(1)
Calculus Exam 2017 model
page(1)
Model Answer
Answer the following questions 20 questions
Q(1) If F(X ) =
c e
X , then F /// (3 ) equals……………..…… X−2
d f
-36 6
-12 4
/ // /// F (X ) = X − 2 − 2X = −2(X − 2)−2 ∴ F (X ) = 4(X − 2)− 3 ∴ F (X ) = −12(X − 2)− 4 = −12 (X − 2 )
Q(2) If
c e
(
)
2
F(X ) = ln X2 + 1 + e sin X
then F / (0 ) = .. ……
d f
1+e e
1 0
/ sin X / sin 0 F (X ) = 42 X + cos Xe ∴ F (0 ) = 42× 0 + cos 0 × e =1 X +1 0 +1
Q(3) If F(X ) = X X then F / (1) = ………….…….
c e
e -1
Y = X
X
∴ lnY = XlnX
/
∴ Y = ln X + 1 Y
d f
1
/
X
0
∴ Y = X lnX + X
X
= 1
Q(4) A body moves on the curve Y 2 = X 3 If dX = 1 unit / sec when Y=-1 then dY at this moment equals…….. dt
c e 2Y
− 3 4 3 4 dY dX = 3X 2 dt dt
dt
2
unit/s
d f Y = -1∴ X = 1 ∴ 2(- 1) ×
− 3 8 3 2
dY dY 1 3 = 3(1)2 × ∴ =− dt dt 2 4
Calculus Exam 2017 model
(1)
(
page(2)
)
Q(5) The slope of tangent to the curve Y = ln 2 − 2 cos X at X = π equals ……
c e
d f
1 -1 2 sin X
/
Y =
2−
2 cos X
Q(6) If Y =
c
3 2
e
0
atX = π 4
e
-2
2−
o
2 cos 45
o
= 1
+ ln( X + 1) then dY =…….. when X=0 dX
2X
(
2
2 sin 45
/
∴Y =
4
)
d
1 2
f
1
(
1
)
1
(
)
dY = 1 e2 X + ln (X + 1) − 2 × ⎛ 2e2 X + 1 ⎞ at X = 0 dY = 1 e 0 + ln 1 − 2 × 2e 0 + 1 = 3 ⎜ ⎟ 2 X + 1⎠ 2 2 dX dX ⎝
Q(7)
lim h→ 0
)
(
cos π + h − 1 3 2 h
c 21 e
d
− 1 2
f
3 2 −
3 2
cos⎛⎜ π + h ⎞⎟ − 1 cos⎛⎜ π + h ⎞⎟ − cos π 3 3 ⎝3 ⎠ 2 ⎠ ⎝3 = lim = d (cos X ) = − sin X = − sin 60 o = − lim 2 h h dX h→ 0 h→ 0
Q(8)
lim ⎛⎜⎝ 1 − X→∞
1 ⎞⎟ X⎠
2X
=
c
e2
d
1
e
e −2
f
0
let - 1 = Y ∴ X = - 1 X Y
2 1 ∴ lim (1 + Y )− Y = lim ⎡(1 + Y ) Y ⎤ ⎥⎦ ⎣ Y →0 Y →0 ⎢
−2
=e
−2
(1) 2 Q(9) If F( X ) = X then ∫− 2 f(X )dX = ..... Calculus Exam 2017 model
c e
d f
-1 0
2 XdX 0
2∫
2
2
2 − X
4
∫
2e
1 dX = ……. X
e
c
1
d
e
e
0
f
ln2
d f
2 4
2e
e
1 dX = [ln X]2 e = ln 2e − ln e = ln 2 e X
∫
Q(11)
c e
π 2
sin X dX =
−π 2
0 1 π
π
A = 2∫ 2 sin X dX = 2[− cos X ]02 = 2 ⎡⎛⎜ − cos π ⎞⎟ − (− cos 0 )⎤ = 2 − −1 = 2 0 ⎢⎣⎝ ⎥⎦ 2⎠ e2
Q(12)
∫1
2
ln X dX = X
c
7 3e 2
d
7 3
e
8 3
f
4 e2
e
∫1
2
e
2
(
2 3 ln X dX = ⎡ (ln X ) ⎤ = ln e ⎢ ⎥ 3 X ⎢⎣ 3 ⎥⎦1 2
)
3
−
(ln 1)3 3
=
(2 ln e )3 3
= 8 3
X
1
−2 −1 −1
2
⎡ ⎤ = 2⎢ X ⎥ = 2 × ⎡ 4 − 0 ⎤ = 4 ⎢⎣ 2 ⎦⎥ ⎣ 2 ⎦0
Q(10)
∫
page(3)
1
2
Calculus Exam 2017 model
(2)
page(1)
Answer the following questions 20 questions
Model Answer
Q(1) If Y = sin 2X cos 2X , then Y // ⎛⎜ π ⎞⎟ equals :……….. ⎝3⎠
c e
d f
-4 4 3
Y = 1 sin 4 X 2
/
Y = 2 cos 4 X
Q(2) If F(X ) = Xe −2 X
c e
∴Y
//
8
= −8 sin 4 X
- 8sin4 × 60 = 4 3
then F / ⎛⎜ 1 ⎞⎟ =……….. ⎝2⎠
d f
0 e2
F / (X ) = e − 2 X − 2 Xe − 2 X
0
e e-1
∴ F / ⎛⎜ 1 ⎞⎟ = e − 1 − e − 1 = 0 ⎝2⎠
3 Q(3) If Y = X 2 ln X then d Y3 =……….. at X=4
a
c e
dX
1 4
dY = 2 X ln X + ⎛ a × 1 ⎞ X2 = 2 X ln X + X ⎜ ⎟ a ⎝ X a⎠ a dX 2 d Y = 2⎛ ln X + a × 1 X ⎞ + 1 = 2⎛ ln X + 1⎞ ⎜ ⎟ ⎜ ⎟ X a ⎠ ⎝ a ⎝ a ⎠ dX2
d f
2 1 2
3 ∴ d Y3 = 2⎛⎜ a × 1 ⎞⎟ = 2 = 2 = 1 X 4 2 ⎝ X a⎠ dX
Q(4) The sides of a right triangle with legs X and Y and hypotenuse Z dZ = 1 and dX = 3 dY at the instant when X=4 increase in such a way that dt dt dt and Y=3 what is dX dt
c e
1 2
d f
5 1 3
∴ 2X dX + 2Y dY = 2Z dZ ∴ X dX + Y dY = Z dZ dt dt dt dt dt dt dX dX dX dX 1 4 + 3× = 5×1 ∴5 = 5 ∴ =1 3 dt dt dt dt X2 + Y 2 = Z 2
Calculus Exam 2017 model
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page(2)
Q(5) Slope of tangent of Y = 1 + 2 csc X + cot X At ⎜⎛ π ,4 ⎞⎟ ⎝4 ⎠ c -2 + 2 d -4
e
f
2
-1
dY = − 2 csc X cot X − csc 2 X at X = π = - 2 × 2 × 1 − dX 4
( 2)
2
= −2 − 2 = −4
Q(6) If The line Y=X is tangent to the curve Y = X + K then K=…… 2
c e
c e
2 1 4
2
2X = 1 ∴ X = 1 ∴ Y = 1 ∴ 1 = ⎛⎜ 1 ⎞⎟ + K ∴ K = 1 2 2 2 ⎝2⎠ 4
Q(7)
lim h→0
)
(
2 1 4
( )
tan 2 π + h − tan π 8 4 = h
c
2
d
4
e
3 2
f
2 3
/ 2 = F (tan 2X ) = 2 sec 2X at X = π 8
Q(8) If X = t 3 − t
(
and Y = 3t + 1 then dY at t=1 is dX
c
1 8
d
8 3
e
3 8
f
3 4
dX = 3t 2 − 1 dt
)
/ 2 o ∴ F ⎛⎜ π ⎞⎟ = 2 sec 45 = 4 ⎝8 ⎠
dY = 3 dt 2 3t + 1
3 1 ∴ dY = × = 3×1 = 3 2 4 dX 2 3t + 1 3t − 1 2 2
Calculus Exam 2017 model
(2)
page(3)
Q(9) If the function f is continuous and even on R and 2
g(X )dX = 2 then
and
∫4
c e
-18
4
∫2 [2F(X) − 3g(X) − 5]dX
d f
10
4
∫ g(X)dX = −2 2
4
∫2 f(X)dX
= 7
= …..
-8 14
∴ 2 × 7 − 3 × −2 − [5 X]24 = 14 + 6 − (20 − 10 ) = 10
( )
Q(10) If F / (X ) = 2 and F e = 5 then F( e ) = …… X
c
5
e
6
d
ln 25
f
ln 5
1 2
F(X ) = 2 ln X + C ∴ 5 = 2 ln e + c ∴ 5 = 1 + C ∴ c = 4 ∴ F(X ) = 2 ln X + 4 ∴ F(e ) = 2 ln e + 4 = 6
Q(11)
c e =
∫ [X 2
2
0
] dX =…..
− X−1
0
d
7 3
5 3
f
1
∫ [X 1
0
2
]
− (1 − X ) dX + ∫
Q(12) If
2
1
[X
2
]
− (X − 1) dX = 5 3
n −1 ∫0 n(X + 1) dX = 15 then the value of n=….. 3
c1
d
2
e4
f
3
[(X + 1) ]
n 3 0
= 15
∴ 4 n − 1n = 15
∴ 4 n = 16
∴ n = 2
Calculus Exam 2017 model
Q(16) find
∫
3X + 5 dX e2X
(2)
−2 X −2 X = − 1 e (3X + 5 ) + ∫ 3 e 2 2 −2 X −2 X 3 1 = − e (3X + 5 ) − e 2 4
page(5)
D I −2X 3X + 5 e 3 − 1 e −2 X 2
Q(17) Using one of the integration techniques to find
∫
2X + 1 dX = e5X
D
I
∫ (2Xe
−5 X
+ e -5X ) dX =
∫
2X + 1 dX e5X
∫ (2Xe ) dX + ∫ (e ) dX −5 X
−5 X
2 2 −5 X 2 2 e −5 X ∴ − Xe −5 X + e dx = − Xe − 5 X − e − 5 X 5 5 5 25 1 − e −5 X 5 2X + 3 2 2 −5 X 1 −5 X dX = − Xe − 5 X − e − e 5X e 5 25 5 10 2 −5 X 5 −5 X 1 −5 X ( − Xe − 5 X − e − e =− e 10 X + 7 ) 25 25 25 25
2X 2
∫
∫
Q(18) If you know that F(0 ) = 5 Use the opposite figure which represent The graph of F / (X ) to find F(6)
6 6 / ∫0 F (X )dX = [F(X )]0 F(6 ) − F(0 ) = 1 × 6 × 5 = 15 2 ( ) ∴ F 6 = 15 − 5 = 10
Calculus Exam 2017 model
(2)
page(6)
Q(19) If f (x) = 4 + Cot X - Sec2 X , find the equation of the normal to the curve of the function f at a point lying on the curve and its x-coordinate equals π 4
When X= π Y = 4 + cot 45 o − sec2 45 o = 4 + 1 − 2 = 3 4 / F (X ) = − csc2 X − 2 sec2 X tan X at X = 45 o / 2 o 2 o o F ⎛⎜ π ⎞⎟ = − csc 45 − 2 sec 45 tan 45 ⎝4⎠ = − 2 − 2 × 2 = −6 Equation of normal Y − 3 = 1 6 X− π 4
Q(20) A rectangle of perimeter 30cm is revolved about one of its side to form a cylinder what is the maximum possible volume that could be generated X
∴ X + Y = 15 ∴ Y = 15 - X Y 2πr = X ∴r = X 2π 2 2 2 2 3 V = πr × Y = π⎛⎜ X ⎞⎟ × (15 − X ) = 1 X (15 − X ) = 1 15 X − X 4π 4π ⎝ 2π ⎠ 2 2 dV = 1 30 X − 3X = 0 ∴ 30 X − 3X = 0 4π dX
r
2X + 2 Y = 30
(
(
)
Y
)
∴ Y = 15 - 10 = 5 ∴ 3X(10 - X ) = 0 ∴ X = 10 2 3 V = 1 15 × 10 − 10 = 500 4π 4π
(
)
If the tangent to the curve
Y = X passes through the point (-1,0) find
the equation of the tangent and the normal at the point of tangency
Slope of tangent
1 (X )− 21 2
Slope of tangent
X 1 1 = ∴ 2X = X + 1 ∴ X = 1 ∴ point 2 X X+1 Equation of the tangent Y − 1 = 1 X−1 2 ∴
X −0 X+1
of tangency is
(1,1)
(3)
Calculus Exam 2017 model
Q(5) lim h→ 0
c e
)
(
tan π + 2h − 3 h
3
page(2)
=………..
d f
-8 8
F (tan 2X ) = 2 sec 2X when X = π 6 2
/
4 -4
⎛ ⎞ 1 2⎜ ⎟ o ⎝ cos 60 ⎠
2
=8
Q(6) The ratio between the slopes of the two curves Y = ln 3 X + 1 and Y = ln 5 X + 1 when X=a is
c e
3:5 5:3
Y1 = ln 3 + 1 ln(X + 1) 2
e
∫
1:1
f
ln3:ln5
1:1
Y2 = ln 5 + 1 ln(X + 1) 2
∫ X ln1X dX = ….
Q(7)
c
,
d
3
3 ln X 1 ln X 3
d
3 ln ln X
f
1 ln ln X 3
1 1 dX = 1 1 dX = 1 X dX = 1 ln ln X 3 3 X ln X 3 ln X X ln X 3
∫
∫
Q(8) lim ⎛⎜ X + 2 ⎞⎟ X→∞ ⎝ X − 1 ⎠
c e
X+ 4
=……
e2
d
1
3
f
e
e
lim ⎛⎜⎝ XX +− 21 ⎞⎟⎠ X→∞
X+ 4
= lim ⎛⎜ X − 1 + 3 ⎞⎟ X→∞
= lim ⎛⎜ 1 + 3 ⎞⎟ X − 1⎠ X→∞ ⎝
X −1
⎝
X−1 ⎠
X+ 4
= 5
4
lim ⎛⎜⎝ 1 + X 3− 1 ⎞⎟⎠ X→∞
× lim ⎛⎜ 1 + 3 ⎞⎟ = e 3 × 1 = e 3 X − 1⎠ X→∞ ⎝
X − 1+ 5