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) =
sin1 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
x2
1 – x > 0 and x + 2 0 x (– , 1) – {0} and x – 2
(vi)
f(x) =
and
1–x1 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 x1 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
–
x2 – 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
D0 (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) = enx, Domain x R+ Domain x R non-identical function f(x) = ex and g(x) = n x fog(x) = en 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 x0 f(x) = 3 0 x 1 x x
RESONANCE
x x x x0 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
fx = 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:DR 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 ey 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 x2 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 xR
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) = en(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) =
0x1
n I x R – {n },
n I
0 |2x + 5| 7 –6x1
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) = +
xR
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 2x + sin (3x) + cos (4x) + ........ + sin (2n – 1) + cos (2px) f(x) = {x + b} + a – b + sin (x) + cos (2x) + sin (3x) + cos (4x) + .... + sin (2n – 1) + cos (2nx) 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 ex 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:xR
(iv)
f(x) =
or
(2x)2 – 3.2x + 2 0 2x [1, 2] x [0, 1]
1 1 x2
x2 + x2
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
x2 1 x 0 and 0 x2 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) =
x5 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)
x2
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 D0
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 – fx = 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) = en 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
x0 x0
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
–1x 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 xR 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)) > –
2y>– Hence range is y (– 2] 11.
(A)
f(x) = e1/2 n x = g(x) =
12.
D:x>0
x,D:x0
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:x0 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:NI
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
xR
[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) + 2h(–2h) = 0
r2 = 3h2
h=
< 0 at h =
Vmax = 2
RESONANCE
= S OLUTIONS (XII) # 103
D-10.
2 + 2r = 440 A = 2r = – 2r2 + 440r = – 4r + 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
xR
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 = 2rh = 2H
= 2H
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
,nN
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)
asf ’(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 xR 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 1k+2k+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 =
= 4r2
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 =
AD
= =
= 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 =
AD
=
14.
Area of PQR = Area of PGQ + Area of QGR + Area of RGP ...(1)
Area of PGQ = = =
RESONANCE
×
×
PG.GQ.sin(PGQ) AD ×
×
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 – ON = R –
ZO = ZM + = from
RcosA +
OZO, using Pythagorous theorem,
we get (R –
)2 = (RcosA +
)2 +
=
PART - II 1.
from
from
ABC ,
=
AB = 2Rsin(A + )
ACB,
=
AC’ = 2Rsin( – A)
=
4RcossinA = 2acos
similarly CA = 2bcos =
RESONANCE
area
BC = 2R(sin (A + ) – sin( – A))
ABC = =
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
radius = 10
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 ABAC, AA = 21 , from
AAC, AA < b + c 21 < b + c
similarly 22 < c + a
...(1) ...(2)
and 23 < 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
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