Sponsor Area
Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm (b) r = 4 cm
(a) When r = 3, rate of change of area = 2
× 3 = 6 cm2/cm.
(b) When r = 4, rate of change of area = 2 
× 4 = 8 
cm2/cm.
Find the rate of change of the volume of a ball with respect to its radius r. How fast is the volume changing with respect to the radius when the radius is 2 m?
(2)2 = 16 
m3/m.How fast is the volume of a ball changing with respect to its radius when the radius is 3 m?
A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is 10 cm.
Let V be volume of balloon of radius r
Rate of increase of volume w.r.t radius
when r = 10 cm, rate of increase of volume = 
The cost function C(x), in rupees, of producing x items (x ≥ 15) in a certain factory is given by 
Determine the marginal cost function and the marginal cost of producing 100 items.
Here, 
Marginal cost function 
When x = 100, marginal cost = 20 (100) - 
The total cost C(x) in Rupees, associated with the production of x units of an item is given by C(x) = 0.005 x3 - 0.02 x2 + 30 x + 5000.
Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.
Here C(x) = 0.005 x3 - 0.02 x2 + 30 x + 5000
Marginal cost = 
When x = 3, Marginal cost = 3 (0.005) (3)2 - 2 (0.02) (3) + 30
= 27(0.005) - 6 (0.02) + 30
=0.0135 - 0.12 + 30 = 30.015 = Rs. 30.02 nearly
Here 
Marginal cost 
When x = 17, marginal cost = (0.021) (17)2 - (0.006) (17) + 15
The total revenue in Rupees received from the sale of x units of a product is given by
R(x) = 13x2 + 26x + 15.
Find the marginal revenue when x = 7.
R(x) = 13x2 + 26x + 15 
Marginal revenue = 
When x = 7, marginal revenue = 26(7) + 26 = 182 + 26 = Rs. 208
A ballon which always remain spherical, has a variable diameter 
. Determine the rate of charge of its volume with respect to x.
Let r be the radius of the balloon.
From given condition, diameter =
Let V be the volume of spherical balloon.
The radius of a circle is increasing at 0.7 cm/sec. What is the rate of increase of its circumference when r = 4.9 cm?
Let C be the circumference of the circle whose radius is r at any time t.
The radius of a circle is increasing uniformly at the rate of 3 cm per second. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Let r cm be the radius of the circle.
From given condition, rate of increase = 3 cm per second
Let A be the area of circle, 
When r = 10, rate of increase of area = 
A stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm per second. At the instant when the radius of the circular waves is 10 cm, how fast is the enclosed area increasing?
Let r be the radius of the circular wave and A the enclosed area at time t.
From given condition, 
Sponsor Area
The radius of a balloon is increasing at the rate of 10 cm per second. At what rate is the surface area of the balloon increasing when its radius is 15 cm?
Let r be the radius of the balloon.
. ...(1)
Let S be the surface area of the balloon.
When r = 15, 
The radius of an air-bubble is increasing at the rate of 
At what rate is the volume of the bubble increasing when the radius is 1 cm?
An edge of a variable cube is increasing at the rate of 3 cm per second. How fast is the volume of the cube increasing when edge is 10 cm long?
(given)
Let S be the surface of the spherical bubble of radius r.
From given condition,
Let V be the volume of the spherical bubble.
Rate of increase of volume = 
The volume of a spherical balloon is increasing at the rate of 25 cm3/sec. Find the rate of change of its surface area at the instant when its radius is 5 cm.
Let r be the radius of spherical soap bubble
(i) Now, 
Let r be radius of spherical soap bubble.
(ii) Again, 
A ballon which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate of which the radius of the balloon is increasing when the radius is 15 cm.
Let r be the radius of the ballon and V be its volume at any time t. Then
But 
(given)
The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 10 centimeters?
Let V be volume of cube of side x
Let S be surface area of cube
Rate of increasing of surface area = 
When x = 10, rate of increase of surface area = 
The volume of a cube is increasing at a rate of 7 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 12 centimeters?
Tips: -
Since the rate of change is represented by derivative with respect to time, it is taken to be positive if the quantity is increasing and negative if the quantity is decreasing.The length x of a rectangle is decreasing at the rate of 2 cm/s and the width y is increasing at the rate of 2 cm/s. When x = 12 cm and y = 5 cm, find the rate of change of (a) the perimeter and (b) the area of the rectangle.
Since the length x is decreasing and the width y is increasing
...(1)
(a) The perimeter P of the rectangle is given by
(b) The area A of the rectangle is given by
The length x of a rectangle is decreasing at the rate of 3 cm/minute and the width y is increasing at the rate of 2 cm/minute. When x = 10 cm and y = 6 cm, find the rate of change of (a) the perimeter and (b) the area of rectangle
Since the length x is decreasing and the width y is increasing
...(1)
(a) The perimeter P of the rectangle is given by
(b) The area A of the rectangle is given by
A = xy
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minutc. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Since the length x is decreasing and the width y is increasing
...(1)
(a) The perimeter P of the rectangle is given by
(b) The area A of the rectangle is given by
The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?
so that 
be area of 
A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Here, 6y = x3 + 2. ...(1)
Differentiating both sides, w.r.t 't'
But 
(given)
At what points of the ellipse 16x2 + 9y2 = 400, does the ordinate decrease at the same rate at which the abscissa increases?
Here, 16x2 + 9y2 = 400 ...(1)
Differentiating both sides w.r.t. 't', we get,
But 
Putting 
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4m away from the wall?
Let the foot A of the ladder be at a distance x metres from the wall and y metres be the height of the wall at any time t.
Differentiating both sides w.r.t. 't', we get,
(given)
Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way what the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand-cone increasing when the height is 4 cm?
(given)
Let r be the radius, h be the height and a be semi-vertical angle of right circular cone. 
But 
(given)
Let V be the volume of the cone.
A man 2 metres high walks at a uniform speed of 6 metre/sec away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.
Let AB be the lamp-post and PQ the man, CP be his shadow at time t. Let AP = x, PC = y. Also AB = 6 m, PQ = 2 m.
Now ∆CAB and ∆CPQ are equiangular and hence similar.
Sponsor Area
A man of height 2 metres walks at a uniform speed of 5 km/h away from a lamp post which is 6 metres high. Find the rate at which the length of his shadow increases.
Let AB be the lamp-post and PQ the man, CP be his shadow at time t.
Let AP PC = y. Also AB = 6 m, PQ = 2 m. Now ∆CAB and ∆CPQ are equiangular and hence similar.
Let AB = a metres be the lamp-post and PQ = b metres the boy, CP = y be his shadow at time t. Let AP = x.
Now ∆CAB and ∆CPQ are equiangular and hence similar
Water is dripping out from a conical funnel, at the uniform rate of 2 cc/sec through a tiny hole at the vertex of the funnel. When the slant height of water is 5 cm, find the rate of decrease of the slant height of the water.
Let V be the volume of the water in the cone i.e. the volume of the water cone CA 'B' at any time t.
Let CO' = h, O' A' = r and CA' = I.
Let α be the semi-vertical angle of the cone. CAB where CO = 15 cm, OA = 5 cm
CO = 15 cm
OA = 5 cm
Then, 
Also, 
...(2)
From (1) and (2), we get,
...(3)
Now, 
(ii) Let A be the water surface area at any time t. Then, A = 
(iii) Let S be the wetted surface area of the vessel at any time t. Then. S = 
Now, 
An inverted cone has a depth of 10 cm and a base of radius 5 cm. Water is poured into it at the rate of 
Find the rate at which the level of water in the cone is rising when the depth is 4 cm.
...(1)
B.
Let A be area of circle of radius r
Rate of change of area w.r.t. r = 
When r = 6, rate of change of area = 
The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x 2 + 36x + 5.
96
90
126
D.
126
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic metre per hour. Then the depth of the wheat is increasing at the rate of
1 m3/h
A.
1 m3/h
Let h be the height of the cylindrical tank at any time t.
The equation of the curve is
y = x2 – 5x + 6
Let 
be slopes of tangents to the curve at the points (2, 0), (3, 0).
Show that the tangent to the curve y = 7x3 + 11 at the points where x = 2 and x = – 2 are parallel.
The equation of curve is
y = 7x3 + 11
At 
At 
Find the points on the curve y = x3 at which the slope of the tangent is equal to the y-coordinate of the point.
The equation of curve is 
From given condition,
...(1)
Also, 
From (1) and (2), 
From (1), 
The curve is x = at2, y = 2at
Find the slope of the tangent to the curve y = x3 – x at x = 2.
The equation of curve is
y = x3 – x 
Find the slope of the tangent to the curve y = 3x4 – 4x at (i) x = 1 (ii) x = 4.
The equation of curve is
(i) At x = 1, 
which is required slope of tangent to the curve.
(ii) 
which is required slope of tangent to the curve
Find the slope of the tangent to the curve 
.
The equation of curve is
which is required slope of tangent to the curve.
The equation of curve is
y = 3x2 + 4 x
At x = 2, 
which is required slope of tangent to the curve.
The equation of curve is
Find the slope of the tangent to the curve y = x3 – 3x + 2 at the point whose. x-coordinate is 3.
The equation of curve is
y = x3 – 3x + 2
Find the slope of the normal to the curve 
Find the slope of the normal at the point (am3 , am2 ) to the curve a x2 = y3.
The equation of the curve is a x2 = y3
Differentiating both sides w.r.t. x, we get,
Find a point on the curve y = (x – 2)2 at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).
The equation of curve is
Also slope of line joining (2, 0) and (4, 4) 
Find the point at which the tangent to the curve 
has its slope 
The equation of curve is
which is slope of tangent to the curve.
From given condition.
Find the equation of the normal to the curve y2 = 4x at the point (1, 2).
The equation of curve is
y2 = 4x
Differentiating both sides w.r.t. x, we get,
Find the equation of tangent to the ellipse 
The equation of ellipse is 
Differentiating both sides w.r.t.x,
Find the equation of tangent and normal to the hyperbola 
The equation of hyperbola is 
Differentiating both sides w.r.t. x,
The equation of the curve is a y2 = x3
Differentiating both sides w.r.t. x,
Find the equation of the tangent and the normal to the curve 
The equation of curve is
Differentiating both sides w.r.t.x, we get.
At (1, 1), slope of tangent 
Find the equation of the tangent and normal to the curve y2 = 4ax at the point (at2, 2at).
The equation of curve is y2 = 4ax 
The equations of curve are x = at2 , y = 2at
Find the equation of tangent to the curve given by x = a sin3 t, y = b cos3 t at a point where 
Solution not provided.
Find the equation of tangent to the curve:
The equations of curve are 
which is slope of tangent
or 
Find the equation of the tangent and normal to the curve
The equations of the curve are 
Also, 
or 
Also slope of normal = 
The equation of curve is 
...(1)
Since tangent is parallel to x-axis.
Sponsor Area
Find the equations of the tangent and normal to the curve y = x2 + 4x + 1 at the point whose abscissa is 3.
The equation of curve is y = x2 + 4x + 1 ...(1)
When x = 3, y = 9 + 12 + 1 = 22
point with abscissa 3 is (3, 22)
Diff. (1) w.r.t.x, 
At (3, 22), 
Show that the normal at any point 0 to the curve
x = a cosө + a ө sinө. y = a sin ө – a ө cos ө
is at a constant distance from the origin.
The equation of the curve are 
Find the equation of the tangent and normal to the given curves at the points given:
y = x2 at (0,0).
The equations of curve is y = x2 
Find the equation of the tangent and normal to the given curves at the points given:
y2 = 4ax at (0, 0).
The equation of curve is y2 = 4ax
Differentiating both sides w.r.t.x, 
At (0, 0) 
does not exist
Find the equation of the tangent and normal to the given curves at the points given:
y = x3 at (1, 1)
The equation of curve is y = x3
At (1, 1), 
Find the equation of the tangent and normal to the given curves at the points given:
y = x2 at (2, 8)
The equation of curve is y = x2
At (2, 8), 
which is the slope of tangent
The equation of normal at (2, 8) is
Find the equation of tangent and normal to the hyperbola 
The equation of hyperbola is 
Differentiating both sides w.r.t.x,
At 
Find the equation of the tangent and the normal to the curve 
The equations of curve is
Differentiating both sides w.r.t. x, we get, 
The equation of curve is 
Find the equation of the tangent and normal to the curve
The equations of the curve are 
Also 
At 
The equations of curve is y = x4 – 6x3 + 13x2 – 10x + 5
At (0, 5). 
The equation of curve is
y = x4 – 6x3 + 13x2 – 10x + 5
The equation of the curve is 
Differentiating both sides w.r.t x,
The equation of curve is 
The equation of curve is
Differentiating both sides w.r.t x,
The equation of curve is 
...(1)
Now, 
lies on it
Put 
or 
Slope of normal = 
the equation of normal at 
The tangent 
cuts x-axis where y = 0, So putting y = 0, we get,
The normal 
cut x-axis where y = 0. So putting y = 0, we get.
The equation of curve is 
Differentiating both sides w.r.t. x, we get,.
The equation of given curve is
It meets x-axis where y = 0
Putting y = 0, we get,
At (1, 0), slope of tangent = 4 (1)3 – 6 (1)2 –1 = 4 – 6 – 1 = – 3
At (2, 0), slope of tangent = 4 (2)3 – 6 (2)2 – 1 = 4 × 8 – 6 × 4 – 1 = 7
∴ equation of tangent at (1, 0) with slope – 3 is
y – 0 = – 3(x – 1) or y = – 3x + 3 or 3x + y – 3 = 0
The equation of tangent at (2, 0) with slope 7 is
y – 0 = 7(x – 2), or y = 7x – 14 or 7x – y - 14 = 0
at the point where it cuts the x-axis.
The equation of curve is
...(1)
It meets x-axis where y = 0
Putting y = 0 in (1), we get
Prove that the line 
is a tangent to the curve 
at the point where the curve cuts y-axis.
The equation of given curve is 
...(1)
It cuts y-axis where x = 0
Differentiating (1), w.r.t. x, we get,
The equation of tangent (0, b) with 
The equation of the curve is y = x3 – 11 x + 5 ...(1)
Find the point on the curve y = x3 – 2x2 – 2x at which the tangent lines are parallel to the line y = 2x – 3.
The equation of curve is y = x3 – 2x2 – 2x 
Find the points on the curve y = x3 – 2x2 – x at which the tangent lines are parallel to the line y = 3x – 2.
The equation of curve is 
...(1)
Find the equation of the tangent to the curve x2 + 3y = 3, which is parallel to the line y – 4x + 5 = 0.
The equation of curve is x2 + 3y = 3 ...(1)
Differentiating w.r.t.x, we get,
Find the equations of all lines having slope -1 that are tangents to the curve 
The equations of the curve is 
∴ there are two tangents to the given curve with slope – 1 and passing through the points (0, – 1) and (2, 1).
The equation of tangent through (0, – 1) is
y – (– 1) = – 1 (x – 0) or y + 1 = – x or x + y + 1 = 0
The equation of tangent through (2, 1) is
y – 1 = – 1 (x – 2) or y – 1 = – x + 2 or x + y – 3 = 0
Find the equation of all lines having slope 2 and being tangent to the curve 
.
The equation of curve is
or 
there are two tangents to the given curve with slope 2 and passing through the points (2, 2) and (4, -2).
The equation of line through (2, 2) is
y – 2 = 2 (x – 2) or y – 2 = 2x – 4 or 2 x – 2 = 0
The equation of line through (4, – 2) is
y – (– 2) = 2 (x – 4) or y + 2 = 2 x – 8 or 2 x – y – 10 = 0
Find the equation of all lines having 0 slope and that are tangent to the curve 
The equation of the curve is 
Find the equation of the tangent to the curve 
which is parallel to the line 4x - 2y + 5 = 0
The equation of curve is 
...(1)
Find the equation of tangents to the curve
y = cos (x + y), – 2 
≤ x ≤ 2 
that are parallel to the line x + 2y = 0.
The equation of curve is
y = cos (x + y) ...(1)
Differentiating both sides w.r.t.x, we get,
or 
The equation of tangent at 
is
or 
The equation of tangent at 
is
or 
Find the point on curve 4x2 + 9y2 = 1, where the tangents are perpendicular to the line 2y + x = 0.
The equation of curve is 4x2 + 9y2 = 1 ...(1)
Diff. w.r.t. x, 
Find the equation of the tangent line to the curve y = x2 – 2x + 7 which is
(a) parallel to the line 2x – y + 9 = 0
(ii) perpendicular to the line 5y – 15y = 13
The equation of curve is y = x2 – 2x + 7 ...(1)
(i) consider the line 
...(2)
Its slope 
Since tangent to the curve is parallel to line (2)
(ii) Consider the line 
...(3)
Its slope 
Since target to the curve is perpendicular to line (3)
The equations of curve is 
...(1)
Differentiating w.r.t.x, 
The equation of curve is y = x3 + 2x + 6 ...(1)
Let 
be slope of line 
...(2)
At what point will the tangent to the curve y = 2x3 – 15x2 + 36x – 21 be parallel to x-axis ? Also, find the equations of tangents to the curve at those points.
The equation of curve is
y = 2x3 – 15x2 + 36x – 21
Differentiating both sides w.r.t.x, we get,
For the points on the curve, where tangents are parallel to x-axis
Find the points on the curve x2 + y2 -2x – 3 = 0 at which the tangents are parallel to the x-axis.
The equation of curve is x2 + y2 -2x – 3 = 0 ...(1)
Differentiating both sides, w.r.t.x, we get,
The equation of curve is
For the points on the curve, where tangents are parallel to x-axis
When 
When 
Find point on the curve y = 3x2 – 12x + 6 at which tangent is parallel to x-axis. Also find the equation of tangent line.
The equation of curve is
y = 3x2 – 12x + 6 ...(1)
Find out on the curve y = x2 – x – 8 at which tangent is parallel to x-axis. Also find the equation of tangent line.
The equation of curve is
y = x2 – x – 8 ...(1)
Find out on the curve y = 2 x2 – 6x – 4 at which tangent is parallel to x-axis. Also find the equation of tangent line.
The equation of curve is
y = 2x2 – 6x – 4 ...(1)
Sponsor Area
Find out the points on the curve 
at which the tangents are (i) parallel to the x-axis and (ii) parallel to the y-axis.
The equation of curve is 
Differentiating both sides w.r.t.x, we get,
(i) For the points on the curve, where tangents are parallel to x-axis.
Putting x = 0 in (1), we get,
there is no real point at which the tangent is parallel to x-axis.
(ii) For the points on the curve, where tangents are parallel to y-axis.
Putting y = 0 in (1), we get,
Find points on the curve 
at which the tangents are (i) parallel to the x-axis (ii) parallel to the y-axis.
The equation of curve is 
...(1)
Differentiating both sides w.r.t. x, we get,
(i) For the points on the curve, where tangent are parallel to x-axis,
Putting x = 0 in (1), we get,
(ii) For the points on the curve, where tangents are parallel to y-axis.
Putting y = 0 in (1), we get,
Find points on the curve 
at which the tangents are (i) parallel to the x-axis (ii) parallel to the y-axis.
The equation of curve is
...(1)
Differentiating both sides w.r.t.x, we get,
(i) For the points on the curve, where tangents are parallel to x-axis
Putting x = 0 in (1), we get,
(ii) For the points on the curve, where tangents are parallel to y-axis.
Putting y = 0 in (1), we get
For the curve y = 4x3 – 2x5, find all the points at which the tangent passes through the origin.
The equation of curve is
y = 4x3 – 2x5 ...(1)
Let the tangent to the curve (1) at (h, k) pass through origin (0, 0)
...(2)
Differentiating (1) w.r.t.x, 
The equation of tangent of (h, k) is
or 
Now this tangent passes through (0, 0).
or 
or 
.
A.
The equation of curve is 
...(1)
Differentiating both sides w.r.t.x, we get,
When x =4, from (1), 
The equation of curve is
y = ax3 + bx2 + cx + 5 ...(1)
It meets y-axis at Q where x = 0
putting x = 0 in (1), we get,
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...(2)
Also (-2, 0) lies on (1)
or 
Subtracting (3) from (2), 4 a +2 = 0 
From (2), -6 - 4b + 3 = 0 
Show that the curves 2x = y2 and 2xy = k cut at right angles if k2 = 8
The equation of two curves are
2x = y2 ...(1)
and 
...(2)
From (1) and (2), 
point of intersection of curves (1) and (2) is 
From (1), 
From (2), 
At 
Curves (1) and (2) cut at right angles if 
i.e., if 
i.e. if 
Hence the result.
The equation of two curves are
x = y2 ...(1)
and xy = k ...(2)
From (1) and (2), 
From (2), 
At 
Curves (1) and (2) cut at right angles if 
i.e., 
Hence the result.
Show that the curve xy = a2 and x2 + y2 = 2a2 touch each other.
The given curves are
...(1)
x2 + y2 = 2a2 ...(2)
Now 
...(3)
Also 
=0
...(4)
Adding (3) and (4), we get,
If the curve αx2 + βy2 = 1 and α' x2 + β'y2 = 1 intersect orthogonally, prove that (α – α') β β') = (β – β') α α'.
The equations of two curves are
...(1)
and 
...(2)
Let curves (1) and (2) intersect at 
...(3)
and 
...(4)
Subtracting (4) from (3), we get,
At 
Similarly for second curve,
Since the two curves (1) and (2) intersect orthogonally,
B.
The equation of curve is
...(1)
The equation of line is
...(2)
Differentiation (1), w.r.t.x, we get,
From (2), slope of tangent = m
This point 
The slope of the normal to the curve y = 2 x2 + 3 sin x at x = 0 is
3
-3
D.
The equation of curve is
y = 2x2 + 3 sin x
The line y = x + 1 is a tangent to the curve y 2 = 4x at the point
(1, 2)
(2, 3)
(1, -2)
(-1, 2)
A.
(1, 2)
The equation of curve is 
Consider the tangent line y =x+1
Its slope = 1
From given condition,
When y = 2, 4 = 4x 
x = 1
B.
x – y = 0The equation of curve is
A.
x + y = 3The equation of the curve is 
...(1)
Let normal at (h, k) pass through (1, 2).
Since (h, k) lies on (1)
...(2)
Slope of tangent at (h, k) = 
Equation of normal at (h, k) is 
or 
From (2) and (3), we get, 
Let x1 , x2 ∊ R and let x1 < x2
Now x1, < x2
⇒ 2x1 < 2x2
⇒ 2x1 + 3 < 2x2 + 3
⇒ f (x1) < f (x2)
⇒ f is an increasing function on R.
Let x1, x2 ∊ R and let x1< x2
Now x1 < x2 ⇒ 7 x1, < 7 x2 ⇒ 7 x1 – 3 < 7 x2, – 3
⇒ f (x1) < f (x2)
∴ f is strictly increasing function in R.
Let x1 , x2 ∊ (0, ∞) and let x1, < x2.
Now, 
...(1)
...(2)
Show that the function f(x) = x2 is a decreasing function in (– ∞ 0).
...(1)
...(2)
Construct an example of a functions which is strictly increasing but whose derivative vanishes at a point in the domain of definition of the function.
Let (x) = x3
It is strictly increasing in [– 2, 2] but f '(x) = 3 x2 ⇒ f '(0) = 0
i.e. f ' (x) vanishes at a point x = 0 ∊ [– 2, 2].
Prove that the exponential function ex is strictly increasing on R.
Let f (x) = ex ∴ Df = R
We are to prove that function f (x) = ex is strictly increasing. Here interval is not given. So we will prove that function ex is strictly increasing in its domain.
Now f ' (x) = ex
Three cases arise:
Case I.
Case II.
x = 0
∴ f ' (x) = ex = I > 0
Case III.
Show that the function given by f(x) = e2x is strictly increasing on R.
Here f (x) = e2 x ⇒ f ' (x) = 2 e2x
Three cases arise:
Case I.
Case II. x = 0
Case III,
Prove that f (x) = ax + b, where a and b are constants and a > 0 is an strictly increasing function for all real values of x. without using the derivative.
Let x1 , x2 x ∊ R and let x1 < x2
Now x1 < x2
⇒ a x1 < a x2: [ ∵ a > 0]
⇒ a x1 + b < a x2+ b ⇒ (x1) < f (x2)
∴ x1 < x2 ⇒ f (x1) < f (x2)
⇒ f is a strictly increasing function in R.
Prove that the logarithmic function is increasing wherever it is defined.
Let 
Now, 
Logarithmic function is increasing wherever it is defined.
Show that the function f given by f (x) = 10x is increasing for all x.
f (x) = 10x
∴ f '(x) = 10x log 10 > 0 ∀ x.
∴ f is increasing ∀ x.
Prove that the function f(x) = x3 – 3x2 + 3x – 100 is increasing on R.
Here f(x) = x3 – 3x2 + 3x – 100
Df = R
f ' (x) = 3x2 – 6 x + 3 = 3 (x2 – 2 x + 1) = 3 (x – 1)2 ≥ 0 ∀ x ∊ R
∴ f (x) is increasing on R.
Prove that the function f(x) = x3 – 3x2 + 3x – 100 is increasing on R.
Here (x) = x3 – 6x2 + 12x – 18 ∴ Df = R
f ' (x) = 3 x2 – 12x + 12 = 3 (x2 – 4 x + 4) = 3 (x – 2)2 ≥ 0 ∀ x ∊ R
∴ (x) is increasing on R.
Here f (x) = 4x3 – 6x2 + 3x + 12 12 ∴ Df = R
f ' (x) = 12x2 – 12x + 3 = 3(2x – 1)2 4x + 1) = 3(2 x – 1)2 ≥ 0 ∀ x ∊ R
∴ f (x) is increasing on R.
Here f (x) = x3 – 3x2 + 4x
Df = R
f '(x) = 3x2 – 6x + 4 = 3 (x2 – 2x + 1) + 1 = 3 (x – 1)2 + 1 > 0 ∀ x ∊ R
∴ f is strictly increasing on R.
Prove that the function f (x) = sinx is
(i) strictly increasing in 
(ii) strictly decreasing in 
(iii) neither increasing nor decreasing in 
.
Here 
(i) Since for each 
(ii) Since for each 
(iii) f(x) is strictly increasing in 
and strictly decreasing in 
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Prove that the function f (x) = cos x is
(i) strictly increasing in 
(ii) strictly decreasing in 
(iii) neither increasing nor decreasing in 
Here f (a) = cos x ⇒ f ' (x) = – sin x
(a) In 
, f '(x) = – sin x > 0
∴ f (x) is strictly increasing in (– 
, 0)
(b) In 
f ' (x) = – sin x < 0
∴ f (x) is strictly decreasing in (0, 
)
(c) Now f (x) is strictly increasing in (– 
, 0) and strictly decreasing in 
∴ f (x) is neither increasing nor decreasing in 
.
(a) Let 
In 
(b) Let 
In 
is strictly decreasing on 
(c) Let f(x) = cos 3x, 
Now 
and 
(d) Let 
Now, 
Here f (x) = x7 + 8 x5 + 1
∴ f ' (x) = 7x6 + 40 x4 ≥ 0 ∀ x
∴ f (x) is increasing function for all values of x.
and strictly decreasing in 
Let 
(i) Now, 
(iv) 
Let f (x) = x100 + sin x – 1 ∴ f ' (x) = 100 x99 + cos x
(i) For – 1 < x < 1, f (x) > 0 is not necessarily true
∴ f (x) is not strictly increasing on (– 1, 1)
(ii) For 0 < x < 1, f ' (x) > 0 ∴ f (x) is strictly increasing on (0, 1).
(iii) 
(iv) 
Find the value of a for which the function f (x) = x3 – 2ax + 6 is increasing for x > 0.
Here f (x) = x2 – 2ax – 6
∴ f ' (x) = 2x – 2a
f (x) is increasing when f ' (x) > 0
⇒ 2 (x – a) > 0 ⇒ x – > 0
⇒ x > a
But x > 0
∴ a ∊ (– ∞ , 0)
Find the values of x for which y = [x ( x – 2)]2 is an increasing function.
The points x = 0, x = 1, x = 2 divide the real line into four disjoining intervals 
Now 
is increasing in (0, 1) and 
Find the intervals in which the function f is given by
is (i) increasing (ii) decreasing
Here 
(i) For f(x) to be increasing.
(ii) For f(x) to be decreasing,
Determine the values of x for which the function 
is increasing and for which it is decreasing.
Here 
For 
to be increasing, 
Find the intervals in which the following function is decreasing:
f (x) = x3 – 12x
Here f(x) = x3 – 12x
∴ f ' (x) = 3x2 – 12 = 3 (x2 – 4) = 3 (x – 2) (x + 2)
For f (x) to be decreasing, f ' (x) < 0
∴ 3 (x – 2) (x + 2) < 0 or (x – 2) (x – 2) < 0
∴ – 2 < x < 2
f (x) is decreasing when – 2 < x < 2.
Find the intervals in which the function f given by f(x) = 2x2 – 3x is
(a) strictly increasing (b) strictly decreasing
f(x) = 2x2 – 3x 
Now 
The point 
divides the real line into two disjoint intervals 
.
Find the intervals in which the function f given by f(x) = x2 – 4x+6 is
(a) strictly increasing (b) strictly decreasing
Here f (x) = x2 - 4x + 6
∴ (x) = 2x – 4
f '(x) = 0 gives us 2x – 4 = 0 or x = 2
The point x = 2 divides the real line into two disjoint intervals (– ∞, 2). (2, ∞).
(a) In the interval (2, ∞), f ' (x) > 0
∴ f is strictly increasing in (2, ∞).
(b) In the interval (– ∞, 2), (x) < 0
∴ f is strictly decreasing in (– ∞, 2).
Note: Given function f is continuous at x = 2. which is the point joining the two intervals (– ∞, 2) and (2, ∞). Therefore f is decreasing in (– ∞, 2) and increasing in (2, ∞).
Find the intervals in which the following functions are strictly increasing or decreasing:
6 – 9x – x 2
Let f(x) = 6 – 9x – x 2 
Now, 
Find the intervals in which the following function is increasing or decreasing:
x3 – 6x2 + 9x + 15.
Let f (x) = x3- 6x2 + 9x + 15
∴ f '(x) – 3x2 – 12x + 9 = 3 (x2 – 4 x + 3) = 3 (x – 1) (x – 3)
f ' (x) = 0 gives us 3(x – 1) (x – 3) = 0
∴ x = 1, 3
The points x = 1, 3 divide the real line into three intervals (– ∞, 1), (1, 3), (3, ∞).
(1) In the interval (– ∞ , 1), f ' (x) > 0
∴ f (x) is increasing in (– ∞, 1)
(2) In the interval (1, 3), f ' (x) < 0
∴ f ' (x) is decreasing in (1, 3).
(3) In the interval (3, ∞), f ' (x) > 0
∴ f (x) is increasing in (3, ∞).
Find the intervals in which the following function f(x) is
(a) increasing (b) decreasing:
f (x) = 2x3 – 9x2 + 12x + 15
Let f (x) = 2x3 – 9x2 + 12x + 15
∴ (x) = 6x2 – 18x + 12 = 6(x2 – 3x + 2) = 6 (x – 1) (x – 2)
f ' (x) = 0 gives us 6 (x – 1) (x – 2) = 0 ⇒ x = 1 , 2
The points x = 1, 2 divide the real line into three intervals (– ∞ , 1), (1, 2), (2, ∞)
1. In the interval (– ∞ , 1), f ' (x) > 0
∴ f (x) is increasing in (– ∞ , 1)
2. In the interval (1, 2), f ' (x) < 0
∴ f (x) is decreasing in (1, 2)
3. In the interval (2, ∞), f ' (x) > 0
∴ f (x) is increasing in (2, ∞)
Determine the values of x for which the function f(x) = 2x3 – 24x + 5 is increasing or decreasing.
f (x) = 2 x3 – 24 x + 5
⇒ f ' (x) = 6 x2 – 24 = 6 (x2 – 4) = 6 (x – 2) (x + 2)
(i) For f (x) to be increasing, f ' (x) > 0
∴ 6 (x – 2) (x + 2) > 0 i.e. (x + 2) (x + 2) > 0
∴ either x < – 2
or x > 2
f (x) is increasing for x > 2 or x < – 2
(ii) For f (x) to be decreasing. f ' (x) < 0
∴ 6 (x – 2) (x + 2) < 0 or (x – 2) (x + 2) < 0
⇒ – 2 < x < 2
∴ f (x) is decreasing when – 2 < x < 2.
Find the intervals in which the functions f (x) = 2x3 – 15x2 + 36x + 1 is strictly increasing or decreasing. Also find the points on which the tangents are parallel to the x-axis.
f (x) = 2x3 –15x2 + 36x + 1
∴ f ' (x) = 6x2 – 30x + 36 = 6 (x2 – 5x + 6) = 6 (y – 2) (x – 3)
f ' (x) = 0 gives us 6 (r – 2) (x – 3) ⇒ x = 2, 3
The points x = 2, 3 divide the real line into three intervals (– ∞, 2), (2, 3), (3, ∞)
(i) In the interval (– ∞, 2), f ' (x) > 0
∴ f (x) is increasing in (2)
(ii) In the interval (2, 3), f ' (x) < 0
∴ f (x) is decreasing in (2, 3)
(iii) In the interval (3, ∞) f ' (x) > 0
∴ f (x) is increasing in (3, ∞)
∴ we see that f (x) increases in (– ∞, 2) ∪ (3, ∞) and decreasing in (2, 3).
Tangent is parallel to x-axis when f ' (x) = 0 i.e. 6 (x – 2) (x – 3) = 0 i.e. x = 2, 3
When x = 2, y = f (2) = 2 (2)3 – 15 (2)2 + 36 (2) + 1 – 16 –60 + 72 + 1 = 29
When x = 3, y = f (3) = 2(3)3 – 15 (3)2 + 36(3) + 1 = 54 – 135 + 108 + 1 = 28
∴ points are (2, 29), (3, 28).
Find the intervals in which the function
f(x) = x3 – 12x2 + 36x + 17 is
(a) strictly increasing (b) strictly decreasing
(i) Here f (x) = x3 – 12x2 + 36x + 17
∴ f '(x) = 3x2 – 24x + 36 = 3 (x2 – 8x + 12) = 3 (x – 2) (x – 6)
f ' (x) = 0 gives us 3 (a – 2) (a – 6)
∴ x = 2, 6
The points x = 2, 6 divide the real line into three intervals (– ∞ , 2), (2, 6), (6, ∞)
(1) In the interval (– ∞, 2), f ' (x) > 0
∴ f (x) is increasing in (– ∞, 2)
(2) In the interval (2, 6), f ' (x) < 0
∴ f ' (x) is decreasing in (2, 6).
(3) In the interval (6, ∞), f ' > 0
∴ f (x) is increasing in (6, ∞).
Find the intervals in which the function f (x) = 2x3 – 15x2 + 36x + 1 is
(a) strictly increasing (b) strictly decreasing
Let f (x) = 2x3 – 15x2 + 36x + 1
f ' (x) = 6x2 – 30x + 36 = 6 (x2 – 5 x + 6) = 6 (x – 2) (x – 3)
(a) For f (x) to be increasing, f' (x) > 0
i.e., 6 (x – 2) (x – 3) > 0 or (x – 2) (x – 3) > 0
⇒ either x < 2 or x > 3
∴ f (x) is increasing in x < 2 or x > 3.
(b) For f (x) to be decreasing, f ' (x) < 0
i.e. 6 (x – 2) (x – 3) < 0. or (x – 2) (x – 3) < 0
⇒ 2 < x < 3
∴ f (x) is decreasing in 2 < x < 3
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
x3 – 6x2 – 36x +4
Let f (a) = x3 – 6 x2 – 36x + 4
f '(x) = 3x2 – 12x – 36 = 3 (x2 – 4x – 12) = 3 (x + 2) (x – 6)
(a) For f (x) to be increasing, f ' (x) > 0
i.e. 3 (x + 2) (x – 6) > 0 or (x + 2) (x – 6) > 0
∴ either x < – 2 or x > 6
∴ f (x) is increasing in x > 6 and x < – 2.
(b) For f (x) to be decreasing, f ' (x) < 0
i.e. 3 (x + 2) (x – 6) < 0 or (x + 2)(x – 6) < 0
⇒ – 2 < x < 6
∴ f (x) is decreasing in – 2 < x < 6.
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
2x3 – 15x2 + 36x + 6
Let f (x) = 2 x3 – 15x2 + 36x + 6
f ' (x) = 6x2 – 30x + 36 = 6 (x2 – 5 x + 6) = 6 (x – 2) (x – 3)
(a) For f (x) to be increasing, f'(x) > 0
i.e., 6 (x – 2) (x – 3) > 0 or (x – 2) (x – 3) > 0
⇒ either x < 2 or x > 3
∴ f (x) is increasing in x < 2 or x > 3.
(b) For f (x) to be decreasing, f ' (x) < 0
i.e. 6 (x – 2) (x – 3) < 0 or (x – 2) (x – 3) < 0 ⇒ 2 < x < 3
∴ f (x) is decreasing in 2 < x < 3
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
6 + 12x + 3x2 – 2x3
Let f (x) = 6 + 12x + 3x2 – 2x3 = – 2x3 + 3x2 + 12x + 6
∴ f ' (x) = – 6x2 + 6x + 12 = – 6 (x2 – x – 2) = – 6 (x + 1 ) (x – 2)
(a) For f (x) to be increasing, f ' (x) > 0
or – 6 (x + 1) (x – 2) > 0 or (x + 1) (x – 2) < 0
⇒ – 1 < x < 2
∴ f (x) is increasing for –1 < x < 2
(b) For f (x) to be decreasing , f ' (x) < 0
or – 6 (x + 1) (x – 2) < 0 or (x + 1) (– 2) > 0
⇒ either x < – 1 or x > 2
∴ f (x) is decreasing for x < – 1 or x > 2.
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
2x3 – 8x2 + 10x + 5
Let f (x) – 2x3 – 8x2 + 10x + 5
∴ f ' (x) = 6x2– 16x + 10 = 2 (3x2 – 8x + 5) = 2 (x – 1) (3x – 5)
For f (x) to be increasing,
f ' (x) > 0 ⇒ 2 (x – 1) (3x – 5) > 0 ⇒ (x – 1) (3 x – 5) > 0
For f(x) to be decreasing,
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
2x3 – 6x2 – 48x + 17
Let f (x) = 2x3 – 6x2– 48x + 17
∴ f ' (x) = 6x2 – 12x – 48 = 6 (x2 – 2x – 8) = 6 (x + 2) (x – 4)
For f (x) to be increasing,
f ' (x) > 0 ⇒ 6 (x + 2) (x – 4) > 0
⇒ (x + 2) (x – 4) > 0
∴ either x < – 2 or x > 4
∴ f (x) is increasing in (– ∞, 2) ∪ (4, ∞)
For f (x) to be decreasing,
f ' (x) < 0 ⇒ 6 (x + 2) (x – 4) = 0
⇒ (x + 2) (x – 4) < 0 ⇒ – 2 < x < 4
∴ f (x) is decreasing in (– 2, 4)
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
f (x) = 2x3 – 9x2 + 12x + 30
Let f (x) = 2x3 – 9x2 + 12x + 30
∴ f '(x) = 6 x2 – 18x + 12 = 6 (x2 – 3x + 2) = 6 (x –1) (x – 2)
f ' (x) = 0 gives us 6 (x – 1) (x – 2) = 0 ⇒ x = 1, 2
The points x = 1, 2 divide the real line into three intervals (– ∞, 1), (1, 2), (2, ∞)
1. In the interval (– ∞, 1), f ' (x) > 0
∴ f (x) is increasing in ( – ∞, 1)
2. In the interval (1, 2), f ' (x) < 0
∴ f(x) is decreasing in (1, 2)
3. In the interval (2, ∞). f ' (x) > 0
∴ f (x) is increasing in (2, ∞)
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
f (x) = 2x3 – 3x2 – 36x + 7
Here f (x) = 2x3 – 3x2 – 36x + 7
∴ f ' (x) = 6x2 – 6x – 36 = 6 (x2 – x – 6) = 6 (x + 2) (x – 3)
f ' (x) = 0 gives us 6 (x + 2) (x – 3)
∴ x = – 2, 3
The points x = – 2, 3 divide the real line into three intervals (– ∞, – 2), (– 2, 3), (3, ∞).
In the interval (– ∞, – 2), f ' (x) > 0
strictly ∴ f (x) is strictly increasing in (– ∞, – 2)
In the interval (– 2, 3), f ' (x) < 0 ∞ f (x) is strictly decreasing in ( – 2, 3)
In the interval (3, ∞), f ' (x) > 0
∴ f (x) is strictly increasing in (3, ∞)
∴ we see that f (x) is strictly increasing in (– 8, – 2) ∪ (3, ∞) and strictly decreasing in ( – 2, 3)
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
f(x) = 2x3 – 21x2 + 36x – 40
Here f (x) = 2x 3 – 21x2 + 36x – 40
∴ f '(x) = 6x2 – 42x + 36 = 6(x2 – 7 x + 6) = 6 (x – 1) (x – 6)
∴ f '(x) = 0 gives us 6 (x – 1) (x – 6) = 0
∴ x = 1, 6
The points x = 1, 6 divide the real line into three intervals – (∞, 1), (1, 6), (6, ∞).
In the interval (– ∞, 1), f ' (x) > 0
∴ f (x) is strictly increasing in (– ∞, 1)
In the interval (1, 6), f '(x) < 0
∴ f (x) is strictly decreasing (1, 6)
In the interval (6, ∞), f ' (x) > 0
∴ f (x) is strictly increasing in (6, ∞)
∴ we see that f (x) is strictly increasing in (– ∞, 1) ∪ (6, ∞) and strictly decreasing in (1, 6).
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
4x3 – 6x2 – 72x + 30
Here f (x) = 4x3 – 6x2 – 72x + 30
∴ f ' (x) = 12x2 – 12x – 72 = 12 (x2 – x – 6) = 12 (x – 3) (x + 2)
f '(x) = 0 gives us 12 (x – 3) (x + 2) = 0
∴ x = – 2, 3
The points x = – 2, 3 divide the real line into three disjoint intervals (– ∞ – 2), (–2, 3), (3, ∞).
Interval Sign of f' (x) Nature of function f
(–∞, –2) (–) (–) > 0 f is strictly increasing
(– 2, 3) (–) (+) < 0 f is strictly decreasing
(3, ∞) (+) (+) > 0 f is strictly increasing
(a) f is strictly increasing in the intervals (– ∞, – 2) and (3,
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
– 2x3 – 9x2 – 12x + 1
Let f (x) = – 2x3 – 9x2 – 12x + 1
∴ f ' (x) = – 6x2 – 18x – 12 = – 6 (x2 + 3x + 2) = – 6 (x + 1) (x + 2)
f '(x) = 0 gives us – 6 (x + 1) (x + 2) = 0 ⇒ x = – 1, – 2
The points x = – 2, – 1 divide the real line into three intervals (– ∞, – 2), (– 2, – 1),
(1) In the interval (– ∞, – 2), f '(x) < 0
∴ f (x) is strictly decreasing in (– ∞, – 2).
(2) In the interval (– 2, – 1), f ' (x) > 0
∴ f (x) is strictly increasing in (– 2, – 1).
(3) In the interval (– 1, ∞), f '(x) < 0
∴ f (x) is strictly decreasing in (– 1, ∞).
Find the intervals in which the function
, is
(a) increasing (b) decreasing.
(a) For f(x) to be increasing,
Find the intervals in which the function f is given by 
(i) increasing (ii) decreasing
Here 
(i) For 
to be increasing
(ii) For f(x) to be decreasing,
Determine for which values of x, the function f (x) = x4 – 2x2 is increasing or decreasing.
f (x) = x4 – 2x2
∴ f '(x) = 4 x3 – 4x = 4x (x2 – 1 )
(i) For f (x) to be increasing , f ' (x) > 0
Case I. Let x > 0 .
∴ f ' (x) > 0 when x2 – 1 > 0 i.e.. (x – 1) (x + 1) > 0
∴ x does not lie in (– 1, 1)
Also x > 0
∴ we have x > 1
Including end point, f (x) is increasing in x ≥ 1.
Case II. Let x < 0
∴ (x) > 0 when x2 – 1 < 0
i.e., when (x – 1) (x + 1) < 0
i.e., when x lies in (– 1, 1)
But x < 0
∴ we have – 1 < x < 0
∴ f (x) is increasing in – 1 < x < 0 or (– 1, 0)
(ii) For f (x) to be decreasing, f '(x) < 0
Case I. Let x > 0
∴ f ' (x) < 0 when x2 – 1 < 0
ie., (x – 1) (x + 1) < 0 i.e., x lies in (– 1, 1)
But x > 0
∴ we have 0 < x < 1
∴ f (x) is decreasing in ( 0, 1)
Case II. Let x < 0
∴ f ' (x) < 0 when x2 – 1 > 0
i.e. (x – 1) (x + 1) > 0
i.e., x does not lie in (– 1, 1)
Also x < 0
∴ we have x < – 1
∴ f (x) is decreasing in x < – 1.
Find intervals in which the function given by 
is (a) strictly increasing (b) strictly decreasing.
f is strictly decreasing in (1, 3)
Find the intervals in which the function 
is increasing or decreasing.
Here, 
(i) For f(x) to be increasing, 
(ii) For f(x) to be decreasing, 
Separate 
into sub-intervals in which the function f (x) = sin 3x is increasing or decreasing.
Here 
(i) For f(x) to be increasing, f '(x) > 0 
(ii) For f(x) to be decreasing, f '(x) < 0
Find the intervals in which the following function is increasing or decreasing:
f (x) = sinx – cosx, 0 < x < 2
.
f(x) = sinx – cosx
For 
to be increasing, 
For 
to be decreasing, 
to be increasing
Find the intervals in which the following function is increasing or decreasing
f (x) = (x + 2) e–x
f (x) = (x + 2) e–x
∴ f ' (x) = (x + 2) · e–x (– 1) + e–x · 1 = e–x (– x – 2 + 1)
∴ f ' (x) = – (x + 1) e–x
For f (x) to be increasing
f ' (x) > 0 ⇒ –(x + 1) e–x > 0
⇒ (x + 1) e–x < 0 ⇒ x + 1< 0 [∵ e–x > 0]
⇒ x < – 1
∴ f (x) is increasing in ( – ∞ , – 1)
For f (x) to be decreasing,
f ' (x) < 0 ⇒ – (x + 1) e–x < 0
⇒ (x + 1) e–x > 0 ⇒ x + 1 > 0 [ ∵ c–x > 0]
⇒ x > – 1
∴ f (x) is decreasing in (– 1, ∞)
is increasing or decreasing. f (x) = sin x + cos x
Now, 
The points 
divide the interval 
into three disjoint intervals 
In 
In 
In 
, 
Let f be a function defined on [a, b] such that f ' (x) > 0, for all x ∊ (a, b). Then prove that f is strictly increasing function of (a, b).
Since f '(x) > 0 in (a, b).
By Lagrange's mean value theorem,
But 
On which of the following intervals is the function f given by f (x) = x100 + sin x – 1 strictly decreasing?
(A) (0, 1) (B) 
(C) 
(D) None of these
Let 
(A) For 0 < x < 1, f '(x) >0![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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(B) For 
f '(x) >0
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The interval in which y = x2e–x is increasing is
D.
(0, 2)Here, 
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For y to be increasing,
Find the maximum and the minimum values, if there be any, of f given by f(x) = 9x2 – 6x + 1, x ∊ R.
The given function is
f (x) = 9x2 – 6x + 1, x ∊ R
= (3x – 1) 2 ≥ 0, ∀ x ∊ R.
Also, f(x) = 0, if 
Therefore, the minimum value of f is 0 and the point of minimum value of f is Futher,f has no maximum value and hence no point of maximum value of in R.
The given function is non-positive for all x ∊ R
Also, f (x) = 0 if x = 0. Therefore, the maximum value of f is 0 and the point of maximum value of f is x.
Further, note that f has no minimum value in R and hence no point of minimum value of f in R
Find the maximum and the minimum values, if there be any, of f given by f(x) = x, x ∊ (0, 1).
Find the maximum and minimum values, if any, of the following function without using the derivatives:
9x2 – 12x + 4
Let f (x) = 9x2 – 12x + 4 = (3 x – 2)2 ≥ 0
∴ f (x) ≥ 0 ∀ a ∊ R
Minimum value of f (x) is 0 and is obtained at 
There is no maximum value of the function.
Find the maximum and minimum values, if any, of the following function without using the derivatives:
4x 2 + 28x + 49
Let f (x) = 4x2 + 28x + 49
= (2x + 7)2 ≥ 0 f (x) ≥ 0 ∀ x ∊ R
Find the maximum and minimum values, if any, of the following function without using the derivatives:
x + 1, x ∊ (– 1, 1)
Let f (x) = x + 1
Here f (x) can assume any positive value depending upon x. Also f (x) can assume any negative value depending upon x.
∴ there is neither maximum nor minimum value.
Find the maximum and minimum values, if any, of the following functions without using the derivatives:
x2
Let f (x) = x2
Now x2 ≥ 0 ∀ a∊ R
∴ minimum value of F (x) = 0. It has no maximum value.
Find the maximum and minimum values, if any, of the following functions without using the derivatives:
– (x – 2 )2 + 4
Let f (x) = – (x – 2)3 + 4
Now ( x – 2)2 ≥ 0 ∀ x ∊ R
⇒ – (x - 2)2 ≥ 0 ∀ x ∊ R ⇒ – (x – 2)2 + 4 ≤ 4 ∀ x ∊ R
∴ maximum value of f (x) is 4. It has no minimum value.
Find the maximum and minimum values, if any, of the following functions without using the derivatives:
– (x – 1)2 + 10
Let f (x) = – (x – 1)2 + 10
Now (x – 1 )2 ≥ 0 ∀ x ∊ R
⇒ – (x – 1)2 ≤ 0 ∀ x ∊ R ⇒ – (x – 1)2 + 10 ≤ ∀ x ∊ R
∴ maximum value of f (x) is 10. It has no minimum value.
Find the maximum and minimum values, if any, of the following functions without using the derivatives:
(2 x – 1)2 + 3
Let f (x) = (2 x – 1)2 + 3
Now (2 x – 1)2 ≥ 0 ∀ x ∊ R
⇒ (2 x – 1)2 + 3 ≥ 3 ∀ x R
∴ minimum value of f (x) is 3. It has no maximum value.
Find the maximum or minimum values, if any, of the following functions without using the derivatives:
x3 + 1
Let f (x) = x3 + 1
Now f (x) → when x → ∞ and f (x) → – ∞ when x → – ∞
∴ f (x) has neither maximum nor minimum.
Find the maximum or minimum values, if any, of the following functions without using the derivatives:
| x + 2 |
Let f(x) = | x + 2 |
Now | x + 2 | ≥ 0 ∀ x ∊ R
∴ f (x) ≥ 0 ∀ x∊ R
⇒ minimum value of f (x) is 0 and it has no maximum value.
Find the maximum or minimum values, if any, of the following functions without using the derivatives:
– | x + 1 | + 3
Let f (x) = – | x + 1 | + 3
Now | x + 1 | ≥ 0 ∀ x ∊ R
⇒ – | x + 1 | ≥ 0 ∀ x ∊ R ⇒ – | x + 1 | + 3 ≤ 3 ∀ a ∊ R
∴ maximum value of f (x) is 3 and it has no minimum value.
Find the maximum or minimum values, if any, of the following functions without using the derivatives:
sin (2x + 3)
Let f (x) = sin (2 x + 3)
We know that – 1 ≤ sin ≤ for all 
∴ maximum value of sin (2 x + 3) is 1 and minimum value of sin (2 x + 3) is – 1.
Find the maximum or minimum values, if any, of the following functions without using the derivatives:
sin 2x + 5
Let f (x) = sin 2x + 5
We know that – 1 ≤ sin 2x ≤ 1
⇒ – 1 + 5 ≤ sin 2x + 5 ≤ 1 + 5 ⇒ 4 ≤ f (x) ≤ 6
∴ maximum value of f (x) is 6 and minimum value of f (x) is 4.
Find the maximum or minimum values, if any, of the following functions without using the derivatives:
|sin 4x + 3|
Let f(x) = |sin 4x + 3|
∵ – 1 ≤ sin 4x ≤ 1
∴ – 1 + 3 ≤ sin 4x + 3 ≤ 1 + 3
⇒ 2 < sin 4x + 3 ≤ 4 ⇒ 2 ≤ | sin 4x + 3 | ≤ 4
⇒ 2 ≤ f (x) ≤ 4
∴ maximum value of f (x) is 4 and minimum value of f (x) is 2.
Find the maximum or minimum values, if any, of the following functions without using the derivatives:
sin (sin x)
Let f (x) = sin (sin x)
For any x, – 1 ≤ sin x ≤ 1
⇒ sin (– 1) ≤ sin x ≤ sin 1 {∵ in [– 1, 1] , sin function is increasing}
⇒ – sin 1 ≤ f (x) ≤ sin 1
∴ maximum value of f (x) is sin 1 and minimum value of f (x) is – sin 1.
Find the maximum or minimum values, if any, of the following functions without using the derivatives:
16x2 – 16x + 28
Let f(x) = 16x2 – 16x + 28 = 16(x2 - x) + 28
Now, 
minimum value of f(x) is 24. It has no maximum value.
Prove that the following functions do not have maxima or minima:
f(x) = ex
Let f (x) = ex
∴ f ' (x) = ex
Now f ' (x) ≠ for any x ∊ R
∴ f (x) has neither maximum nor minimum value.
Prove that the following functions do not have maxima or minima:
g(x) = log x
g(x) = log x, x>0, 
Now, 
Prove that the following functions do not have maxima or minima:
h(x) = x3 + x2 + x + 1
Let 
Now, 
These values of x are not real.
∴ there is no real value of x for which f ' (x) = 0
∴ f (x) has neither maxima nor minima.
Find the absolute maximum value and the absolute minimum value of 
The function 
is differentiable for all x in 
and 
Now, 
i.e., when (x - 5) (x - 1) = 0
i.e., when x = 1 or x = 5
But only 
Now, 
Hence absolute maximum value is 
and absolute minimum value is 
. These are attained at 1 and 4 respectively.
Find the absolute maximum and minimum values of a function f is given by
f (x) = 2x3 – 15x2 + 36x + 1 on the interval of [1, 5].
f (x) = 2x3 – 15x2 + 36x + 1
∴ f ' (x) = 6x2 – 30x + 36 = 6 (x2 – 5x + 6) = 6 (x – 2) (x – 3)
f ' (x) = 0 ⇒ 6 (x – 2) (x – 3) = 0 ⇒ x = 2, 3 ∊ [1,5]
Now f (1) = 2 – 15 + 36 + 1 = 24
f (2) = 2 × 8 – 15 x 4 + 36 x 2 + 1 = 16 – 60 + 72 + 1 = 29
f (3) = 2 × 27 – 15 x 9 + 36x3 + 1 = 54 – 135 + 108 + 1 = 28
f (5) = 2 × 125–15 × 25 + 36 x 5 + 1 = 250–375+ 180+ 1 = 56
∴ absolute maximum value = 56 at x = 5
and absolute minimum value = 24 at x = 1.
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
Here f (x) = x3
The given function is differentiate for all x in [– 2, 2],
f ' (x) = 3x2
Now f ' (x) = 0 ⇒ 3x2 = 0 ⇒ x = 0 ∊ [– 2, 2]
f (0) = 0, f (– 2) = – 8, f (2) = 8
∴ absolute maximum value = 8 and absolute minimum value = – 8.
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
Here 
It is differentiable for all x in 
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
Let 
It is differentiable for all x in 
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
Here f (x) = (x – 1 )2 + 3
The given function is differentiable for all x in [– 3, 1],
f '(x) = 2 (x – 1)
Now f ' (x) = 0 ⇒ 2 (x – 1) = 0 ⇒ x = [∊ [– 3,1]
f (– 3) = (– 3 – 1)2 + 3 = 19, f (1) = (1 – 1 )2 + 3 = 3
∴ absolute maximum value = 19 and absolute minimum value = 3.
Find the absolute maximum value and the absolute minimum value of
Let 
The given function is differentiable for all x in [-2, 2.5]
absolute maximum value = 19.625 and absolute minimum value = -1.75
Find the absolute maximum and minimum values of the function f given by
f (x) = cos2x + sinx, x ∊ [0, 
].
.
Find the points at which the function f given by f (x) = (x – 2)4 (x + 1 )3 has
(i) local maxima (ii) local minima (iii) point of inflexion .
Here, 
When x < 2 slightly, 
When x > 2 slightly, 
When x < -1 slightly f ' (x) = (-) (+) (-) = +ve
When x > -1 slightly, f ' (x) = (-) (+) (-) = +ve
at x = -1, f ' (x) does not change sign
x = -1 is a point of inflexion.
Examine the following function for extreme values:
f(x) = (x – 3)5 (x + 1)6
Here f (x) = (x – 3)5 (x + 1)6
Differentiating (I) w.r.t. x, we get
f ' (x) = 5 (x – 3)4 (x + 1)6 + (x – 3)5 (x + 1 )5
∴ f ' (x) = (x – 3)4 (x + 1)5 [5 (x + 1) + 6 (x – 3)]
= (x – 3)4 (x + 1)5 (11 x – 13)
Now, 
(a) When x = 3
If x < 3 (slightly), f ' (x) = (+) (+) (+) = + ve
If x > 3 (slightly), f ' (x) = (+) (+) (+) = + ve
Hence f ' (x) does not change sign as x passes through 3.
∴ x = 3 is neither a point of maxima, nor a point of minima. 3 is a point of inflexion.
(b) When x = – 1
If x < – 1 (slightly), f ' (x) = (+) (–) (–) = + ve
If x > – 1 (slightly), f ' (x) = (+) (+) (–) = – ve
∴ f ' (x) changes from positive to negative as a passes through – 1
Hence x = – 1 is a point of local maxima and maximum value of the function at x = – 1 is f (– 1) = 0.
(c) When 
Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:
The constant function 
Let f (x) = α
∴ f ' (x) = 0
∴ there is no point at which f (x) has extreme value
⇒ f (x) has neither local maxima nor local minima.
Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:
f(x) = x2
Let f (x) = x2
∴ f ' (x) = 2x
f ' (x) = 0 gives us x = 0
When x < 0 slightly, f ' (x) = 2 (– ve) = – ve
When x > 0 slightly, f ' (x) = 2 ( + ve) = +ve
∴ at x = 0, f ' (x) changes from – ve to + ve
∴ f (x) has local minimum value at x = 0
and this local minimum value = 0.
Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:
Let f (x) = x3 + 3 x.
∴ f ' (x) = 3x2 – 3 = 3 (x – 1) (x + 1)
f ' (x) = 0 ⇒ 3 x2 – 3 = 0
∴ x2 – 1 = 0 ⇒ x2 = 1 ⇒ x = – 1, 1
When x < – 1 slightly, f ' (x) = 3 (– ve) (– ve ) = + ve
When x > – 1 slightly, f ' (x) = 3 (– ve) (+ ve) = – ve
∴ at x = – 1 , f ' (x) changes from + ve to – ve
∴ f (x) has local maximum value at x = – 1
and this local maximum value = (– 1 )3 – 3 (– 1) = – 1 + 3 = 2
When x < 1 slightly, f ' (x) = 3 (– ve) ( + ve) = – ve
When x > 1 slightly, f ' (x) = 3 ( + ve) (+ ve) = + ve
∴ at x = 1, f ' (x) changes from – ve to + ve
∴ f (x) has local minimum value at x = 1
and this local minimum value = (1)3 – 3 (1) = 1 – 3 = – 2
Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:
Let f (x) = cos x
∴ '(x) = – sin x
Now f ' (x) ≠ 0 for any x ∊ (0, 
)
∴ given function has no extreme points
Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:
f(x) = x3 – 6x2 + 9x + 15
Let f (x) = x3 – 6x2 + 9x + 15
∴ f ' (x) = 3x2 - 12x + 9 = 3 (x2 – 4 x + 3) = 3 (x – 1) (x – 3)
f ' (x) = 0 ⇒ 3 (x – 1) (x – 3) = 0 ⇒ x = 1, 3
When x < 1 slightly, f ' (r) = 3 (– 1)(–) = + ve
When x > 1 slightly, f ' (x) = 3 (+)(–) = – ve
∴ at x = 1, f ' (x) changes from + ve to – ve
∴ f (x) has local maxima at x = 1
and local maximum value = 1 – 6 + 9 + 15 = 19
When x < 3 slightly, f ' (x) = 3 (+) (–) = – ve
When x > 3 slightly, f ' (x) = 3 (+) (+) = + ve
∴ at x = 3, f ' (x) changcs from – ve to + ve
∴ f ' (x) has local minimum value at = 3
and local minimum value = (3)3 – 6 (3)2 + 9 (3) + 15 = 27 – 54 + 27 + 15 = 15
Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:
f(x) = (x – 1)(x + 2)2
f ' (x) = 3 x2 + 6 x
= (x – 1), 2 (x + 2) + (x + 2)2, 1 = 2 (x2 + x – 2) + x2 + 4 x + 4
= 2 x2 + 2 x – 4 – x2 + 4 x + 4.
f ' (x) = 3 x2 + 6 x
f ' (x) = 0 gives us 3 x2 – 6 x = 0. or x2 + 2 x = 0
∴ x (x + 2) = 0 ∴ x = 0, – 2
f ' ' (x) = 6 x + 6
At x = 0, f ' ' (x) = 0 + 6 = 6 > 0
∴ f (x) has a local minimum at x = 0
∴ local minimum value = (c – 1) (0 + 2)2 = – 4
At x = – 2, f ' ' (x) = – 12 + 6 = – 6 < 0
∴ f (x) has a local maximum at x = – 2
∴ local minimum value = (– 2 – 1) (– 2 + 2)2 = 0
Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:
Let f (x) = (x – 3)4.
∴ f ' (x) = 4 (x – 3)3
Now f ' (x) = 0 ⇒ 4(x – 3)3 = 0 ⇒ (x– 3)3 = 0 ⇒ x = 3
When x < 3 slightly, f ' (x) = – ve
When x > 3 slightly, f '(x) = + ve
∴ at x = 3, f '(x) changes from – ve to + ve
∴ f (x) has local minima at x = 3
and this local minimum value = (3 – 3)4 = 0
Let 
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#13 {main}</pre>](/application/zrc/images/qvar/MAEN12079610-4.png)
When x<0 slightly, f '(x) = (+) (-) (-) = + ve
When x>0 slightly, f '(x) = (+) (-) (-) = + ve
at x = 0, f '(x) does not change sign
Let f (x) = – (x – 1)3 ( x + 1 )2
∴ f ' (x) = – [(x – 1)3. 2(x + 1) + (x + 1)2. 3(x – 1)2]
= – (x – 1)2 (x + 1) [2 (x – 1) + 3 (x + 1)] = – (x – 1)2 (x + 1) (5 x + 1)
f ' (x) = 0 ⇒ – (x – 1)2 (x + 1) (5 x + 1) = 0 
.
When x < 1 slightly, f ' (x) = – [( + )( + )( + )] = – ve
When x > 1 slightly, f ' (x) = – [( + )( + )( + )] = – ve
∴ at x = 1 , f '(x) does not change sign ,
∴ x = 1 is a point of inflexion.
When x < – 1 slightly. f ' (x) = – [( + )( – )( – )] = – ve
When x > – 1 slightly. f ' (x) = – [( + )( + ) ( – )] = + ve
∴ at x = – 1, f ' (x) changes from – ve to + ve
∴ f (x) has local minimum value at x = – 1 and this local minimum value = 0.
Find all the local maxima or minima of the function
f (x) = x3 – 12x.
Let f (x) = x3 – 12 x
∴ f ' (x) =3 x2 – 12
f ' ' (x) = 0 ⇒ 3 x2 – 12 = 0 ⇒ x2 – 4 = 0 ⇒ x = 2, – 2
f ' ' (x) = 6 x
At x = 2.
f ' ' (x) = 12 > 0
∴ f (x) has local minimum value at x = 2
and this minimum value = (2)3 – 12 (2) = – 16
At x = – 2, f ' ' (x) = – 12 < 0
∴ f (x) has local maximum value x = – 2
and this value = (– 2)3 – 12 (– 2) = – 8 + 24 = 16.
Here f (x) = x3 – 27 x + 3
∴ f ' (x) = 3 x2 – 27
and f ' ' (x) = 6 x
Now f ' (x) = 0 ⇒ 3 x2 – 27 = 0 ⇒ x2 – 9 = 0
⇒ x2 = 9 x = – 3, 3
At x = 3, f ' ' (x) = 18 > 0
∴ f has a local minima at x = 3
and local minimum value = (3)3 – 27 (3) + 3 = 27 – 81 + 3 = – 51
At x = – 3, f ' ' (x) = – 18 < 0
∴ f has a local maxima at x = – 3
and local maximum value = (– 3)3 –21 ( 3) + 3 = 27 + 81 + 3 = 57
f (x) = x3 – 3 x + 3
∴ f ' (x) – 3 x2 3 = 3 (x – 1) ( v + 1)
f ' (x) = 0 ⇒ 3 x2 – 3 = 0 ⇒ x2 – 1 – 0 ⇒ x2 = 1 ⇒ v = – 1, 1
When x < – 1 slightly, f ' (x) = 3 ( ve) ( ve ) = + ve
When x > – 1 slightly, f ' (x) = 3 ( ve) ( + ve) = ve
∴ at x = – 1, f ' (x) changes from + ve to ve
∴ f (x) has local maximum value at x = 1
When x < 1 slightly, f ' (x) = 3 (– ve) ( + ve) = – ve
When x > 1 slightly, f ' (x) = 3 ( + ve) (+ ve) = + ve
∴ at x = 1, f ' (x) changes from ve to + ve
∴ f (x) has local minimum value at x = 1
f (x) = 2 x3 – 6 x2 + 6 x + 5
∴ f ' (x) = 6 x2 – 12 x + 6 = 6 (x2 – 2 x + 1)
∴ f ' (x) = 6 ( x – 1)2
Now f ' (x) = 0 ⇒ 6 (x – 1)2 = 0 ⇒ r = 1
When x < 1 slightly, f ' (x) = 6 (1) = + ve
When x > 1 slightly, f ' (x) = 6 (+) = ve
∴ f ' (x) does not change sign as x passes through I
∴ x = 1 is a point of inflexion.
Find local minimum value of the function f given by f(x) = 3+ | x |, x ∊ R.
f (x) = 3 + | x |, x ∊ R
f is not differentiable at x = 0
∴ 0 is a critical point of f.
Now, 
Find local maximum and local minimum values of the function f given by
f (a) = 3x4 + 4x3 – 12x2 + 12
f (x) = 3 x4 + 4 x3 – 12 x2 + 12
f ' (x) = 12 x3 + 12 x2 – 24 x = 12 x (x2 + x – 2)
f ' (x) = 12 x (x – 1) (x + 2)
f ' (x) = 0 ⇒ 12 a (x – 1) (x + 2) = 0
∴ x = 0, 1, – 2
f ' (x) = 36 x2 + 24 x – 24
At x = 0, f '' (x) = 0 + 0 24 = – 24 < 0
∴ x = 0 is a point of local maximum
and local maximum value = 0 + 0 – 0 + 12 = 12
At x = 1, f ' (x) = 36 + 24 – 24 = 36 > 0
∴ x = 1 is a point of local minima
and local minimum value = 3+ 4 – 12 + 12 = 7
At x = – 2, f '' (x) = 36 (4) + 24 (– 2) – 24 = 72 > 0
∴ x = – 2 is a point of local minima
and local minimum value = 3 (–2)4 + 4 (–2)3 – 12(–2)2 + 12
= 3 (16) + 4 (– 8) – 12 (4) + 12 = 48– 32 – 48 + 12 = – 20
Find both the maximum value and the minimum value of
3x4 – 8x3 + 12x2 – 48x + 25
Let f (x) = 3x4 – 8x3 + 12x2 – 48x + 25
f ' (x) = 12x3 – 24x2 + 24x – 48
f ' (x) = 0 ⇒ 12x3 – 24x2 + 24x – 48 = 0 ⇒ x3 – 2x2 + 2x – 4 = 0
⇒ (x – 2) (x2 + 2) = 0 ⇒ x = 2 , ± i 
.
Now x = 2 is the only real value in [0, 3]
f (0) = 0 – 0 + 0 – 0 + 25 = 25
f (2) = 48 – 64 + 48 – 96 + 25 = – 39
f (3) = 243 – 216 + 108 – 144 + 25 = 16
∴ max. value = 25 and min. value = – 39.
Let f (x) = x4 – 62x2 + ax + 9
f ' (x) = 4x3 – 124x + a
∴ f ' (1) = 4 – 124 + a = – 120 + a
∴ f (x) attains max. value at x = 1
∴ f ' (1) = 0 ⇒ – 120 + a = 0 ⇒ a – 120.
has a turning point at P(2, -1). Find the values of a and b and show that y is maximum at P.
...(1)
...(2)
lies on (1)
...(3)
Find the shortest distance of the point (0, c) from the parabola y = x2 , where 0 ≤ c ≤ 5.
Let x be the distance between (0, c) and any point (x, y) on the parabola y =x2.
Let (x, y) be any point on y = x2 + 7 at which helicopter in at a particular moment.
∴ helicopter is at (x, x2 + 7 ).
Let d be the distance between the jet at (x, x2 + 7 ) and soldier at (3, 7).
∴ d2 = (x – 3)2 + (x2 + 7 – 7)2 = (x – 3)2 + (x2)2
∴ d2 = x4 + x2 – 6 x + 9
Let f (x) = d2 = x4 + x2 – 6 x + 9
f ' (x) = 4 x3 + 2 x – 6 = 2 (2 x3 + x – 3) = 2 ( x – 1) (2 x2 + 2 x + 3)
f ' (x) = 0 ⇒ 2 (x – 1) (2 x2 + 2 x + 3) = 0
⇒ x = 1 as we reject imaginary values of x.
f ' ' (x) = 12x2 + 2
At x = 1, f ' ' (x) = 12 + 2 – 14 > 0 ⇒ f (x) lies a local minimum at x = 1
But x – 1 is only extreme point
∴ f (x) is minimum at x = 1
The point on the curve x2 = 2y which is nearest to the point (0, 5) is
(0, 0)
(2, 2)
A.
Let (x, y) be the point on 
which is nearest to the point (0, 5).
We are not interested in this case.
Determine two positive numbers whose sum is 15 and the sum of whose, squares is minimum.
Let one number = x
Amongst all pairs of positive numbers with product 64, find those whose sum is the least.
Let 
be the two numbers
Determine two positive numbers whose sum is 30 and whose product is the maximum.
Let one number = x.
Find two numbers whose sum is 24 and whose product is as large as possible.
Let two numbers be x and 24 - x.
Let P = x(24 - x) = 24x - x2 
Let two numbers be x and 16 – x
Let f (x) = x3 + (16 – x)3
∴ f (x) = 3 x2 + 3 (16 – x)2 (– 1) = 3 x2 – 3 (16 – x)2
= 3 x2 – 3 (256 + x2 – 32 x) = 96 x – 768 = 96 (x – 8)
f ' (x) = 0 ⇒ 96 (x – 8) = 0 ⇒ x = 8
f ' ' (x) = 96
At x = 8, f ' ' (x) = 96 > 0
∴ f (x) is minimum when x = 8 ⇒ numbers are 8, 8.
Here x + y = 40 ⇒ y = 40 – x ...(1)
Let f (x) = x y3 = x (40 – x)3
f'(x) = 
= x [3(40 – x)2 (–1)] + (40 – x)3
= – 3 x (40 – x)2 +(40 – x) 3
= (40 – x)2 (– 3 a + 40 – x) = (40 – x)2 (– 4 x + 40)
= 4 (40 – x)2 (10 – x)
f ' (x) = 0 ⇒ 4 (40 – x)2 (10 – y) = 0 ⇒ x = 10, 40
Rejecting a = 40 as 0 < x < 40, we get, x = 10
When x < 10 slightly, f ' (x) = 4 (+) (+) = +ve
When x > 10 slightly, f ' (x) = 4 (+)(–) = –ve
∴ at x = 10, f ' (x) changes from +ve to –ve
∴ f (x) has local maximum at x = 10
But f (x) has only one extreme point x = 10
∴ f (x) is maximum when x = 10, y = 40 – 10 = 30
∴ x = 10, y = 30.
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.
= (60 – x)2 [ – 3 x + 60 – x] = (60 – x)2 (– 4 x + 60)
= 4 (60 – x)2 (15 – x)
f '(x) = 0 ⇒ 4 (60 – x)2 (15 – x) = 0 ⇒ x = 15, 60
Rejecting a = 60 as 0 < x < 60, we get, x = 15
When x < 15 slightly, f ' (x) = 4 (+) (+) = + ve
When x > 15 slightly, f ' (a) = 4 (+) (–) = – ve
∴ at x = 15, f ' (x) changes from + ve to – ve
∴ f (x) has local maximum at x = 15.
But f (x) has only one extreme point x = 15
∴ f (x) is maximum when x = 15, y = 60 – 15 = 45
∴ x = 15, y = 45
Find two positive numbers a and y such that their sum is 35 and the product x2y5 is a maximum.
Here x + y = 35 ⇒ y = 35 – x ...(1)
Let f (x) = x2 y5 = x2(35 – x)5
∴ f ' (x) = a2· 5 (35 – x)4 (– 1) + (35 – x)5 · 2x
= x (35 – x)4 [ – 5x + 2 (35 – x)] = x (35 – x)4 (– 7x + 70)
= – 7x (35 – x)4 (x – 10)
f ' (x) = 0 ⇒ – 7x (35 – x)4 (x – 10)
⇒ x (x – 10) (35 – x)4 = 0 ⇒ x = 0, 10, 35
Rejecting x = 0. 35 as 0 < x < 35, we get, x = 10
When x < 10 slightly, f ' (x) = – (+)(+) (–) = + ve
When x > 10 slightly , f ' (x) = – (+) (+) (+) = – ve
∴ at x = 10, f ' (x) changes from +ve to negative
∴ f (x) has maximum value at a = 10, y = 35 – 10 = 25
∴ x = 10, y = 25
Find two positive numbers a and y such that their sum is 35 and the product x2y5 is a maximum.
Here x + y = 35 ⇒ y = 35 – x ...(1)
Let f (x) = x2 y5 = x2(35 – x)5
∴ f ' (x) = a2·5 (35 – x)4 (– 1) + (35 – x)5 · 2x
= x (35 – x)4 [ – 5x + 2 (35 – x)] = x (35 – x)4 (– 7x + 70)
= – 7x (35 – x)4 (x – 10)
f ' (x) = 0 ⇒ – 7x (35 – x)4 (x – 10)
⇒ x (x – 10) (35 – x)4 = 0 ⇒ x = 0, 10, 35
Rejecting x = 0, 35 as 0 < x < 35, we get, x = 10
When x < 10 slightly, f ' (x) = – (+)(+) (–) = + ve
When x > 10 slightly , f ' (x) = – (+) (+) (+) = – ve
∴ at x = 10, f ' (x) changes from +ve to negative
∴ f (x) has maximum value at a = 10, y = 35 – 10 = 25
∴ x = 10, y = 25
Divide a number 15 into two parts such that the square of one multiplied with the cube of the other is a maximum.
Let the two parts be x, 15-x.
Let 
Rejecting x = 0, 15 as far these values of x, we have y = 0, which is not possible. 
How should we choose two numbers, each greater than or equal to -2 whose sum is 
so that the sum of the square of the first and cube of the second is the minimum?
Let x and y be two numbers.
...(1)
Let 
When 
Two sides of a triangle are given. Find the angle between them such that the area shall be maximum.
Let 
be the angle between two given sides of lengths a and b
where 
is area of triangle
Now, 
Show that of all the rectangles with a given perimeter, the square has the largest area.
Let x , y be the lengths of sides of rectangle and 2 k be the perimeter.
...(1)
Let 
∴ area of a rectangle of given perimeter is maximum when its sides are equal i.e., when it is a square.
Find the area of the largest rectangle having the perimeter of 200 metres.
Let x, y be the lengths of sides of rectangle having perimeter of 200 metres.
...(1)
Let 
Show that among rectangles of given area, the square has the least perimeter.
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#13 {main}</pre>](/application/zrc/images/qvar/MAEN12079887-5.png)
Show that, of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Let O be the centre of circle of radius a. Let A BCD be the rectangle inscribed in the circle such that AB = x, AD = y.
Now AB2 + BC2 = AC2
∴ x2 + y2 = 4 a2 ...(1)
Let P be the area of rectangle. 
Find the dimensions of the rectangle of greatest area that can be inscribed in a semi-circle of radius r.
Let ABCD be the rectangle inscribed in the semi-circle of radius r such that OC = r.
Let 
and width of rectangle = BC = 
Find the area of greatest rectangle which can be inscribed in a circle of radius 10 cm.
...(1)
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#6 [internal function]: _hx_lambda->execute('Http Error #404')
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#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
#12 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#13 {main}</pre>](/application/zrc/images/qvar/MAEN12079932-2.png)
Find a point on the curve y2 = 4 x, which is nearest to the point (2, 1).
Any point on the parabola 
Let d be the distance between Q (2, 1) and 
Now d is maximum or minimum when D is maximum or minimum.
Find the point on the curve y2 = 2x which is nearest to the point (1, – 4).
Any point on the parabola 
is 
, Let 
Let d be the distance between Q(1, -4) and 
Find the point on the curve y2 = 2x which is nearest to the point (1, 4).
Any point on the parabola y2 = 2x is 
Let Q be (1, 4)
Let a be the distance between 
Now, d is maximum or minimum when D is maximum or minimum.
Find the point on the curve x2 = 8y which is nearest to the point (2, 4).
Let 
be the point on x2 = 8y which is nearest to the point (2, 4).
Now d is maximum or minimum when D is maximum or minimum. 
Find the point on the curve x2 = 4y which is nearest to the point (–1, 2).
Let (x, y) be the point on 
which is nearest to the point (–1, 2).
At 
Find the points on the curve 
which are nearest from the origin.
The equation of curve is
...(1)
Let D be the distance of any point (x, y) on the curve 
from the origin (0, 0).
Now, 
Rejecting 
When 
Find the points on the curve 
which are nearest to the point (0, 5).
Let (x, y) be the point on 
which is nearest to the point (0, 5).
Let d be the distance between (0, 5) and 
Now d is maximum or minimum when D is maximum or minimum.
Again at 
A jet of an enemy is flying along the curve y = x2 + 2. A soldier is placed at the point (3, 2). What is the nearest distance between the soldier and the jet?
Let (x, y) be any point on y = x2 + 2 at which jet is at a particular moment.
∴ jet is at (x, x2 + 2)
Let d be the distance between the jet at (x, x2 + 2) and solider at (3, 2).
∴ d2 = (x – 3)2 + [ (x2 + 2) – 2]2 = (x – 3)2 + (x2)2
∴ d2 = x4 + x2 – 6 x + 9
Let f (x) = d2 = x4 + x2 – 6 x + 9
f ' (x) = 4 x3 + 2 x – 6 = 2 (2 x3 + x – 3) = 2 (x – 1) (2 x2 + 2 x + 3)
f ' (x) = 0 ⇒ 2 (x – 1) (2 x2 + 2 x + 3) = 0
⇒ x = 1 as we reject imaginary values of x.
f ' ' (x) = 12 x2 + 2
At x = 1, f ' ' (x) = 12 + 2 = 14 > 0 ⇒ f (x) has a local minimum at x = 1
But x = 1 is only extreme point
∴ f (x) is minimum at x = 1
∴ nearest distance = d at x = 1
...(1)
A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.
Total length of wire = 36 cm
Let x cm be length of each side of square and y cm be length of each side of quilateral triangle.
Length of wire used for square = 4x cms
and length of wire used for triangle = 3y cms
∴ 4x + 3y = 36 ⇒ 3y = 36 – 4x
...(1)
Let A denote the sum of the areas of square and equilateral triangle.
Length of piece required for square = 4x = 
and length of piece required for triangle = 3y = 36 - 4x
be the radius of semi-circle with centre at O.
A window consists of a semi-circle with a rectangle on its diameter. If the perimeter of the window is 30 metres, find the dimensions of the window in order that its area may be maximum.
Or
A window is in the form of a rectangle surmounted by a semi-circle. If the total perimeter of the window is 30 m. find the dimensions of the window so that maximum light is admitted.
be the radius of semi-circle with centre at O.
area of figure is maximum i.e. maximum light is admitted when length of rectangle = 
be the radius of the semi-circle with centre at O.
area of figure is maximum i.e., maximum light is admitted when length of rectangle = 
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
An open box with a square base is to be made out of a given quantity of sheet of area c2. Show that the maximum volume of the box is 
...(1)
An open topped box is to be constructed by removing equal squares from each corner of a 3 metre by 8 metre rectangular sheet of aluminum and folding up the sides. Find the volume of the largest such box.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off squares from each corner and folding up the flaps. What should be side of the square to be cut off so that the volume of he box is maximum?
...(1)![<pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('com_wiris_plugi...', Array)
#2 [internal function]: _hx_lambda->execute('Http Error #404')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(532): call_user_func_array(Array, Array)
#4 [internal function]: haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Http Error #404')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('haxe_Http_5', Array)
#6 [internal function]: _hx_lambda->execute('Http Error #404')
#7 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(27): call_user_func_array(Array, Array)
#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
#12 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#13 {main}</pre>](/application/zrc/images/qvar/MAEN12080232-1.png)
A square tank of capacity 250 cubic metres has to be dug out. The cost of the land is Rs.50 per square metre. The cost of digging increases with the depth and for the whole tank is Rs 400 h2, where h metres is the depth of the tank. What should be the dimensions of the tank so that the cost be minimum?
Let x side of square base.
Volume of tank = 250 cubic metres.
...(1)
Cost of land = 
Cost of digging = 
Let C be total cost.
When 
Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius is a square of side 
Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.
Let x, x, y be the length, width and height of the cuboid such that base is a square of side x.
...(1)
where V is volume of cuboid.
Let S be surface area of cuboid.
When 
S is minimum when 
When 
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic
centimetres, find the dimensions of the can which has the minimum surface
area?
Or
Of all the closed cylindrical tin cans (right circular) which enclose a given volume of 100 cubic cm., which has the minimum surface area?
...(1)
Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
Let r be the radius of base of circular cylinder and h be its height. Let V be the volume and S be total surface area.
...(1)
Now, 
V is maximum when h = 2r i.e., height of the cylinder is equal to the diameter of the base.
Show that a cylinder of given volume open at the top has minimum total surface area provided its height is equal to the radius of its base.
...(1)
Show that the height of the cylinder, open at the top, of given surface area and greatest volume is equal to the radius of its base.
...(1)
height = radius of base A rectangle is inscribed in a semi-circle of radius r with one of its sides on the diameter of the semi-circle. Find the dimensions of the rectangle so that its area is maximum Find also this area.
Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm.
Let R be radius and l be the slanted height of cone.
Let OD = x, OA = OB = OC = 12, where 12 is a radius of sphere.
Let S be curved surface area of cone.
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('com_wiris_plugi...', Array)
#2 [internal function]: _hx_lambda->execute('Http Error #404')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(532): call_user_func_array(Array, Array)
#4 [internal function]: haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Http Error #404')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('haxe_Http_5', Array)
#6 [internal function]: _hx_lambda->execute('Http Error #404')
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#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
#12 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#13 {main}</pre>](/application/zrc/images/qvar/MAEN12080316-3.png)
For S to be maximum or minimum, 
...(1)
...(2)
(given)
...(1)
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#1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('com_wiris_plugi...', Array)
#2 [internal function]: _hx_lambda->execute('Http Error #404')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(532): call_user_func_array(Array, Array)
#4 [internal function]: haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Http Error #404')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('haxe_Http_5', Array)
#6 [internal function]: _hx_lambda->execute('Http Error #404')
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#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
#12 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#13 {main}</pre>](/application/zrc/images/qvar/MAEN12080332-5.png)
Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.
...(1)
Prove that the radius of the right circular cylinder of greatest curved surface which can be inscribed in a given cone is half of that of the cone.
are similar
Show that the height of a cylinder of maximum volume that can be inscribed in a sphere of radius R is a sphere of radius R is 
Also, find the maximum volume.
Let h be the height and r be the base radius of the inscribed cylinder in a sphere of radius R.
In rt. angled 
Let V be the volume of the cylinder
Find the height of a right circular cylinder of maximum volume, which can be inscribed in a sphere of radius 9 cm.
Solution not provided.
Show that the volume of the greatest cylinder which can be inscribed in a cone of height h and semi-vertical angle 30° is 
Show that height of the cylinder of greatest volume which can be inscribed in a right ciruclar cone of height h and semi vertical angle 
is one-third that of the cone and the greatest volume of cylinder is 
v is maximum when 
Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius 
of the volume of the sphere.Let R be the radius of cone, r be the radius of sphere and OD = x
∴ height of cone i.e. h = x + r
Now x2 + R2 = r2 ⇒ R2 = r2 – x2 ...(1)
Let V be the volume of cone.
Now x = – r is not possible as x cannot be negative.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is 
Let x be radius, h be the vertical height and y be the slant height of the cone.
∴ y2 = x2 + h2 ...(1)
Let V be volume of cone
For V to be maximum or minimum, 
For a given current surface area of right circular cone when the volume is maximum. Prove that the semi-vertical angle is 
where 
Let x be radius, h be the vertical height and y be the slant height of cone.
∴ y2 = h2+ x2 ...(1)
Let V be volume of cone.
Now total surface area of cone = constant
Now, V will be maximum when 
is maximum.
Prove that a conical tent of given capacity will require the least amount of convas when the height is 
times the radius of the base.
Let x be the radius of the base, h be the height and y the slant height of the conical tent.
Let V be the given capacity (i.e., volume)
Let S be the curved surface.
Find the altitude of a right circular cone of maximum curved surface which can be inscribed in a sphere of radius r.
Let R be radius and l be the slanted height of cone.
Let OD = x, OA = OB = OC = r, where r is radius of sphere.
Let S be curved surface area of cone. 
For S to be maximum or minimum, 
When 
(slightly), 
Find the equation of the line through the point (3, 4) which cuts from the first quadrant a triangle of minimum area.
to be maximum or minimum,
Prove that the area of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles.
Let ABC be right angled triangle in which
Let ∆ denote the area of triangle ABC.
Rejecting 
∴ area of a right-angled triangle of given hypotenuse is maximum when the triangle is isosceles.
Let ABC be right-angled triangle in which ∠ABC = 90°, AB = x, AC = y (constant).
Let S denote the perimeter of the triangle.
Find the largest possible area of a right-angled triangle whose hypotenuse is 5 cm long.
Let ABC be given triangle in which
, so that AC = l is hypotenuse.
Let P be any point on AC and 
Let 
Now, 
Now, 
with its vertex at one end of the major axis.
Let APQ be the isosceles triangle inscribed in the ellipse with centre at C. A is (a, 0).
Let P and Q be (a cos ө , b sin ө) and (a cos ө, – b sin ө) respectively.
Let ∆ be area of ∆ APQ
Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m. BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A such that RP2 + RQ2 is minimum.
Let R be a point on AB such that AR = x metres, RB = (20 – x) metres
∴ RP2 = AR2 + AP2 = x2 + (16)2 = x2+ 256
and RQ2 = RB2 + BQ2
= (20 – x)2 + (22)2
= 400+ x2 – 40 x + 484
= x2 – 40 x + 884
Let y = RP2 + RQ2
If length of three sides of a trapezium other than base are equal to 10 cm. then find the area of the trapezium when it is maximum.
From D, draw DP X ⊥ and from C, draw CQ ⊥ AB
so that PQ = 10 cm.
Now ∆APD ≡ ∆QBC
AP = QB = x cm. say.
In rt. ∠ d ∆APD.
DP2 = AD2 – AP3 = 100 – x2 ⇒ DP =
Also 
Let y be area of trapezium.
Rejecting x = -10 as distance x cannot be negative, we get, x = 5
Use differentials to approximate:
Take 
Then 
...(1)
Now, 
is approximately equal to dy and
is approximately equal to 5.02.
...(1)
is approximately equal to dy and
Take 
...(1)
Now, 
is approximately equal to dy
...(1)
is approximately equal to dy
Use differentials to approximate fourth root of 82.
...(1)
is approximately equal to dy.
Use differentials to approximate fourth root of 81.5.
...(1)
Now 
is approximately equal to dy.
Use differentials, find the approximate value of each of the following upto 3 places of decimal:
Now 
is approximately equal to dy
If y = x4 – 10 and if x changes from 2 to 1.99, what is the approximate change in y?
Now, 
is approximately equal to dy and
Here, 
Here y = x4 + 10
Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2 .
f (x) =4 x2+ 5x + 2
Let x = 2 and ∆x = 0.01
Now ∆y = f (x + ∆x) – f (x)
∴ f (x + ∆x) = f (x) + ∆y
⇒ f (x + ∆x) = f (x) + f ' (x) ∆x [∵ dx = ∆x]
⇒ (x + ∆x) = f (x) + (8 x + 5) ∆x
∴ f (3.2) = f (2) + {8(2) + 5 } (0.01)
= [4 (2)2 + 5(2)+ 2] + (16 + 5) (0.01)
= (16 + 10 + 2) +(21) (0.01)
= 28 + 0.21
∴ f (3.2) = 28.21
Find the approximate value of f (5.001), where f (x) = x3 –7x2 + 15.
f (x) = x3 – 7 x2 + 15
Let x = 5, ∆x = 0.001
Now ∆y = f (x + ∆y) – f (x)
∴ f (x + ∆x) = f (x) + ∆x
⇒ f (x + ∆x) = f (x) + f ' (x) ∆x [∵ dx = ∆x]
∴ f (x + ∆x) = f (x)+ (3 x2 – 14 x) ∆y
∴ f (5.001) = f (5) + {3 (5)2 – 14 (5)} (0.001)
= [(5)3 – 7(5)2 +15] + {3 (25) – 70} (0.001)
= (125 – 175 + 15)+ (75 – 70) (0.001)
= (– 35) + 5 (0.001) = – 35 + 0.005
∴ f (5.001) = – 34. 995
Find the approximate value of f(3.02), where f (x) = 3x2 + 5x + 3 .
f (x) = 3 x2 + 5x + 3
Let x = 3 and ∆x = 0.02.
∆y = f (x + ∆x) – f (x)
∴ f (x + ∆x) = f (x) + ∆y = f (x) + f ' (x) ∆x (as dx = ∆x)
⇒ f (x + ∆x) = f (x) + (6 x + 5) ∆x
∴ f (3.2) = f(3) + {6(3) + 5} (0.02)
= [3 (3)2 + 5(3) + 3] + (18 + 5) (0.02)
= (27 + 15 + 3)+ 23 (0.02)
= 45 + 0.46 = 45.046
∴ approximate value of f (3.02) is 45.46.
If f (x) = 3x2 + 15x + 5, then find the approximate value of f (3.02) is
47.66
57.66
67.66
77.66
D.
77.66
f (x) = 3 x2 + 15 x + 5
Let x = 3 and ∆x = 0.02
∆y = f (x + ∆x) – f (x)
∴ f (x + ∆x) = f (x) + ∆y = f (x) + f ' (x) ∆x [∵ dx = ∆x]
⇒ f (x + ∆x) = f (x) + (6 x + 15) ∆x
∴ f (3.02) = f (3)+ {6 (3) + 15} (0.02)
= [3 (3)2 + 15 (3) + 5]+ (18 + 15) (0.02)1111
= (27+ 45 + 5) + 33 (0.02)
= 77 + 0.66 = 77.66
∴ approximate value of f (3.02) is 77.66.
If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume.
If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.
Let r be the radius of the sphere and ∆r be the error in measuring the radius.
∴ r =7 m. ∆r = 0.02 m
Let V be volume of sphere
approximate error in caluclating the volume = 
If the radius of a sphere is measured as 9 m with an error of 0.03 m. then find the approximate error in calculating its surface area.
Let r be the radius of the sphere and ∆r be the error in measuring the radius.
∴ r = 9 m, ∆r = 0.03 m. Let S be the surface area of sphere.
A circular disc of radius 3 cm is being heated. Due to expanison, its radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm.
0.06 x3 m3
C.
0.09 x3 m3Let V be volume of cube of side x metre.
If the sum of the lengths of hypotenuse and a side of a right angled triangle is given, show that the area is maximum when the angle between them is 60°.
.
...(1)
gives us
Find the equations of the tangent and normal to the curve 
Let 
be the equation of the curve.
Rewriting the above equation as,
Differentiating the above function with respect to x, we get,
Slope of the tangent is 
Equation of the tangent is
Slope of the normal is 
Equation of the normal is
If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum when the angle between them is 
Let ABC be the right angle triangle with base b and hypotenuse h.
Given that b+ h = k
Let A be the area of the right triangle.
Differentiating the above function with respect to be, we have
For the area to be maximum, we have
Again differentiating the function in equation (1), with respect to b, we have
Now substituting 
in equation (2), we have
Thus area is maximum at 
Now, 
Let 
be the angle between the base of the triangle and the hypotenuse of the right angle. 
Show that the height of the cylinder of maximum volume, which can be inscribed in a sphere of radius R is 
Also find the maximum volume.
Given, radius of the sphere is R.
Let r and h be the radius and the height of the inscribed cylinder respectively.
We have:
Let Volume of cylinder = V
Differentiating the above function w.r.t r, we have, 
For maxima or minima, 
Now, when 
When 
Hence, the volume of the cylinder is the maximum when the height of the cylinder is 
Find the equation of the normal at a point on the curve x2 = 4y which passes through the point (1, 2). Also, find the equation of the corresponding tangent.
The equation of the given curve is x2 = 4y.
Differentiating w.r.t. x, we get
Let (h, k) be the co-ordinates of the point of contact of the normal to the curve x2 = 4y.
Now, slope of the tangent at (h, k) is given by
Hence, slope of the normal at (h, k) = 
Therefore, the equation of normal at (h, k) is 
Since, it passes through the point (1, 2) we have
Now, (h, k) lies on the curve x2 = 4y, so, we have:
...(3)
Solving (2) and (3), we get,
h = 2 and k = 1.
From (1), the required equation of the normal is:
Also, slope of the tangent = 1
Equation of tangent at (1, 2) is:
The total cost C(x) associated with the production of x units of an item is given by C(x) = 0.005x3 - 0.02 x2 + 30 x +5000. Find the marginal cost when 3 units are produced, whereby marginal cost we mean the instantaneous rate of total cost at any level of output.
Given,
C(x) = 0.05x3 - 0.02 x2 + 30 x +5000
Marginal cost (CM),
Find the equations of the tangent and the normal, to curve 16x2 + 9y2 = 145 at the point (x1,y1) where x1 = 2 and y1 > 0
16x2 + 9y2 = 145 is the curve and points (x1,y1) where
x1 = 2 and y1 >0
⇒ 16 (2)2 + 9
So, the required point is (2,3). Now 16x2 + 9y2 = 145, on differentiating w.r.t. x gives
Find the intervals in which the function f(x) = is
a) Strictly increasing
b) Strictly decreasing
f'(x) = x3 -3x2 - 10 x +24
f'(x) = (x+ 3)(x-2)(x-4)
f(x) is strictly increasing
if f'(x)>0
An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when the depth of the tank is half of its width. If the cost is to be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question?
Let the length, width and height of the open tank be x, x and y units respectively. Then, its volume is x2Y and the total surface area is x2 + 4xy.
It is given that the tank can hold a given quantity of water. This means that its volume is constant. Let it be V. Then,
V = x2y
The cost of the material will be least if the total surface area is least. Let S denote the total surface area.
Then,
S = x2 + 4xy
We have to minimize S object to the condition that the volume V is constant.
Now,
The critical numbers of S are given by dS/dx = 0
Now, ds/dx = 0
Hence, S is minimum when x =2y i.e the depth (height) of the tank is half of its width.
Show that the rectangle of maximum area that can be inscribed in a circle is a square.
Let a rectangle ABCD be inscribed in a circle with radius r.
Let A be the area of the rectangle ABCD.
Therefore, by the second derivative test, is the point of the local maxima of A.
So, the area of the rectangle ABCD is the maximum at
Now,
Hence, the rectangle of the maximum area that can be inscribed in a circle is a square.
Show that the height of the cylinder of maximum volume that can be inscribed in a cone of height h is h.
Let a cylinder be inscribed in a cone of radius R and height h.
Let a radius of the cylinder be r and its height be h1.
It can be easily seen that AGI and ABD are similar.
It can be noted that if h1 = h, then the cylinder cannot be inscribed in the cone.
Therefore, by the second derivative test, h1 = is the point of local maxima of V.
So, the volume of the cylinder is the maximum when h1 = .
Hence, the height of the cylinder of the maximum volume that can be inscribed in a cone of height h is .
Find the equation of the tangent to the curve which is parallel to the line 4x – 2y + 5 = 0
Since, the tangent is parallel to the line 4x - 2y = - 5
Therefore, slope of tangent can be obtained from equation
Comparing equations (i) and (ii), we have,
Thus, substituting the value of x in the above euation,
Equation of tangent is
Find the intervals in which the function f given by is
(i) increasing
(ii) decreasing.
(i) For an increasing function, we should have,
So, f( x ) is increasing on
(ii) For a decreasing function, we should have f ( x ) 0
So, f ( x ) is decreasing on ( -1, 0 ) ( 0, 1 )
Find the volume of the largest cylinder that can be inscribed in a sphere of radius r.
The given sphere is of radius R. Let h be the height and r be the radius of the cylinder inscribed in the sphere.
Volume of cylinder
In right angled triangle
Putting the value of R2 in equation (i), we get
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs Rs. 70 per square metre for the base and Rs. 45 per square metre for sides, what is the cost of least expensive tank?
Let l, b, and h denote the length breadth and depth of the open rectangular tank.
Given h = 2m
V = 8m3
i.e. 2 l b= 8
Surface area, S, of the open rectangular tank of the depth 'h' = l b + 2( l + b ) x h
In this problem, b = , l b = 4 metre, h = 2 metre
For maxima or minima, differentiating with respect to l we get,
l = 2m for minimum 0r maximum
= 4 + 2 x 8 = 4 + 16 = 20 square metres
Base Area = 4 square metres; Lateral surface area = 16 square metres
Cost = 4 x 70 + 16 x 45
= 280 + 720
= Rs. 1000.
Find the equations of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Equation of the curve is y= x3 + 2x + 6
Slope of the normal at point ( x, y ) =
on substitution, we get
Normal to the curve is parallel to the line x + 14y + 4 = 0,
So the slope of the line is the slope of the normal.
When x = 2, y = 18 and when x = -2, y = -6
Therefore, there are two normals to the curve y = x3 + 2x + 6.
Equation of normal through point ( 2, 18 ) is given by:
Equation of normal through point ( -2, -6 ) is given by:
Therefore, the equation of normals to the curve are x + 14y - 254 = 0 and x + 14y + 86 = 0.
Find the values of x for which f(x) = [x(x - 2)]2 is an increasing function. Also, find the points on the curve where the tangent is parallel to x-axis.
So, the tangents to curve f ( x ) is parallel to the x-axis if x = 0, x = 1 or x = 2.
Now points 0, 1 and 2 will divide the number line into 4 disjoint intervals
Which gives us x = 0, 1, 2
Hence, x = 0, y = 0
x = 1, y = 0
x = 2, y = 0
Required points are ( 0, 0 ), ( 1, 1 ), ( 2, 0 ).
Show that the right circular cylinder, open at the top, and of given surface area and maximum volume is such that its height is equal to the radius of the base.
let r and h be the radius and height of the right circular cylinder with the open top.
So surface area of the cylinder S is given by,
Let V be the volume, so
Using this (i)
And in this case radius of base = height.
Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of t heradius of the base. How fast is the sand cone increasing when the height is 4 cm?
The volume of a cone with radius r and height h is given by the formula
According to the question,
Substituting in the formula,
The rate of change of the volume with respect to time is
Find the points on the curve x2 + y2 – 2x – 3= 0 at whichthe tangents are parallel to x-axis.
Let P ( x, y ) be any point on the given curve x2 + y2 - 2 x - 3 = 0.
Tangent to the curve at the point (x, y ) is given by .
Differentiating the equation of the cueve w.r.t. x we get
Let P ( x1, y1 ) be the point on the given curve at which the tangents are parallel to the x-axis.
To get the value of y1 just substitute x1 = 1 in the equation x2 + y2 - 2 x - 3 = 0, we get
( 1 )2 + ( y1 )2 - 2 x 1 - 3 = 0
So, the points on the given curve at which the tangents are parallel to the x-axis are ( 1, 2 ) and ( 1, - 2 ).
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Let the rectangle of length l and breadth b be inscribed in circle of radius a.
Then, the diagonal of the rectangle passes through the centre and is of length 2a cm.
Now, by applying the Pythagoras Theorem, we have:
( 2a )2 = l2 + b2
Thus, frrom the second derivative test, when l = , the area of the rectangle is maximum.
Since l = b = , the rectangle is square.
Hence, of all the rectangles inscribed in the given circle, the square has the maximum area.
Let y be an implicit function of x defined by x2x – 2xxcoty – 1 = 0. Then y′ (1) equals
-1
1
log 2
-log 2
A.
-1
When x = 1, y=π/2
=(xx– cot y)2= cosec2y
xx = cot y + |cosec y|
when x = 1, y=π/2
⇒ xx = cot y + cosec y
diff. w.r.t. to x
Given P(x) = x4+ ax3 + cx + d such that x = 0 is the only real root of P′ (x) = 0. If P(–1) < P(1),then in the interval [–1, 1].
P(–1) is the minimum and P(1) is the maximum of P
P(–1) is not minimum but P(1) is the maximum of P
P(–1) is the minimum but P(1) is not the maximum of P
neither P(–1) is the minimum nor P(1) is the maximum of P
B.
P(–1) is not minimum but P(1) is the maximum of P
P(x) = x4+ ax3+ bx2+ cx + d
P′(x) = 4x3+ 3ax2+ 2bx + c
As P′(x) = 0 has only root x = 0
⇒ c = 0
P′(x) = x(4x2+ 3ax + 2b)
⇒ 4x3+ 3ax + 2b = 0 has non real root.
and 4x2+ 3ax + 2b > 0 ∀ x ∈ [−1, 1].
As P(−1) < P(1) ⇒ P(1) is the max. of P(x) in [−1, 1]
Suppose the cube x3– px + q has three distinct real roots where p > 0 and q > 0. Then which one of the following holds?
The cubic has minima at 
and maxima at –
The cubic has minima at –
and maxima at
The cubic has minima at both 
and- 
The cubic has maxima at both 
and- 
A.
The cubic has minima at and maxima at –
Let f(x) = x3– px + q
Now for maxima/minima
f′(x) = 0
⇒ 3x2– p = 0
⇒ x2
= p/3
The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is
(x – 2)y′2 = 25 – (y – 2)2
(y – 2)y′2 = 25 – (y – 2)2
(y – 2)2y′2= 25 – (y – 2)2
(x – 2)2y′2= 25 – (y – 2)2
C.
(y – 2)2y′2= 25 – (y – 2)2
(x – h)2 + (y – 2)2 = 25
The normal to the curve x = a(1 + cosθ), y = asinθ at ‘θ’ always passes through the fixed point
(a, 0)
(0, a)No
(0,0)
(a,a)
A.
(a, 0)
Eliminating θ, we get (x – a)2 + y2 = a2 .
Hence normal always pass through (a, 0).
Sponsor Area
Sponsor Area
.
is increasing or decreasing for 
.
and strictly increasing on 
is an increasing function of x throughout its domain.
is an increasing function of 
in 
.
is strictly increasing on I.
.
.
is
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The cost price of x items is Rs. 
Find the number of items he should sell to earn maximum profit. 
maximum value. Find the point on the beam at which the bending moment has th
give us 
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