WorksheetsLevel 11 MCQ
Total questions: 40
Worksheet time: 10hrs 0mins
sin−1(sin4)=?
π−4
4
4−π
4−2π
cos−1(cos8)
8−2π
8
8−π
2π−8
If the equation's solution for tan−12x+tan−13x=4π is ba , find the value of a+b.
(a)
If sin−1(sin x)=π−x , then x belongs to
[2π,23π]
[0,2π]
[23π,2π]
[−2π,2π]
Calculate sin−1(51)+cot−1(3)
4π
3π
1050
6π
Given that ∫26 k(x)dx = −5, find the value of p if ∫26 [2p+3k(x)]dx=33 .
(a)
∫01 (x2+1)2 xdx =a1 ,
find the value of a
(a)
Find the value for the definite integral ∫−42 17+4x dx = ba . What is the value of a+b ?
a+b为多少 (整数)?
(a)
Given that ∫−13f(x)dx=4 and ∫−13 g(x) dx=9 , then find the value of [∫−13 5f(x) dx +∫3−1 g(x) dx]
11
29
20
15
A curve has the gradient function of (5x−3)210 . If the curve passes through the point (1,2) , find the equation of the curve. 某曲线的斜率为 (5x−3)2 10 和经过点 (1,2) . 求该曲线方程式。
y=−(5x−3)2+3
y=(5x+3)2+3
y=−(5x−3)10+c
y=(5x−3)2+c
∫01 x(2x2−1)10 dx
221
111
−111
441
Find the value of
∫−2π0 sinxcosx dx
−21
21
−41
41
Find the value
∫02π cos2x sinx dx
31
21
41
51
Find the area of the region bounded by the given curves. y=16−x2 and y=x2−16
3512unit2
3352unit2
3252 unit2
3 152unit2
A particle moves along a straight line and passes through a fixed point O with a velocity of −12ms−1 . Its acceleration, a ms−2 , t seconds after passing O is given by a=4t−2 . Find the total distance travelled by the particle in the first 5 seconds.
3157m
3 127m
386m
65 m
A particle moves along a straight line and passes through a fixed point O such that its velocity, v ms−1 . t seconds after passing through O is given by v=8−2t . Find the maximum displacement, s of the particle.
16m
12m
20m
24m
A particle moves along a straight line and passes through a fixed point O. The velocity, v ms−1 of the particle is given by v=3t2−2t−21 , where t is the time in seconds, after passing through O. Find the total distance travelled by the particle is the first 5 seconds.
85 m
5 m
40 m
45 m
Given that a=∫02 x2dx , b=∫02 x3 dx , c=∫02sin x dx , which of the following is true?
c<a<b
b<c<a
a<b<c
a<c<b
∫023 9−4x2 dx 可以看成为
半径为 23 的圆的面积的二分之一
半径为 23 的圆的面积的八分之一
半径为 23 的圆的面积的三分之一
半径为 23 的圆的面积的四分之一
The diagram shows part of the curves undefined andundefined which intersect at P. Find the area undefined of the shaded region. (A是整数)
(a)
The diagram shows the curve y=x2 , the straight lines y=16 and x=1 . The volume generated when the shaded region is revolved through 360° about the x-axis is P52π unit3 . What is the value of P? (键入P的整数值)。
(a)
32x23+136x613+c
23x32+613x136+c
34x43+512x125+c
43x34+125x512+c
The diagram show part of the curve y=x28 and the straight line y=x which intersect at p . Find the area, A unit2 of the shaded region.
314 unit2
4 unit2
411 unit2
213 unit2
The diagram shows part of the curve y2=x−3 and the straight line y=2 . Find the volume generated when the shaded region is revolved through 360° about the y-axis.
5202π unit3
5192π unit3
5182π unit3
5172π unit3
Find the area of the shaded region
31
41
21
61
sin (3π−sin−1(−21)) is equal to
1
31
41
21
Find the value of sin−1 (cos 533π)
−10π
57π
−4π
52π
Evaluate sin 21cos−1 54
1010
53
1
−53
Find the value of cos2 21(cos−1 53)
54
−41
−21
31
x→∞lim (x+5x+6)x=
e
e56
e−56
e65
x→∞lim (xx−3)2x=
e−23
e23
e32
e−32
Given y=cosx+cosx+cos x+...∞ , find the dxdy
1−2ysin x
3y+22 sin x
2y+1sin x
sin x2y
Given y=tan x+tan x +tan x+...∞ , find the dxdy
2y−1sec2x
sec2xy
sec2xy+1
y+2sec2x
Given that y3−3xy2=x3+3x2y ,find the value of dxdy at point (1,1)
−34
−21
−32
−41
Differentiate of y=sin−1 3x4
−x2 9x2−164x2
x29x2−16x2
−x 9x2−16x2
x216−9x22x2
Derivative of y=cos−1 (73x+2)
−−9x2−12x+453
−9x2−12x+453
−9x2+12x−453
9x2+12x−453
Derivative of y=(sin−1x)2
1−x22 sin−1x
1+x22 sin−1x
1−x22 sin−1x
1+x22 sin−1x
Derivative of y=(cos−12x)2
−1−4x24cos−12x
1−4x24 cos−12x
1−4x2cos−12x
−1−4x2cos−12x
dxd16−x2⋅ sin−1(4x)
−16−x2xsin(4x)+1
16−x2xsin−1(4x)+1
−16−x2x sin−1(4x)−1
16−x2x sin−1(4x)−1
Given x→∞lim (1−x3)2x=ek1 , what is the value of k?
(a)
