Solutions Hwt Dc 2

Solutions to Home Work Test-4/Mathematics

3.(D) 3 2 10 x y x   





 









2 3 2 3 2 2 3 10 1 30 2 2 10 10         x x x dy x x y dx _{x x} or









2 3 2 15 10 x x dy dx _{x y}     slope of tangent = 2  equation of tangent is



y5

 

2 x5



or y2x5  2





2 2 2 5 4 25 20 y  x  x   x For co-ordinates of Q 3 2 4 25 20 10 x x x x      2 3 2 3 40x 250 200 x4x 25x20x x  3 2 5x 60x 225x2500  3 2 12 45 50 0 x  x  x   x2 and y 1 4.(C) 1 2 2 1 1 2 1 4 1 0 1 0 1 1 2 1 4 0 1 1                     _ _ _  _          _  tan x x x x x y cos ec sec x x tan x x x    2 2 4 1 1 0 1 0 0 4 0 1 1 0 1   _{ }            _{ }    _  x x x y y x x x for x 



1 0,

  

 1, 5.(BC) For f (x) to be defined 2 2 0 16 x    or 2 2 16 x  or 4 4 x  , _  

Now, according to the question

2 2 0 16 x ,4  _    _ _ or 2 2 0 1 16 tan  x ,    _ _{ }     or f x

 

A

 

1

max f x_  _ maximum value of _f x

 

_{ is not defined.}1

9.(A) f x

 





x442x280x32



3

 



 

2



4 2 3 3 42 80 32 4 84 80 f x  x  x  x x  x

 

0 f x   4



x321x20



0  4



x1



x4



x 5



0  x   



4, 1

 

5,



So f (x) is monotonically increasing in x   



4, 1

 

5,



Differential Calculus - 2

HWT - 1

10.(ABD) The graph of f (x) is 1.(BC) Let f x

 

ax b  f 1

 

x x b a  _  _ 1

_{ }

x b f x a  _{ } 

Slope of y f x

 

is a and that of y f1

 

x is 1 a

  f (x) and f1

 

 are orthogonal.x

Similarly f

 

 andx 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 x2x3 is a decreasing function as its derivative is always negative

 it will have minimum value at x = 1

 

1 14 10 1 1 4

f     

Now for this value to be minimum



2



10 4 1

log p   (As log₁₀



p24



is an increasing function)

 2 4 10 p   or p214. For domain p2 4 0  p24 1.(C) dy 3x2 4x 1 3 dx     2 2 3 x ,

2.(A) Let line be y = mx.

2 3 dy x m dx    y



2x3



x  2 2 3 y x  x

Solve two equations to get :

2 2 2x 3xx 3x4 2 4 x   x 2

Differential Calculus - 2

HWT - 2

Differential Calculus - 2

HWT - 3

3.(A) f

 

x sin cos x sin x





 In 0

 

2 , f x    _{ }     is negative  In 0, 2 f

 

x   _{ }     is positive.  In

 

2, f x    _{ }  

  is negative.  (A) option is not correct.

7.(A) a b 88  Area



88 b b



Will be maximum when 88 b b

 b44  a44 

 

2 44 121 16 1936 A    8.(C)

 

2 2 1 1 x x f x x      2 1 0 x   x and x2   1 0 x R

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

2.(D) f

 

x 6



x2 x 2



6



x2

 

x1



f (x) has local maximum at x 1

And f (x) is increasing from  to 12  and 2 to 4

 f (x) will have maximum at either 4 or  . Check f (4) and1 f

 

 to get values.1

5.(C)

 





2 ad bc f x c sin x d cos x      f

 

x 0 for bcad 6.(B) f x

 

2n x



 2



x24x1

Domain is : x > 2 (as log takes positive inputs)

 

2



_

3



_

1



2 4 2 2 2 x x f x x x x           For f (x) to increase f

 

x 0 











2 3 1 0 2 x x x      









3



1 0 2 x x x     But x > 2  x  and2 0 x 1 0 



x 3



0  x3 Hence x

 

2 3, 9.(D)

 

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

Differential Calculus - 2

HWT - 4

1.(A) In Module 2.(C) 1 0 2 dy cos x dx x sin x      cos x 1  sin x0  2 0 y x  y x  parabola 7.(B)

 











 





2 2 2 3 3 3 2 1 3 x x x x f x x        

 





2 2 6 7 3 x x f x x      for x > 3

Has local minima at x 3 2



 



1 2

 

2 2



2 2 3 2 3 2 2 2 2 f x   _{ }     8.(B) f

 

x 2x nx  x x



2nx 1



0

 

0 f x  at x 0, x 1 e

  but both are < 1

For x  _1, e_{ f (x) is increasing}

 fmin at x and f1 max at x = e  fmin = 0 fmaxe2

 

1 e2

9.(B) See from graph :

1

cos x

 _{ }

  sin1x 

 x _sin ,1 1_ [As sin 1 > cos 1]

3.(A)

 

3 3

 

3 3 1 1 1 1 1 1 2 1 1 1 1 1 dy dy dx x dx x x          4.(CD) x is not in domain .0 5.(B) f

 

x 5x420x315x2











2 2 2 5x x 4x 3 5x  x 3 x 1     _   _

Maximum at 1 and minimum at 3.

 p1, q3

Differential Calculus - 2

HWT - 5

6.(C) By graph 7.(C)

 

 

1 1 1 1 x x lim f x lim f x         discontinuous 8.(D) 3 1 2 x   a cos x And derivative also equal.

 3 2 sin x x 3      1 2 2 3  a  1 2 2 3 a   10.(C) f x

 

x e2 2x

 

2 2 2 2 2 2





2 x 2 x 2 x 2 x 1 f x  xe  x e  e _xx _ e x x

 f

 

x  at0 x0, x1  f (x) is increasing in 0  and decreasing inx 1 x 



,0

  

 1,

 Maximum at x = 1

 

2 2 1  2 1 1 max f e e    1.(C) g x

 

 f



4x



f



2x



 

0



4





2



g x   f x  f x  4  x 2 x [As f

 

x is increasing   asx R f

 

x   0 x R]  x 1  x  



1,



2.(D) Graph is  6 solutions

3.(B) Do by graph. (Discontinuity at only x  ).1

4.(C) f (x) is continuous but not diff. as

 

 

0 0 x x f x f x        5.(A) 4x y xy2yy0



2



4 y x y  xy 





   

    

4 1 2 4 6 1 2 2 2 1 2 2 1 4 x y y x , y x y                 

Differential Calculus - 2

HWT - 7

6.(D) Make it

 

1  form  a1 Now check it 2 1 x x x lim x     _{ }        1 _{   } 1 1 1 1 1 1 x x x x x x x lim lim e x e               _  _        None of these

7.(AC) Let P x , y



_{1 1}



be the required point. The given curve is

3 2 2 yx  x x . . . .(i)  1 1 2 2 1 1 3 4 1 3 4 1 x , y dy dy x x x x dx dx       _{ }     

Since the tangent at



x , y1 1



is parallel to the liney3x .2  Slope of the tangent at



x , y1 1



= slope of the liney3x2  1 1 1 2 1 3 3 4 1 3 x , y dy x x dx   _{    }     . 







2 1 1 1 1 1 2 3 4 4 0 2 3 2 0 2 3 x  x    x  x    x  ,

Since



x , y_{1 1}



lies on (i), therefore y₁x₁32x₁2 .x₁

When x12; y1232 2

 

2  2 2 When 3 2 1 1 2 2 2 2 14 2 3 3 3 3 27 x   , y _ _  _ _       

Thus, the required points are



2,2



and 2 14 3 , 27

 

 

 

 

8.(B) Let



x , y_{1 1}



be the point of intersection of the curves.

Then ax₁2by₁2 1 . . . .(iii) a x₁2b y₁2 1 . . . .(iv)

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

2ax 2bydy 0 dy ax dx dx by        1 1 1 1 1 x , y ax dy m dx by   _{ }     . . . .(v)

Differentiating (ii) w.r.t. x, we obtain 2a x 2b ydy 0 dy a x dx dx b y         



1 1



1 2 1 x , y a x dy m dx b y     _{ }      . . . .(vi)

The two curves will intersect orthogonally, if

1 1 1 2 1 1 1 ax a x 1 m m by b y           2 2 1 1 aa x bb y    . . . .(vii)

Subtracting (iv) from (iii), we obtain





2





2

1 1

aa x   b b y . . . .(viii)

Dividing (viii) by (vii), we get : a a b b 1 1 1 1

aa bb a b a b

 

 _  _{   }

9.(D) 4 2 9 2 36 8 18 0 4 9 dy dy y x y x y dx dx y       

 Slope of the tangent 4 9

x y



For this tangent to be perpendicular to the straight line 5x2y10 , we must have0

4 5 10 1 9 2 9 x x y y     

Putting this value of y in 4x29y2 36, we get 64x2 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 :





2

1 0 0 0 1

x yxy  xy x   x , y , x .

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 1

2

y .

Thus, the two curves intersect at (0, 1) and 1 1 2 ,       Now, x y2  1 y  2 2 2 2 1 dy dy dy xy x xy dx  dx  dx x   _{ }0 1 0 , dy dx   _     and _1 1 2_ 1 2 , dy dx   _{ }    

The equations of the required tangents are





1 0 0

y  x and 1 1



1



2 2

y  x  y = 1 and x2y 2 0

These two tangents intersect at (0, 1).

1.(C) 1 3 1 3 1 3 1 2 3 1 2 3 0 3 2 dy x y a x y dx           2 3 2 3 2 3 2 3 8 8 1 a , a dy x y dy dx y x dx          _{ }    

The equation of the tangent at 8 8 a a ,       is given by 1 8 8 a a y   _x _   or 4 0 a x y 

The x and y intercepts of this line on the coordinate axes are each equal to 4 a . So, we have 2 2 2 4 4 4 a a a   _  _{  }        

2.(A) We have x33xy2 2 0 . . . .(i) and 3x y2 y3 2 0 . . . .(ii)

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

1 2 2 2 c dy x y dx xy   _{ }     and ₂ 2 2 2 c dy xy dx x y   _{ }      1 2 1 c c dy dy dx dx   _  _{ }    

    Hence, the two curves cut at right angles.

3.(C) We have, 0 dy dx at (0, 1) and



1 0,



 c and 40  a3b c 0  4a3b and0 c0  3 4 b a and c0 . . . .(i)

Also, the curve passes through (0, 1) and



1 0,



 d and 0 a b c d1    

 a   b c 1 0 . . . .(ii)

From (i) and (ii), we get : a3, b4, c and d = 10

 4 3 3 2 3 4 1 dy 12 12 y x x x x dx       Now, dy 0 12x3 12x2 0 x 1 0 x 1 dx         

4.(A) Let P x , y



1 1



be any point on the curve x yn an.

Then, x y1n 1an . . . .(i) Now, x yn an  n xn 1y xn dy 0 dx  _{ }  dy ny dx  x  _{ } 1 1 1 1 x , y ny dy dx x    _      1 1 1 1 n n x , y dy a n dx x    _{ }     [Using (i)]

Then equation of the tangent at P x , y



1 1



is 1 ₁



1



1 n n n a y y x x x     

This meets the coordinate axes at

1 1 1 1 0 n n x y A x , na           and 1 1 0 n n n a B , y x           Area of 1





2 AOB OA OB    1 1 1 1 1 1 1 2 n n n n x y na x y n a x         _  _  _     1 1 1 1 1 2 n n n n x a na x n x x    _{ }  _  _  _  _{ } [Using (i)]





2 1 1 1 1 2 n n n a x n   

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

5.(B) We have,

xt cos t and yt sin t

 dx cos t t sin t

dt   and

dy

sin t t cos t

dt  

At the origin, we have x0, y0

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

So, the equation of the tangent at the origin is y 0 0



x0



 y0

6.(A) Let f x

 

ex1  . Then,x 2

 

0

1 1 2 0

f e   

 x is a real root of the equation1 f x

 

0 Let x be a real root of f x

 

 such that0   .1 Now,   1  or1  1

Let us assume that  1 Consider the interval 1,_ _

Clearly, f

 

1  f x

 

 . So, by Rolle’s theorem0 f

 

x  has a root in0

 

1, . But f

 

x ex1  for all x.1 1

 f

 

x  for any0 x

 

1,

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

7.(BC) Any point on xy is1 t ,1 , t 0 t   _     Now, xy 1 xdy y 0 dy y dx dx x         ₂ 1 1 t , t dy dx _ _ t     _{ }  

   Slope of the normal at

1 t , t       is 2 t .

Since axby  is normal to xy = 1, therefore slope of the normal isc 0 b

a



Thus, t2 b b

a

   and a are of opposite signs.

 either b > 0 and a < 0 or b < 0 and a > 0. 8.(A) Since (2, 3) lies on y2  px3 , thereforeq

98 pq . . . .(i)  2 3 2 3 2 2 3 2 dy dy px y px q y px dx dx y         2 3 12 2 6 , dy p p dx   _{ }    

Since y4x is tangent to5 y2  px3 at (2, 3), thereforeq

 2 3,

dy dx

   

  = slope of the liney4x5  2p4  p2

Putting p = 2 in (i), we get : q 7

9.(D) Differentiating yexy  w.r.t. x, we obtainx 0 1 0 xy dy dy e x y dx dx    _  _     1 1 1 1 xy xy xy xy dy ye dx xe dx xe dy ye       

If the curve has a vertical tangent, then dx 0 1 xexy 0 exy 1

dy      x

Clearly y = 0, x = 1 satisfies the equations yexy  andx 0 exy 1 x



10.(B) By the algebraic meaning of Rolle’s theorem between any two roots of a polynomial there is always a root of its derivative.

1.(C) We have f x

 

x sin x  and g x

 

x tan x   f

 

x sin x ₂x cos x sin x    and

 

2 2 tan x x sec x g x tan x   

Let 

 

x sin xx cos x and

 

x tan xx sec x2

Then,

 

 

2 x f x sin x   and

 

 

2 x g x tan x   

Now, 

 

x cos xcos xx sin xx sin x and  

 

x sec x2 sec x2 2x sec x tan x2  2x sec x tan x2

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

 

 

x x sin x and0  

 

x  and 00  x 1  

 

x is increasing on (0, 1] and

 

x is decreasing on (0, 1]

 

 

x 

 

0 and

 

x 

 

0  

 

x  and0 

 

x 0  f

 

x

 

₂x 0 sin x     and g

 

x

 

₂x 0 tan x   

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

2.(B) Let f x

 

log



1x



 . Clearly, f (x) is defined forx x  .1

Now, f x

 

log



1x



x 

 

1 1 1 1 x f x x x       

 f

 

x  for x > 0 and0 f

 

x  for 10   x 0

 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



  for 1 xx 0      log



1x



 forx x  



1,



3.(BC) f x

 

2log x



2



x24x1 

 

2 2 4 2 f x x x      

 





2







1 2 1 3 2 2 2 2 x x x f x x x  _{ }  _{ }             

 













2 2 1 3 2 2 x x x f x x        f

 

x 0  2



x1



x3



x2



0 



x1



x2



x3



0  x 



,1

  

 2 3,

Thus, f (x) is increasing on



,1

  

 2 3, . Clearly, it includes answers (B) and (C)

4.(C) f x

 

x3ax2bx5sin x2 is increasing on R.

 f

 

x  for all x0 R  3x22ax b 5sin x2  for x0 R

 2





3x 2ax b5  for all x0 R 

 

2a 2  4 3



b5



0  2 3 15 0 a  b  5.(BC) We have,

 

 

1 0

 

 

f x  f  x log   f   x f x

So, f (x) is an odd function

Now, f x

 

log x_ 3 x61_ 

 

2 5 2 3 6 6 6 1 6 3 3 0 1 2 1 1 x x f ' x x x x x x      _  _         f (x) is increasing. 6.(B) 7.(AB) 8.(A) We have yx33x26x17  2





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

Hence, y increases for all values of x.

9.(C) The function f x

 

x3 increases for all x and the function g x

 

6x215x increases if5

 

5

0 12 15 0

4

g x   x   x 

Thus, f (x) and g (x) both increase for 5

4

x 

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

 2

3x 12x150  x24x 5 0





x5



x 1



0    1 x 5

Hence, x3 increases less rapidly than 6x215x on the interval5



1 5,



.

10.(C) 1.(C) Let 1 x x y x x    _{ } 

  . Then log y x log x

 1dy



1 log x



y dx    or



1



dy y log x dx    And,









2 2 2 1 1 d y dy y y log x y log x dx x x dx        









2 2 2 1 2 1 1 x x x d y x log x x log x x dx          

For maximum and minimum, dy 0

dx

 dy 0 y



1 log x



0 1 log x 0

dx       

 1 1 1 log x x e e       B e log A B A e  _{  }    Also, 1 2 1 1 2 2 1 1 1 1 1 e e x e d y log e e e dx       _{     }       _{     }   _{     }  

 

₁ 1





₂

 

₁ 1 1 1 e e e  log e e       1





2 1 1 1 1 1 1 0 e e e e e  e         2.(D) If f (x) has an extremum at 3 x , then :

 

0 f x  at 3 x Now,

 

1 3 3

f x a sin x sin x  f

 

x a cos xcos x3

 0 0 2 3 3 f_{ }    a cos _{ } cos   a     3.(C) We have,

 

2 2 2 1 1 1 1 1 x , x x _x f x x x _{, x} x   _   _    _  _ 

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

 

3 3 2 1 2 1 x , x x f x x , x x    _   _{ }   _ 

Clearly, f

 

x  at x = 2. So, x = 2 is also a critical point.0 Hence, f (x) has three critical points, viz. 0, 1 and 2.

4.(B) We have, f x

  

 x1



1 3



x1



1 3 

 





















2 3 2 3 2 3 2 3 ₂ 2 3 1 1 1 1 1 3 _{1 1} 3 1 x x f x x x _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  for any other value of0 x _0 1, _{ . The value of f (x) at} x is 2.0 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,

 



2 3



f x   sin xsin xsin x 2 3 2 3 3

2 2 2 2

x x x x

sin cos sin cos

   _  _   3 3 2 2 2 2 x x x

sin cos cos 

  _  _   3 4 2 2 x x

sin sin sin x



 

0 0 2 3 f x   x ,  , and 2 Now,

 

0 1 1 1 7 2 13

 

1 2 3 6 3 12 6 f     , f_ _  , f      and

 

7 2 6 f  

The largest and the smallest of these values are 7 6 and

13 12

 respectively.

Hence, the required difference 7 13 9

6 12 4    _{ }    6.(B) Let f x

 

x25



1x



75. Then, f

 

x x24



1x

 

74 1 4 x



Now,

 

0 0 1 1 4 f x   x , , .

Clearly, f

 

x  in the left neighborhood of0 1

4and f

 

x  in the right neighborhood of0 1

4. So, f

 

x changes its sign from positive to negative in the neighbourhood of 1

4. Hence, it attains maximum at 1

4

x

7.(C) We have, P x

 

a0a x1 2a x2 4. . .a xn 2n

For maximum or minimumP x

 

0





2 2 2



1 2

2x a 2a x . . .na x_n n 0  x0 Now, P

 

x 2a₁12a x₂ 2 . . . 2n



2n1



a x_n 2n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 0 8 2x 0 x 4

dx       . Clearly

2

2 0

d P

dx  . Hence, P is maximum

for x  . The maximum value of P is 16.4 y

9.(D) Let A be the area of the rectangle shown in figure.

Then A2x .2 r2x2 4x r2x2 





2 2 2 2 2 4 r x dA dx _{r x}     0 2 dA r x dx   

It can be easily checked that

2

2 0

d A

dx  for this value of x. Hence, A is maximum for 2

r

x and the maximum value of A is

Given by 2 2 2 4 2 2 2 r r A r   r 10.(D)