f (x) is defined for x  1 . f  x 

1 x 1  1 x 1 x

Now,

f  x   log 1  x   x



f   x   0 for x > 0 and f   x   0 for 1  x  0



3.(BC)

f  x 

HWT - 9



f (x) is decreasing on 0,   and increasing on  1, 0



f  x   f  0  for 0  x   f  x   f  0  for 1  x  0



f  x   f  0  for 1  x  



log 1  x   x  0 for 1  x  



log 1  x   x for x   1,  



 1   x  2 2  x  1 x  3   2  f  x  2   x2  x2  

f  x   2 log  x  2   x 2  4 x  1 2  2x  4 x2



f  x 



f  x 



f   x   0   2  x  1 x  3 x  2   0



 x  1 x  2  x  3  0

2  x  1 x  3 x  2 

 x  2 2 

x    , 1   2 , 3

Thus, f (x) is increasing on   , 1   2 , 3 . Clearly, it includes answers (B) and (C)

VMC/Differential Calculus - 2

42

Solutions to HWT-4/Mathematics

Vidyamandir Classes 4.(C)

5.(BC)

f  x   x3  ax 2  bx  5 sin 2 x is increasing on R. 

f   x   0 for all x  R



3 x 2  2ax   b  5   0 for all x  R 



a 2  3b  15  0



3x 2  2ax  b  5 sin 2 x  0 for x  R

 2a 2  4  3   b  5   0

We have, f  x   f   x   log 1  0  f   x    f  x 

So, Now, 

f (x) is an odd function f  x   log  x3  x 6  1   



 6 x5   3x 2   x3  x 6  1  2 x 6  1  1

3x 2 x6  1

0

f (x) is increasing.

6.(B) 8.(A)

f '  x 

7.(AB) We have y  x3  3 x 2  6 x  17 

dy 2  3 x 2  6 x  6  3  x  1  1  0 for all x.   dx

Hence, y increases for all values of x. 9.(C)

The function f  x   x3 increases for all x and the function g  x   6 x 2  15 x  5 increases if g   x   0  12 x  15  0  x  

Thus,

f (x) and g (x) both increase for x  

5 4

5 4

It is given that f (x) increases less rapidly than g (x), therefore the function   x   f  x   g  x  is decreasing function, which implies that    x   0 . 

3 x 2  12 x  15  0



x2  4 x  5  0



 x  5 x  1  0



1  x  5

Hence, x3 increases less rapidly than 6 x 2  15 x  5 on the interval  1, 5  . 10.(C)

Differential Calculus - 2

HWT - 10

x

1.(C)

1 Let y     x  x . Then log y   x log x x



And,

dy 1 dy   y 1  log x    1  log x  or dx y dx

d2y dx

2



dy y y 1  log x    y 1  log x 2  dx x x

For maximum and minimum,





d2y dx 2

  x  x 1  log x   x  x 1  log x   x  x 1 2

2

dy 0 dx

dy  0   y 1  log x   0  1  log x  0 dx

VMC/Differential Calculus - 2

43

Solutions to HWT-4/Mathematics

Vidyamandir Classes

Also,

1 e 2 1 e 1  d2y  1 1 1     1  log      2    dx  e e e    x 1 e

 

If f (x) has an extremum at x  f   x   0 at x 

3.(C)

 log e A  B  A  e B   

log x  1  x  e 1 

  e1 2.(D)

1 e



1 e

1  log e 2   e1 

1 e 1

 3

f  x   a sin x 



    f     0  a cos    cos   0  a  2 3 3

We have,

f  x 

1 sin 3x 3

x2

2

 , then : 3

Now,

x 1

 e1 e 1  1  e1 e 1  e1 e 1  0



f   x   a cos x  cos 3 x

 x 1  2 , x 1  x  1  x , x  1  x 2

Clearly, f (x) is not differentiable at x = 0 and x = 1. So, by definition, these are two of the critical points. For points other than those two, we have

 x  2  3 , x 1  x f  x    x 2 , x 1  x3 Clearly, f   x   0 at x = 2. So, x = 2 is also a critical point. Hence, 4.(B)

f (x) has three critical points, viz. 0, 1 and 2.

We have, f  x    x  1

13

  x  1

13

   x  12 3   x  12 3 1 1 1   23 3   x  12 3  x  12 3  3 x2  1  



f  x 

Clearly,

f   x  does not exist at x  1

Now, f   x   0 



 x  12 3   x  12 3



 x0

Clearly, f   x   0 for any other value of x  0 , 1 . The value of f (x) at x  0 is 2. Hence, the greatest value of f (x) is 2. 5.(C)

Clearly, f (x) is a periodic function with period 2 . So, the difference between the greatest and the least values of the function f (x) is the difference between the greatest and the least values on the interval 0 , 2  . Now, we shall find these values on the interval 0 , 2  .

We have, 3x x 3x 3x   f   x     sin x  sin 2 x  sin 3x     2 sin cos  2 sin cos  2 2 2 2    2 sin

x 3x 3x  x 3x  cos  cos   4 sin sin sin x 2 2 2  2 2 

VMC/Differential Calculus - 2

44

Solutions to HWT-4/Mathematics

Vidyamandir Classes 

2 , 3 1 1 7  2 f  0  1    , f  2 3 6  3

f   x   0  x  0,

and 2

13 1  and    12 , f     6  7 13 The largest and the smallest of these values are and  respectively. 6 12

Now,

Hence, the required difference 6.(B)

7 6

7  13  9   6  12  4

Let

f  x   x 25 1  x 

Now,

f   x   0  x  0 , 1,

75

f  2  

. Then,

f   x   x 24 1  x 

74

1  4 x 

1 . 4

1 1 and f   x   0 in the right neighborhood of . So, f   x  changes its sign 4 4 1 from positive to negative in the neighbourhood of . 4 1 Hence, it attains maximum at x  4

Clearly,

7.(C)

f   x   0 in the left neighborhood of

We have, P  x   a0  a1 x 2  a2 x 4  . . .  an x 2 n For maximum or minimum P   x   0







2 x a1  2a2 x 2  . . .  nan x 2 n  2  0  x  0

Now,

P   x   2a1  12a2 x 2  . . .  2n  2n  1 an x 2 n  2



P   0   2a1  0

 a1  0

Hence, P(x) has only one minimum at x = 0. 8.(B)

Let P = xy. Then x  y  8  P  x  8  x  . Now,

dP d2P  0  8  2 x  0  x  4 . Clearly  0 . Hence, P is maximum dx dx 2

for x  4  y . The maximum value of P is 16. 9.(D)

Let A be the area of the rectangle shown in figure. Then

A  2x . 2 r 2  x2  4x r 2  x2



dA r 0  x dx 2

It can be easily checked that

Given by A  4

r 2

r2 

d2A dx 2





r 2  2 x2 dA 4 dx r 2  x2



 0 for this value of x. Hence, A is maximum for x 

r 2

and the maximum value of A is

r2  2r 2 2

10.(D)

VMC/Differential Calculus - 2

45

Solutions to HWT-4/Mathematics