# 2nd-Dispatch DLPD IIT-JEE Class-XII English PC(Maths)

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CONTENT

MATHEMATICS CLASS : XII Preface

1.

Functions Exercise

2.

79 - 100

Application of derivatives Exercise

6.

54 - 78

Method of differentiation Exercise

5.

29 - 53

Continuity and derivability Exercise

4.

01 - 28

Limits Exercise

3.

Page No.

101 - 126

Solution of triangle Exercise

127 - 149

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FUNCTIONS EXERCISE # 1 PART - I Section (A) : A–3.

(i)

f(x) =

f(x) =

x 3  5x  3 x2  1 x 3  5x  3 ( x  1)( x  1)

Division by zero is undefined Domain x  R – {1, –1}

 

x±1 x  (–, –1)  (–1, 1)  (1, )

(ii)

f(x) =

sin1 x x For sin–1x, x  [–1, 1] and division by zero is undefined x  0

Domain x  [–1, 0)  (0, 1]

1 (iii)

(iv)

(v)

f(x) =

x | x |

for function to be defined x + |x| > 0 for x > 0, x + |x| = 2x > 0 for x  0, x + |x| = 0  Domain is x  (0, ) f(x) = ex + sin x Domain x  R as there is no restriction for exponent of e.

1 f(x) = log (1  x ) + 10

x2

1 – x > 0 and x + 2  0 x  (– , 1) – {0} and x  – 2

(vi)

f(x) =

and 

1–x1 x  [–2, 0)  (0, 1)

x

 3x  1 –1 1 2x + 3 sin  2 

1 – 2x  0 and – 1 

3x  1 1 2

1 2

and –

1 x1 3

Taking intersection 

 1 1 Domain x   ,   3 2

(vii)

f(x) = 2 sin

1

x

x

1

1 

x

x  (0, 1)  (1, ) and 0 < x –

  1  f(x) = logx  log 2  x  1/ 2      In case of composite function in log. We start with outer log. x > 0, x  1 and

1/2

x2 – 1  x  1 and x > 2 (viii)

–1/3

1

1 >1 x 2 1 3 x  (0, ) – {1} and 0 Range 0 < sin–1 x 

(iv)

1

>1+

x 1

Range y  (0, 1]

x  (0, 1]

 2

 –  < n (sin–1x)  n   2 Inequality doesn't change as n is increasing function f(x) = 2 – 3x – 5x2 Domain x  R Method 1 y = – 5x2 – 3x + 2 opening downward parabola 0

 D  Range y     , 4a  

49   y    ,  20   Method 2 5x2 + 3x + (y – 2) = 0

D0 (v)

– D/4a 2

9 – 20 (y – 2)  0

20y – 49  0

y

49 20

f(x) = 3 |sin x| – 4|cos x| f(x) is a periodic function with period . So analysis is limited in [0, ]

 , |sin x| = 1, |cos x| = 0 2 fmin = 3.0 – 4.1 = – 1 at x = 0, |sin x| = 0, |cos x| = 1 fmax = 3.1 – 4.0 = + 3 at x =

(vi)

f(x) =

sin x

Range y  [–4, 3]

cos x

1  tan x 1  cot 2 x f(x) = sin x |cos x| + cos x |sin x| periodic period = 2   sin 2x    0 f(x) =   sin 2x    0 

+

2

  x   0,  2    , x   ,  2    , x   ,  2   3  , x   , 2  2   ,

Range y  [–1, 1]

RESONANCE

S OLUTIONS (XII) # 2

Section (B) : B–1.

(ii)

2 x 2 and g(x) = ( x ) Domain x  R, Domain x  [0, ) non-identical functions f(x) = sec(sec –1x) and g(x) = cosec (cosec –1x) Domain x  (–, –1]  [1, ), Domain x  (–, –1]  [1, ) f(x) = x g(x) = x Identical functions

(iii)

f(x) =

(i)

(iv)

B–5.

(i)

(ii)

(iii)

(iv)

f(x) =

1 cos x and g(x) = cos x 2 f(x) = |cos x| non-identical function f(x) = x and g(x) = enx, Domain x  R+ Domain x  R non-identical function f(x) = ex and g(x) = n x fog(x) = en x = x, x > 0 gof(x) = n ex = x, x  R f(x) = |x| and g(x) = sin x fog(x) = f(sin x) = |sin x| gof (x) = g(|x|) = sin |x| f(x) = sin–1x and g(x) = x2 fog(x) = sin–1(g(x)) = sin–1x2 gof(x) = (f2(x)) = (sin–1x)2 f(x) = x2 + 2, g(x) =

fog(x) = g2(x) + 2 =

gof(x) =

B–6.

x x 1

x2 ( x  1)2

+2=

3x 2  4x  2 ( x  1)2

f (x) x2  2 = 2 f (x)  1 x 1

1  x 2 , x 1 f(x) =   1  x , 1  x  2 g(x) = 1 – x, – 2  x  1

 1  g2 , g( x )  1  x  [0,1] fog(x) =  1  g( x ) , 1  g( x )  2  x  [–1,0)

1  (1 – x )2 fog (x) =   1  (1 – x )

,

x  [0,1]

, x  [–1, 0)

2 – 2x  x 2 , x  [0,1] fog (x) =   2 – x , x  [–1, 0)

Section (C) : C–1.

(i)

y = |(x + 2) (x + 3)| many - one function

(ii)

y = |nx| many - one function

RESONANCE

S OLUTIONS (XII) # 3

(iii)

   f(x) =sin 4x, x   – ,   8 8

 2 one-one function period =

1 1 , x  (0, ) x x many one function

(iv)

f(x) = x +

(v)

f(x) =

1   –1 

1 – e x 1   –1 .

e x f =

2 1– e

1 x2  0

1   –1 x 

increasing function

Hence one - one

C–5.

3x 2 – cos( x ) 4 Hence many - one

(vi)

f(x) =

(vii)

f(x) = sin–1 x – cos–1 x = 2 sin–1 x –

(i) (ii) (iii)

f(x) = sin (x2 +1)  f(x) = x + x2  f(x) = x – x3 

f(– x) = f (x) = even function f(– x) = x2 – x  f (x) or – f(x) Neither even nor odd function f(–x) = – x + x3 = – f(x) odd function

(iv)

 a x – 1   f(x) = x  x   a  1

 ax – 1    f(–x) = – x  – x   a  1

even function

 monotonically increasing. 2

 a x – 1   = f(x) even function f(–x) = x  x   a  1

(v)

f(x) = log (x +

(vi)

2 2   f(x) + f(– x) = log ( x  x  1)(– x  x  1) = log [(x2 + 1) – x2] = 0 hence odd function   f(x) = sin x + cos x  f(– x) = – sin x + cos x  f(x) or – f(x) Neither even nor odd. f(x) = (x2 – 1) |x|  f(–x) = f(x) even function.

(vii)

(viii)

x2  1 )

f(–x) = log (–x +

x2  1 )

 | tan(tan 1 x ) | x    2 x [ 2 x ] 1        x 1 f(x) =   sec(sec 1 x ) x   x   | x |     x  x0   f(x) =  3 0  x 1   x x 

RESONANCE

x    x     x  x0   f(x) =  3 0  x 1   x x 

S OLUTIONS (XII) # 4

C–6.

(i)

x 2 – sin x , – 1  x  0 even extension of f(x) = f(–x) =   – x  e x , x  –1

(ii)

– x 2  sin x , – 1  x  0 odd extension of f(x) = – f(– x) =   x – e x , x  –1

Section (D) : D–2.

(i)

f(x) = 2 + 3 cos (x – 2)

(ii)

f(x) = sin 3x + cos2 x + |tan x|

fundamental period = 2 

 2  period of f(x) = L.C.M.  , ,   = 2   3  f(x + ) = – sin x + cos2 x + |tan x|  f(x)

f(x) = sin

(iv)

f(x) = cos

for fundamental period 

fundament period = 2

3x sin 2x – 5 7

period 

10 ,7 3

 10  , 7  = 70  period of f(x) = L.C.M.  3  

Fundament period = 70

f(x) = [sin 3x] – |cos 6x|

period 

 2   2 period of f(x) = L.C.M.  ,  =  3 3 3

Fundamental period =

(vi)

f(x)=

(vii)

f(x) =

2 3

 3 2 3

1 fundamental period = 2 1  cos x sin12x

   period of f(x) = L.C.M.  ,  = 6 3 3

2

1  cos 6 x for fundamental period

  fx   = 6 

(viii)

2 , , 3

x x + sin 4 3 period  8, 6 period of f(x) = L.C.M. (8, 6) = 24 fundamental period = 24

(iii)

(v)

period 

  sin12 x   6    1  cos 2 6 x   6 

= f(x)

f(x) = sec3 x + cosec3 x  period  2 Fundamental period = L.C.M. (2, 2) = 2

Fundament period =

 6

2

Section (E) : E–1.

(i)

f:DR f(x) = 1 – 2–x

f (x) = 2– x n2 > 0 increasing function  one one function

D : [x  R), Range : (–, 1)  codomain 

function is not bijective

RESONANCE

f –1 does not exist S OLUTIONS (XII) # 5

(ii)

f(x) = (4 – (x – 7)3)1/5 f (x) =

1 (4 – (x – 7)3) 5

– 4/5

. (– 3 (x – 7)2)  0 decreasing function  one one function

Lim f ( x )  –  x 

Lim f ( x )  

x – 

D:R

Range : R = codomain

 onto function

function is bijective (invertible)

y = (4 – (x – 7)3)1/5 4 – y5 = (x – 7)3 x = 7 + (4 – y5)1/3 (iii)

f –1 (x) =

E–6.

f –1(x) = 7 + (4 – x 5)1/3

or

x=

f(x) = 1 ± xn

 or

f(x) = 1 – x3 f(1) = – 3

f(x) = n  x  1  x 2    D : x  R, Range : R y = n  x  1  x 2   

E–5.

or

e y  ey 2

e x  e x 2

 1  1 f(x) . f   = f(x) + f   x x f(3) = – 26 f(x) = – 3x2

f(x + y) = f(x) . f(y) and f(1) = 2 10

 f (n) = f(1) + f(2) + ........... + f(10)

n 1

 210  1    = 21 + 22 + 23 + ....... + 210 = 2  2  1  = 2046  

PART - II Section (A) : A–2.

For domain – log0.3(x – 1)  0  log0.3(x – 1)  0  (x – 1)  1  x2 Taking intersection x  [2, )

A–3.

f(x) = cot–1

x( x  3 ) + cos –1

for domain x(x + 3)  0  x  (–, –3]  [0, ) Taking intersection x  {–3, 0}

RESONANCE

and and and

x2 + 2x + 8 > 0  (x + 1)2 + 7 > 0  xR

x 2  3x  1

and and

0  x 2 + 3x + 1  1 x 2 + 3x + 1  0 and

x 2 + 3x  0  x  [–3, 0]

S OLUTIONS (XII) # 6

A–5.

f(x) = 4x + 2x + 1 Let 2x = t > 0,  x  R

f(x) = g(t) = t2 + t + 1,

t>0

2

1  3 g(t) =  t   + 2 4  2

1  1 t   > 2  2

1  1 t   > 2 4 

2

1  3 t   + >1 2 4 

Range is (1, )

Section (B) : B–2.*

–1

f(x) = en(sec x ) = sec–1x, x  (–, – 1]  (1, ) –1 g(x) = sec x, x  (–, – 1] [1, ) non-identical functions f(x) = tan (tan–1 x) = x, x  R g(x) = cot (cot–1 x) = x, x  R identical functions

(A)

(B)

 1 x 0  f(x) = sgn (x) =  0 x  0 – 1 x  0 

(C)

 1 x 0  g(x) = sgn(sgn x) =  0 x  0 – 1 x  0 

Identical functions f(x) = cot2 x . cos2 x, x  R – {n }, g(x) = cot2 x – cos2 x = cot2 x (1 – sin2 x) = cot2 x. cos2 x Identical functions

(D)

B–4.

B–6.*

Domain of f(g(x)) Range of g(x)  Domain of f(x)  – 5  |2x + 5|  7  – 12  2x  2 f(x) =

1– x , 1 x fog(x) =

0x1

n I x  R – {n },

n I

 

0  |2x + 5|  7 –6x1

g(x) = 4x (1 – x),0  x  1

–7  2x + 5  7

1 – g( x ) 1 – 4 x(1 – x ) 1 – 4 x  4x 2 = = 1  g( x ) 1  4 x(1 – x ) 1  4x – 4x 2

 1– x   1– x  gof(x) = 4f(x) . (1 – f(x)) = 4  1  x  1 –  1  x      

=

8 x(1 – x ) (1  x )2

Section (C) : C–2.

One One / Many One f(x) =

f(x) =

f(x) =

2x 2  x  5 7 x 2  2x  10

, Domain x  R

( 4x  1)(7 x 2  2x  10)  (14 x  2)(2x 2  x  5) (7 x 2  2x  10)2 11x 2  30x  20 2

(7 x  2x  10)

2

 30  > 0  x  (– , 0)   ,    11   30   f (x) < 0  x   0,  11 

f(x) = 0

RESONANCE

x = 0,

30 11 S OLUTIONS (XII) # 7

Function is increasing and decreasing in different intervals, so non monotonic  Many one function. Onto / Into f(x) =

2x 2  x  5

7 x 2  2x  10 2x 2 – x + 5 > 0,  x  R and 7x 2 + 2x + 10 > 0  x  R a = 2 > 0 and  a = 7 and D = 4 – 280 < 0 D = 1 – 40 = – 39 < 0  f(x) > 0  x  R Also f(x) never tends to ± as 7x 2 + 2x + 10 has no real roots, Range  Codomain so into function.

C–3.

f(x) = x 3 + x 2 + 3x + sin x, f(x) = 3x 2 + 2x + 3 + cos x 

3x 2 + 2x + 3 

–1  cos x  1

lim f(x) = + 

xR

32 as a = 3 > 0 and D < 0 12 so f(x) > 0  x  R

lim f(x) = – 

x 

x  

Hence f(x) is one-one and onto function (as f(x) is continuous function) C–6.

f(g(x1)) = f(g(x2)) as f is one - one function hence f(g(x1)) = f(g(x2))  x1 = x2

 

g(x1) = g(x2) x1 = x2

f(g(x)) is one - one function

as

g is one - one function

Section (D) : 2 D–2.

f(x) = sin

 [a] x  .

[a] = 4 D–3.

Period =

[ a]

=

a  [4, 5)

f(x) = x + a – [x + b] + sin x + cos 2x + sin (3x) + cos (4x) + ........ + sin (2n – 1) + cos (2px) f(x) = {x + b} + a – b + sin (x) + cos (2x) + sin (3x) + cos (4x) + .... + sin (2n – 1) + cos (2nx) Period of f(x) = L.C.M (1, 2,

2 2 2 2 , , ........., , )=2 3 4 2n  1 2n

 period of f(x) = 2 since f(1 + x)  f(x) , hence fundamental period is 2 D–7.*

(A) (B)

f(x) = cos (cos–1 x) = x, x  [–1, 1] odd function f (x + ) = cos (sin (x +)) + cos (cos (x + )) f (x + ) = cos (sin x) + cos (cos x) = f(x)

          f  x   = cos  sin x    + cos  cos x    2  2 2       = cos (cos x) + cos (sin x) = f(x) fundamental period = (C)

 2

f(x) = cos (3 sin x), x  [–1, 1] – 3 sin1 3 sin x 3 sin 1  cos (3 sin 1) cos (3 sin x) 1

Range is [cos (3 sin1), 1]

 1 x   f–1(x) = n   1 x 

Section (E) : E–1.

ex  ex y = x 1 e  e x By compnendo and dividendo

1 y 2e x = 1 y 2

RESONANCE

 1 y  x = n  1  y   

S OLUTIONS (XII) # 8

E–7.

E–8.

f(1) = 1 = 2 – 1 f(n + 1) = 2f(n) + 1  f(3) = 7 = 23 – 1 f(4) = 15 = 24 – 1 Similarly f(n) = 2n – 1

f(2) = 2f(1) + 1 = 2. 1 + 1 = 3 = 22 – 1

Method 1 : (usual but lengthy) x2 f(x) + f(1 – x) = 2x – x4 .....(1) replace x by (1 – x) in equation (1) (1 – x)2 f(1 – x)+ f(x) = 2 (1– x) – (1 – x)4 .....(2) eliminate f(1 – x) by equation (1) and (2) we get f(x) = 1 – x2 Method 2 : Since R.H.S. is polynomial of 4th degree and also by options consider f(x) = ax2 + bx + c x2 f(x) + f(1 – x) = 2x – x4  x 2 (ax2 + bx + c) + a (1 – x)2 + b (1 – x) + c = 2x – x4 by comparing coefficients a=–1 b=0 c=1  f(x) = – x2 + 1

EXERCISE # 2 PART - I 1.

(i)

f(x) =

or

3  2 x  2 .2  x 3 – 2x – 2 .2–x  0 (2x – 1) (2x – 2)  0

(ii)

f(x) =

(iii)

1 – 1 x 2  0 f(x) = (x2 + x + 1)–3/2 D:xR

(iv)

f(x) =

or  

(2x)2 – 3.2x + 2  0 2x  [1, 2] x [0, 1]

1 1 x2

x2 + x2

1 x 2 1

0  1 – x2  1

or

x 

or

x  2n

or

x  (0, 1]  [4, 5)

x [– 1, 1]

1 x 1 x

x2 1 x 0 and 0 x2 1 x x  (– , –2)  [2, ) and x  (–1, 1] D: (v)

f(x) =

tan x  tan 2 x

tan x – tan2x  0

or

0  tan x  1

 n 

  n , n  4   

1

(vi)

(vii)

f(x) = 2 sin x 2

f(x) =

sin

x 0 2

 5x  x 2   log1 / 4    4  

5x  x 2 1 4

RESONANCE

and

5x – x2 > 0

S OLUTIONS (XII) # 9

(viii) or 2.

(i)

f(x) = log10 (1 – log10(x2 – 5x + 16)) 1 – log10 (x2 – 5x + 16) > 0 x  (2, 3) f(x) = 1 – |x – 2| |x – 2|  [0, )

or

x2 – 5x + 6 < 0

f(x)  (– , 1]

1  f(x)   , 1 3 

1 (ii)

(iii)

f(x) =

x5 D : x  (5, ) R : f(x)  (0, ) 1 f(x) = 2  cos 3 x range of cos 3x is [–1, 1] cos 3x [–1, 1]

(iv)

x2

f(x) =

=y x  8x  4 x + 2 = yx2 – 8yx – 4y for x to be real D  0 (8y + 1)2 + 4y (4y + 2)  0 64y2 + 16y + 1 + 16y2 + 8y  0 2

80y2 + 24y + 1  0 (v)

f(x) =

(viii) (ix)

f(x) = 3 sin

(i)

1   1  ,  y     ,     4   20 

or

1  y   , 3 3 

2  x2 16

   D : x   ,   4 4

f(x) = x3 – 2x2 + 5 = (x2 – 1)2 + 4 R : [4, ) f(x) = x3 – 12x , x  [–3, 1] = x (x2 – 12) f(x) = 3x2 – 12 = 0 or f(x) = sin2x + cos4x = sin2x + 1 + sin4x – 2 sin2x = sin 4x – sin2x + 1 1  2 3 =  sin x   + 2  4

7.

or

=y x 2  2x  4 x2 – 2x + 4 = yx2 + 2xy + 4y x2 (1 – y) – 2x(1 + y) + 4(1 – y) = 0 D0

  2  x 2  0 ,   4 16

(vii)

yx2 – x (8y + 1) – (4y + 2) = 0

x 2  2x  4

4(1 + y)2 – 16(1 – y)2  0

(vi)

or

f(–x) =

 3  f(x)  0 ,  2 

x=±2

R : [–11, 16]

3  R :  , 1 . 4 

(2 x  1)7 (2 x )6

neither even non add (ii) (iii)

sec x  x 2  9 = f(x) x sin x f(–x) = – f(x) odd f(–x) =

RESONANCE

even

S OLUTIONS (XII) # 10

(iv)

 x  1  x2  2 [ x ] [ x ] 1      x  1 , even by graph of function f(x) =  2  x x 1 

(v)

f(x) =

2x(sin x  tan x ) x 2   1 

if x = n, f(n) = 0 if x  n

8.

 x   x    = – 1        

f(–x) = – f(x) odd function

(i)

f(x) = 1 –

  cos2 x sin2 x – fx   = 1 –  f(x) 2  1 – tan x 1 – cot x fundamental period = 

(ii)

(iii)

(iv)

sin2 x cos2 x – 1  cot x 1  tan x period of f(x) = L.C.M. (, ) =  For fundamental period

  f(x) = tan  [ x ]  : [x]  2n + 1 2  f (x) = 0 By graph fundamental period = 2 f(x) = log (2 + cos 3x) fundamental period of f(x) = fundamental period of (2 + cos 3x) (as log is a monotonic function) f(x) = en sin x + tan3x – cos (3x – 5) f(x) = sin x + tan3x – cos (3x – 5), sin x > 0

period  2 ;  ,

2 3

2    = 2 Period of f(x) = L.C.M.  2 , , 3  

(v)

x x x   x x x x   f(x) =  sin x  sin 2  sin 4  ....  sin n –1    tan  tan 3  tan 5  ...  tan n  2 2 2 2   2 2 2  

Period of f(x) = L.C.M. of 2, 23 , 25 ,......2n , 2, 23 .....2n  = 2 n

(vi)

f(x) =

sin x  sin 3 x cos x  cos 3 x

2 2   period of f(x) = L.C.M.  2. , 2,  = 2 3 3   For fundamental period

f(x + ) = 11.

f(x) =

sin ( x  )  sin(3 x  3 ) – sin x – sin 3 x = cos( x  )  cos(3 x  3) – cos x – cos 3 x

1 x

1 x2

f(x) = 0 at x = 1 ±

Fundamental period = 

2

for x   2  1, 1  2 f is bijective function hence f is invertible.

RESONANCE

S OLUTIONS (XII) # 11

1 x 1 x2 or

or

=y x2y + x + (y – 1) = 0 x=

 1  1  4 y( y  1) 2y

=

 1  4y  4 y 2  1

  1  4x  4x 2  1  , f–1(x) =  2x  , 1

2y

x0 x0

as f (1)  0

n

 f (a  k ) = 16 (2

n

13.

k 1

or

or Now or 15.

(i)

 (ii)

– 1)

f(a + 1) + f(a + 2) + ......... + f(a + n) = 16 (2n – 1) f(x + y) = f(x) . f(y) f(0) = 1, f(1) = 2 f(x) = 2x f(a + 1) + f(a + 2) + ........ + f(a + n) = 2a [2 + 4 + ......... + 2n] = 2a . 2(2n – 1) 16 = 2a + 1 or a=3 f(x) = Ax2 + Bx + C x   and f(x)   at x = 0, f(0) = C  at x = 1, f(1) = A + B + C C is integer  at x = –1, f(–1) = A – B + C f(1) + f(–1) = 2A + 2C  2A is also integer f(x) = A x(x – 1) + (A + B) x + C f(x) = 2A

C is integer A + B is also integer C is integer

x( x  1) + (A + B)x + C 2

x( x  1) is also an integer and 2A, (A+ B), C   2 f(x) is also an integer.

If x is an integer then 

PART - II 2.

f(x) = here

1

x

 1 cos 1 (2 x  1) tan 3 x

– 1  2x + 1 < 1

– 2  2x < 0

–1x 3x > –

 2

or

   x   , 0  6 

      Domain :   , 0   (–1, 0)    , 0  6    6 

RESONANCE

S OLUTIONS (XII) # 12

3.

 1 x3    f(x) = sin  3 / 2  + sin(sin x ) + log(3{x} + 1) (x 2 + 1)  2x  Domain : 3{x} + 1  1 or 0  x   –1

and

6.

1 x 3

1 2x 3 / 2 – 2x 3/2  1 + x 3  2x 3/2 1 + x 3 + 2x 3/2  0 (1 + x 3/2)2  0  xR 1 + x 3 – 2x 3/2  0 or (1 – x 3/2)2  0 3/2 or 1–x =0 or x=1 Hence domain x f(x) = (sin–1x + cos –1x)3 – 3 sin–1x cos –1x (sin–1x + cos –1x) –1

=

 3 3  1   3 2 – 3 sin–1x   cos x  = – sin–1 x + 3 (sin–1x)2 2  2 8 8 2 4

3 3 = + 8 2

2   2   2 1 1 3 3 3 3  1 sin x (sin x )  sin x   –    = + 2 16  4  32 32 2 

maximum value of f(x) at x = – 1 8.

f maximum =

9 3 7 3 3 3 + × = 32 2 16 8

Here (2 – log2 (16 sin2x + 1) > 0 

0 < 16 sin2x + 1 < 4

1  16 sin2x + 1  4

3 16 0  log2 (16 sin2x + 1) < 2

log

2

2  2 – log2 (16 sin x + 1) > 0

0 sin2x <

2

2  log

2

(2 – log2 (16 sin2x + 1)) > – 

 2y>– Hence range is y  (– 2] 11.

(A)

f(x) = e1/2 n x = g(x) =

12.

D:x>0

x,D:x0

D : x  ± (2n +1)

 2

(B)

tan–1 (tan x) = x

(C)

cot–1 (cot x) = x D : x  ± n f(x) = cos 2x + sin4x = cos 2x + (1 – cos 2x)2 = 1 – cos 2x + cos 4x = sin2x + cos 4x g(x) = sin2x + cos 4x

(D)

f(x) =

|x| , D:x0 x g(x) = sgn (x), D : x  R

f(6{x}2 – 5{x} + 1) (3{x} – 1) (2{x} – 1)  0

13.

x,

f((3{x} – 1) (2 {x} – 1)) or

 1 1 {x}  ,  3 2

x

n 

1 1  n  3 , n  2   

  R+   0 ,   2 g(x) = 2x – x2 R  R f(g(x)) = cot–1 (2x – x2), where x  (0, 1]

f(x) = cot–1x

   hence f(g(x))   ,  4 2 

RESONANCE

S OLUTIONS (XII) # 13

20.

21.

25.

1 1 2 1     f  x   =  x   +  x   + [x + 1] – 3  x   + 15 3 3 3 3         1 2   =  x   +  x   + [x] – 3x + 15 = f(x) 3 3     f(x) = |x – 1| f : R+  R x g(x) = e , g : [–1, )  R fog(x) = f[g(x)] = |ex – 1| D : [–1, ) R : [0, )

f(x) = (A) (B) (C)

(D)

fundamental period is 1/3

sin( [ x ]) =0,x { x} By graph fundamental period is one f(–x) = 0 = f(x)  even function Range y  {0}  { x }  y = sgn  sgn – 1, x  { x }   y = sgn (1) – 1  y=1–1 y = 0, x   Identical to f(x)

28.

f(x) = sin x + tan x + sgn (x2 – 6x + 10) f(x) = sinx + tan x + sgn ((x – 3)2 +1) f(x) = sin x + tan x + 1 period = L.C.M. (2, ) = 2 fundamental period = 2

29.

f:NI

n – 1 , n  odd  f(n) =  2 n – , n  even  2 For For

n  odd numbers f(n)  0, 1, 2, 3, ...... n  even numbers f(n)  –1, –2, –3, ...... range I

 

f(n) is one -one onto function.

EXERCISE # 3 2.

 x = 2

(A)

sin–1 x + cos–1

(B)

2   sin–1 x + cos–1 1  x  = 0  

 (C)

x  [0, 1]

x  [–1, 0]

x [0, )

 1 x   tan–1 x + tan–1= tan–1   1– x 

 1 – x2    g 2  = 2h(x)  1 x 

(D)

cos–1 1 x 2 = – sin–1(x)

cos

–1

 1 – x2    –1  1  x 2  = 2 tan x  

 1 x   h(x) + h (1) = h   1– x   x (–, 1)

RESONANCE

S OLUTIONS (XII) # 14

Comprehension # 2 (6, 7, 8) Period of e

x tan   4

is 4

 (1 2 [ x])   =0 2  

cos 

xR

 [x] 

Period of sin  2  is 4   then y =

8  2[ x ]  [ x ] 2

p=4

   

[x]2 – 2[x] – 8  0 – 2  [x]  4 q=–2 , r=5 r–q–1=5+2–1=6 x  2 , f2 (x) =  2  x ,

Period of f(x) is 4

– [x]2 + 2 [x] + 8  0

i.e., 

([x] – 4) ([x] + 2)  0 –2  xx–1 f(x + 1) < f(x – 1) as f(x) is M.D. g(f(x + 1) < g(f(x – 1)) as g(x) is M..

Let

f(x) =

, x

f(x) = Let

f(x)  0

.

g(x) = x sec2x – tan x g(x) = 2x sec2x tanx > 0 x>0 g(x) > g(0), g(x) > 0 x1 < x2 f(x1) < f(x2) <

RESONANCE

f(x) > 0

f(x) is M.I.

< S OLUTIONS (XII) # 102

Section (D) : D-2.

f(x) = 3x 3 f(x) = 0  x=0 x = – 2, f(–2) = – 8 x = 0, f(0) = 0 x = 2, f(2) = 8 Minimum = – 8, maximum = 8 (ii) f(x) = cos x – sin x (i)

f(x) = 0

 x=

x = 0, x=

f(0) = 1 ,

x = ,

f

=

f() = – 1

Minimum = –1, Maximum = (iii) f(x) = 4 – x f(x) = 0  x=4 x = – 2,f(–2) = – 10 x = 4, f(4) = 8 x=

,f

=

Minimum = –10, (iv)

Maximum = 8

f(x) = cos x – sin 2x f(x) = 0

x = 0, f(0) = Minimum = D-6_.

cos x = 0, sin x =

x=

, x=

x=

f

=

 x=

, f

=

, Maximum =

Let No. of children of john & anglina = y  x + (x + 1) + y = 24  y = 23 – 2x Number of fights F = x(x + 1) + x(23 – 2x) + (x + 1) (23 – 2x) F = – 3x2 + 45x + 23  But 'x' wil be integral.

D-8.

,

2

2

= 0  – 6x + 45 = 0

 x = 7.5

check x = 6 or x = 7, F = 191

2

R +r =h R2 = h2 – r2 volume of cylinder , V = R2 (2h) =  (2h) (

)2

= 2 (r2 – h2) + 2h(–2h) = 0 

r2 = 3h2

h=

< 0 at h =

Vmax = 2

RESONANCE

= S OLUTIONS (XII) # 103

D-10.

2 + 2r = 440 A =  2r = – 2r2 + 440r = – 4r + 440 = 0

D-12.

at

r=

Let  Let

x, y be dimensions of rectangle. 2x + 2y = 36. V be volume sweeped V = x 2y V = x 2(18 – x) = x.3.(12 – x)

At

x = 12, V has maximum value 

y=6

Section (E) : E-2.

Let (h, k) be point of inflection h sin h = k y = sin x + x cos x y = cos x + cos x – x sin x y = 0  2 cos h – h sin h = 0  2 cos h = k sin2 h + cos 2 h = 1 = 1  4k 2 + h2k 2 = 4h2

+

...(1)

...(2)

locus y2 (4 + x 2) = 4x 2

Section (F) : F-2.

F-4.

Let

f(x) = 3x2 + px –1

f(x) = x3 +

 

f(x) satisfies conditions in Rolle's theorem 3x2 + px –1 = 0 has atleast one root in (–1,1).

–x+c

f(–1) =

+ c = f(1) 

f(c) = 0 for atleast one c (–1,1)

Let h(x) = f(x) g(x) h(a) = 0 = h(b) By Rolle’s theorem on [a, b] h(x) = 0, for at least one c  (a, b).  f (c) g(c) + f(c) g(c) = 0

PART - II Section (A) : A-1.

V=

V=

=

77 × 103 =

× 70 × 70 ×

( 1 litre = 103 c.c.)

= 20 cm/min.

RESONANCE

S OLUTIONS (XII) # 104

A-3_.

= Let x = 25 and x = 0.2 such that f(x) =  f (x) =

 f(x + x) = f(x) + f(x). x

=

+

.x

= A-4_.

+

× 0.2 =

=5+

= 5 + 0.02 = 5.02

V = x3 x = (3x 2) x = (3x 2) (0.04x) = 0.12x 3m 3

V =

Section (B) : B-5*.

2y3 = ax 2 + x 3 6y2

= 2ax + 3x 2

=

=

Tangent at (a, a) is 5x – 6y = – a =

,

2 +  2 = 61

= = 61

a2 = 25.36 a = ± 30 B-8.

P1 : y2 = 8x C1 : x 2 + (y + 6)2 = 1

Equation of normal of parabola y = mx – 2am – am 3 if passes through (0,–6) –6 = – 2am – am 3  a=2  3 = 2m + m 3 3 m + 2m – 3 = 0 m = 1. Point on parabola (am 2 , – 2am)  (2, –4).

Section (C) : C-2.

5x 4 – 3

f(x) =

It is sufficient to solve for p, the condition f(x)  0

xR

5x 4 – 3  0  x  R

RESONANCE

S OLUTIONS (XII) # 105

Case - 

1–p1 Inequality holds true. 1–p>0 p 0 and for x  cos x – x < 0, 

RESONANCE

f(x) has maxima at x 0

S OLUTIONS (XII) # 106

D-8*.

f(x) =

,x>0

=

x > 0.

f(x) is decreasing On

x > 0.

, greatest value is

f(

)=

n

f

=

n

and least value is

.

D-10. S = 2rh = 2H

= 2H

Maximum at r =

D-13.

Let d be distance between (k, 0) and any point (x, y) on curve. d= ( y2 = 2x – 2x 2).

d=

Maximum d =

Maximum d =

Section (E) : E-3.

x=1

a=–

E-4.

3=a+b

=0

,

b=

3a + b = 0

f(x) = n(x – 2) –

f(x) =

=

=

=

.

As n(x – 2) is defined when x > 2  f(x) is M.. for x  (2, )

f –1(x) is M.. wherever defined

Also

f(x) is always concave downward

f(x) =

RESONANCE

0

If x > 0, (x) > 0 Hence (x) is increasing As we know ex  x + 1  (ex)  (x + 1) ex + 13.

x+1+

Let x > –1 Consider f(x) = (1 + x)n(1 + x) – tan–1 x f(x) = n(1 + x) + 1 – f(x) =

>0

 f(x) is increasing  f(x) < 0

For x < 0, f(x) < f(0)  f(x) decreasing  f(x) > f(0)

 (1 + x)n(1 + x) – tan–1 x > 0 For x > 0,

f(x) > f(0)  f(x) > 0

 f(x) > 0

n(1 + x) >  f(x) is increasing

 f(x) > f(0)

 f(x) > 0

 n(1 + x) > Hence larger of these is n(1 + x). 15.

f(x) = a0 + a1x + a2x 2 + a3x 3 + a4x 4 + a5x 5 + a6x 6 a0 = a1 = a2 = a3 = 0 = e2

a4 = 2

f(x) = 2x 4 + a5x 5 + a6x 6 f(x) = x 3 ( 8 + 5a5x + 6a6x 2) f(1) = 0, f(2) = 0 , 17.

f(x) = 2x 4

xy = 18 Area of printed space

=

=

Maximum when x=

RESONANCE

y= S OLUTIONS (XII) # 109

21.

f(x) = 0  sin x=

=0

= n

,nN

x = .......,

,

, .......

,

Consider interval

, 1.

=0=

By Rolle’s theorem f(x) vanishes at least once in Infinite number of such intervals are there. Hence f(x) vanishes at infinite number of points in (0, 1) 24.

Let g(x) =

, x  [a, b]

By Rolle’s theorem, g’(x0) = 0 =0

 25.

f(x0) =

Let f(x) = (x + a) – (x)  f(x) = (x + a) – (x) + k f(0) = (a) – (0) + k f(2a) = (3a) – (2a) + k  f(0) = f(2a) By Rolle's theorem on [0, 2a], f(c) = 0 for at least one c  (0, 2a)  (x + a) = (x) has at least one root in (0, 2a)

PART - II 3.

f(x) =

>0

 f(x) increasing hence g(x) is also increasing function 7.

Let

 be quantity y=

y2 = y

=

=

=a 9.

= = a

f(x) = ; g(x) = for a > 1, a  1 and x  R n a h(x) = n f(x) + ng(x) 

(n a) h(x) =

h(x) =

na +

n a

+ |x|

 h(x) = a sgn x Now h(–x) = a|–x| sgn (–x) = –h(x)  h(x) is an odd function Also graph of h(x) is It is clear from the graph that h(x) is an increasing function

RESONANCE

S OLUTIONS (XII) # 110

12.

f(x) =

f(x) = For 13.

16.

0 < x < 1,

tan

f(1–) f(1) and f(1+)  f(1) – 2 + log2 (b2 – 2)  5 

f (x) > 0

0 < b2 – 2  128

f(x) is increasing.

2 < b2  130

2 = h2 + x 2 Area of base (triangle) is

3x =

a

Volume V = h

. 4 . 3.x 2 = 3

=h

=3 18.

a2

(2 – 3h2)

h (2 – h2)

V is maximum when h =

Maximum of f(x) is Given expression is f(x 1) + f(x 2)

20.

.

f(x 1) 

f(x 2) 

f(x 1) + f(x 2) 



f (x) = 1 +1– 2– At x = 0, f(x) is least. Least value = f(0) = 1

23.

f(x) = m – n is odd. f (x) < 0

25.

>0

x  (–, 0) 

f (x) > 0

x  (0, )

(x) = (3(f(x))2 – 6(f(x)) + 4)f(x) + 5 + 3 cos x – 4 sin x 5–

5 + 3 cosx – 4 sin x  5 +

adding (3(f(x))2 – 6(f(x)) + 4)f(x) (3(f(x))2 – 6(f(x)) + 4)f(x)  (x)  (3(f(x))2 – 6(f(x)) + 4)f(x) + 10  3(f(x))2 – 6f(x) + 4 = 3 (f(x) – 1)2 + 1 > 0 2 (3(f(x)) – 6(f(x)) + 4)f(x)  0 when ever f(x) is increasing.  (x) 0  (x) is increasing, when ever f(x) is increasing. If f(x) = – 11 then (3(f(x))2 – 6f(x) + 4) f(x) + 10 = – 33 (f(x) – 1)2 – 1 < 0  (x) < 0  (x) is decreasing.

RESONANCE

S OLUTIONS (XII) # 111

28.

f(x) = f(x) = 0 at x = 0, x = – 3, x = 1 so at x = 0, f(x) has local minima. and at x = –3, x = 1 ; f(x) has local maxima f(1) = 

,

f(– 3) =

. f(–3) < 0, f(1) > 0 and f(x)  0

f(x) is undefined at point(s) in (–3, 1). Hence f(x) has no absolute maxima.

EXERCISE # 3 PART - I 2.

(A) (B)

f(x) is continuous and differentiable f(0) = f() Hence condition in Rolle’s theorem and LMVT are satisfied. f(1–) = –1, f(1) = 0, f(1+) = 1 f(x) is not continuous at x = 1, belonging to Hence, atleast one condition in LMVT and Rolle’s theorem is not satisfied

(C)

f’(x) =

(x – 1)–3/5, x  1

(D)

At x = 1, f(x) is not differentiable. Hence at least one condition in LMVT and Rolle’s theorem is not satisfied. At x = 0

L.H.D. =

3.

=

= –1

R.H.D. = 1 At x = 0, f(x) is not differentiable Hence at least one condition in LMVT and Rolle’s theorem is not satisfied. (A) Let PQ = x Then BP =

PS =

tan60º =

area A of rectangle =

=

=–

(4 – x) x

(4 – 2x) = 0

x=2

0

g(x) = –

× f(x).g(x)

g(x) > 0 g(x) = f –1 (x) concave upward

15.

f(x) =

 17.

(1 – nx) f(x)  0, when x  e  f(x) is decreasing function, when x  e >e  f() < f(e) 1/ < e1/e  e > e Statement-1 is True, Statement-2 is False

f ’ (x) = 50x49 – 20x19 = 10x19(5x30 – 2) x = 0 is stationary point. Statement-2 is ture. f(0) = 0 =

0 increasing

gmax = g(1) = e +

h(x) = x2 

+

+

hmax = h(1) = e +

RESONANCE

h(x) = 2x ,

+ 2x3 so

– 2x

= 2x

>0

a=b=c S OLUTIONS (XII) # 120

26.

(A)

Re

= Re

= –1  y  1 =

1 or

= Re

= Re (–1/y) =

– 1

Alternate

Re

= Re

= Re

= Re

= Re

= Re as –1  sin  1 (– , 0 )  (0, ) (B)

(C) (D)

–1 

1

–1 

0

1

–1 

0

0 

0

1

– 1 0

 t  (– , –9]  [–1 , 1]  [9, )  x  (– , 0)  [2 , ) f() = 2 sec2  f()  2  f(x) = x3/2 (3x – 10) 

f ’(x) = x3/2 3 +

0

f()  [2, )

x1/2 (3x –10)

asf ’(x)  0 0

 27.

– 15  0

x 2

3x +

– 15  0

x  [2, )

f(x) = x4 – 4x3 + 12x2 + x – 1 f(x) = 4x3 – 12x2 + 24x + 1 f(x) = 12x2 – 24x + 24 = 12 (x2 – 2x + 2) > 0 xR  f(x) is S.I. function Let is a real root of the eqution f(x) = 0  f(x) is MD for x (– , ) and M.I. for x (, ) where < 0  f(0) = – 1 and < 0  f() is also negative  f(x) = 0 has two real & distinct roots.

RESONANCE

S OLUTIONS (XII) # 121

28.

p = (x – 1) (x – 3) = (x2 – 4x + 3) p(x) = (x3/3 –2x2 + 3x) +  p(1) = 6 6 = (1/3 – 2 + 3) +  6 = (1/3 + 1) +  18 = 4 + 3 ...(i) p(3) = 2 2 = (27/3 – 2 × 9 + 9) +  2= =2=3 p(x) = 3(x – 1) (x – 3) p(0) = 3(–1)(–3) =9

29.

f(x) = |x| + |x2 – 1|

f(x) =

f(x) =

PART - II 1.

Let

f(x) = x +

f(x) = Minimum occures at x = 1. 2.

3.

f(x) = 6x2 – 18ax + 12a2 f(x) = 0  x = a, 2a f(x) = 12 x – 18a f(a) = – 6a < 0 and f (2a) = 6a > 0  p = a, q = 2a  p2 = q a = 2 (a > 0) Let

f (x) = ax2 + bx + c

f(x) =

a2 – 2a = 0

a = 0, 2

f(0) = d = f(1) (2a + 3b + 6c = 0) By Rolle's theorem, at least one root of f(x) = 0 lies in (0, 1) 4.

y2 = 18x Differentiating w.r.t. t

...... (1)

From equation (1), x =

RESONANCE

2y.2 = 18

y=

Required point is S OLUTIONS (XII) # 122

5.

x = a (1 + cos ), y = a sin 

Equation of normal at point (a(1+ cos ), a sin ) is y – a sin =

(x – a (1 + cos ))

y cos = (x – a)sin It is clear that normal passes through fixed point (a, 0) 6.

y = x2 – 5x + 6 = 2x – 5

= – 1,

product of slopes = –1

=1

angle between

tangents is 7.

Let f1(x) = x3 + 6x2 + 6 Increasing in (–, – 4] Let

f2 (x) = 3x2 – 2x + 1

f1(x) = 3x (x + 4)

f2(x) = 6x – 2

Not increasing in Let

f3(x) = 2x3 – 3x2 – 12x + 6

f3(x) = 6x2 – 6x – 12 = 6(x +1) (x – 1)

Increasing in [2, ) Let f4(x) = x3 – 3x2 + 3x + 3 Increasing in (–, ) 8.

V=

 4 (10 + r)2

, 0  r  15 = – 50. = – 50 

=

9.

Any point on ellipse is P(x, y) = (a cos , b sin ) Area of rectangle = 4ab cos sin  = 2ab sin 2   Maximum area = 2ab.

10.

By LMVT and f(x) 2

f(x) = 3(x –1)2 0

= f(x),

x (1, 6)

11*.

(where r = 5)

2

2

f(6) 8.

 Equation of normal is y – a(sin – cos ) =

(x – a (cos  + sin ))

 x cos + y sin = a. Distance from origin = |a| (constant) Slope of normal = –

RESONANCE

= tan

angle made with x-axis is

+ . S OLUTIONS (XII) # 123

12.

f (x) = f (x) changes sign as x crosses 2. f(x) has minima at x = 2.

13.

= f(c)  1 < c < 3 

c =

14.

f(x) =

15.

Graph of y = x3 – px + q cuts x-axis at three distinct pints

(cos x – sin x)

=0 16.

= 2 log3e

x = 

Maxima at x = –

, minima at x =

Condition for shortest distance is slope of tangent to x = y2 must be same as slope of line y = x +1. 

=1

y=

,

x=

,x–y+1=0

Shortest distance =

17.

P (x) = 4x 3 + 3ax 2 + 2bx + c and P(0) = 0  c=0  P (x) = x(4x 2 + 3ax + 2b) D = 9a2 – 32b < 0

b>

>0

(P(x) = 0 has only one root x = 0)

P (– 1) < P (1)  a>0 P(x) has only one change of sign.  x = 0 is a point of minima. P(–1) = 1 – a + b + d, P(0) = d P(1) = 1 + a + b + d  P(–1) < P(1), P (0) < P(1), P (–1) > P(0)  P(–1) is not minimum but P(1) is maximum.

18.

y=x+ y = 1 –

=0

y=2+

=3

x3 = 8

x=2

(2, 3) is point of contact Thus y = 3 is tangent Hence correct option is (3)

RESONANCE

S OLUTIONS (XII) # 124

19.

f(x) = 1 f(–1) = k + 2 f(x) = k + 2 f has a local minimum at x = – 1 1k+2k+2 possible value of k is – 1 Hence correct option is (3)

 

20.

ex + 2e–x  2

f(–1+)  f(–1)  f(–1–) k–1

(AM  GM)

 f(x) > 0

so statement- 2 is correct

As f(x) is continuous and

f(c) =

 21.

belongs to range

for some C.

of f(x),

Hence correction option is (4).

y–x=1 y2 = x 2y

=1

=

  y =

= 1 , x=

tangent at

y=

 y=x+

y–x=

distance =

22.

=

=

f(x) =

In right neighbourhood of ‘0’ tan x > x

RESONANCE

S OLUTIONS (XII) # 125

In left neighbourhood of ‘0’ tan x < x as(x < 0) at x = 0 , f(x) = 1  x = 0 is point of minima so statement 1 is true. statement 2 obvious

23.

r3

V=

4500  =

= 4r2

45 × 25 × 3 = r3

r = 15 m after 49 min = (4500 – 49.72) =972 m3 972 =

r3

r3 = 3×243 = 3× 35 r=9 72  = 4 × 9 × 9

=

24.

f '(x) = at x = – 1

+ 2bx + a –1 – 2b + a = 0 a – 2b = 1

at x = 2

+ 4b + a = 0

a + 4b =

On solving (i) and (ii)

f '(x) =

...(i)

=

a=

...(ii)

, b=

=

So maxima at x = – 1, 2

RESONANCE

S OLUTIONS (XII) # 126

SOLUTION OF TRIANGLE EXERCISE # 1 PART - I Section (A) : A-1.

(i)

L.H.S. = a sin (B – C) + b sin (C – A) + c sin (A – B) = k sin A sin (B – C) + k sin B sin (C – A) + k sin C sin (A – B) = k (sin2 B – sin2 C) + k (sin2C – sin2 A) + k (sin2 A – sin2 B) = 0 = R.H.S.

(ii)

L.H.S. =

first term =

= = k 2 sin (B + C) sin (B – C) = k 2 (sin2 B – sin2 C) = k 2 (sin2 C – sin2 A)

Similarly

= k 2 (sin2 A – sin2 B)

and

L.H.S. = k 2 (sin2 B – sin2C + sin2C – sin2A + sin2 A – sin2 B) = 0 = R.H.S. (iii) L.H.S. = 2bc cos A + 2ca cos B + 2ab cos C = b2 + c 2 – a2 + a2 + c 2 – b2 + a2 + b2 – c 2 = a2 + b2 + c 2 = R.H.S 

(iv) L.H.S. = a2

– 2ab

= a2 + b2 – 2ab cos C = a2 + b2 – (a2 + b2 – c 2) = c 2 = R.H.S. (v)  L.H.S. = b2 sin 2C + c2 sin 2B = 2b2 sin C cos C + 2c 2 sin B cos B = 2k 2 sin2 B cos C sin C + 2k 2 sin2 C sin B cos B = 2k 2 sin B sin C [sin B cos C + cos B sin C] = 2(k sin B) (k sin C) sin (B + C) = 2bc sin A (vi)  R.H.S =

(b = ksin B, c = ksin C)

c = a cos B + b cos A, b = c cos A + a cos C

=

=

=

A–4.

    

= L.H.S.

= sin(B + C) sin(B – C) = sin(A + B) sin(A – B) sin2 B – sin2 C = sin2 A – sin2 B 2 sin2 B = sin2 A + sin2 C 2b2 = a2 + c 2  a2, b2, c 2 are in A.P.

RESONANCE

S OLUTIONS (XII) # 127

A–7.

x 3 – Px 2 + Qx – R = 0

a2 + b2 + c 2 = P a 2b 2 + b 2c 2 + c 2a 2 = Q a2b2c 2 = R  abc =

+

+

=

[a2 + b2 + c 2] =

Section (B) : B–1.

(i)

L.H.S. = 2a sin2 = = = =

(ii)

+ 2 c sin2

a(1 – cos c) + c(1 – cos A) a + c – (a cos C + c cos A) a+c–b R.H.S.

 L.H.S. =

+

=

.

+ +

= (iii)

.

+

=

.

.

L.H.S. = 2bc(1 + cos A) + 2ca(1 + cos B) + 2ab(1 + cos C) = 2bc + 2ca + 2ab + 2bc cos A + 2ca cos B + 2 ab cos C + a2 + b2 + c 2 = (a + b + c)2

=2 = R.H.S. (iv)

 L.H.S. = (b – c)

 (b – c) cot

+ (c – a)

+ (a – b)

= k(sin B – sin C)

= 2k cos

sin

= 2k sin

sin

= k [cos C – cos B] similarly (c – a) cot and (a – b) cot

= k[cos A – cos C]

= k[cos B – cos A]

 L.H.S. = k[cos C – cos B + cos A – cos C + cos B – cos A] =0 = R.H.S.

RESONANCE

S OLUTIONS (XII) # 128

(v) L.H.S. = 4 (cot A + cot B + cot C) = 4 = 2bc cos A + 2 ca cos B + 2ab cos C = a2 + b2 + c 2 = R.H.S. (vi) L.H.S. =

cos

. cos

. cos

= =

=  = R.H.S.

B–3.

Let ADB =   we have to prove that tan = if we aply m – n rule, then (1 + 1) cot= 1.cot C – 1.cotA. =

=

= =

[2(a2 – c 2)]

2cot =

tan =

Section (C) : C–2.

(i)

r. r1 .r2 .r3 =

(ii)

r1 + r2 – r3 + r = 4R cosC

= 2

L.H.S. = =

=

=

RESONANCE

S OLUTIONS (XII) # 129

=

=

=c

=

L.H.S. =

(iii)

=

   

= R.H.S.

=

(s – a + s – b + s – c) =

·

=

=

=

=

=

=

=

 = 24 sq. cm 2s = 24  r1, r2, r3 are in H.P.

=

=

=r

=r

.... (i) s = 12 .... (ii)

are in A.P..

[4s 2 – 2s(a + b + c) +a2] =

R.H.S. =

similarly we can show that 

= 4RcosC

L.H.S. =

=

C–4.

=

(s + s – a + s – b + s – c)2 = 4

(v)

=

L.H.S. =

=

cos C =

[s 2 + (s – a)2 + (s – b)2 + (s – c)2] =

(iv)

=

=





a, b, c are in A.P. 2s = 24 a + b + c = 24 3b = 24 b=8

are in A.P.. 

2b = a + c

a + c = 16

But  =   = 24 × 24 = 12 × (12 – a) × 4 × (12 – c)  2 × 6 = 144 – 12 (a + c) + ac 12 = 144 – 192 + ac  ac = 60 and a + c = 16  a= 10, c = 6 or a = 6, c = 10 and b = 8

RESONANCE

S OLUTIONS (XII) # 130

Section (D) : D–1.

(i) 

,=

=

, =

=

R.H.S. =

=

=

L.H.S. = R.H.S.

(ii)

=

=

=

R.H.S. =

=

= L.H.S.= R.H.S.

PART - II Section (A) : A–4.

(b + c)2 – a2 = kbc

 (a + b + c) (b + c – a) = kbc

b2 + c2 – a2 = (k – 2) bc

 In a ABC –1 < cos A < 1

–1 <

b

=

= cos A

0, c = 5x + 5y is the largest side   C is the largest angle. Now cos C =

=

r2 > r3

>

>

 s – a < s – b < s – c  –a < –b < –c;

3.

tan

=

; sin

r+R=

RESONANCE

a>b>c

=

r+R=

.cot

S OLUTIONS (XII) # 143

4.

a

=

=

  a + c = 2b 5.

a + b + c = 3b.

a, b, c are in A.P.

AD = 4 

AG =

Area of ABG =

=

Area of  ABC = 3(Area of  ABG)

6.

cos =

7.

C = /2

×4=

×

×

× AB × AG sin 30º

×

=

Sin 60º =

AB =

=

 = 120º

=

=–

r = (s – c) tan

C = 90º

r = s – 2R  2r + 2R = 2 (s – 2R) + 2R. = 2s – 2R = (a + b + c) –

C = 90º

=a+b+c–c =a+b

8.

are in H.P.

are in A.P.

9.

a,b,c are in A.P.

= cos

Let cos

=

As

for some n  3, n  N

cos

 cos

 cos

 3  n < 4, which is not possible so option (2) is the false statement so it will be the right choice Hence correct option is (2)

RESONANCE

S OLUTIONS (XII) # 144

ADVANCE LEVEL PROBLEMS PART - I 1.

From figure, AD = c sin B Hence number of triangle is 0 if b < c sin B one triangle for b = c sin B two triangles for b > c sin B

2.

C = 60° Hence c2 = a2 + b2 – ab

3.

=

= 2 cos

Using properties of pedal triangle, we have

 MLN = 180° – 2A  LMN = 180° – 2B  MNL = 180° – 2C

Hence the required sum =

sin2A + sin2B + sin2C

=

4.

5.

4sinA sinB sinC

From figure, we can observe that

OGD is directly similar to PGA

BD = s – b, CE = s – c and AF = s – a Hence BD + CE + AF = s

6.

, as cos

= cos

A = B, in either case

 7.

, Using cosine rule in

ABO, we get

h=

8.

In

ABD,

RESONANCE

S OLUTIONS (XII) # 145

Comprehension # 1

9.

10.

+

+

= b sin B + c sin C + a sin A =

k = 2R

cot A + cot B + cot C =

=

(b2 + c2 + a2) =

(b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2)

=

=

.

=

11.

=

k=

=6

Comprehension # 2 (12 to 14) 12.

 PG =

= =

=  PG = =

.ab sin C

or

b sin C

(  =

ac sin B)

ac sin B c sin B

13.

 Area of GPL =

and

Area of ALD =

(PL) (PG) (DL) (AD)

=

 PL =

=

DL and PG =

=

14.

 Area of PQR = Area of PGQ + Area of QGR + Area of RGP ...(1)

Area of PGQ = = =

RESONANCE

×

×

×

BE sin ( – C) sin C S OLUTIONS (XII) # 146

=

×

=

bc sin A ×

ac sin B × sin C

sin A.sin B.sin C

Similarly Area of QGR =

sin A.sin B .sin C and Area of RGP =

sin A.sin B.sin C

 From equation (1), we get (a2 + b2 + c 2) sin A.sin B.sin C

Area of PQR = 15.

In

CDB ,

=

 Also from same triangle 16.

17.

 BD =

cosAcosB + sinAsinBsinC = 1

(cosA – cosB)2 + (sinA – sinB)2 + 2sinAsinB(1 – sinC) = 0

a:b:c=1:1:

A = B & C = 90°

We have

 18.

=

a:b:c=5:4:3

from figure, OO = ON – ON = R –

ZO = ZM + = from

RcosA +

OZO, using Pythagorous theorem,

we get (R –

)2 = (RcosA +

)2 +

=

PART - II 1.

from

 from

ABC ,

=

AB = 2Rsin(A + )

ACB,

=

AC’ = 2Rsin( – A)

=

4RcossinA = 2acos

similarly CA = 2bcos =

RESONANCE

 area

BC = 2R(sin (A + ) – sin( – A))

ABC = =

4cos2. S OLUTIONS (XII) # 147

2.

c2 – 2bc cosA + (b2 – a2) = 0 c1 & c2 are roots of this quadratic equation Hence (c1 – c2)2 + (c1 + c2)2tan2A = 4a2

3.

Area =

=

= =

2Rs

= 4.

We know that OA = R, HA = 2RcosA and applying Appoloneous theorem to

AOH, we get

2.(AQ)2 + 2(OQ)2 = OA2 + (HA)2

2.(AQ)2 = R2 + 4R2cos2A –

5.

=

+

using sine rule, diameter of required circle =

 6.

=

= 20

L.H.S. =

(a2 (b + c – a) + b2 (c + a – b) + c2 (a + b – c))

=

=

=

abc

=

4R

RESONANCE

S OLUTIONS (XII) # 148

7.

from the parellelogram ABAC, AA = 21 , from

AAC, AA < b + c 21 < b + c

similarly 22 < c + a

...(1) ...(2)

and 23 < a + b ...(3) (1) + (2) + (3) gives 1 + 2 + 3 < 2s 8.

ZXY =

and

 Area of

=

 Area of

 Area of XYZ = 2R2 cos

9.

cos

cos

=

Drop a perpendicular from the apex P to the base

ABC.

The foot of perpendicular is at circum centre O of

ABC

Using given data, we get from, right angle

POB, we get

= = 8.83 m 10.

from cyclic quadrileteral CQFP, we get

from cyclic quadriletral AQMF, we get  FQM =  FAM = 90º – B

 AQM = 90º + 90º – B = 180º – B

 

P, Q, M are collinear

similarly P, Q, N are collinear hence, P, Q, M, N are collinear

RESONANCE

S OLUTIONS (XII) # 149