- (8) : 2 ≥ max.{|x – y|, |x + y|}
⇒ |x – y| ≤ 2 and |x + y| ≤ 2, which forms a square
of diagonal length 4 units.
2
–2 2 –=–2+= 2 –= 2+=–2
⇒ The area of the region =
1
2××= 448 sq. units.- (4) : Area = 4
0
1
∫|ln |xdx=−^4 ∫
0
1
lnxdx=− 4 []xxxln − 01 = 4 sq. units- (7) : Then required area will be equal to area
enclosed by y = f (x), the y-axis between the abscissa
at y = –2 and y = 6.
= 6
(0, 2)
–1 =–2Required area = { 62 ( )} [ ( ) ( )]
0
110
∫∫− + −−
−fx dx fx dx⇒ 9
2
=m ⇒− 27 =
nm- (0) : I= ∫ 2 t tdt
4
2
sin cos
//ππ- ∫{( sin cos ) (sin cos )}− ++∫ − sin cos
//tt ttdt 2 ttdt542 ππππ= ∫∫sin − sin =
///
220
4254
tdt tdt
ππππ- (6) :
Area=∫()()6543−− − −−^22 ∫
1
315
xx dx xx dx+∫() 43 −−^2 +∫() 315 −
3445
xx dx xdx=^73
6sq. units- (8) : ′ =
+ − +
→fx fx h fx
h h( ) lim ()()
00= ++ −−−
→lim () () () () () ()
hfx fh h fx fx f fx
0 h00= −
−⎛
⎝⎜⎞
lim→ ⎠⎟+
() () ()
hfh f
hfx
00
0⇒ fx′()= fx() ⇒′ =
∫∫fx
fx()dx dx
()⇒ (^2) fx x c()=+ ⇒ fx()=x
2
4
When α = 0, area is minimum.
Required minimum area = 22
0
9
∫ ydy
= ⎛
⎝⎜
⎞
⎠⎟
4 =
32
72
32
0
9
y/
/
sq. units.
- (8) : x^2 + y^2 ≤ 144 and sin(x + y) ≥ 0
⇒ 2 nπ ≤ x + y ≤ (2n + 1)π ; n ∈I
Hence, we get the area
S=π⋅ ⇒ S=
π144
298- (1) : Area = n + 1
+ 1
–11
+ 1