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WorksheetsDifferentiation and AOD
Total questions: 50
Worksheet time: 32mins
. y = x3 – 2x2 + 5x + 8. The gradient of the curve at x = 2 is
9
8
15
25
The two natural numbers whose sum is 30 and product is a minimum are,
10, 20
12, 18
15, 15
14, 16
The length of seg AB is 12 cm. The point P on seg AB such that AP2 + BP2 is minimum is,
AP = 5, BP = 7
AP = 6, BP = 6
AP = 8, BP = 4
AP = 9, BP = 3
Consider the function
P (x) = 2 + 3x2 + 5x4 + x6. Then P (x) has
Neither a maximum nor a minimum
only one maximum
Only one minimum
Only one maximum and one minimum
A man is walking at the rate of 6.5km/hr.toward the foot of a tower 120m. high at what rate is he approaching the top of the tower when he is 50m. away from the tower ?
2.5 km/hr.
0.25km/hr.
2.5m/sec.
none
The volume of a cube is increasing at a rate of 7cm3/sec. How fast is the surface area increasing when the length of an edge is 12cm.?
7/3 cm2/sec.
3/7 cm2/sec.
7/3 cm/sec.
7/3 cm3/sec.
Approximate value of 6x3-7x2+x-3 when x =2.01 is
20
19.50
19.45
19.54
The function f (x) = - x2 – 2x + 15 is increasing in the interval
(- ∞, - 1)
(- 1, ∞)
(- 1, 1)
none
Local maximum value of the function logx/x is
1
e
1/e
none
Approximate value of log10(103)
2.00121
2.001302
2.00421
2.00142
The approximate value of tan(29030’)
0.6005
0.6007
0.4005
none
Equation of normal to the curve
2x2 – y2=14 which is parallel to x +3y = 4 is
x +3y=7
x+3y-9=0
2x +3y=4
none
The tangent to the parabola x2 = 2y at the point (1, 1/2) makes with the x –axis an angle of
00
450
600
900
A stone thrown vertically upwards satisfies the equation x = 80t – 16t2. The time required to reach the maximum height is
2
3
4
2.4
A manufacturer can sell x items at a price of Rs. (330 – x) each. The cost of production of x items is Rs. (x2 + 10x + 12). For maximum profit the number of items to be sold is
20
40
60
80
Let f be the function with derivative given by f'(x) = x2 - 2/x. On which of the following intervals is f decreasing?
(-∞, 0]
(-1, 0]
(0, ∛2]
(∛2, ∞)
A solid sphere has radius r cm, surface area A cm2 and volume V cm3 . The radius is increasing at a rate of 5π1 cm s−1 . Find the rate of increase of the surface area when r=3.
2.4 cm s−1
24 cm s−1
4.8 cm s−1
48 cm s−1
y=5x2e3x
Find dxdy for y=tan4(2x+1) .
4(tan(2x+1))4⋅sec2(2x+1)(2)
(tan(2x+1))3⋅sec2(2x+1)(2)
4(tan(2x+1))3⋅sec2(2x+1)
4(tan(2x+1))3⋅sec2(2x+1)(2)
If 2x+2y=2x+y, then at x=y=1 dy/dx
1
2
-1
0
If log(x+y) = 2xy, then y’ (0)=
1
-1
2
0
If y=1+x.ey, then
ye4
(2+y)ey
ey/(2−y)
Ifx=cosθ , y=sinθ, then dxdy=?
cotθ
−cotθ
tanθ
−tanθ
x=a(cost+logtan2t), y=asint,then dxdy=
tan t
-tan t
cot t
-cot t
If y = f (x3), z = g (x5), f’ (x) = tan x and
g’ (x) = sec x, then the value of dy/dz is
(5x2)3 sec〖x5〗⋅tan〖x3〗
3(5x2) tan〖x3〗⋅sec〖x5〗
5sec〖x5〗(3x2)⋅tan〖x3〗
none of these
y = ax + xa + xx , find dy/dx
ax log a + axa-1 – x.xx-1
xax-1 + axa-1 + xx.log x
ax log a + a.xa-1 + xx (1 + log x)
ax + a.xa-1 + xx (1 + log x)
If cos (x + y) = y sin x, then find dy/dx
−((sinx+sin(x+y))(sin(x+y)+ycosx)
((sinx+sin(x+y))(sin(x+y)+ycosx)
(sinx−sin(x+y))(ycosx−sin(x+y))
none of these
If x3=(x+y)n.y2 and dx dy=xy then n=
1
2
3
5
If y=(x−c)(x−d)(x−a)(x−b) dxdy=
2y (x−a1+x−b1−x−c1−x−d1)
1y (x−a1+x−b1−x−c1−x−d1)
21 (x−a1+x−b1−x−c1−x−d1)
none of these
Ify=logx+logx+logx+.......∞ dy/dx=
2y−1x
2y+1x
x(2y−1)1
x(2y+1)1
Ifx32+y32=a32 then dxdy=
(xy)3 1
−(xy)3 1
(yx)3 1
−(yx)3 1
x2+y2 =t−t 1 , x4 +y4 =t2 +t21, find x3y dxdy=
0
1
-1
-3
Verify Rolle’s theorem for each of the following functions on the indicated intervals :f(x) = x(x−2)2 in [0,2].Find c
2
1
1.5
0
Iff(x)=exsinx in [0,π],then c(degree) in Rolle's theorem is
30
45
135
90
Given an interval[a, b] that satisfies hypothesis of Rolle's theorem for the function f (x) = x^3 − 2x^2 + 3. It is known that a = 0. Find the value of b
2
1
0
3
If f(x)=x+4 Verify Lagrange's mean value theorem for the function on [0,5].
Write the answer as yes or no
(a)
A Rectangular sheet of paper has it area 24 sq. meters. The margin at the top and the bottom are 75 cm each and at the sides 50 cm each. What are the dimensions of the paper, if the area of the printed space is maximum ?
Hint answer as (2,3)
(a)
An open box is to be cut out of piece of square card coard of side 18 cm by cutting of equal squares from the corners and turning up the sides. Find the maximum volume of the box.
(a)
The normal to the curve x^2 + 2xy − 3y^2 = 0 at (1, 1)
Meets the curve again in second quadrant
Does not meet the curve again
Meets the curve again in third quadrant
Meets the curve again in fourth quadrant
x+11
−(x+1)21
−x+11
(1+x)2
Ify=sin−1 ((a2+b2)(asinx+bcosx) ) dxdy=
1
0
a2+b2
a2 −b2
If y = a sin ( logx) + bcos( logx), then
……
x2 dx2(d2y)−x dxdy−y=0
x2 dx2(d2y)−x dxdy+y=0
x2 dx2(d2y)+x dxdy−y=0
x2 dx2(d2y)+x dxdy+y=0
Ify=(tan−1x)2,then (1+x2)2y22+2xy1(1+x2)=…
1
0
2
4
[Ify=cos21(cos−1x)],then
4y
y4
4y
4y1
dxd (x2+x+1)(x4+x2+1)=ax+b then (a,b)
2,-1
2,1
1,2
-3,1
The derivative ofsec−1((2x2−1)1)w.r.t. √(1−x2) at x=½ is
2
4
1
-2
If y = tan-1( secx – tanx) , then dy/dx =
1/2
1
-1/2
-1
If f(x) = logx (log x), then f '(x) at x = e is
e
1/e
1
none
The function x5 – 5x4 – 10 has a maximum when x =
3
2
4
0
A square plate is contracting at the uniform rate at 2 cm2/sec. The rate at which the perimeter is decreasing when the side of the square is 16 cm long, is
1/2
1/4
1
none of these
