NCERT Solutions for Class 12 Subject-wise · A balloon, which always remains spherical has a...

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NCERT Solutions for Class 12 Subject-wise ● Class 12 Mathematics

● Class 12 Physics

● Class 12 Chemistry

● Class 12 Biology

● Class 12 Accountancy

● Class 12 Business Studies

● Class 12 English

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#418988

Topic: Rate Measurement

Find the rate of change of the area of a circle with respect to its radius r when

(i) r = 3 cm

(ii) r = 4 cm

Solution

(i)

The area of a circle (A) with radius (r) is given by,

A = πr2

Now, the rate of change of the area with respect to its radius is given by,

dA

dr =d

dr (πr2) = 2πr

At r = 3 cm,

dA

dr = 2π(3) = 6π

(ii)


A = πr2

Now, the rate of change of the area with respect to its radius is given by,

dA

dr =d

dr (πr2) = 2πr

At r = 4 cm,

dA

dr = 2π(4) = 8π

#418990


The volume of a cube is increasing at the rate of 8cm3 /s. How fast is the surface area increasing when the length of an edge is 12cm?

Solution

Let x be the length of a side, V be the volume, and S be the surface area of the cube.

Then, V = x3 and S = 6x2

It is given that dV

dt = 8cm3 /s.

Then, by using the chain rule, we have:

∴ 8 =dV

dt =d

dt (x3) ⋅d

dx = 3x2 ⋅dx

dt

⇒dx

dt =8

3x2 . . . . . . . . . (1)

Now, dS

dt =d

dt (6x2) ⋅d

dx = (12x) ⋅dx

dt [By chain rule]

= 12x ⋅dx

dt = 12x ⋅

8

3x2 =32

x

Thus, when x = 12 cm, dS

dt =32

12 cm2 /s =8

3 cm2 /s.

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of 8

3 cm2 /s.

#418995


The radius of a circle is increasing uniformly at the rate of 3cm/s. Find the rate at which the area of the circle is increasing when the radius is 10cm.

Solution

( )




A = πr2

dA

dt =d

dt (πr2) ⋅dr

dt = 2πrdr

dt, [By chain rule]

It is given that,

dr

dt = 3cm/s.

∴dA

dt = 2πr(3) = 6πr

Thus, when r = 10cm,

∴dA

dt = 6π(10) = 60πcm2 /s.

#419005


An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?

Solution

Let x be the length of a side and V be the volume of the cube.

Then, V = x3

dV

dt = 3x2 ⋅dx

dt , (By chain rule)

It is given that,

dx

dt = 3cm/s

∴dV

dt = 3x2(3) = 9x2

Thus, when x = 10cm,

∴dV

dt = 9(10)2 = 900cm3 /s

#419011

Topic: Maxima and Minima

A stone is dropped into a quiet lake and waves move in circles at the speed of 5cm/s. At the instant when the radius of the circular wave is 8cm, how fast is the enclosed area

increasing?

Solution

The area of a circle (A) with radius (r) is given by A = πr2.

Therefore, the rate of change of area (A) with respect to time (t) is given by,

dA

dt =d

dt (πr2) =d

dr (πr2)dr

dt = 2πrdr

dt , [By chain rule]

It is given that dr

dt = 5cm

Thus, when r = 8cm,

dA

dt = 2π(8)(5) = 80πcm2 /s

#419021


The radius of a circle is increasing at the rate of 0.7cm/s. What is the rate of increase of its circumference?

Solution



The circumference of a circle (C) with radius (r) is given by C = 2πr.

Therefore, the rate of change of circumference (C) with respect to time (t) is given by,

dC

dt =dC

dr ⋅dr

dt (By chain rule)

=d

dr (2πr)dr

dt

= 2π ⋅dr

dt

It is given that dr

dt = 0.7cm/s

Hence, the rate of increase of the circumference is 2π(0.7) = 1.4πcm/s.

#419032


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/minute. When x = 8cm and y = 6cm find the rates of change

of (a) perimeter, and (b) the area of the rectangle.

Solution

(i)

It is given that length (x) is decreasing at the rate of 5 cm/min

and the width (y) is increasing at the rate of 4 cm/min,

⇒dx

dt = − 5 cm/min and dy

dt = 4 cm/min

Thus the perimeter (P) of a rectangle is, P = 2(x + y)

dP

dt = 2

dx

dt +dy

dt = 2( − 5 + 4) = − 2 cm/min

Hence, the perimeter is decreasing at the rate of 2 cm/min.

(ii)

It is given that length (x) is decreasing at the rate of 5 cm/min

and the width (y) is increasing at the rate of 4 cm/min,

⇒dx

dt = − 5 cm/min and dy

dt = 4 cm/min

Thus the area (A) of a rectangle is, A = xy

∴dA

dtx = 8 , y = 6

= ydx

dt + xdy

dtx = 8 , y = 6

= 6( − 5) + 8(4) = 2 cm2/min

Hence, the area is increasing at the rate of 2 cm2/min.

#419045


A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon

increases when the radius is 15cm.

Solution

( )

( ) ( )



The volume of a sphere (V) with radius (r) is given by,

V =

4

3 πr3

∴ Rate of change of volume (V) with respect to time (t) is given by,

dV

dt =dV

dr ⋅dr

dt = 4πr2 ⋅dr

dt

It is given that dV

dt = 900cm3 /s.

∴ 900 = 4πr2 ⋅dr

dt

⇒dr

dt =900

4πr2 =225

πr2

Therefore, when radius = 15 cm,

dr

dt =225

π(15)2 =1

π cm/s

#419049


A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10cm.

Solution

The volume of a sphere (V) with radius (r) is given by V =4

3 πr3.

Rate of change of volume (V) with respect to its radius (r) is given by,

dV

dr =d

dr

4

3 πr3 =4

3 π(3πr2) = 4πr2

Therefore, when radius = 10 cm,

dV

dr = 4π(10)2 = 400π.

#419058


A ladder 5m 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?

Solution

Let ym be the height of the wall at which the ladder touches. Also, let the foot of the ladder be xm away from the wall.

Then, by Pythagoras theorem, we have:

x2 + y2 = 25 [Length of the ladder = 5m]

⇒ y = √25 − x2

Then, the rate of change of height (y) with respect to time (t) is given by,

dy

dt =

−x

√25 − x2 ⋅dx

dt

It is given that dx

dt = 2cm/s

∴dy

dt =

−2x

√25 − x2

Now, when x = 4m, we have:

dy

dt =

−2 × 4

√25 − 42 = −8

3

Hence, the height of the ladder on the wall is decreasing at the rate of 8

3 cm/s.

#419296


( )



The radius of an air bubble is increasing at the rate of 1

2cm/s. At what rate is the volume of the bubble increasing when the radius is 1cm?

Solution

The air bubble is in the shape of a sphere.

Now, the volume of an air bubble (V) with radius (r) is given by,

V =4

3 πr3

The rate of change of volume (V) with respect to time (t) is given by,

dV

dt =4

3 πd

dr (r3) ⋅dr

dt

4

3 π(3r2) ⋅dr

dt = 4πr2dr

dt

It is given that dr

dt =1

2 cm/s

Therefore, when r = 1 cm,

dV

dt = 4π(1)21

2 = 2πcm3 /s

Hence, the rate at which the volume of the bubble increases is 2πcm3 /s.

#419303


A balloon, which always remains spherical, has a variable diameter 3

2(2x + 1). Find the rate of change of its volume with respect to x.

Solution

The volume of a sphere (V) with radius (r) is given by,

V =4

3 πr3

It is given that,

d =

3

2 (2x + 1)

∴ r =

3

4 (2x + 1)

∴ V =4

3π

3

43(2x + 1)3 =

9

16π(2x + 1)3

Hence, the rate of change of volume with respect to x is given as,

dV

dx =9

16 πd

dx (2x + 1)3 =9

16 π × 3(2x + 1)2 × 2 =27

8 π(2x + 1)2.

#419312


Sand is pouring from a pipe at the rate of 12cm3 /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 the radius of the

base. How fast is the height of the sand cone increasing when the height is 4cm?

Solution

( )

( )



The volume of a cone (V) with radius (r) and height (h) is given by,

V =1

3 πr2h

It is given that,

⇒ h =1

6 r

⇒ r = 6h

∴ V =1

3 π(6h)2h = 12πh3

The rate of change of volume with respect to time (t) is given by,

dV

dt = 12xd

dh (h3) ⋅dh

dt = 12π(3h2)dh

dt = 36πh2dh

dt

It is also given that dV

dt = 12cm3 /s

Therefore, when h = 4 cm, we have:

12 = 36π(4)2dh

dt

⇒dh

dt =12

36π(16) =1

48π

Hence, when the height of the sand cone is 4cm, its height is increasing at the rate of 1

48πcm/s

#419317


The total cost C(x) in Rupees associated with the production of x units of an item is given by C(x) = 0.007x3 − 0.003x2 + 15x + 4000.

Find the marginal revenue when x = 17.

Solution

Marginal cost is the rate of change of total cost with respect to output.

∴ Marginal cost (MC) =

dC

dx = 0.007(3x2) − 0.003(2x) + 15

= 0.021x2 − 0.006x + 15

Now When x = 17,

MC = 0.021(172) − 0.006(17) + 15 = 0.021(289) − 0.006(17) + 15

= 6.069 − 0.102 + 15 = 20.967

Hence, when 17 units are produced, the marginal cost is Rs. 20.967.

#419324


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.

Solution

Marginal revenue is the rate of change of total revenue with respect to the number of units sold.

⇒ Marginal Revenue (MR)=

dR

dx = 13(2x) + 26 = 26x + 26

Thus when x = 7, MR = 26(7) + 26 = 182 + 26 = 208

Hence, the required marginal revenue is Rs 208.

#419327


The rate of change of the area of a circle with respect to its radius r at r = 6cm is.

A 10π



B 12π

C 8π

D 11π

Solution

The area (A) of circle with radius r is, A = πr2

Thus rate of change of area of circle with radius is,

⇒dA

dr = 2πr.

Hence at r = 6 cm ,dA

dr = 12π cm

#419344

Topic: Nature of the Function

Show that the function given by f(x) = sinx is (a) strictly increasing in 0,π

2 (b) strictly decreasing in π

2, π (c) neither increasing nor decreasing in (0, π).

Solution

The given function is f(x) = sinx.

∴ f ′ (x) = cosx

(a) Since for each x ∈ 0,π

2, , cosx > 0 ⇒ f ′ (x) > 0.

Hence, f is strictly increasing in 0,π

2 .

(b) Since for each x ∈π

2, π , cosx < 0 ⇒ f ′ (x) < 0.

Hence, f is strictly decreasing in π

2, π .

(c) From the results obtained in (a) and (b), it is clear that f is neither increasing nor

decreasing in (0, π).

#419353


Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is (a) strictly increasing (b) strictly decreasing.

Solution

The given function is f(x) = 2x3 − 3x2 − 36x + 7

f ′ (x) = 6x2 − 6x − 36 = 6(x2 − x − 6) = 6(x + 2)(x − 3)

∴ f ′ (x) = 0 ⇒ x = − 2, 3

The points x = 2 and x = 3 divide the real line into three disjoint intervals

i.e., ( − ∞, − 2), ( − 2, 3), and (3, ∞).

In intervals ( − ∞, − 2) and (3, ∞), f ′ (x) > 0

while in interval ( − 2, 3), f ′ (x) < 0

Hence, the given function (f) is strictly increasing in intervals

( − ∞, − 2) and (3, ∞) while function (f) is strictly decreasing in interval ( − 2, 3).

#420089


( ) ( )

( )( )

( )( )



Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x2 + 2x − 5

(b) 10 − 6x − 2x2

(c) −2x3 − 9x2 − 12x + 1

(d) 6 − 9x − x2

(e) f(x) = (x + 1)3(x − 3)3

Solution

(a)

We have,

f(x) = x2 + 2x − 5

∴ f ′ (x) = 2x + 2

Now, f ′ (x) = 0 ⇒ x = − 1

The point x = − 1 divides the real line into two disjoint intervals

i.e., ( − ∞, − 1) and ( − 1, ∞).

In interval ( − ∞, − 1), f ′ (x) = 2x + 2 < 0

∴ f is strictly decreasing in interval ( − ∞, − 1).

And in interval ( − 1, ∞), f ′ (x) = 2x + 2 > 0

Thus, f is strictly increasing for x > 1.

(b)

f(x) = 10 − 6x − 2x2

∴ f ′ (x) = − 6 − 4x

Now, f ′ (x) = 0 ⇒ x = −3

2

The point x = −3

2 divides the real line into two disjoint intervals i.e., − ∞, −

3

2 and −3

2, ∞ .

In interval − ∞, −3

2 i.e., when x < −3

2, f ′ (x) = − 6 − 4x < 0.

∴ f is strictly decreasing for x < −3

2 and in interval −

3

2, ∞ , f ′ (x) > 0

Thus f is strictly increasing in this interval.

(c)

f(x) = − 2x3 − 9x2 − 12x + 1

∴ f ′ (x) = − 6x3 − 18x2 − 12 = − 6(x2 + 3x + 2) = − 6(x + 1)(x + 2)

Now,

f ′ (x) = 0 ⇒ x = − 1 and x = − 2

Points x = − 1 and x = − 2 divide the real line into three disjoint intervals

i.e., ( − ∞, − 2)( − 2, − 1) and ( − 1, − ∞).

In intervals ( − ∞, − 2) and (1, − ∞) i.e., when x < − 2 and x > − 1,

f ′ (x) = − 6(x + 1)(x + 2) < 0

∴ f is strictly decreasing for x < − 2 and x > − 1.

Now, in interval ( − 2, − 1) i.e., when −2 < x < − 1, f ′ (x) = − 6(x + 1)(x + 2) > 0

∴ f is strictly increasing for −2 < x < − 1.

(d)

f(x) = 6 − 9x − x2

f ′ (x) = − 9 − 2x

For f to be strictly increasing f ′ (x) > 0 ⇒ 2x + 9 < 0 ⇒ x < −9

2

And for f to be strictly decreasing, f ′ (x) < 0 ⇒ 2x + 9 > 0 ⇒ x > −9

2

( ) ( )( )

( )



(e)

We have,

f(x) = (x + 1)3(x − 3)3

f ′ (x) = 3(x + 1)2(x − 3)3 + 3(x − 3)2(x + 1)3

= 3(x + 1)2(x − 3)2[x − 3 + x + 1]

= 3(x + 1)2(x − 3)2(2x − 2)

= 6(x + 1)2(x − 3)2(x − 1)

Now,

f ′ (x) = 0 ⇒ x = − 1, 3, 1

The points x = − 1, x = 1, and x = 3 divide the real line into four disjoint intervals

i.e.,( − ∞, − 1), ( − 1, 1), (1, 3) and (3, ∞).

In intervals ( − ∞, − 1) and ( − 1, 1), f ′ (x) = 6(x + 1)2(x − 3)2(x − 1) < 0

∴ f is strictly decreasing in intervals ( − ∞, − 1) and ( − 1, 1).

In intervals (1, 3) and (3, ∞)f ′ (x) = 6(x + 1)2(x − 3)2(x − 1) > 0

∴ f is strictly increasing in intervals (1, 3) and (3, ∞).

#420133


Find the values of x for which y = [x(x − 2)]2 is an increasing function.

Solution

y = [x(x − 2)]2 = [x2 − 2x]2

∴

dy

dx = y ′ = 2(x2 − 2x)(2x − 2) = 4x(x − 2)(x − 1)

∴

dy

dx = 0 ⇒ x = 0, x = 2, x = 1

The points x = 0, x = 1, and x = 2 divide the real line into four disjoint intervals

i.e., ( − ∞, 0), (0, 1), (1, 2) and (2, ∞).

In intervals ( − ∞, 0) and (1, 2) , dy

dx < 0

∴ y is strictly decreasing in intervals ( − ∞, 0) and (1, 2)

However, in intervals (0, 1) and (2, ∞),

dy

dx > 0.

∴ y is strictly increasing in intervals (0, 1) and (2, ∞).

∴ y is strictly increasing for 0 < x < 1 and x > 2.

#420143


Prove that the function f given by f(x) = x2 − x + 1 is neither strictly increasing nor strictly decreasing on ( − 1, 1).

Solution



The given function is f(x) = x2 − x + 1.

∴ f ′ (x) = 2x − 1

Now, f ′ (x) = 0 ⇒ x =

1

2

.

The point 1

2

divides the interval (1, 1) into two disjoint intervals i.e., − 1,1

2 and 1

2, 1

Now, in interval − 1,1

2 , f ′ (x) = 2x − 1 < 0.

Therefore, f is strictly decreasing in interval − 1,1

2 .

However, in interval 1

2, 1 , f ′ (x) > 0 ⇒ . f is increasing.

Hence, f is neither strictly increasing nor decreasing in interval ( − 1, 1).

#420526


Which of the following functions are strictly decreasing on 0,π

2 ?

A cosx

B cos2x

C cos3x

D tanx

Solution

A function f(x) is said to be strictly decreasing on (a, b) if

f ′ (x) < 0 for all x ∈ (a, b)

Option A

f(x) = cosx

f ′ (x) = − sinx

For x ∈ (0,

π

2 ), sinx is positive

i.e. sinx > 0 for x ∈ (0,

π

2 )

−sinx < 0 for x ∈ (0,

π

2 )

⇒ f ′ (x) < 0 for x ∈ (0,

π

2 )

Hence, f(x) = cosx is strictly decreasing on (0,

π

2 )

Option B,

f(x) = cos2x

f ′ (x) = − 2sin2x

Since, 0 < x <

π

2

⇒ 0 < 2x < π

We know that sint > 0 for t ∈ (0, π)

So, sin2x > 0 for 2x ∈ (0, π)

( ) ( )( )

( )( )

( )



So, sin2x > 0 for x ∈ (0,

π

2 )

⇒ − 2sin2x < 0 for x ∈ (0,

π

2 )

⇒ f ′ (x) < 0 for x ∈ (0,

π

2 )

Hence, f(x) = cos2x is strictly decreasing on (0,

π

2 )

Option C,

f(x) = cos3x

f ′ (x) = − 3sin3x

Since, 0 < x <

π

2

⇒ 0 < 3x <

3π

2

⇒ 3x ∈ (0, π) ∪ π ∪ (π,

3π

2 )

We know that sint > 0 for t ∈ (0, π) and sint < 0 for t ∈ (π,

3π

2 )

So, sin3x > 0 for 3x ∈ (0, π) and sin3x < 0 for 3x ∈ (π,

3π

2 )

−3sin3x < 0 for x ∈ (0,

π

3 ) and −3sin3x > 0 for

3x ∈ (π,

3π

2 )

Hence, f(x) is strictly decreasing in x ∈ (0,

π

3 ) and increasing in

(π

3 ,

π

2 )

Hence, f(x) = cos3x is not strictly decreasing on (0,

π

2 )

Option D,

f(x) = tanx

f ′ (x) = sec2x

For 0 < x <

π

2

, sec2x > 0

⇒ f ′ (x) > 0

Hence, f(x) is strictly increasing on (0,

π

2 )

#420540


Find the least value of a such that the function f given by f(x) = x2 + ax + 1 is strictly increasing on (1, 2).

Solution



We have,

f(x) = x2 + ax + 1

∴ f ′ (x) = 2x + a

Now, function f will be increasing in (1, 2), if f ′ (x) > 0 in (1, 2).

⇒ 2x + a > 0

⇒ 2x > − a

⇒ x >−a

2

Therefore, we have to find the least value of a such that

⇒ x >−a

2, when x ∈ (1, 2).

Thus, the least value of a for f to be increasing on (1, 2) is given by,

−a

2 = 1 ⇒ a = − 2

Hence, the required value of a is −2.

#421163


Prove that the function given by f(x) = x3 − 3x2 + 3x − 100 is increasing in R.

Solution

f(x) = x3 − 3x2 + 3x − 100

f ′ (x) = 3x2 − 6x + 3 = 3(x2 − 2x + 1)

= 3(x − 1)2 ≥ 0, ∀x ∈ R.

Hence, the given function (f) is increasing in R.

#422230

Topic: Approximations and Differentials

Find the approximate value of f(2.01), where f(x) = 4x2 + 5x + 2.

Solution

Let x = 2 and Δx = 0.01.

Now, Δy = f(x + Δx) − f(x)

f(x + Δx) = f(x) + Δy

≃ f(x) + f ′ (x) ⋅ Δx, as (dx ≃ Δx)

⇒ f(2.01) ≈ (4x2 + 5x + 2) + (8x + 5)Δx

= [4(2)2 + 5(2) + 2] + [8(2) + 5](0.01)

= 28 + (21)(0.01) = 28 + 0.21 = 28.21

Hence, the approximate value of f(2.01) is 28.21.

#422255


Find the approximate value of f(5.001), where f(x) = x3 − 7x2 + 15.

Solution



Let x = 5 and Δx = 0.001

Now, Δy = f(x + Δx) − f(x)

∴ f(x + Δx) = f(x) + Δy

≈ f(x) + f ′ (x) ⋅ Δx (as dx = Δx)

∴ f(5.001) ≈ (x3 − 7x2 + 15) + (3x2 − 14x)Δx

= [(5)3 − 7(5)2 + 15] + [3(5)2 − 14(5)](0.001)

= (125 + 175 + 15) + (75 − 70)(0.001)

= − 35 + (5)(0.001) = − 35 + 0.005 = − 34.995

Hence, the approximate value of f(5.001) is 34.995

#422286


If the radius of a sphere is measured as 7m with an error of 0.02m, then find the approximate error in calculating its volume.

Solution

Let r be the radius of the sphere and Δr be the error in measuring the radius.

We have given, r = 7 m and Δr = 0.02 m

Now, the volume V of the sphere is given by,

V =4

3 πr2

⇒dV

dr = 4πr2

∴ ΔV ≈ dV =

dV

dr Δr

= (4πr2)Δr

= 4π(7)2(0.02)m3 = 3.92π m3

Which is approximate error in measuring volume

#422298


If the radius of a sphere is measured as 9m with an error of 0.03m, then find the approximate error in calculating in surface area.

Solution

Let r be the radius of the sphere and Δr be the error in measuring the radius.

Then, r = 9 m and Δr = 0.03 m

Now, the surface area of the sphere (S) is given by,

S = 4πr2

⇒dS

dr= 8πr2

∴ dS =dS

dr Δr

= (8πr2)Δr

= 8π(9)2(0.03) m2 = 2.16π m2

Hence, the approximate error in calculating the surface area is 2.16π m2

#422708


If f(x) = 3x2 + 15x + 5, then the approximate value of f(3.02) is.

A 47.66

B 57.66

( )

( )



C 67.66

D 77.66

Solution

We have f(x) = 3x2 + 15x + 5

⇒ f ′ (x) = 6x + 15

Let x = 3, and Δx = .02

Thus using approximation,

f(x + Δx) ≈ f(x) + f ′ (x)Δx = (3x2 + 15x + 5) + (6x + 15)Δx

∴ f(3 + .02) = f(3.02) ≈ (3(3)2 + 15(2) + 5) + (6(3) + 15)(.02)

= 77 + 33(.02) = 77 + .66 = 77.66

#422735


Using differentials, find the approximate value of each of the following up to 3 places of decimal.

(i) √25.3

(ii) √49.5

(iii) √0.6

(iv) (0.009)13

(v) (0.999)1

10

(vi) (15)14

(vii) (26)13

(viii) (255)14

(ix) (82)14

(x) (401)12

(xi) (0.0037)12

Solution

(i)

Consider y = √x. Let x = 25 and Δ x = 0.3.

Then,

Δy = √x + Δx − √x = √25.3 − √25 = √25.3 − 5

⇒ √25.3 = Δy + 5

Now, dy is approximately equal to Δy and is given by,

dy =dy

dx Δx =1

2√x(0.3) [as y = √x]

=1

2√25(0.3) = 0.03

Hence, the approximate value of √25.3 is 5 + .03 = 5.03

(ii)

Consider y = √x.

Let x = 49 and Δx = 0.5.

Then,

Δy = √x + Δx − √x = √49.5 − √49 = √49.5 − 7

⇒ √49.5 = 7 + Δy


dy =dy

dx Δx =1

2√x(0.5) [as y = √x]

=1

2√49(0.5) = (0.5) =

1

14(0.5) = 0.035

Hence, the approximate value of √49.5 is 7 + .035 = 7.035

( )

( )



(iii)

Consider y = √x.

Let x = 1 and Δx = − 0.4.

Then,

Δy = √x + Δx − √x = √0.6 − 1

⇒ √0.6 = 1 + Δy


dy =dy

dx Δx =1

2√x(Δx) [as y = √x]

=1

2√1(2) = ( − 0.4) = − 0.2

Hence, the approximate value of √0.6 is 1 − .02 = .98

(iv)

Consider y = x13 .

Let x = .008, Δx = .001

Δy = (x + Δx)13 − x

13 = (0.009)

13 − 0.2

⇒ (0.009)13 = 0.2 + Δy


dy =dy

dx Δx =1

3 ( x )23

(Δx) [as y=x13 ]

=1

3 × 0.04(0.001) =

0.001

0.12= 0.008

Hence, the approximate value of (0.009)13 is 0.2 + 0.008 = 0.208.

(v)

Consider y = x1

10 . Let x = 1 and Δx = 0.001.

Then,

Δy = (x + Δx)1

10 − x1

10 = (0.009)1

10 − 1

⇒ (0.009)1

10 = 1 + Δy


dy =dy

dx Δx =1

3 ( x )9

10(Δx) [as y = x

110 ]

=1

10( − 0.001) = − 0.0001

Hence, the approximate value of (0.999)1

10 is 1 + ( − 0.0001) = 0.9999.

(vi)

Consider y = x14 .

Let x = 16 and Δx = − 1.

Then,

Δy = (x + Δx)14 − x

14 = (15)

14 − (16)

14 = (15)

14 − 2

⇒ (15)14 = 2 + Δy


dy =dy

dx Δx =1

4 ( x )3

4(Δx) [as y = x

14 ]

=1

4 ( 16 )3

4( − 1) =

−1

4 × 8=

−1

32= − 0.03125

Hence, the approximate value of (15)14 is 2 + ( − 0.03125) = 1.96875.

(vii)

Consider y = x13 .

( )

( )

( )

( )



Let x = 27and Δx = − 1.

Then,

Δy = x + Δx13 − x

13 = (26)

13 − (27)

13 = (26)

13 − 3

⇒ (26)13 − ) = 3 + Δy


dy =dy

dx Δx =1

3(x)23

(Δx) [as y = x13 ]

=1

3(27)2 / 3 ( − 1) = −1

27 = − .037


(viii)

Consider y = x14 . Let x = 256 and Δx = − 1.

Then,

Δy = (x + Δx)14 − x

14 = (255)

14 − (256)

14 = (255)

14 − 4

⇒ (255)14 = 4 + Δy


dy =dy

dx Δx =1

4(x)34

(Δx) [as y = x14 ]

=1

4(256)3 / 4 ( − 1) =1

4(64) ( − 1) = −1

256 = − .0039


(ix)

Consider y = x14 . Let x = 81 and Δx = 1.

Then,

Δy = (x + Δx)14 − x

14 = (82)

14 − (81)

14 = (82)

14 − 3

⇒ (82)14 = Δy + 3


dy =dy

dx Δx =1

4(x)34

(Δx) [as y = x14 ]

=1

4(81)3 / 4 ( − 1) =1

4(27) ( − 1) = −1

108 = − .009


(x)

Consider y = x12 .

Let x = 400 and Δx = 1.

Then,

Δy = √x + Δx − √x = √401 − √400 = √401 − 20

⇒ √401 = 20 + Δy


dy =dy

dx Δx =1

2√x(Δx) [as y = x

12 ]

=1

2√400 (1) =1

40 = .025

Hence, the approximate value of √401 is 20 + .025 = 20.025

(xi)

Consider y = x12 . Let x = 0.0036 and Δx = 0.0001.

Then,

Δy = (x + Δx)12 − (x)

12 = (0.0037)

12 − (.0036)

12 = (0.0037)

12 − 0.06

( )

( )

( )

( )



⇒ (0.0037)12 = 0.06 + Δy


dy =dy

dx Δx =1

2√x(Δx) [as y = x

12 ]

=1

2√.0036 (.0001) =1

2 × .06 (.0001) = .00083

Hence, the approximate value of (0.0037)12 is .06 + .00083 = .06083

#422936


Minimise Z = x + 2y

subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0

Solution

The feasible region determined by the constraints, 2x + y ≥ 3, x + 2y ≥ 6, x ≥ 0, y ≥ 0 is as shown.

The corner points of the feasible region are A(6, 0) and B(0, 3)

The values of Z at these corner points are as follows.

Corner point Z = x + 2y

A(6, 0) 6

B(0, 3) 6

It can be seen that the value of Z at points A and B is same. If we take any other point such as (2, 2) on line x + 2y = 6, then Z = 6

Thus, the minimum value of Z occurs for more than 2 points, and is equal to 6

#422953


Minimise and Maximise Z = x + 2y

subject to x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200; x, y ≥ 0

Solution

( )



The feasible region determined by the constraints x + 2y ≥ 100, 2x − y ≤ 0, 2x + y ≤ 200; x ≥ 0, y ≥ 0 is as shown.

The corner points of the feasible region are A(0, 50), B(20, 40), C(50, 100) and D(0, 200)


Corner point Z = x + 2y

A(0, 50) 100 → Minimum

B(20, 40) 100 → Minimum

C(50, 100) 250

D(0, 200) 400 → Maximum

The maximum value of Z is 400 at (0, 200) and the minimum value of Z is 100 at all the points on the line segment joining the points (0, 50) and (20, 40).

#422978


Maximise Z = x + y, subject to x − y ≤ − 1, x − y ≥ 0, x, y ≥ 0

Solution

The region determined by the constraints Z = x + y, subject to x − y ≤ − 1, x − y ≥ 0, x, y ≥ 0 is as shown.

There is no feasible region and thus, Z has no maximum value.

#422998


One kind of cake required 200 g flour and 25 g of fat, and another kind of cake required 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made

from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes?

Solution



Let there be x cakes of first kind and y cakes of second kind. Therefore, x ≥ 0 and y ≥ 0

The given information can be complied in a table as follows.

Flour (g) Fat (g)

cakes of first kind, x 200 25

cakes of second kind, y 100 50

Availability 5000 1000

∴ 200x + 100y ≤ 5000

⇒ 2x + y ≤ 50

25x + 50y ≤ 1000

⇒ x + 2y ≤ 40

Total numbers of cakes, Z, that can be made are, Z = x + y

The mathematical formulation of the given problem is

Maximise Z = x + y. . . . . . . (1)

subject to the constraints,

2x + y ≤ 50........(2)

x + 2y ≤ 40........(3)

x, y ≥ 0............(4)

The feasible region determined by the system of constraints is as shown.

The corner points are A(25, 0), B(20, 10), O(0, 0) and C(0, 20)


Corner point Z = x + y

A(25, 0) 25

B(20, 10) 30 → Maximum

C(0, 20) 20

O(0, 0) 0

Thus, the maximum numbers of cakes that can be made are 20 of one kind and 10 of the other kind.

#423010


Passage

A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman's time in its making while a cricket bat takes 3 hour of

machine time and 1 hour of craftsman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman's time.

What number of rackets and bats must be made if the factory is to work at full capacity?

Solution



Let the number of rackets and the number of bats to be made be x and y respectively.

The machine time is not available for more than 42 hours.

∴ 1.5x + 3y ≤ 42.........(1)

The craftsman's time is not available for more than 42 hours.

∴ 3x + y ≤ 24.........(2)

The factory is to work at full capacity. Therefore,

1.5x + 3y = 42

3x + y = 24

On solving these equations, we obtain

x = 4, y = 12

Thus, 4 rackets and 12 bats must be made.

#423083


A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A required 5 minutes each for cutting and 10 minutes each for assembling.

Souvenirs of type B required 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours of assembling. The

profit is Rs.5 each for type A and $Rs.6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?

Solution



Let the company manufacture x souverins of type A and y souvenirs of type B.

Therefore,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as follows.

Type A Type B Availability

Cutting (min) 5 8 3 × 60 + 20 = 200

Assembling (min) 10 8 4 × 60 = 240

The profit on type A souvenirs is Rs.5 and on type B souvenirs is Rs.6.Therefore, the constraints are

5x + 8y ≤ 200

10x + 8y ≤ 240 i.e., 5x + 4y ≤ 120

Total profit, Z = 5x + 6y


Maximise Z = 5x + 6y. . . . . . . . . (1)

subject to the constraints

5x + 8y ≤ 200........(2)

5x + 4y ≤ 120........(3)

x, y ≥ 0........(4)


The corner points are A(24, 0), B(8, 20) and C(0, 25)


Corner point Z = 5x + 6y

A(24, 0) 120

B(8, 20) 160 → Maximum

C(0, 25) 150

The maximum value of Z is 200 at (8, 20).

Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs.160

#423199


The corner points of the feasible region determined by the following system linear inequalities:

2x + y ≤ 10, x + 3y ≤ 15, xy ≥ 0 are (0, 0), (5, 0), (3, 4) and (0, 5).

Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is

A p = q

B p = 2q

C p = 3q

D q = 3p

Solution



The maximum value of Z is unique.

It is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5)

∴ Value of Z at (3, 4) = Value of Z at (0, 5)

⇒ p(3) + q(4) = p(0) + q(5)

⇒ 3p + 4q = 5q ⇒ q = 3p

#459555


Show that y = log(1 + x) −

2x

2 + x , x > − 1, is an increasing function of x throughout its domain.

Solution

By watching slope of a function, it can be observed that a function is increasing or decreasing.

If slope is always positive then it means that the function is always increasing.

Here f′ (x) =

1

1 + x−

4

(2 + x)2

=x2

(x + 2)2. (1 + x)

f′ (x) = 0 at x = 0 and is not defined at x = − 1, − 2; it is also clear that it is > 0 for all x > − 1.

As function is always positive in given domain, so this function is increasing in its domain.

#459561


Find the maximum and minimum values, if any of the following function given by:

f(x) = (2x − 1)2 + 3

Solution

Maximum or minimum can be seen by using derivatives.

Steps1: First find first derivative of the function

Step2: Put it equal to zero and find x were first derivative is zero

Step3: Now find second derivative

Step4: Put x for which first derivative was zero in equation of second derivative

Step5: If second derivative is greater than zero then function takes minimum value at that x and if second derivative is maximum then function will take maximum value at that x.

Here f′ (x) = 8x − 4

Putting this equal to 0

f′ (x) = 0

⇒ 8x − 4 = 0

⇒ x =1

2.

Second derivative of this function

f″ (x) = 8 which is positive.

Which means given function will take minimum value at x =1

2

And that is given by

f1

2 = 2 ×1

2− 1 + 3

= 3

Maxima does not exist.

#459562


( ) ( )



Find the maximum and minimum values, if any of the following function given by: f(x) = 9x2 + 12x + 2

Solution






Step5: If second derivative is greater than zero then function takes minimum value at that x and if second derivative is negative then function will take maximum value at that x.

So here f′ (x) = 18x + 12

Putting this equal to 0 we get

18x + 12 = 0

⇒ x =−2

3.

Let's look at the second derivative of this function

f″ (x) = 18

Which is positive, this means that the given function will take minimum value at x =−2

3

Value is given by

f−2

3 = 9−2

32

+ 12 ×−2

3+ 2

⇒ f−2

3 = − 2

#459563



f(x) = − (x − 1)2 + 10

Solution






Step5: If second derivative is greater than zero then function takes minimum value at that x and if second derivative is less than then function will take its maximum value at that

x.

f′ (x) = − 2x + 2

Putting this equal to 0

−2x + 2 = 0

⇒ x = 1.

Now let's look at the second derivative of this function

f″ (x) = − 2

is −2 which is negative.

Which means given function will take maximum value at x = 1 and that value can be found out putting x = 1 in given function

f(1) = − (1 − 1)2 + 10

which is equal to 10

( ) ( )( )



#459564



g(x) = x3 + 1

Solution






Step5: If second derivative is greater than zero then function takes minimum value at that x and if second derivative is negative then function will take maximum value at that x. If

Second derivative is zero them it means that this is the point of inflection.

So here g′ (x) = 3x2

Putting this equal to 0, we get

g′ (x) = 3x2 = 0

⇒ x = 0.

Second derivative of this function is

f″ (x) = 6x

At x = 0

f″ (0) = 6 ∗ 0 = 0

Which means this function has no maxima or minima but will show an inflection i.e slope will change from negative to positive or from positive to negative at x = 0

#459571


Passage

Find the local maxima and local minima, if any of the given functions. Find also the local maximum and local minimum values, as the case may be:

f(x) = x2

Solution

Here f′ (x) = 2x

Putting this equal to zero, we get

2x = 0

⇒ x = 0.

Now let's see the double derivative of this function.

f″ (x) = 2

Which is constant and always positive. So function will take a minimum value at x = 0

Value is f(0) = 02 = 0

#459572


Passage


g(x) = x3 − 3x

Solution









If Second derivative is zero them it means that this is the point of inflection.

f′ (x) = 3x2 − 3


3x2 − 3 = 0

⇒ x = ± 1.


f″ (x) = 6x

At x = 1

f″ (1) = 6

Which is positive, this means function will take minimum value at x = 1

Minimum value of the function is f(1) = − 2

At x = − 1

f″ ( − 1) = − 6

Which is negative, this means function will take maximum value at x = − 1

Maximum value of the function is f(1) = 2

#459573


Find the local maxima and local minima, of the given functions. Also find the local maximum and local minimum values:

h(x) = sinx + cosx, 0 < x <

π

2

Solution

h(x) = sin(x) + cos(x)

Dividing and multiplying by √2

= √2(1

√2sinx +

1

√2cosx)

= √2(cos45sinx + sin45cosx)

= √2(sin(x + 45))

⇒ h(x) = √2(sin(x + 45))

We know that

−1 ≤ sin(x) ≤ 1

⇒ − √2 ≤ √2sin(x) ≤ √2

Here x can take any value

⇒ − √2 ≤ √2sin(x + 45) ≤ √2

So maximum and minimum value of the function h(x) is √2 and −√2

But since domain has been constrained to 0 < x <π

2, therefor h(x) will attain a minimum value of 1 at x = 0 and x =

π

2

#459574


Passage




f(x) = sinx − cosx, 0 < x < 2π

Solution

h(x) = sin(x) − cos(x)


= √2(1

√2sinx −

1

√2cosx)

= √2(cos45sinx − sin45cosx)

= √2(sin(x − 45))

⇒ f(x) = √2(sin(x − 45))

We know that

−1 ≤ sin(x) ≤ 1

⇒ − √2 ≤ √2sin(x) ≤ √2


⇒ − √2 ≤ √2sin(x − 45) ≤ √2

So maximum and minimum value of the function f(x) is √2 and −√2

#459575


Passage


f(x) = x3 − 6x2 + 9x + 15

Solution










f′ (x) = 3x2 − 12x + 9


f′ (x) = 0

3x2 − 12x + 9 = 0

⇒ (x − 1)(x − 3) = 0

⇒ x = 1, 3


f″ (x) = 6x − 12

At x = 1

f″ (1) = 6 ∗ 1 − 12 = − 6

So function will take maximum value at x = 1, which is given by

f(1) = 19

At x = 3

f″ (3) = 6 ∗ 3 − 12 = 6

This is positive at x = 3, so function will take a minimum value at x = 3.

Minimum value is given by f(3) = 33 − 6 ∗ 32 + 9 ∗ 3 + 15 = 15

Minimum value of the function is 15

Maximum value of the function is 19

#459576



g(x) =

x

2 +

2

x , x > 0

Solution




Step 1: First find first derivative of the function

Step 2: Put it equal to zero and find x were first derivative is zero

Step 3: Now find second derivative

Step 4: Put x for which first derivative was zero in equation of second derivative

Step 5: If second derivative is greater than zero then function takes minimum value at that x and if second derivative is negative then function will take maximum value at that x.


g′ (x) =

1

2−

1

x2


g′ (x) =

1

2−

1

x2= 0

⇒ x = ± 2.

Please note that −2 is not in the domain of given function so it is of no use to us


f″ (x) =

4

x3

At x = 2

f″ (2) =

4

23=

1

2

Which is positive, so function will take minimum value at x = 2

Minimum value is given by

f(x) =2

2+

2

2= 2

#459577



g(x) =

1

x2 + 2

Solution

Given, g(x) =

1

x2 + 2

g′ (x) =

−2x

(x2 + 2)2


g′ (x) =

−2x

(x2 + 2)2 = 0

⇒ x = 0.


f″ (x) =

−1

2

which is negative so function will take maximum value at x = 0

Maximum value of the function is

f(0) =

1

0 + 2 =

1

2

#459578




Passage


f(x) = x√1 − x, 0 < x < 1

Solution

Given, f(x) = x√1 − x

f′ (x) =

−x

2√1 − x+ √1 − x

Putting this equal to zero we get

f′ (x) =

−x

2√1 − x+ √1 − x = 0

⇒ − 3x + 2 = 0

⇒ x =2

3.


f″ (x) = −2√1 − x −

2x

√1 − x

4(1 − x)

−1

√1 − x

At x =2

3

f″

2

3 =

− 2 1 −23 −

2 ∗2

3

1 −2

3

4 ( 1 −23 )

−

1

1 −2

3

⇒ f″

2

3 = − 4√3

This is negative so function will take maximum value at x =2

3

Maximum value of the function is

f2

3 =2

3× 1 −

2

3

=2

3√3

#459584


Passage

Find the absolute maximum value and the absolute minimum value of the given functions in the given intervals:

f(x) = x3, x ∈ ( − 2, 2)

Solution

Given: f(x) = x3

f′ (x) = 3x2


f′ (x) = 0 at x = 0

Now lets look at second derivative:

f″ (x) = 6x

At x = 0

f″ (0) = 6 ∗ 0

This is also 0 at x = 0 which means given function will not have a maxima or minima in given domain. It will just have a point of inflection at x = 0

#459586


( )√ √

√( )

( ) √



Passage


f(x) = 4x −

1

2 x2, x ∈ − 2,

9

2

Solution

f′ (x) = 4 − x

Putting this equal to zero

f′ (x) = 4 − x = 0

⇒ x = 4

Now let's look at second derivative of the given function.

f″ (x) = − 1 and as we can see, this is negative for all x; hence f(x) will have it's maxima at x = 4.

Maximum value of the given function is

f(4) = 4 × 4 −

1

2 × 42 = 8

#459587


Passage


f(x) = (x − 1)2 + 3, x ∈ [ − 3, − 1]

Solution

Given, f(x) = (x − 1)2 + 3

f′ (x) = 2x − 2


f′ (x) = 2x − 2 = 0

∴ x = 1


f″ (x) = 2 and as we can see, this is positive for all x

hence f(x) will have it's minima at x = 1 but as you can see that 1 is not in domain of the function so function will have maxima and minima at two extream of domain i.e at − 1 and

−3 respectively.

Minimum value of the given function is 7

Maximum value of the given function is 19

#459588


Find the maximum profit that a company can make if the profit function is given by

p(x) = 41 − 72x − 18x2

Solution

[ ]



Given, p(x) = 41 − 72x − 18x2

p′ (x) = − 72 − 36x

Putting this equal to zero we get x = − 2


f″ (x) = − 36 and as we can see, this is negative for all x; hence f(x) will have it's maxima at x = − 2.

Maximum profit which the company is going to make is

p( − 2) = 41 − 72 × ( − 2) − 18 × (−22)

= 113.

#459593


It is given that at x = 1, the function x4 − 62x2 + ax + 9 attains its maximum value, on the interval [0, 2]. Find the value of a.

Solution

If function is attaining maximum at x = 1, then f′ (1) = 0.

Here f′ (x) = 4x3 − 124x + a

f′ (1) = 4 − 124 + a.

Putting this to zero,we get

4 − 124 + a = 0.

∴ a = 120

#459594


Find the maximum and minimum values of x + sin2x on [0, 2π].

Solution



Let

f(x) = x + sin2x

f ′ (x) = 1 + 2cos2x

For critical points,

1 + 2cos2x = 0

Or

2cos2x = − 1

cos2x =−1

2

2x =2π

3,

4π

3

x =π

3,

2π

3

Now

f(π

3) =

π

3+ sin(

2π

3)

=π

3+ √3

2 ...(i)

f(2π

3) =

2π

3+ sin(

4π

3)

=2π

3−

√3

2 ...(ii)

Hence

Maximum value is

π

3+

√3

2 and Minimum Value is

2π

3−

√3

2

#459599


A square piece of tin of side 18cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of

the square to be cut off so that the volume of the box is the maximum possible.

Solution



let the side of square be x

Then remaining dimensions of cuboid for volume is

length18 − 2x, width 18 − 2x and height x

Volume(v) = (18 − 2x)2xdv

dx= (18 − 2x)2 − 2x(18 − 2x)

dv

dx= (18 − 2x)(18 − 2x − 2x) = (18 − 2x)(18 − 4x)

Nowdv

dx= 0 at x = 9,

9

2

Now,

d2v

dx2= − 4(18 − 2x) − 2(18 − 4x)

at x =9

2= 4.5

d2v

dx2< 0 ∴ at x = 4.5 that is side of square for which we will get maximum volume

#459600


A rectangular sheet of tin 45cm by 24cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the

square to be cut off so that the volume of the box is maximum?

Solution

let the side of square be x

Then remaining dimensions of cuboid for volume is

length45 − 2x, width 24 − 2x and height x

Volume(v) = (45 − 2x)(24 − 2x)xdv

dx= (45 − 2x)(24 − 2x) − 2x(24 − 2x) − 2x(45 − 2x)

dv

dx= 12(x2 − 23x + 90) = 12(x − 5)(x − 18)

Nowdv

dx= 0 at x = 5, 18

Now,

d2v

dx2= 12(2x − 23)

at x = 5d2v

dx2< 0 ∴ at x = 5

that is side of square for which we will get maximum volume

#459601


Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Solution



The diagonal of the rectangle will be diameter of the circle, since the rectangle has all four co-ordinates inscribed on the circumference of the circle.

Hence let the sides of the rectangle be x and y.

Now applying Pythagoras theorem gives us

x2 + y2 = 4r2 where 'r' is the radius.

Now area of the rectangle is

A = xy

= x√4r2 − x2

dA

dx= √4r2 − x2 −

x2

√4r2 − x2

=4r2 − 2x2

√4r2 − x2 = 0

implies 4r2 = 2x2

2r2 = x2

√2r = x

Hence

y = √4r2 − x2

= √4r2 − 2r2

= √2r

Hence

x = y = √2r.

Thus it is a square. Hence the square has the maximum surface area out of all the rectangles inscribed by the circle.

#459613


Using differentials, find the approximate value of

17

81

14

Solution

Here we will assume a function f(x) = x14

Formula for approximation is given by

f(x + △x) = f(x) + f′ (x). △x

f′ (x) =

1

4 (x)− 34

Now most important step is to find △x and for this we need to refer to question.

We need to choose x and one important thing which we need to keep in mind is that x when inserted in f(x) should give a perfect outcome, here x should be the nearest number

to 17

81 and should give a perfect fourth root.

Best number which can be seen is 16

81 so from here we can conclude that △x =

1

81

Plugging all the value in formula f(x + △x) = f(x) + f′ (x). △x

We get 16

81

14 +

1

4

16

81

− 34 .

1

81

This is equal to 65

96.

( )

( ) ( )



#459614


Using differentials, find the approximate value of (33)15

Solution

Here we will assume a function f(x) = x15

Formula for approximation is given by

f(x + △x) = f(x) + f′ (x). △x

f′ (x) =

1

5 (x)− 45

Now most important step is to find △x and for this we need to refer to question.

We need to choose x and one important thing which we need to keep in mind is that x when inserted in f(x) should give a perfect outcome, here x should be nearest number to

33 and should give a perfect fifth root.

Best number which can be seen is 32 so from here we can conclude that △x = 1

Plugging all the value in formula f(x + △x) = f(x) + f′ (x). △x

We get (32)15 +

1

5 (32)− 45 .1

This is equal to 161

80

#459620


Find the intervals in which the function f given by f(x) = x3 +

1

x3, x ≠ 0 is

(i) increasing (ii) decreasing.

Solution

A function f(x) is increasing if f ′ (x) > 0 and decreasing if f ′ (x) < 0

f(x) = x3 +1

x3 f ′ (x) = 3x2 −3

x4 =3x6 − 3

x4

Now,

f ′ (x) > 0 x ∈ ( − 1, 0) and (1, ∞)f ′ (x) < 0 x ∈ ( − ∞, − 1) and x ∈ (0, 1)

#459629

Topic: Turning Points and Points of Inflexion

Passage

Find the points at which the function f given by f(x) = (x − 2)4(x + 1)3 has

point of inflexion

Solution



Derivative of given function is

f′ (x) = 4(x − 2)3(x + 1)3 + 3(x + 1)2(x − 2)4

= (x + 1)2(x − 3)3[7x − 2]

Putting this to zero we get x = − 1, 2,2

7

Double derivative of this function is given by

f″ (x) = 12(x − 2)2(x + 1)3+12(x − 2)3(x + 1)2+6(x − 2)4(x + 1)+12(x − 2)3(x + 1)2

At x = − 1

i. e f″ ( − 1)

f″ ( − 1) = 12( − 1 − 2)2( − 1 + 1)3+12( − 1 − 2)3( − 1 + 1)2+6( − 1 − 2)4( − 1 + 1)+12( − 1 − 2)3( − 1 + 1)2

= 0

Let's do a first derivative test at x =2

7

Value close to the it, to the left f′ (x) > 0 and to the right f

′ (x) < 0 Which shows that this is the point of Maxima.

At x = 2

f″ (2) = 12(2 − 2)2(2 + 1)3+12(2 − 2)3(2 + 1)2+6(2 − 2)4(2 + 1)+12(2 − 2)3(2 + 1)2

= 0

So both at x = − 1 and x = 2 double derivative is 0 and hence function will not have any maxima or minima at these points.

Rather such a point is called as point of inflection.

#459632


Let f be function defined on [a, b] such that f ′ (x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b).

Solution

Let us take any 2 points c1 and c2 such that {c1, c2} ∈ (a, b) and c2 = c1 + h, where h → 0

Now, f ′ (c1) = limh → 0

f(c1 + h) − f(c1)

h =

f(c2) − f(c1)

c2 − c1

Now, it is given that f ′ (x) > 0 ∀x ∈ (a, b).

Therefore, f ′ (c1) > 0

⇒

f(c2) − f(c1)

c2 − c1> 0

From the above fraction, we can conclude that:

1. For c2 > c1, f(c2) > f(c1)

2. For c1 > c2, f(c1) > f(c2)

Hence, f(x) is an increasing function in (a, b).

#459635


A cylindrical tank of radius 10m 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

A 1m3 /h

B 0.1m3 /h

C 1.1m3 /h

D 0.5m3 /h

E None of these

Solution



dv

dt= 314 given

Now V = πr2hdV

dt= πr2

dh

dt314 = 3.14(10)2

dh

dt∴

dh

dt= 1m3 /h

#459783


The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Solution

Volume of sphere is directly proportional to radius

Now,

dr

dt= kr = kt + c

Initially r = 3 at t = 0

3 = 0 + cc = 3r = kt + 3

Now at t = 3 r = 66 = k × 3 + 3 ∴ k = 1r = t + 3

#459784


In a bank, principal increase continuously at the rate of r% per year. Find the value of r if Rs. 100 double itself in 10 years (log_e 2=0.6931)

Solution

dp

dt=

r

100 p100dp

p= rdt100lnp = rt + cp = e

rt

100+ k

at t = 0 p = 100

100 = ek

at t = 10

2p = 200 = er.10

100+ k = e

r10 . ek

2 = er

10

log2 =r

10r = 0.6931 × 10 = 6.9

#459785


In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).

Solution

dp

dt=

5

100 p20.dp

p= dt20lnp = t + cp = e

t

20+ k

at t = 0 p = 1000

1000 = ek

at t = 10

p = e1

2+ k = e0.5. ek

Given that e0.5 = 1.648

p = 1.648 × 1000 = 1648

#459786


In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is

proportional to the number present?

Solution

( ) ( )

( )

( ) ( )

( )



Let the number of bacteria at a given times 't' be N

Then,

dN

dt= kN where k is the proportionality constant.

dN

N= kdt

∫NN0

dN

N= ∫t

0kdt

ln(N

N0) = kt

N = N0ekt ...(i)

If N0 = 105 and t = 2hours and

N = (1 +10

100) × 105

= 1.1 × 105

Hence

lnN

N0= kt

ln(1.1 × 105

105) = 2k

ln(1.1) = 2k

k =1

2ln(1.1)hours − 1

= ln(√1.1)hours − 1

Now

ln(N

N0) = kt

N is now 2 × 105

Hence

N

N0

=2 × 105

105

= 2

Thus

ln(N

N0) = kt implies

ln2 = ln√1.1t



ln2 =ln(1.1)

2t

2ln2 = ln(1.1)t

t =ln4

ln(1.1)

= 14.54 hours

#459857


The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20, 000 in

1999 and 25000 in the year 2004, what will be the population of the village in 2009?

Solution

Let the number of people at a given year 't' be N

Then,

dN

dt= kN where k is the proportionality constant.

dN

N= kdt

∫NN0

dN

N= ∫t

0kdt

ln(N

N0) = kt

N = N0ekt ...(i)

If N0 = 2 × 104 and t = 2004 − 1999

t = 5years and

N = 2.5 × 104

Hence

lnN

N0= kt

ln(2.5 × 104

2 × 104) = 5k

ln(1.25) = 5k

k =1

5ln(1.25)years − 1

Now

t = 2009 − 1999

= 10years



Hence

ln(N

2 × 104 ) =1

5ln(1.25)t

ln(N

2 × 104) =

1

5ln(1.25) × 10

ln(N

2 × 104 ) = ln(1.25) × 2

ln(N

2 × 104) = ln(1.252)

N

2 × 104= 1.252

N = 2 × (1.25)2 × 104

= 3.125 × 104

= 31250



#419292


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.

Solution

The equation of the curve is given as,

6y = x3 + 2

Differentiating both sides w.r.t. t

6dy

dt = 3x2dx

dt + 0

⇒ 2dy

dt = x2dx

dt

It is given that, dy

dt = 8dx

dt ,

Thus we have,

2 8dx

dt = x2dx

dt

⇒ 16dx

dt = x2dx

dt

⇒ (x2 − 16)dx

dt = 0

⇒ x2 − 16 = 0 ⇒ x = ± 4

When x = 4, y =

43 + 2

6 =66

6 = 11

when x = ( − 4), y =

( − 4)3 + 2

6 = −62

6 = −31

3

Hence, the points required on the curve are (4, 11) and − 4,−31

3 .

#419328


The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5. The marginal revenue, when x = 15 is.

A 116

B 96

C 90

D 126

Solution

We have, R(x) = 3x2 + 36x + 5

⇒ Marginal revenue =dR(x)

dx = 6x + 36

Thus at x = 15, Marginal revenue = 6(15) + 36 = 126

#419347


Find the intervals in which the function f given by f(x) = 2x2 − 3x is (a) strictly increasing (b) strictly decreasing.

Solution

( )

( )

( )



The given function is f(x) = 2x2 − 3x.

⇒ f ′ (x) = 4x − 3

∴ f ′ (x) = 0 ⇒ x =3

4

Now, the point 3

4 divides the real line into two disjoint intervals i.e., − ∞,

3

4 and 3

4, − ∞

In interval − ∞,3

4 , f ′ (x) = 4x − 3 < 0.

Hence, the given function (f) is strictly decreasing in interval − ∞,3

4 In interval 3

4, ∞ , f ′ (x) = 4x − 3 > 0.

Hence, the given function (f) is strictly increasing in interval 3

4, ∞ .

#420558


On which of the following intervals is the function f given by f(x) = x100 + sinx − 1 strictly decreasing ?

A (0, 1)

B π

2, π

C π

2, π

D None of these

Solution

f(x) = x100 + sinx − 1

f ′ (x) = 100x99 + cosx

Option A,

For 0 < x < 1 , both 100x9 and cosx are greater than 0.

So,f ′ (x) > 0

Hence, f(x) is strictly increasing on (0, 1)

Option C,

For x ∈ (

π

2 , π],

cos(

π

2 ) = 0, cos(π) = − 1

i.e cosx decreases from 0 to −1 in (0,

π

2 )

At x =

π

2

, 100(

π

2 )99 and x = π ,100(π)99

i.e. 100x99 increases from (

π

2 , π]

Hence, 100x99 + cosx > 0

⇒ f ′ (x) > 0 in (

π

2 , π)

Hence, f(x) is strictly increasing on (

π

2 , π)

( ) ( )( )

( ) ( )( )

( )

( ]



#421159


Let I be any interval disjoint from [ − 1, 1]. Prove that the function f given byf(x) = x +1

x is strictly increasing in interval disjoint from I.

Solution

f(x) = x +1

x

∴ f ′ (x) = 1 −1

x2

f ′ (x) = 0

⇒1

x2 = 1

⇒ x = ± 1

The points x = 1 and x = − 1 divide the real line in three disjoint intervals i.e., ( − ∞, − 1), ( − 1, 1), and (1, ∞)

In interval ( − 1, 1), it is observed that:

⇒ x2 < 1

⇒ 1 −

1

x2 < 0, x ≠ 0

Thus f is strictly decreasing on ( − 1, 1)

and f ′ (x) = 1 −

1

x2 > 0 on ( − ∞, − 1) and (1, ∞)

∴ f is strictly increasing on ( − ∞, − 1) and (1, ∞)

Hence, function f is strictly increasing in interval I disjoint from ( − 1, 1).

Hence, the given result is proved.

#421160


Prove that the function f given by f(x) = logsinx is strictly increasing on 0,π

2 and strictly decreasing on π

2, π

Solution

We have, f(x) = logsinx

∴ f ′ (x) =1

sinx cosx = cotx

In interval 0,π

2 , f ′ (x) = cotx > 0.

∴ f is strictly increasing in 0,π

2 .

In interval π

2 , π , f ′ (x) = cotx < 0.

∴ f is strictly decreasing in π

2 , π .

#421161


Prove that the function f given by f(x) = logcosx is strictly decreasing on 0,π

2 and strictly increasing on π

2, π .

Solution

( ) ( )

( )( )

( )( )

( ) ( )



We have f(x) = logcosx

Differentiate both sides wrt x to get

f′ (x) =

−sinx

cosx = − tanx

comparing it to the graph of h(x) = tanx in the intervals 0,π

2 andπ

2 , π

we can say that the given function is strictly decreasing on 0,π


2 , π

#421162


Prove that the function f given by f(x) = logcosx is strictly decreasing on 0,π


2, π .

Solution

We have, f(x) = logcosx

∴ f ′ (x) =

1

cosx ( − sinx) = − tanx

In the interval 0,π

2 , tanx > 0 ⇒ − tanx < 0.

∴ f ′ (x) < 0 on 0,π

2

∴ f is strictly decreasing on 0,π

2 .

In interval π

2, π , tanx < 0 ⇒ − tanx > 0.

∴ f ′ (x) > 0 on π

2, π

∴ f is strictly increasing on π

2, π

#421164


The interval in which y = x2ex is decreasing is.

A ( − ∞, ∞)

B (2, 0)

C (2, ∞)

D (0, 2)

Solution

y = x2ex

dy

dx = x2ex − 2xex = ex(x2 − 2x)

Now for given function to be decreasing,

dy

dx < 0 ⇒ ex(x2 − 2x) < 0

⇒ x(x − 2) < 0 ⇒ x ∈ (0, 2)

#421173

Topic: Applications on Geometrical Figures

( ) ( )( ) ( )

( ) ( )

( )( )

( )

( )( )

( )



Find points at which the tangent to the curve y = x3 − 3x2 − 9x + 7 is parallel to the x-axis.

Solution

The equation of the given curve is y = x3 − 3x2 − 9x + 7 .

dy

dx = 3x2 − 6x − 9

Now, the tangent is parallel to the x-axis if the slope of the tangent is zero.

∴ 3x2 − 6x − 9 ⇒ x2 − 2x − 3 = 0

⇒ (x − 3)(x + 1) = 0

⇒ x = 3 or x = − 1

When x = 3, y = (3)3 − 3(3)2 − 9(3) + 7 = 27 − 27 − 27 + 7 = − 20.

When x = 1, y = (1)3 − 3(1)2 − 9(1) + 7 = 1 − 3 − 9 + 7 = − 4.

Hence, the points at which the tangent is parallel to the x-axis are (3, − 20) and (1, − 4).

#421174


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).

Solution

If a tangent is parallel to the chord joining the points (2, 0) and (4, 4), then,

The slope of the tangent = the slope of the chord.

The slope of the chord is 4 − 0

4 − 2 =4

2 = 2.

Now, the slope of the tangent to the given curve at a point (x, y) is given by, dy

dx = 2(x − 2)

Since the slope of the tangent = slope of the chord, we have:

2(x − 2) = 2 ⇒ x − 2 = 1 ⇒ x = 3

When x = 3, y = (3 − 2)2 = 1

Hence, the required point is (3, 1).

#421175


Find the point on the curve y = x3 − 11x + 5 at which the tangent is y = x − 11.

Solution

The equation of the given curve is y = x3 − 11x + 5.

The equation of the tangent to the given curve is given as y = x − 11

(which is of the form y = mx + c).

∴ Slope of the tangent = 1

Now, the slope of the tangent to the given curve at the point (x, y) is given by,

dy

dx = 3x2 − 11

Then, we have:

3x2 − 11 = 1 ⇒ 3x2 = 12

⇒ x2 = 4

⇒ x = ± 2

When x = 2, y = (2)3 − 11(2) + 5 = 8 − 22 + 5 = − 9.

When x = − 2, y = ( − 2)3 − 11( − 2) + 5 = − 8 + 22 + 5 = 19.

Hence, the required points are (2, 9) and (2, 19).



#421858


Find points on the curve x2

9 +y2

16 = 1 at which the tangents are

(i) Parallel to x-axis .

(ii) Parallel to y-axis.

Solution

The equation of the given curve is x2

9 +y2

16 = 1

On differentiating both sides with respect to x, we have:

2x

9 +2y

16

dy

dx = 0

⇒dy

dx =−16x

9y

(i)

The tangent is parallel to the x-axis if the slope of the tangent is 0.

i.e.,−16x

9y = 0, which is possible if x = 0..

Then, x2

9 +y2

16 = 1

for x = 0, ⇒= y2 = 16 ⇒ y = ± 4

Hence, the points at which the tangents are parallel to the x-axis are (0, 4) and (0, − 4).

(ii)

The tangent is parallel to the y-axis if the slope of the normal is 0,

which gives

−1− 16

9y =9y

16x = 0 ⇒ y = 0

Then, x2

9 +y2

16 = 1 for y = 0, ⇒ x = ± 3

Hence, the points at which the tangents are parallel to the y-axis are (3, 0) and ( − 3, 0).

#421885


Find the points on the curve y = x3 at which the slope of the tangent is equal to the y coordinate of the point.

Solution

The equation of the given curve is y = x3.

∴dy

dx = 3x2

When the slope of the tangent is equal to the y-coordinate of the point,

then dy

dx = y = 3x2 .

Also, we have y = x3 .

3x2 = x3 ⇒ x2(x − 3) = 0 ⇒ x = 0, x = 3

When x = 0, then y = 0 and when x = 3, then y = 3(3)2 = 27.

Hence, the required points are (0, 0) and (3, 27).

#422024


Find the equation of the normal to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Solution

Given equation of curve is

( )



y = x3 + 2x + 6 .....(1)

Differentiating w.r.t x, we get

dy

dx = 3x2 + 2

Let P(x1, y1) be any point on the curve.

Slope of the tangent to the given curve at P(x1, y1) is

dy

dx ( x1 , y1 )= 3x2

1 + 2

Slope of the normal to the given curve at point P(x1, y1) is

−1

Slope of the tangent at the point(x1, y1)

=

−1

3x2 + 2

The equation of the given line is x + 14y + 4 = 0 which can be written as

y = −

1

14 x −

4

14

(which is of the form y = mx + c)

So, Slope of this line = −

1

14

Since, the normal is parallel to this line.

So, slope of normal = slope of the given line.

∴

−1

3x21 + 2 =

−1

14

⇒ 3x21 + 2 = 14

⇒ 3x21 = 12

x21 = 4

x1 = ± 2

Since, P(x1, y1) lies on the curve

y1 = x31 + 2x1 + 6 (by (1))

When x1 = 2,

⇒ y1 = 8 + 4 + 6 = 18

When x1 = − 2,

⇒ y1 = − 8 − 4 + 6 = − 6

Therefore, there are two normals to the given curve with slope −1

14

and passing through the points (2, 18) and ( − 2, − 6).

Thus, the equation of the normal through (2, 18) is given by,

y − 18 =

−1

14 (x − 2)

⇒ 14y − 252 = − x + 2

⇒ x + 14y − 254 = 0

And, the equation of the normal through ( − 2, − 6) is given by,

y − ( − 6) =

−1

14 [x − ( − 2)]

⇒ y + 6 =−1

14(x + 2)

⇒ 14y + 84 = − x − 2

( )



⇒ x + 14y + 86 = 0

Hence, the equations of the normals to the given curve (which are parallel to the given line) arex + 14y − 254 = 0 and x + 14y + 86 = 0.

#422091


Find the equation of the tangent to the curve y = √3x − 2 which is parallel to the line 4x − 2y + 5 = 0 .

Solution



Given equation of curve is y = √3x − 2

Differentiating w.r.t. x, we get

dy

dx =

1

2√3x − 2 × 3

Let P(x1, y1) be any point on the curve.

Slope of the tangent to the curve at point P(x1, y1) is

dy

dx ( x1 , y1 )=

3

2 3x1 − 2

Given, equation of line is 4x − 2y + 5 = 0.

y = 2x +

5

2

(which is of the form y = mx + c)

Slope of the given line = 2

Since, the tangent to the curve is parallel to the line 4x − 2y + 5 = 0

So, slope of the tangent is equal to the slope of the line.

3

2 3x1 − 2 = 2

⇒ 3x1 − 2 =

3

4

⇒ 3x1 − 2 =

9

16

⇒ 3x1 =

9

16 + 2

⇒ 3x1 =

41

16

⇒ x1 =

41

48

When x1 =

41

48

, y1 = 3

41

48− 2 =

41

16− 2 =

41 − 32

16

=9

16

=

3

4.

Equation of tangent passing through the point 41

48 ,

3

4 is

y −

3

4 = 2 x −

41

48

⇒4y − 3

4 = 2

48x − 41

48

⇒ 4y − 3 =

48x − 41

6

⇒ 24y − 18 = 48x − 41

⇒ 48x − 24y = 23

Hence, the equation of the required tangent is 48x − 24y = 23.

( ) √

√

√

√ ( ) √( ) √ √

( )

( )

( )



#422264


Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1 percent.

Solution

The volume of a cube (V) with side x is given by V = x3 .

∴ dV =dV

dx Δx = (3x2)Δx

= (3x2)(0.001x) [Since given Δx = 1% of x, = 0.01x]

= 0.03x3

Hence, the approximate change in the volume of the cube is 0.03x3 m3

#422273


Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1 percent.

Solution

The surface area of a cube (S) with side x is given by S = 6x2 .

∴ dV =

dS

dxΔx = (12x)Δx

= (12x)( − 0.01x) [Since given as Δx = -1% of x, = − 0.01x]

= − 0.12x2

#422716


The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is :

A 0.06x3m3

B 0.6x3m3

C 0.09x3m3

D 0.9x3m3

Solution

Volume (V) of a cube with side x is given by, V = x3

⇒dV

dx = 3x2 ≡ ΔV = 3x2Δx,

Now it is given that Δx = .03x

∴ ΔV = .09x3 m3

#422900


Maximize Z = 3x + 4y

subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0

Solution

( )

( )



The feasible region determined by the constraints x + y ≤ 4, x ≥ 0, y ≥ 0 is as follows.

The corner points of the feasible region are O(0, 0), A(4, 0) and B(0, 4).

The values of z at these points are as follows.


O(0, 0) 0

A(4, 0) 12

B(0, 4) 16 → Maximum

Therefore, the maximum value of Z is 16 at the point B(0, 4)

#422905


Minimize Z = − 3x + 4y, subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Solution

The feasible region determined by the system of constraints x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 is as follows.

The corner points of the feasible region are O(0, 0), A(4, 0), B(2, 3) and C(0, 4)


Corner point Z = − 3x + 4y

O(0, 0) 0

A(4, 0) −12 → Minimum

B(2, 3) 6

C(0, 4) 16

Therefore, the minimum value of Z is −12 at the point (4, 0)

#422921


Maximize Z = 5x + 3y

subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

Solution



The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0 are as shown.

The corner points of the feasible region are O(0, 0), A(2, 0), B(0, 3) and C(

20

19 ,

45

19 )



O(0, 0) 0

A(2, 0) 10

B(0, 3) 9

C(

20

19 ,

45

19 )

235

19→ Maximum

Therefore, the maximum value of Z is 235

19

at the point (

20

19 ,

45

19 )

#422924


Minimise Z = 3x + 5y

such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0

Solution



The feasible region determined by the system of constraints x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0, and x, y ≥ 0 is as follows.

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are A(3, 0), B(

3

2 ,

1

2 ) and C(0, 2)



A(3, 0) 9

B(

3

2 ,

1

2 )7 → Smallest

C(0, 2) 10

As the feasible region is unbounded, therefore, 7 may or may not be the minimum value of Z

For this, we draw the graph of the inequality, 3x + 5y < 7 and check whether the resulting half plane has points in common with the feasible region or not It can be seen that the

feasible region has no common point with 3x + 5y < 7

Therefore, the minimum value of Z is 7 at (

3

2 ,

1

2 )

#422926


Maximise Z = 3x + 2y

subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0

Solution

The feasible region determined by the constraints x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0 is as shown.

The corner points of the feasible region are A(5, 0), B(4, 3) and C(0, 5).


Corner points Z = 3x + 2y

A(5, 0) 15


C(0, 5) 10

Therefore, the maximum value of Z is 18 at the point (4, 3)



#423017


A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour

on machine B to produce a package of bolts. He earns a profit, of Rs.17.50 per package on nuts and Rs.7.00 per package on bolts. How many packages of each should be

produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?

Solution

Let the manufacturer produce x package of nuts and y packages of bolts.

Therefore, x ≥ 0 and y ≥ 0

The given information can be compiled in a table as follows:

Nuts Bolts Availability

Machine A (h) 1 3 12

Machine B (h) 3 1 12

The profit on a package of nuts is Rs. 17.50 and on a package of bolts is Rs. 7.

Therefore, the constraints are

x + 3y ≤ 12

3x + y ≤ 12

Total profit, Z = 17.5x + 7y


Maximise Z = 17.5x + 7y .........(1)


x + 3y ≤ 12 .......(2)

3x + y ≤ 12 ......(3)

x, y ≥ 0 .....(4)


The corner points are A(4, 0), B(3, 3) and C(0, 4).

The values of Z at these corner points are as follows:

Corner point Z = 17.5x + 7y

O(0, 0) 0

A(4, 0) 70

B(3, 3) 73.5 → Maximum

C(0, 4) 28

The maximum value of Z is Rs.73.50 at (3, 3).

Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of Rs.73.50.

#423026




A factory manufactures two types of screws, A and B. Each type screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic

and 6 minutes on hand operated machines to manufacture a package of screws A, while takes 6 minutes on automatic and 3 minutes on the hand operated machines to

manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs.7 and

screws B at a profit of Rs.10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to

maximise profit? Determine the maximum profit.

Solution

Let the factory manufacture x screws of type A and y screws of type B on each day.

Therefore,

x ≥ 0 and y ≥ 0


Screw A Screw B Availability

Automatic Machine (min) 4 6 4 × 60 = 120

Hand Operated Machine (min) 6 3 4 × 60 = 120

The profit on a package of screw A is Rs.7 and on the package of screws B is Rs.10.

Therefore, the constraints are

4x + 6y ≤ 240

6x + 3y ≤ 240

Total profit, Z = 7x + 10y


Maximise Z = 7x + 10y. . . . . . . . (1)


4x + 6y ≤ 240........(2)

6x + 3y ≤ 240........(3)

x, y ≥ 0



The values of Z at these corner points are as follows


A(40, 0) 280

B(30, 20) 410 → Maximum

C(0, 40) 400

The maximum value of Z is 410 at (30, 20).

Thus, the factory should produce 30 package of screws A and 20 packages of screws B to get the maximum profit of Rs.410

#423041




A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting

machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any

day, the sprayer is available for at least for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs.5 and that from

a shade is Rs.3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?

Solution

Let the cottage industry manufacture x pedestal lamps and y wooden shades. Therefore, x ≥ 0 and y ≥ 0


Lamps Shades Availability

Grinding/Cutting Machine (h) 2 1 12

Sprayer 3 2 20

The profit on a lamp is Rs.5 and on the shades is Rs.3. Therefore, the constraints are

2x + y ≤ 12

3x + 2y ≤ 20

Total profit Z = 5x + 3y


Maximise Z = 5x + 3y. . . . . . . . . (1)


2x + y ≤ 12.......(2)

3x + 2y ≤ 20.....(3)

x, y ≥ 0........(4)



The values of Z at these corner points are as follows


A(6, 0) 30


C(0, 10) 30

The maximum value of Z is 32 at (4, 4)

Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximise his profits.

#459556


Prove that y =

4sinθ

(2 + cosθ) − θ is an increasing function of θ in 0,

π

2

Solution

[ ]



Given, y =4sinθ

(2 + cosθ)− θ

A derivative of a function can be used to see if a function is increasing or decreasing in a given domain.

Here slope of the function dy

dx=

8cosθ + 4

(2 + cosθ)2 and this is always positive in given interval so given function is increasing in this domain.

#459559


Find the equation of the tangent to the curve y = √3x − 2 which is parallel to the line 4x − 2y + 5 = 0

Solution

Slope of line 4x − 2y + 5 = 0 is 2; which means slope of required tangent is also 2.

Curve: y2 = (3x − 2); which is a parabola

Equation of a tangent to parabola with slope m is given by (y − y1) = m(x − x1) +a

m; here (x1, y1) is the vertex of parabola.

Given: y1 = 0; x1 =2

3 and a =

3

8

So equation of tangent to the curve is y = 2 x −2

3 +3

8

#459568



f(x) = |sin4x + 3|

Solution

f(x) = | sin4x + 3 |

We know that, − 1 ≤ sinθ ≤ 1, ∀θ ∈ R

Hence fmax = | 1 + 3 | = 4 and fmin = | − 1 + 3 | = 2

#459569



h(x) = x + 1, x ∈ ( − 1, 1)

Solution

h(x) = x + 1

h ′ (x) = 1 > 0 ⇒ h(x) is strictly increasing function

Hence in the given interval ( − 1, 1)

hmax = h(1) = 1 + 1 = 2 and hmin = − 1 + 1 = 0

#459580


Prove that the following functions do not have maxima or minima:

f(x) = ex

Solution

f(x) = ex

f′ (x) = ex and this is positive for all values of x.

Which means this function will always be increasing function and hence will never have a minimum or maximum.

#459581


( )




g(x) = logx

Solution

Domain for logx is x > 0

Here f′ (x) =

1

x which is always positve in given domain. Which means logx is a increasing function and hence will not have any maxima or minima.

#459582



h(x) = x3 + x2 + x + 1

Solution

Given, h(x) = x3 + x2 + x + 1

h′ (x) = 3x2 + 2x + 1

Here h ′ (x) > 0 for all real x which means, h(x) is a continuous increasing function and will have no maxima or minima.

#459585


Passage


f(x) = sinx + cosx, x ∈ [0, π]

Solution

f(x) = sin(x) + cos(x)


= √2(1

√2sinx +

1

√2cosx)


= √2(sin(x + 45))

⇒ f(x) = √2(sin(x + 45))

We know that

−1 ≤ sin(x) ≤ 1

⇒ − √2 ≤ √2sin(x) ≤ √2


⇒ − √2 ≤ √2sin(x + 45) ≤ √2

But as it has been mentioned that the domain of x is [0, π]

So maximum and minimum value of the function h(x) is √2 and −1

#459589


Find both the maximum value and the minimum value of

3x4 − 8x3 + 12x2 − 48x + 25 on the interval [0, 3]

Solution



f′ (x) = 12x3 − 24x2 + 24x − 48


12x3 − 24x2 + 24x − 48 = 0

Using hit and trial, we get

x = 2 as one of the root

So this can be written as (x − 2)(12x2 + 24) = 0

Putting (12x2 + 24) = 0

We can see that the equation has two imaginary root

So only real root or point or critical point is x = 2


f″ (x) = 36x2 − 48x + 24;

putting 2 in this we get

f″ (2) = 3622 − 48 × 2 + 24

= 72, which is positive.

Hence, f(x) will have it's minima at x = 2.

Minimum value of the function will be

f(2) = 3 × 24 − 8 × 23 − 48 × 2 + 25

= − 39

We also have to look for maxima, so let's look at two extreme of domain so at x = 0, function will take a value of 25 and at x = 3 it will take a value of 16. Therefor maxima is at

x = 0

Maximum value will be,

f(0) = 3 × 04 − 8 × 03 − 48 × 0 + 25

= 25

Minimum value of the given function is −39

Maximum value of the given function is 25

#459590


At what points in the interval [0, 2π], does the function sin2x attain its maximum value?

Solution

Maximum value of function sin2x is 1 and that happens when x =(2n + 1)π

2{where n is (0, 1, 2...)

So number of places where sin(2x) will attain maxima in given domain will be at x =π

2,

3π

2 i.e. for n = 0 and 1.

#459591


What is the maximum value of the function sinx + cosx?

Solution



f(x) = sin(x) + cos(x)


= √2(1

√2sinx +

1

√2cosx)


= √2(sin(x + 45))

⇒ f(x) = √2(sin(x + 45))

We know that

−1 ≤ sin(x) ≤ 1

⇒ − √2 ≤ √2sin(x) ≤ √2


⇒ − √2 ≤ √2sin(x + 45) ≤ √2

Which clearly shows that maximum and minimum value of the function is √2 and −√2

#459602


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.

Solution

Total surface area of the cylinder (S) = 2πrh + 2πr2

∴ h =

S − 2πr2

2πr

...(i)

Volume of cylinder (V) = πr2h

⇒ V = πr2

S − 2πr2

2πr

dV

dr =

1

2 (S − 6πr2)

For maximum volume, dV

dr = 0.

∴ S − 6πr2 = 0

∴ S = 6πr2

Substitiuting the value of S in eq(i), we get

h =

6πr2 − 2πr2

2πr = 2r

Now, d2V

dr2 = − 6πr ⇒ Negative

Hence, the volume of the cylinder is maximumwhen h = 2r or h = d., i.e. diameter of the base.

#459603


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?

Solution

( )



Volume = πr2h = 100

Total surface area is given by 2πrh + 2πr2

= 200

r+ 2πr2

Now let's maximise this

ds

dr=

−200 + 4πr3

r2

Putting this to zero we get

4πr3 − 200 = 0

r =50

π

13

Now lets look at the second derivative

d2s

dr2=

400

r3+ 4π

And this is positive at critical r so we will have minimum area at this r

So height is 100

π

π

50

23

#459604


A wire of length 28m 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?

Solution

( )

( )



Let one part be of length x, then the other part will be 28 − x.

Let the part of the length x be covered into a circle of radius.

2πr = x

⇒ r =x

2π

Area of circle = πr2

= πx

2π2

= x2

4π

Now second part of length 280 − x is covered into a square.

Side of a square =28 − x

4

Area of square =28 − x

42

Thus total area =x2

4π+

28 − x

42

dA

dx=

2x

4π+

2

16(28 − x)( − 1)

=x

2π−

28 − x

8

Lets take dA

dx= 0

Thus x

2π−

28 − x

8= 0 ....(1)

4x = 28π − πx

4x + πx = 28π

x[4 + π] = 28π

x =28π

4 + π

Other part = 28 − x = 28 −28π

4 + π

=112 + 28π − 28π

4 + π

=112

4 + π

Now again differentiating, we get

d2A

dx2=

1

2π+

1

8= + ve

A is minimum.

When x =28π

4 + π and 28 − x =

112

4 + π

#459605


Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8

27 of the volume of the sphere.

Solution

( )

[ ][ ]



Let the centre of the sphere be O and radius be R. Let the height and radius of the variable cone inside the sphere be h and r respectively.

So, in the diagram, OA = OB = R, AD = h, BD = r

OD = AD − OA = h − R

Using Pythagoras Theorem in △OBD,

OB2 = OD2 + BD2

⇒ R2 = (h − R)2 + r2

⇒ R2 = h2 + R2 − 2hR + r2

⇒ r2 = 2hR − h2

Volume of the cone V =

1

3 πr2h =

1

3 π(2hR − h2)h =

2πh2R

3 −

πh3

3


dh= 0

⇒ 0 =

4πhR

3 − πh2

⇒ h =

4R

3

∴ V =

2πR

3

16R2

9 −

64πR3

81 =

(96 − 64)πR3

81 =

32πR3

81 =

8

27 ×

4

3 πR3

We know that the volume of the sphere is Vs =

4

3 πR3

Therefore, V =

8

27 Vs

#459606


Show that the right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base.

Solution

It is given that

V = consant =π

3 r2h

Then

h =3V

π. r2 ...(i)

Now

CSA = πrl

= πr(√h2 + r2)



= πr9V2

π2r4+ r2

=π. r

π. r2√9V2 + π2r6

= √9V2 + π2r6

r

Let A = CSA2

=9V2 + π2r6

r2

=9V2

r2+ π2r4

dA

dr=

−18V2

r3+ 4π2r3 = 0

Hence

18V2

r3= 4π2r3

r6 =18V2

4π2

r3 =3√2V

2π

r2. r =3√2V

2π...(ii)

Now

h =3V

π. r2 Or

r2 =3V

πh

Substituting in ii gives us

r2. r =3√2V

2π

3V

πh. r =

3√2V

2π

r

h=

√2

2

r

h=

1

√2

h = √2r.

Hence proved.

√



#459607


Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan − 1√2

Solution

The slant height is given by

l = √h2 + r2

Now volume is

V =π

3(r2)h

=π

3(l2 − h2)h

dV

dh=

π

3[l2 − h2 − h(2h)]

Now for critical points

dV

dh= 0

Or

l2 − h2 − 2h2 = 0

l2 = 3h2

l = √3h

h

l=

1

√3= cosθ where θ is the semi vertical angle.

Hence

tanθ = √1 − cos2θ

cosθ

=

√2

√3

1

√3

= √2

Therefore

tanθ = √2

Or

θ = tan − 1(√2)

#459608


Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin − 1

1

3.

Solution

( )



The surface area of the cone will be S = πr2 + πrl, where r is the radius and l is the slant height of the cone.

⇒ l =

S − πr2

πr

Also, we know that h2 + r2 = l2, where h is the verical height of the cone.

⇒ h2 + r2 =

S − πr2

πr

2

⇒ h2 =

S2 + π2r4 − 2πSr2

π2r2 − r2 =

S2 + π2r4 − 2πSr2 − π2r4

π2r2 =

S2 − 2πSr2

π2r2

⇒ h =

√S2 − 2πSr2

πr

Now, volume of the cone V =

1

3 πr2h =

1

3 πr2

√S2 − 2πSr2

πr =

r

3√S2 − 2πSr2


dr= 0

⇒

√S2 − 2πSr2

3 +

r

3

−4πSr

2√S2 − 2πSr2 = 0

⇒

S2 − 2πSr2

3 =

2πSr2

3

⇒ S2 = 4πSr2

Since, S ≠ 0, we have

S = 4πr2

Now, l =

S − πr2

πr =

4πr2 − πr2

πr = 3r

Let α be the semi-vertical angle of the cone.

Then sinα =

r

l =

r

3r =

1

3

∴ α = sin − 11

3

#459610


For all real values of x, the minimum value of 1 − x + x2

1 + x + x2 is

A 0

B 1

C 3

D 1

3

E 2

Solution

( )

( )



Let's find the first derivative of this function

(2x − 1)(x2 + x + 1) − (2x + 1)(x2 − x + 1)

(x2 + x + 1)2

Putting this to zero, we get

x2 − 1 = 0

Which gives us x = ± 1

Since we can see that the double derivative is quite difficult to solve, so lets put the value of x in main function and see the result.

So for x = 1 the function is 1

3 and for x = − 1 function is 2

So minimum value is 1

3 at x = 1

#459615


Show that the functions given by f(x) =

logx

x

has maximum at x = e.

Solution

Given, f(x) =logx

x

f′ (x) =

1 − logx

x2


x = e

Which means the given function will have maximum at x = e

#459616


The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3cm per second. How fast is the area decreasing when the two equal sides are equal

to the base?

Solution

Given, side is decreasing at the rate of 3cm per second.

Let's assume that the two equal sides are of length x at any point of time and it has been given that dx

dt= − 3cm per second.

So area of this triangle is b

2x2 −

b2

4

dA

dt=

bx

2 x2 −b2

4

At x = b

dA

dt=

b

√3

dx

dt

And as given in question dx

dt= − 3, so

dA

dt= − b√3

#459622


A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2m and volume is 8m3. If building of tank costs Rs.70 per sq metres

for the base and Rs.45 per square metre for sides. What is the cost of least expensive tank.

Solution

√

√



Let l, b and h = 2 m be the length, breadth and depth of the tank respectively.

Volume of the tank = 8 m3

⇒ lbh = 8

⇒ lb(2) = 8

⇒ lb = 4 m2

⇒ b =4

l

Area of the base = lb = 4 m2

Total area of the sides = 2lh + 2bh = 2(l + b)h = 4(l + b) m2

Total cost of building the tank = 70 × 4 + 45 × 4(l + b) = 280 + 180(l + b) = 280 + 180 l +

4

l

Now, applying AMGM property, we get that

l +4

l≥ 2 l.

4

l

⇒ l +

4

l ≥ 4

Therefore, for the cost to be minimum, l +

4

l

should be minimum, i.e, equal to 4.

Therefore, minimum cost = 280 + 180(4) = 280 + 720 = 1000

Hence, minimum cost of building the tank is Rs. 1000

#459623


The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the

circle.

Solution

Let the radius of circle be r and sides of square be a

According to question, we have

2πr + 4a = ka =k − 2πr

4

Sum of area will be

A = πr2 + a2A = πr2 +k − 2πr

42

Now , we will calculate dA

dr

dA

dr= 2πr + 2

k − 2πr

16 ( − 2π)dA

dr= 0

at

r =k

2(π + 4)

Now,

d2A

dr2= 2π +

π2

2> 0

∴ at r =k

2(π + 4) A has least value

then a =k

(π + 4) putting the value of r in a =

k − 2πr

4

#459624


( )

√

( )

( )



A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10m. Find the dimensions of the window to admit maximum

light through the whole opening.

Solution

Let the length and breadth of the rectangular part of the window be 2l and 2b respectively. Thus the radius of the semicircular arc will be l. The total perimeter of the window will

be 1 length of rectangle, 2 breadths of rectangle and the length of the semicircular arc.

⇒ l + 2b + πl = 10

⇒ (1 + π)l + 2b = 10

⇒ b =1

2[10 − (1 + π)l]

To admit maximum light through the window implies that the area of the window is maximum.

Area of the window will be A = (2l)(2b) +

πl2

2 = 4lb +

πl2

2 = 2l[10 − (1 + π)l] +

1

2 πl2 = 20l − 2l2 − 2πl2 +

1

2 πl2 = 20l − 2l2 −

3

2 πl2

For area to be maximum, dA

dl= 0

⇒ 20 − 4l − 3πl = 0

⇒ l =

20

4 + 3π

b =

1

2 [10 − (1 + π)l] =

1

210 − (1 + π)

20

4 + 3π =

1

2

40 + 30π − 20 − 20π

4 + 3π=

10 + 5π

4 + 3π

Hence, the breadth of the rectangular window is 2b =

20 + 10π

4 + 3π

, the length of the window is 2l =

40

4 + 3π

and the radius of the semicircular arc is l =

20

4 + 3π

#459625


A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle.

Show that the maximum length of the hypotenuse is a23 + b

23

32.

Solution

[ ] [ ]

( )



Let P be a point on the hypotenuse AC of right △ABC such that

PL ⊥ AB = a

PM ⊥ BC = b

Let ∠APL = ∠ACB = θ

AP = asecθ and PC = bcosecθ

Let l be the length of the hypotenuse, then l = AP + PC

⇒ asecθ + bcosecθ, 0 < θ <π

2

Now differentiate l with respect to θ, we get

dl

dθ= asecθ. tanθ − bcosecθ. cotθ

For maxima and minima dl

dθ= 0

Thus asecθ. tanθ = bcosecθ. cotθ

⇒ asin3θ = bcos3θ

a

b=

cos3θ

sin3θ

Thus b

a= tan3θ

⇒ tanθ =b

a

13

Now d2l

dθ2= a(secθ. sec2θ + tanθ. secθ. tanθ)

= − b(cosecθ( − cosec2θ) + cotθ( − cosecθ. cotθ))

= asecθ(sec2θ + tan2θ) + bcosecθ × bcosecθ + cot2θ

Since 0 < θ <π

2, so all t ratios of θ are positive.

Also a > 0 and b < 0

Therefore, d2l

dθ2 is positive.

⇒ l is least when tanθ =b

a

13

Least value of l = asecθ + bcosecθ

=a. a

23 + b

23

a13

+b. a

23 + b

23

b13

= a23 + b

23 (a

23 + b

23 )

= (a23 + b

23 )

32

Hence, proved.

#459630


Find the absolute maximum and minimum values of the function f given by

f(x) = cos2x + sinx, x ∈ [0, π]

Solution

[ ]

[ ]

√ √

√



f′ (x) = − 2cosxsinx + cosx

Putting this to zero, we get

f′ (x) = − 2cosxsinx + cosx

⇒ − 2cosxsinx + cosx = 0

⇒ cosx(2sinx − 1)

⇒ x =π

6 or

π

2

Now let's evaluate the value of the function at critical points and at extreme points of domain.

fπ

6 = cos2π

6 + sinπ

6 =5

4

fπ

2 = cos2π

2 + sinπ

2 = 1

f(0) = cos2(0) + sin(0) = 1

f(π) = cos2(π) + sin(π) = 1

And we can see that Function will have maxima at x =π

6 and will have minima at x =

π

2, 0, π

#459631


Show that the altitude of the right circular cone of maximum volume that can be inscribed on a sphere of radius R is 4R

3

.

Solution

Let the centre of the sphere be O and radius be R. Let the height and radius of the variable cone inside the sphere be h and r respectively.

So, in the diagram, OA = OB = R, AD = h, BD = r

OD = AD − OA = h − R

Using Pythagoras Theorem in △OBD,

OB2 = OD2 + BD2

⇒ R2 = (h − R)2 + r2

⇒ R2 = h2 + R2 − 2hR + r2

⇒ r2 = 2hR − h2

Volume of the cone V =

1

3 πr2h =

1

3 π(2hR − h2)h =

2πh2R

3 −

πh3

3


dh= 0

⇒ 0 =

4πhR

3 − πh2

⇒ h =

4R

3

#459633


( ) ( ) ( )( ) ( ) ( )



Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R

√3. Also find the maximum volume.

Solution

Radius of the sphere = R

Let r and h be the radius and height of the cylinder

Then using Pythagoras theorem:

h=2\sqrt{{R}^{2}-{r}^{2}}

Volume of the cylinder will be V=\pi {r}^{2}h

\Rightarrow V=2\pi {r}^{2}\sqrt{{R}^{2}-{r}^{2}}

\dfrac{dV}{dr}=\dfrac{4\pi r{R}^{2}-6\pi {r}^{3}}{\sqrt{{R}^{2}-{r}^{2}}}

Putting this to zero

\dfrac{dV}{dr}=0

\Rightarrow 4\pi r{R}^{2}-6\pi {r}^{3}=0

\Rightarrow {r}^{2}=\dfrac{2{R}^{2}}{3}

\dfrac{{d}^{2}V}{d{r}^{2}}=\dfrac{4\pi {r}^{2}-22\pi {r}^{2}{R}^{2}+12\pi {r}^{4}+4\pi {R}^{2}{r}^{2}}{{({R}^{2}-{r}^{2})}^{\dfrac{3}{4}}}

It can be seen at {r}^{2}=\dfrac{2{R}^{2}}{3} double derivative is less that zero

So the volume is maximum when {r}^{2}=\dfrac{2{R}^{2}}{3}

Height at maximum volume will be \dfrac{2R}{\sqrt3}

#459634


Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle \alpha is one-third that of the cone and the

greatest volume of cylinder is \cfrac{4}{27}\pi {h}^{3}\tan ^{ 2 }{ \alpha } .

Solution

Let VAB be the cone of height h, semi vertical angle \alpha and let xbe the radius of the base of the cylinder A'B'DC which is inscribed in the cone VAB.

Then OO' height of the cylinder =VO-VO'=(h-x\cot\alpha)\pi x^2

Now differentiating with respect to x,

\dfrac {dV}{dx}=2\pi xh -3\pi x^2\cot \alpha

For maxima or minima V,\dfrac {dV}{dx}=0

2\pi xh-3\pi x^2\cot \alpha=0

\Rightarrow 2\pi x h=3\pi x^2\cot \alpha

\Rightarrow x=\dfrac {3\pi x^2\cot \alpha}{2\pi h}

\Rightarrow 2hx=3\pi x^2\cot \alpha

\Rightarrow 2h=3\pi x\cot \alpha

\Rightarrow x=\dfrac {2h}{3\pi \cot \alpha}

We know \tan \alpha=\dfrac {1}{\cot \alpha}

Thus x=\dfrac {2h}{3}\tan \alpha

Now \dfrac {d^2V}{dx^2}=2\pi h-6\pi x\cot \alpha

When x=\dfrac {2h}{3}\tan \alpha, we have

\dfrac {d^2V}{dx^2}=\pi(2h-4h)=-2\pi h<0

\Rightarrow V is maximum when x=\dfrac {2h}{3}\tan \alpha

\Rightarrow OO'=h-x\cot \alpha

= h-\dfrac {2h}{3}

=\dfrac {h}{3}

The maximum volume of cylinder is V=\pi\left (\dfrac {2h}{3}\tan \alpha\right)^2\left(h-\dfrac {2h}{3}\right)

\Rightarrow V=\dfrac {4}{27}\pi h^3\tan^2\alpha



#459638


The normal at the point (1,1) on the curve 2y+{x}^{2}=3 is

A x+y=0

B x-y=0

C x+y+1=0

D x-y=1

E answer required

Solution

Equation of the curve is {x}^{2}=3-2y

Here slope of tangent is given by m=\dfrac{dy}{dx}=-x=-1

So slope of normal will be \dfrac{-1}{m}=1

Hence, equation is given by (y-1)=1(x-1)

\Rightarrow y=x

#459639


The normal to the curve {x}^{2}=4y passing (1,2) is

A x+y=3

B x-y=3

C x+y=1

D x-y=1

E answer required

Solution

Given, x^2=4y

Slope of tangent to the curve will be 4\dfrac{dy}{dx}=2x

\Rightarrow \dfrac{dy}{dx}=\dfrac{x}{2}

So slope of normal will be -\dfrac{dx}{dy}=\dfrac{-2}{x}

Let the point on the curve be (h,k)

So slope of normal will be \dfrac{-2}{h}

Equation of normal is given by (y-k)=\dfrac{-2}{h}(x-h)

It is given that the normal passes through (1,2)

\Rightarrow (2-k)=\dfrac{-2}{h}(1-h).............(1)

Also, (h,k) lies on the curve so {h}^{2}=4k

From (1) we have

\dfrac{{h}^{3}}{4}=2h+2-2h=2

\Rightarrow {h}^{3}=8

\Rightarrow h=2

k=\dfrac{{h}^{2}}{4} \Rightarrow k=1

So the equation will be

y-1=\dfrac{-2}{2}(x-2)

x+y=3

#459805




Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent

to the curve at that point by 5.

Solution


\dfrac{dy}{dx}+5=x+y

\dfrac{dy}{dx}-y=x-5

IF = e^{\int { -1dx}}=e^{-x}\\ y\times e^{-x}=\int{ e^{-x} \times (x-5) dx}\\ y\times e^{-x}=-(x-4)e^{-x}+c\\ y=-(x-4)+ce^x

and it is passing through (0,2)

2=4+c\\ c=-2

Equation of curve is y=-(x-4)-2e^x



#419332


Show that the function given by f(x) = 3x + 17 is strictly increasing on R.

Solution

Given, f(x) = 3x + 17

f ′ (x) = 3 > 0, ∀x ∈ R.

Thus, the given function is strictly increasing on R.

#419334


Show that the function given by f(x) = e2x is strictly increasing on R.

Solution

Given, f(x) = e2x

f ′ (x) = 2e2x > 0, ∀x ∈ R

Hence f is strictly increasing on R

#420135


Prove that the logarithmic function is strictly increasing on (0, ∞).

Solution

The given function is f(x) = logx, with domain = (0, ∞)

⇒ f ′ (x) =

1

x

It is clear that for x > 0, f ′ (x) =

1

x > 0.

Hence, f(x) = logx is strictly increasing in interval (0, ∞).

#421165


Find the slope of the tangent to the curve y = 3x4 − 4x at x = 4.

Solution

The given curve is y = 3x4 − 4x.

Thus, the slope of the tangent to the given curve at x = 4 is given by,

dy

dxx = 4

= 12x3 − 4x = 4=12(4)3 − 4 = 12(64) − 4 = 764

#421168


Find the slope of the tangent to the curve y =

x − 1

x − 2 , x ≠ 2 at x = 10.

Solution

[ ] [ ]



The given curve is y =

x − 1

x − 2 , x ≠ 2.

∴dy

dx =(x − 2)(1) − (x − 1)(1)

(x − 2)2

=x − 2 − x + 1

(x − 2)2 =−1

(x − 2)2

Thus, the slope of the tangent at x = 10 is given by,

dy

dxx = 10

=

−1

(x − 2)2

x = 10

= −1

64

#421169


Find the slope of the tangent to curve y = x3 − x + 1 at the point whose x-coordinate is 2.

Solution

The given curve is y = x3 − x + 1

∴dy

dx = 3x2 − 1

The slope of the tangent to the curve at x = 2 is given by,

dy

dxx = 2

= 3x2 − 1x = 2 = 3(2)2 − 1 = 11

#421170


Find the slope of the tangent to the curve y = x3 − 3x + 2 at the point whose x coordinate is 3.

Solution

The given curve is y = x3 − 3x + 2

⇒dy

dx = 3x2 − 3

Thus slope of the tangent to the given curve at x = 3 is

dy

dxx = 3

= 3x2 − 3x = 3 = 3(3)2 − 3 = 24

#421172


Find the slope of the normal to the curve x = 1 − asinθ, y = bcos2θ at θ =π

2.

Solution

( ) ( )

( ) ( )

( ) ( )



It is given that x = 1 − asinθ and y = bcos2θ.

∴

dx

dθ = − acosθ

and dy

dθ= 2bcosθ( − sinθ) = − 2bsinθcosθ

dy

dx =

dy

dθ

dx

dθ =−2bsinθcosθ

−acosθ =2b

a sinθ

Therefore, the slope of the tangent at θ =π

2 is given by,

dy

dxθ=

π4

=2b

a

Hence, the slope of the normal at θ =π

2 is given by,

−

1

slope of the tangent atθ =π

4=

−12b

a= −

a

2b

#421176


Find the equation of all lines having slope -1 that are tangents to the curve y =1

x − 1, x ≠ 1.

Solution

The equation of the given curve is y =1

x − 1, x ≠ 1.

The slope of the tangents to the given curve at any point (x, y) is given by,

dy

dx=

−1

(x − 1)2

If the slope of the tangent is −1, then we have:

− 1

(x − 1)2= − 1

⇒ (x − 1)2 = 1

⇒ x − 1 = ± 1

⇒ x = 2, 0

When x = 0, y = − 1 and when x = 2, y = 1.

Thus, there are two tangents to the given curve with the slope −1. These are passing through the points (0, − 1) and (2, 1).

∴ The equation of the tangent through (0, − 1) is given by,

y − ( − 1) = − 1(x − 0) ⇒ x + y + 1 = 0

And equation of the tangent through (2, 1) is given by,

y − 1 = − 1(x − 2)

⇒ y + x − 3 = 0

Hence, the equations of the required lines are y + x + 1 = 0 and y + x − 3 = 0

#421851


Find the equation of all lines having slope 2 which are tangents to the curve y =

1

x − 3 , x ≠ 3.

Solution

( )( )

( )



The equation of the given curve is y =

1

x − 3 , x ≠ 3.

The slope of the tangent to the given curve at any point (x, y) is given by,

dy

dx =

−1

(x − 3)2

If the slope of the tangent is 2, then we have:

−1

(x − 3)2 = 2

⇒ 2(x − 3)2 = − 1

⇒ (x − 3)2 =

−1

2

This is not possible since the L.H.S. is positive while the R.H.S is negative.

Hence, there is no tangent to the given curve having slope 2.

#421852


Find the equations of all lines having slope 0 which are tangent to the curve y =

1

x2 − 2x + 3

.

Solution

The equation of the given curve is y =

1

x2 − 2x + 3

.

The slope of the tangent to the given curve at any point (x, y) is given by,

dy

dx =−(2x − 2)

(x2 − 2x + 3)3 =−2(x − 1)

(x2 − 2x + 3)3

If the slope of the tangent is 0, then we have:

−2(x − 1)

(x2 − 2x + 3)3 = 0

When x = 1, y =1

1 − 2 + 3 =1

2

∴ the equation of the tangent through 1,

1

2 is given by,

y −

1

2 = 0(x − 1) ⇒ y =

1

2

#421862


Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x4 − 6x3 + 13x2 − 10x + 5 at (0, 5).

(ii) y = x4 − 6x3 + 13x2 − 10x + 5 at (1, 3)

(iii) y = x3 at (1, 1)

(iv) y = x2 at (0, 0)

(v) x = cost, y = sint at t =π

4

Solution

(i)

We have

y = x4 − 6x3 + 13x2 − 10x + 5

⇒

dy

dx = 4x3 − 18x2 + 26x − 10

( )



∴dy

dx( 0 , 5 )

= − 10

Thus, the slope of the tangent at (0, 5) is −10. Ergo equation of the tangent is given as,

(y − 5) = − 10(x − 0) ⇒ 10x + y − 5 = 0

The slope of the normal at (0, 5) is −1

Slope of the tangent at(0, 5) =

1

10

.

Therefore, the equation of the normal at (0, 5) is given as,

(y − 5) =

1

10 (x − 0) ⇒ 10y − 50 = x ⇒ x − 10y + 50 = 0

(ii)

We have

y = x4 − 6x3 + 13x2 − 10x + 5

⇒

dy

dx = 4x3 − 18x2 + 26x − 10

∴dy

dx( 1 , 3 )

= 4 − 18 + 26 − 10 = 2

Thus, the slope of the tangent at (1, 3) is 2. Ergo equation of the tangent is given as,

(y − 3) = 2(x − 1) ⇒ 2x − y + 1 = 0


Slope of the tangent at(1, 3) = −

1

2

.


(y − 3) = −

1

2 (x − 1) ⇒ x + 2y − 7 = 0

(iii)

On differentiating with respect to x, we get:

dy

dx = 3x2

dy

dx( 1 , 1 )

= 3(1)2 = 3

Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as,

y − 1 = 3(x − 1) ⇒ y = 3x − 2

The slope of the normal at (1, 1) is − 1

Slope of the tangent at(1, 1) =−1

3.


y − 1 = −1

3 (x − 1) ⇒ x + 3y − 4 = 0

(iv)

We have

y = x2

⇒

dy

dx = 2x

∴dy

dx( 0 , 0 )

= 0

Thus, the slope of the tangent at (0, 0) is 0. Ergo equation of the tangent is given as,

(y − 0) = 0(x − 0) ⇒ y = 0


Slope of the tangent at(0, 0) = − ∞.

( )

( )

( )

( )




(y − 0) = − ∞(x − 0) ⇒ x = 0

#421875


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.

(b) perpendicular to the line 5y − 15x = 13.

Solution



(a)

The equation of the given curve is y = x2 − 2x + 7.

On differentiating with respect to x, we get:

dy

dx = 2x − 2

The equation of the line is 2x − y + 9 = 0. ⇒ y = 2x + 9

This is of the form y = mx + c.

Slope of the line = 2

If a tangent is parallel to the line 2x − y + 9 = 0, then the slope of the tangent is equal to the slope of the line.

Therefore, we have: 2 = 2x − 2

⇒ 2x = 4 ⇒ x = 2

Now, at x = 2

⇒ y = 22 − 2 × 2 + 7 = 7

Thus, the equation of the tangent passing through (2, 7) is given by,

y − 7 = 2(x − 2)

⇒ y − 2x − 3 = 0

Hence, the equation of the tangent line to the given curve (which is parallel to line (2x − y + 9 = 0) is y − 2x − 3 = 0.

(b)

The equation of the line is 5y − 15x = 13.

Slope of the line = 3

If a tangent is perpendicular to the line 5y − 15x = 13,

then the slope of the tangent is −1

Slope of the line=

−1

3.

⇒dy

dx= 2x − 2 =

−1

3

⇒ 2x =−1

3+ 2

⇒ 2x =5

3

⇒ x =5

6

Now, at x =5

6

⇒ y =25

36−

10

6+ 7 =

25 − 60 + 252

36=

217

36

Thus, the equation of the tangent passing through 5

6,

217

36 is given by,

y −217

36 = −1

3x −

5

6

⇒=36y − 217

36−

1

18(6x − 5)

⇒ 36y − 217 = − 2(6x − 5)

⇒ 36y − 217 = − 12x + 10

⇒ 36y + 12x − 227 = 0

Hence, the equation of the tangent line to the given curve

(which is perpendicular to line 5y − 15x = 13) is 36y + 12x − 227 = 0

#421883


Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = − 2 are parallel.

Solution

( )( ) ( )



The equation of the given curve is y = 7x3 + 11.

∴

dy

dx = 21x2

Thus slope of the tangent to a curve at x = 2 is =

dy

dx x = 2= (21x2)x = 2 = 84

and slope of the tangent to a curve at x = − 2 is =

dy

dx x = − 2= (21x2)x = − 2 = 84

It is observed that the slopes of the tangents at both the are equal.

Hence, the two tangents are parallel.

#421956


For the curve y = 4x3 − 2x5, find all the points at which the tangent passes through the origin.

Solution

The equation of the given curve is y = 4x3 − 2x5.


⇒dy

dx = 12x2 − 10x4

Let P(x1, y1) be the point on the curve at which tangent passes through the orgin.

Slope of the tangent at P(x1, y1) is dy

dx x1 , y1

= 12x12 − 10x1

4.

Equation of the tangent at (x1, y1) is given by,

y − y1 = (12x21 − 10x4

1 )(x − x1) ....(1)

Since, the tangent passes through the origin (0, 0)

So, equation (1) becomes

−y1 = (12x21 − 10x4

1 )( − x1)

y1 = 12x31 − 10x5

1 .....(2)

Since, P(x1, y1) lies on the curve

So ,y1 = 4x31 − 2x5

1 ......(3)

From eqn (2) and (3), we get

12x31 − 10x5

1 = 4x31 − 2x5

1

⇒ 8x51 − 8x3

1 = 0

⇒ x51 − x3

1 = 0

⇒ x31 (x2

1 − 1)

⇒ x1 = 0, ± 1

When x1 = 0, y1 = 4(0)3 − 2(0)5 = 0.

When x1 = 1, y1 = 4(1)3 − 2(1)5 = 2.

When x1 = − 1, y1 = 4( − 1)3 − 2( − 1)5 = − 2.

Hence, the required points are (0, 0), (1, 2) and ( − 1, − 2).

#421977


Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are parallel to the x-axis.

Solution

( )

( )

( )



The equation of the given curve is x2 + y2 − 2x − 3 = 0.

On differentiating with respect to x, we have:

2x + 2ydy

dx − 2 = 0

⇒ ydy

dx = 1 − x

⇒dy

dx =1 − x

y

Now, the tangents are parallel to the x-axis if the slope of the tangent is 0.

∴1 − x

y = 0 ⇒ 1 − x = 0 ⇒ x = 1

Putting x = 1 in x2 + y2 − 2x − 3 = 0

⇒ y2 = 4

⇒ y = ± 2

Hence, the points at which the tangents are parallel to the x-axis are (1, 2) and (1, 2).

#421986


Find the equation of the normal at the point (am2, am3 ) for the curve ay2 = x3 .

Solution

Given equation of curve is

ay2 = x3


2ay

dy

dx = 3x2

⇒

dy

dx =

3x2

2ay

Slope of the tangent to the curve at (am2, am3) is

dy

dx( am2 , am3 )

=

3(am2)2

2a(am3) =

3a2m4

2a2m3 =3m

2

Slope of normal at (am2, am3)

=

−1

slope of the tangent at(am2, am3) =

−2

3m

Equation of the normal at (am2, am3) is

y − am3 =

−2

3m (x − am2)

⇒ 3my − 3am4 = − 2x + 2am2

⇒ 2x + 3my − am2(2 + 3m2) = 0

#421989


The slope of the normal to the curve y = 2x2 + 3sinx at x = 0 is.

A 3

B 1

3

C −3

D −1

3

( )



Solution

y = 2x2 + 3sinx

dy

dx = 4x + 3cosx

Thus slope of tangent at x = 0 is 4(0) + 3cos0 = 3

Hence slope of normal at the same point is −

1

m = −

1

3

#421992


The line y = x + 1 is a tangent to the curve y2 = 4x at the point.

A (1, 2)

B (2, 1)

C (1, 4)

D (2, 2)

Solution

We have y = x + 1, y2 = 4x

Solving these, ⇒ (x + 1)2 = 4x ⇒ x2 − 2x + 1 = 0

⇒ (x − 1)2 = 0 ⇒ x = 1 ∴ y = (x + 1)x = 1 = 2

Thus required point of contact is (1, 2)

#459557


Prove that the curves x = y2 and xy = k cut at right angles, if 8k2 = 1.

Solution

Put x = y2 in xy = k

we get y3 = k

⇒ y = k13

⇒ x = k23

x = y2 is a parabola with a =1

2

Equation of a tangent to a parabola at (x1, x2) is given by yy1 =1

2(x + x1)

So slope of the tangent to parabola is given by 2ydy

dx= 1

⇒dy

dx=

1

2y

⇒dy

dx= k

− 13

2

m1 = k− 13

2

Slope of tangent to the curve xy = k is given by

dy

dx= −

k

x2 = − k− 13 .

m2 = − k− 13

Angle between the curve is same as angle between the tangents to the curve at the point of intersection.

If the two curves make right angle then it means that their tangents make a right angle. Which means m1m2 = − 1

⇒ k− 13

2. − k

− 13 = − 1

Which given us k− 23 = 2

8k2 = 1



#459592


Find the maximum value of 2x3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [ − 3, − 1].

Solution

Let's solve for Domain [1, 3]

f′ (x) = 6x2 − 24

Putting this equal to zero we get

f′ (x) = 6x2 − 24 = 0

∴ x = ± 2

Note: −2 not in domain


f″ (x) = 12x

At x = 2, f″ (2) = 12 × 2 = 24

This is positive for x = 2

So function will take minima at x = 2

value will be

f(2) = 2 × 23 − 24 × 2 + 107

= 75

Now let's work for domain [ − 3, − 1]

As we have seen above that the function will have it's critical point at x = − 2

f″ ( − 2) = 12 × ( − 2) = − 24

So function will take maxima at x = − 2

Value will be

f( − 2) = 2 × −23 − 24 × ( − 2) + 107

= 135

We can check at both the extreme of domain to find minima and minima will occur at x = − 3 and the value is 125

#459595


Find the two numbers whose sum as 24 and whose product is as large as possible.

Solution

Let's assume that one number is x

then other number is 24 − x.

Now we have to maximise their product.

So product is given by −x2 + 24x

First derivative of this function is given −2x + 24

The second derivative of the function is given by −2.

Putting 1st derivative to zero we get, x = 12

Which means product is zero only if one number is 12

then other number is 24 − 12 = 12

NOTE: If sum of two number is constant then product is maximum when both the numbers are equal.

#459596




Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.

Solution

We have to maxims xy3 and we have x + y = 60

So put x = 60 − y in xy3

we get

f(y) = 60y3 − y4

First derivative is given by

f′ (y) = 180y2 − 4y3


180y2 − 4y3 = 0

y = 0, y = 45

Now lets look at the second derivative

f″ (y) = 360y − 12y2

f″ (0) = 0

and f″ (45) < 0

So given function will have maxima when y = 45

So x = 60 − 45 = 15

#459597


Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum.

Solution

We have to maxims x2y5 and we have x + y = 35

So put x = 35 − y in x2y5

we get (1225 + y2 − 70y)(y5)

f(y) = 1225y5 + y7 − 70y6

f′ (y) = 6125y4 + 7y6 − 420y5


y = 35, y = 25

f″ (y) = 24500y3 + 42y5 − 2100y4

f″ (35) = 24500 × 353 + 42 × 355 − 2100 × 354

f″ (35) > 0

At y = 25

f″ (25) = 24500 × 253 + 42 × 255 − 2100 × 254

f″ (25) < 0

So given function will have maxima when y = 25

So x = 10

#459598


Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Solution



We have to maximise x3 + y3 and we have x + y = 16

Put x = 16 − y in x3 + y3

we get f(y) = 48y2 − 768y + 4096

First derivative is given by

f′ (y) = 96y − 768

Putting this to zero we get, y = 8

f″ (y) = 96

Given that second derivative is positive

So given function will have minima when y = 8

So x = 16 − 8 = 8

#459611


The maximum value of [x(x − 1) + 1]13 , 0 ≤ x ≤ 1

A 3

4

13

B 1

2

C 1

D 0

E answer required

Solution

Given, [x(x − 1) + 1]13

Derivative of this function will be

1

3 [x(x − 1) − 1]− 23 (2x − 1)

Now let's put this equal to zero, so we get x =1

2.

Now you can find double derivative of this function and see it's value at x =1

2 and you will see that it is negative.

So, function will take maxima at x =1

2 and the maximum value is

3

4

13 .

#459619


Find the intervals in which the function f given by

f(x) =

4sinx − 2x − xcosx

2 + cosx

is

(i) increasing (ii) decreasing.

Solution

( )

( )



a function f(x) is increasing if f'(x)>0 and decreasing if f'(x)<0

f(x) =4sinx − 2x − xcosx

2 + cosxf ′ (x) =

xsinx + 3cosx − 2

2 + cosx+

(sinx)(4sinx − 2x − xcosx)

(2 + cosx)2=

4sin2x + 3cos2x + 4cosx − 4

(cosx + 2)24(1 − cos2x) + 3cos2x + 4cosx − 4

(cosx + 2)2=

4cosx − cos2x

(cosx + 2)2=

cosx(4 − cosx)

(2 + cosx)2

f ′ (x) =cosx(4 − cosx)

(2 + cosx)2

Sign f ′ (x) only depends on cosx because (4 − cosx) and (2 + cosx)2 are always greater than 0.

f ′ (x) when cosx > 0 and f ′ (x) < 0 when cosx < 0

#459627


Passage

Find the points at which the function f given by f(x) = (x − 2)4(x + 1)3 has

local maxima

Solution

f(x) = (x − 2)4(x + 1)3


f′ (x) = 4(x − 2)3(x + 1)3 + 3(x + 1)2(x − 2)4

= (x + 1)2(x − 3)3(7x − 2)


7

duble derivative of this function is given by

f″ (x) = 12(x − 2)2(x + 1)3 + 12(x − 2)3(x + 1)2 + 6(x + 1)(x − 2)4 + 12(x + 1)2(x − 2)3

At x = − 1 and 2 double derivative is 0 and hence function will not have any maxima or minima at these points.

At x =2

7

f″

2

7 = 122

7− 2 2 2

7+ 1 3

+ 122

7− 2 3 2

7+ 1 2

+ 62

7+ 1 (

2

7− 2)

4+ 12

2

7+ 1 2 2

7− 2 3

⇒ f″

2

7 < 0

So function will have local maxima

#459628


Find the points at which the function f given by f(x) = (x − 2)4(x + 1)3 has local minima.

Solution

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )




f′ (x) = 4(x − 2)3(x + 1)3 + 3(x + 1)2(x − 2)4

= (x + 1)2(x − 3)3[7x − 2]


7

Double derivative of this function is given by

f″ (x) = 12(x − 2)2(x + 1)3+12(x − 2)3(x + 1)2+6(x − 2)4(x + 1)+12(x − 2)3(x + 1)2

At x = − 1

i. e f″ ( − 1)

f″ ( − 1) = 12( − 1 − 2)2( − 1 + 1)3+12( − 1 − 2)3( − 1 + 1)2+6( − 1 − 2)4( − 1 + 1)+12( − 1 − 2)3( − 1 + 1)2

= 0

Let's do a first derivative test at x =2

7

Value close to the it, to the left f′ (x) > 0 and to the right f

′ (x) < 0 Which shows that this is the point of Maxima.

At x = 2

f″ (2) = 12(2 − 2)2(2 + 1)3+12(2 − 2)3(2 + 1)2+6(2 − 2)4(2 + 1)+12(2 − 2)3(2 + 1)2

= 0

So both at x = − 1 and x = 2 double derivative is 0 and hence function will not have any maxima or minima at these points.

#459636


The slope of the tangent to the curve x = t2 + 3t − 8, y = 2t2 − 2t − 5 at the point (2, − 1) is

A 22

7

B 6

7

C 7

6

D −6

7

E answer required

Solution

Slope to any curve is given by dy

dx

Here, dx

dt = 2t + 3 ...(1)

And dy

dt = 4t − 2 ...(2)

Dividing (2) by (1), we get dy

dx =

dydt

dxdt

⇒dy

dx=

4t − 2

2t + 3

At point (2, − 1), t = 2

So slope is given by,

⇒dy

dx=

4 × 2 − 2

2 × 2 + 3=

6

7

#459637




The line y = mx + 1 is a tangent to the curve y2 = 4x, if the value of m is

A 1

B 2

C 3

D 1

2

E answer required

Solution

Lets put y = mx + 1 in y2 = 4x we get

(mx + 1)2 = 4x

⇒ m2x2 + 1 + 2mx − 4x = 0

Tangent touches a curve at one point so determinant of the equation should be zero.

⇒ (2m − 4)2 − 4 × 1 × m2 = 0

⇒ 4m2 + 16 − 16m − 4m2 = 0

⇒ m = 1

#459640


The points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes are

A

4, ±

8

3

B

4,

−8

3

C

4, ±

3

8

D

± 4,

3

8

E answer required

Solution

Let the point on curve be (h, k)

y2 = x3

92y.

dy

dx=

1

9.3x2 ∴

dy

dx= x2

6y

dy

dx gives slope of curve at given point

dy

dxat (h, k) =

h2

6k

but this is the slope of tangent and we need slope of normal

As we know that slope of normal × slope of tangent is −1

∴ slope of normal is −6k

h2

Given in question that normal makes equal intercepts on axes which means slope of normal is either 1 or −1

−6k = h2 and 6k = h2

Now go through the option you will that option A satisfy our equation

( )

( )

( )

( )



#459782


At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4, -3). Find the equation of the curve given

that it passes through (-2, 1).

Solution


dy

dx= 2

y + 3

x + 4

dy

(y + 3)= 2

dx

(x + 4)ln(y + 3) = 2ln(x + 4) + lncln(y + 3) = ln(x + 4)2 + lnc(y + 3) = c. (x + 4)2

and it passes through ( − 2, 1)

(1 + 3) = c( − 2 + 4)24 = c4c = 1

equation of curve is (y + 3) = (x + 4)2

#459804


Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Solution


dy

dx= x + y

dy

dx− y = x

IF = e ∫ − 1dx = e − xy × e − x = ∫e − x × xdxy × e − x = ( − x − 1)e − x + cy = ( − x − 1) + cex

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