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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

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(

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

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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

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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

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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)

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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