- (a) : f ′(x) = f (x) + k
′- ∫ fx =∫
fx k
- ∫ fx =∫
()dx dx
()
⇒ log(f (x) + k) = x + C ⇒ f(x) = k 1 ex – k
∴ (^) fkk() 0 4 e
(^13)
2
= − = −^ .... (1)
Also, kfxdx=∫ ()
0
2
⇒ 3 k = k 1 (e^2 – 1) .... (2)
Solving (1) and (2), we get
fx e
()= x−⎛e −
⎝⎜
⎞
⎠⎟
(^21)
3
- (a, b, c) : 0
2
<<x ⇒ x
xπ sinis decreasing and sinx < x < tanx
⇒ sin(sin )>>
sinsin sin(sin )
tanx
xx
xx
x⇒ I 1 > I 2 > I 3- (a, d) : Let f–1(x) = y ⇒ x = f(y)
⇒ dx = f ′(y)dy
Ifybyfydy
ab
=∫2()(− ) ()′=bfyfydy yfyfydy∫∫′ − ′
abab
22 () () () ()=bf b f a−−bf b af a++∫f ydy
ab
(() ())22 2 2 2() () ()=+∫fxdx abfa− =∫ fx fadx−
abab 2222
() ( ) () ( () ())- (a, b, c, d) : dx
x
x
xxdx
() 11 +^2010 ()^2010=
∫ ∫ += (^2) ∫t 2010 −^1 =+ 1
t
dt ()wheret x
= ⎡ −
⎣
⎢
⎤
⎦
(^2) ⎥+
1
2009
1
tt^200920082008
c
⇒ α = 2009, β = 2008
- (a, c, d) : Let I x
x
x
x= + dx
−− −
+⎛
⎝⎜⎞
−∫ ⎠⎟1
11
121212//=
−=−
− −∫∫
4
12 4
12 2 1120 2x^12
xdx x
xdx
///
()=−− (^414) {}=− ⎝⎜⎛^3 ⎠⎟⎞
4
2
0
12
ln | | ln
/
x
= (^4) ⎝⎜⎛^4 ⎟⎠⎞= ⎛⎝⎜ ⎟⎠⎞=− ⎛⎝⎜ ⎞⎠⎟
3
256
81
81
256
ln ln ln
- (a, b, c, d) :
′ =×
+
∫ +
fx x x
xx
tdtx
() sin
coscos
/ cos2
11(^22241)
2
π
- (b, d) : Let I dt
e tt
x
=
∫/ ()+cot1 1 2Put t = 1/z ⇒dt=−
z(^12) dz
∴ =
−
⎛⎝⎜ + ⎞⎠⎟
- =
∫ ∫ +
I z
dz
z z
zdz
z
tdt
x t
e
x
e
e
1
(^11111)
2
2
() ()^22
tan tan
tanxx
∫
∴
∫ ∫ +
tdt
t
dt
e tt
x
e
x
1 11221 / ()
cot
/
tan
=
∫∫t =+
t
dt t
t
dt tdt
t
t
e
x
x
e
e
e
(^111) e
1
2
2 1
1
22
2
1
/ 1
tan
tan
/
/
[ln( )]
ee
∫
=+⎧⎨ − ⎛⎝⎜ + ⎞⎠⎟
⎩
⎫
⎬
⎭
(^1) ==
2
1 1 1 1
2
ln( e^2 ) ln 2 (ln^2 ) 1
e
e
Also,^2
1
4
1
(^41)
22
1
0
1
1
1
πππ
dt
t
dt
t
= ⋅ −
−
∫ ∫ tan = ⋅ =
4
4
1
π
π
- (a) : − + − + − + −
−
−−∫∫∫∫^5432
112120012121
dx dx dx dx////
=−−−−^5 =−
22 3
217- (c) : 01501
0
1150
dx dxtantan∫∫+=−tan
π
π- (c) : 01
2 2
32
dx dx
ππππ π//
∫∫+ − =−- A (Q), B (P), C (R), D (S)
(A) Let I x
xxdxn
= nn+−
∫sin
sin cos/απα 2
... (i)Idxnnn=⎛⎝ − ⎞⎠⎝⎛ − ⎞⎠+ ⎛⎝ − ⎞⎠−
∫sinsin cos/ π αα π α π απα 2222=
+−
∫cos
cos sin/ n
nnα
α ααπα 2
... (ii)