NCERT Solutions for Class 12 Maths Chapter 6—Application of Derivatives

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,

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

 

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,
=

(a) When r = 3 cm, then

sq. cm

(b) When r = 4 cm, then

sq. cm

 

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

Given: Rate of increase of volume of cube = 8 cm3/sec

Then, V = x3 and S = 6x2 where x is a function of time t

It is given tha

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.

 

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,
=

It is given that,

Hence, the rate at which the area of the circle is increasing when the radius is 10 cm is 60π cm2/s.

 

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

V = x3.

Hence, the volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm long.

 

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,

Hence, when the radius of the circular wave is 8 cm, the enclosed area is increasing at the rate of 80π cm2/s.

 

Since the length (x) is decreasing at the rate of 5 cm/minute and the width (y) is increasing at the rate of 4 cm/minute, we have:

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

 

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

V = 4/3πr3

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

Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is 1/π cm/s.

 

Since, V = 4/3πr3

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

Hence, the volume of the balloon is increasing at the rate of 400π cm2.

 

Let y m be the height of the wall at which the ladder touches. Also, let the foot of the ladder be x maway from the wall.

Then, by Pythagoras theorem, we have:

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

⇒ y = √25 – x2

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

Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/s.

 

Given: Equation of the curve  6y = x3 + 2……….(i)

The rate of change of the position of the particle with respect to time (t) is given by,

 

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 = 4/3πr3

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

Hence, the rate at which the volume of the bubble increases is 2π cm3/s.

 

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

V = 4/3πr3

It is given that:

Diameter 3/2 (2x + 1)

 

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

V = 4/3πr3

It is given that,

h = 1/6r ⇒r = 6h

Hence, when the height of the sand cone is 4 cm, its height is increasing at the rate of 1/48π cm/s.

 

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

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

 

Marginal Revenue (MR) =

When x = 7,

MR = 26(7) + 26 = 182 + 26 = 208

Hence, the required marginal revenue is Rs 208.

 

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

A = πr2

Therefore, the rate of change of the area with respect to its radius r is

Hence, the required rate of change of the area of a circle is 12π cm2/s.

The correct answer is B.

 

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

∴ Marginal Revenue (MR)= dR/dx= 3(2x) + 36 = 6x + 36

∴ When x = 15,

MR = 6(15) + 36 = 90 + 36 = 126

Hence, the required marginal revenue is Rs 126.

The correct answer is D.

 

Given: f'(x) = e2x

Now,

x ∈ R

Since the value of  e2x   is always positive for any real value of x, e2x > 0.

⇒2e2x > 0

⇒f'(x) > 0

So f(x) is incerasing on R.

 

Given:

The given function is f(x) = sin x.

 

Given:

 

(a)Given:

     

 

(a) Given:

 

Given:

∴dydx=11+x-(2+x)(2)-2x(1)(2+x)2=11+x-4(2+x)2=x2(1+x)(2+x)2

Now, dydx=0

⇒x2(1+x)(2+x)2=0⇒x2=0  [(2+x)≠0 as x>-1]⇒x=0

Since x > -1 , point = 0 divides the domain (−1, ∞) in two disjoint intervals i.e., −1 < x  < 0 and x > 0

When −1 < x < 0, we have:

x<0⇒x2>0x>-1⇒(2+x)>0⇒(2+x2)>0

∴ y’=x2(1+x)(2+x)2>0

Also, when x > 0

x>0⇒x2>0, (2+x)2>0

∴ y’=x2(1+x)(2+x)2>0

Hence, function f  is increasing throughout this domain

 

 

 

Given: The given function is f(x) =  logx

∴ f'(x) = 1/x

It is clear that for x > 0,  f'(x) = 1/x > 0.

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

 

Given: The given function f (x) = x2 – x + 1

hence,f is neither strictly increasing nor decreasing on the interval (-1,1)

 

 

 

Given:

 

 

Given:

∴ 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.

 

Given:

 

Given:

 

Given:

 

Given:

 

It is given that x =a cos3θ and y =a sin3θ.

 

The equation of the given curve is y =x3 + 2x + 6.

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

∴ Slope of the normal to the given curve at any point (x,y) =

The equation of the given line is x + 14y + 4 = 0.

x + 14y + 4 = 0⇒(which is of the form y =mx +c)

∴Slope of the given line = -1/14

If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.

When x = 2, y = 8 + 4 + 6 = 18.

When x = −2, y = − 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,

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

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

 

The equation of the given parabola is y2 = 4ax.

On differentiating y2 = 4ax with respect to x, we have:

 

The given curve isy = 3x4 − 4x.

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

 

The given curve is .

Hence, the slope of the tangent at x = 10 is -1/64

 

The given curve is y =x3 −x + 1

 

The given curve is y =x3 − 3x + 2

 

It is given thatx = 1 −a sinθ and y =b cos2θ.

 

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

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 = 12.

Hence, the points at which the tangent is parallel to thex-axis are (3, −20) and (−1, 12).

 

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,

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

 

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:

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

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

Hence, the required points are (2, −9) and (−2, 19). But, both these points should satisfy the equation of the tangent as there would be point of contact between tangent and the curve.

∴ (2, −9) is the required point as (−2, 19) is not satisfying the given equation of tangent.

 

The equation of the given curve is y = 1/x-1 ≠ 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:

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

Thus, there are two tangents to the given curve having 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)

⇒y +1 = −x

⇒y +x +1 = 0

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

y − 1 = −1 (x − 2)

⇒y − 1 = −x + 2

⇒y +x − 3 = 0

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

 

The equation of the given curve is y = 1/x-3 ≠ 3

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

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

 

The equation of the given curve is

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

 

The equation of the given curve is

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

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

 

(i) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5.

On differentiating with respect tox, we get:

(ii) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5.

On differentiating with respect tox, we get:

(iii) The equation of the curve isy =x3.

On differentiating with respect tox, we get:

(iv) The equation of the curve isy =x2.

On differentiating with respect tox, we get:

(v) The equation of the curve is x = cost, y = sint.

 

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

On differentiating with respect tox, we get:

dy/dx = 2x – 2

(a) The equation of the line is 2x −y + 9 = 0.

2x −y + 9 = 0⇒y = 2x+ 9

This is of the formy = 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,x = 2

y = 4 − 4 + 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.

5y − 15x = 13⇒ y = 3x + 13/5

This is of the formy =mx+c.

∴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

Thus, the equation of the tangent passing throughis given by,

Hence, the equation of the tangent line to the given curve (which is perpendicular to line 5y − 15x = 13) is 36y + 12x – 227 = 0

 

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

It is observed that the slopes of the tangents at the points where x = 2 and x = −2 are equal.

Hence, the two tangents are parallel.

 

The equation of the given curve isy =x3.

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

When the slope of the tangent is equal to they-coordinate of the point, theny = 3x2.

Also, we havey =x3.

∴3x2 =x3

⇒x2 (x − 3) = 0

⇒x = 0,x = 3

Whenx = 0, theny = 0 and whenx = 3, theny = 3(3)2 = 27.

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

 

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

When x = 1,y = 4 (1)3 − 2 (1)5 = 2.

When x = −1,y = 4 (−1)3 − 2 (−1)5 = −2.

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

 

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

On differentiating with respect to x, we have:

But, x2 +y2 − 2x − 3 = 0 for x = 1.

y2 = 4⇒ y = ± 2

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

 

The equation of the given curve is ay2 =x3.

On differentiating with respect tox, we have:

 

The equations of the given curves are given as x =y2 and xy = k

Putting x =y2 in xy =k, we get:

 

Differentiatingwith respect to x, we have:

Therefore, the slope of the tangent at

Then, the equation of the tangent at (x0 ,y0) is given by,

 

The equation of the given curve is y = √3x – 2

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

The equation of the given line is 4x − 2y + 5 = 0.

4x − 2y + 5 = 0⇒y = 2x + 5/2 (which is of the form y = mx + c)

∴Slope of the line = 2

Now, the tangent to the given curve is parallel to the line 4x − 2y − 5 = 0 if the slope of the tangent is equal to the slope of the line.

∴Equation of the tangent passing through the point(41/48, 3/4) is given by,

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

 

The equation of the given curve is y = 2x2 + 3 sinx

Slope of the tangent to the given curve at x = 0 is given by,

Hence, the slope of the normal to the given curve at x = 0 is

The correct answer is D.

 

The equation of the given curve is y2 = 4x

Differentiating with respect tox, we have:

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

dy/dx = 2/y

The given line isy = x + 1 (which is of the form y =mx +c)

∴ Slope of the line = 1

The liney = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve.

Thus, we must have:

Hence, the line y =x + 1 is a tangent to the given curve at the point (1, 2).

The correct answer is A.

 

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

Therefore, option (C) is correct.

 

(i) √25.3

(ii) √49.5

(iii)  √0.6

\

(iv) (0.009)1/3

(v)(0.999)1/10

(vi) (15)1/4

(vii) (26)1/3

(viii) (255)1/4

(ix)(82)1/4

(x) (401)1/2

(xi) (0.0037)1/2

(xii) (26.57)1/3

(xiii) (81.5)1/4

(xiv)(3.968)3/2

(xv) (32.15)1/5

 

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

 

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

 

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

 

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

Hence, the approximate change in the surface area of the cube is 0.12x2 m2.

 

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

Then,

r = 7 m and Δr = 0.02 m

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

 

 

Hence, the approximate error in calculating the volume is 3.92 π m3.

 

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

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

 

Therefore, option (D) is correct.

 

(i) The given function is f(x) = (2x − 1)2 + 3.

It can be observed that (2x− 1)2 ≥ 0 for every x∈R.

Therefore, f(x) = (2x − 1)2 + 3 ≥ 3 for every x∈R.

The minimum value of f is attained when 2x − 1 = 0.

2x − 1 = 0⇒ x = 1/2

∴Minimum value of f == 3

Hence, functionf does not have a maximum value.

(ii) The given function is f(x) = 9x2 + 12x + 2 = (3x + 2)2 − 2.

It can be observed that (3x + 2)2 ≥ 0 for every x∈R.

Therefore, f(x) = (3x + 2)2 − 2 ≥ −2 for every x∈R.

The minimum value of f is attained when 3x + 2 = 0.

3x + 2 = 0⇒ x = -2/3

∴Minimum value off =

Hence, functionf does not have a maximum value.

(iii) The given function is f(x) = − (x − 1)2 + 10.

It can be observed that (x − 1)2 ≥ 0 for every x∈R.

Therefore,f(x) = − (x − 1)2 + 10 ≤ 10 for every x∈R.

The maximum value is attained when (x − 1) = 0.

(x − 1) = 0⇒x = 1

∴Maximum value off = f(1) = − (1 − 1)2 + 10 = 10

Hence, functionf does not have a minimum value.

(iv) The given function isg(x) =x3 + 1.

Hence, function neither has a maximum value nor a minimum value.

 

We have,

f(x) = ex

∴ f'(x) = ex

Now, if f'(x) = 0. But, the exponential function can never assume 0 for any value ofx.

Therefore, there does not exist∈R such that f'(c) = 0

Hence, function f does not have maxima or minima.

We have,

g(x) = logx

∴ g'(x) = 1/x

Since log x is defined for a positive number x, g'(x) > 0 for any x

Therefore, there does not existc∈R such that g'(c) = 0

Hence, function does not have maxima or minima.

We have,

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

h'(x) =3x2 +2x + 1

Now,

h(x) = 0 ⇒ 3x2 + 2x + 1 = 0⇒

Therefore, there does not exist c∈R such that  h'(c) = 0

Hence, functionh does not have maxima or minima.

 

Letf(x) = sin 2x.

 

Let f(x) =x + sin 2x.

Hence, we can conclude that the absolute maximum value off(x) in the interval [0, 2π] is 2π occurring atx = 2π and the absolute minimum value off(x) in the interval [0, 2π] is 0 occurring atx = 0.

 

Let one number be x. Then, the other number is (24 −x).

Let P(x) denote the product of the two numbers. Thus, we have:

∴By second derivative test,x = 12 is the point of local maxima of P. Hence, the product of the numbers is the maximum when the numbers are 12 and 24 − 12 = 12.

 

The two numbers arex andy such that x +y = 60.

⇒y = 60 −x

Let f(x) =xy3.

∴By second derivative test,x= 15 is a point of local maxima off. Thus, functionxy3 is maximum whenx = 15 andy = 60 − 15 = 45.

Hence, the required numbers are 15 and 45.

 

(i) The given function isf(x) = x3.

f'(x) = 3x2

Now, f'(x) = 0 ⇒  x =  0

Then, we evaluate the value off at critical point x = 0 and at end points of the interval [−2, 2].

f(0) = 0

f(−2) = (−2)3 = −8

f(2) = (2)3 = 8

Hence, we can conclude that the absolute maximum value off on [−2, 2] is 8 occurring atx = 2. Also, the absolute minimum value off on [−2, 2] is −8 occurring at x = −2.

(ii) The given function is f(x) = sinx + cosx.

Then, we evaluate the value off at critical point π/4 and at the end points of the interval [0, π].

Hence, we can conclude that the absolute maximum value off on [0, π] is√2  occurring atx = π4 and the absolute minimum value off on [0, π] is −1 occurring atx = π.

Then, we evaluate the value off at critical pointx = 4 and at the end points of the interval.

Hence, we can conclude that the absolute maximum value off onis 8 occurring atx = 4 and the absolute minimum value off onis −10 occurring atx = −2.

Hence, we can conclude that the absolute maximum value off on [−3, 1] is 19 occurring atx = −3 and the minimum value off on [−3, 1] is 3 occurring atx = 1.

 

 

The profit function is given as p (x) = 41 − 72x− 18x2.

∴ p'(x)=−72−36x

⇒ x=−7236 =−2

Also, p”(−2)=−36 <0 By second derivative test, x=−2

is the point of local maxima of p.

∴ Maximum profit=p(−2)

=41−72(−2)−18(−2)2

=41+144−72=113

Hence, the maximum profit that the company can make is 113 units. The solution given in the book has some errors. The solution is created according to the question given in the book.

 

Let f(x) = 3x4 − 8x3 + 12x2 − 48x + 25.

Now,f'(x) = 0 gives x = 2 or x2+ 2 = 0 for which there are no real roots.

Therefore, we consider only x = 2∈[0, 3].

Now, we evaluate the value off at critical pointx = 2 and at the end points of the interval [0, 3].

Hence, we can conclude that the absolute maximum value off on [0, 3] is 25 occurring atx= 0 and the absolute minimum value off at [0, 3] is − 39 occurring atx = 2.

 

Let f(x) = sinx + cosx.

Now,f”(x) will be negative when (sinx + cosx) is positive i.e., when sin xand cosxare both positive. Also, we know that sinx and cosx both are positive in the first quadrant. Then,f”(x) will be negative when x ∊ (0, π/2).

Thus, we consider x = π/4

∴By second derivative test,f will be the maximum at x = π/4 and the maximum value of f is.

 

Let f(x) = 2x3 − 24x + 107.

We first consider the interval [1, 3].

Then, we evaluate the value off at the critical pointx = 2∈ [1, 3] and at the end points of the interval [1, 3].

f(2) = 2(8) − 24(2) + 107 = 16 − 48 + 107 = 75

f(1) = 2(1) − 24(1) + 107 = 2 − 24 + 107 = 85

f(3) = 2(27) − 24(3) + 107 = 54 − 72 + 107 = 89

Hence, the absolute maximum value off(x) in the interval [1, 3] is 89 occurring atx = 3.

Next, we consider the interval [−3, −1].

Evaluate the value off at the critical pointx = −2∈ [−3, −1] and at the end points of the interval [1, 3].

f(−3) = 2 (−27) − 24(−3) + 107 = −54 + 72 + 107 = 125

f(−1) = 2(−1) − 24 (−1) + 107 = −2 + 24 + 107 = 129

f(−2) = 2(−8) − 24 (−2) + 107 = −16 + 48 + 107 = 139

Hence, the absolute maximum value off(x) in the interval [−3, −1] is 139 occurring atx = −2.

 

Let f(x) =x4 − 62x2 +ax + 9.

Hence, the value ofa is 120.

 

Let one number bex. Then, the other number is y = (35 −x).

Let P(x) =x2y5. Then, we have:

∴ By second derivative test, P(x) will be the maximum when x = 10 and y = 35 − 10 = 25.

Hence, the required numbers are 10 and 25.

 

Let one number be x. Then, the other number is (16 −x).

Let the sum of the cubes of these numbers be denoted by S(x). Then,

∴ By second derivative test, x = 8 is the point of local minima of S.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 − 8 = 8.

 

Let the side of the square to be cut off be x cm. Then, the length and the breadth of the box will be (18 − 2x) cm each and the height of the box is x cm.

Therefore, the volume V(x) of the box is given by,

∴ By second derivative test,x = 3 is the point of maxima of V.

Hence, if we remove a square of 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

 

Let the side of the square to be cut off be x cm. Then, the height of the box isx, the length is 45 − 2x, and the breadth is 24 − 2x.

Therefore, the volumeV(x) of the box is given by,

∴By second derivative test,x = 5 is the point of maxima.

Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm.

 

Let a rectangle of length and breadth be inscribed in the given circle of radius.

Then, the diagonal passes through the centre and is of length 2a cm.

∴By the second derivative test, when l = √2a , then the area of the rectangle is the maximum.

Since l = b = √2a, the rectangle is a square.

Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.

 

Letr andh be the radius and height of the cylinder respectively.

Then, the surface area (S) of the cylinder is given by,

Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter.

 

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

Then, volume (V) of the cylinder is given by,

 

Let a piece of length  be cut from the given wire to make a square.

Then, the other piece of wire to be made into a circle is of length (28 −l) m.

Now, side of square = l/4

Let r be the radius of the circle. Then, 2πr = 28 – l ⇒ r = 1/2π(28  – l)

The combined areas of the square and the circle (A) is given by,

 

Letr andh be the radius and height of the cone respectively inscribed in a sphere of radius R.

 

Let randh be the radius and the height (altitude) of the cone respectively.

Then, the volume (V) of the cone is given as: V =13πr2h ⇒ h =3Vπr2

The surface area (S)of the cone is given by,

S = πrl (wherel is the slant height)

Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to√2 times the radius of the base.

 

Letθ be the semi-vertical angle of the cone.

It is clear that

Let r,h, and l be the radius, height, and the slant height of the cone respectively.

The slant height of the cone is given as constant.

Now,r =l sinθ and h =l cosθ

The volume (V) of the cone is given by,

V=1/3πr2h =1/3π (l2sin2θ)(lcosθ)

=1/3πl3 sinθ cosθ

⇒dV/dθ =l3π/3 [sin2θ (−sinθ) + cosθ(2sinθ cosθ)]

=l3π / 3[−sin3θ+2sinθcos2θ]

⇒d2V / dθ2 =l3π / 3 [−3sin2θ cosθ + 2cos3θ− 4sin2θ cosθ ]

=l3π/3 [2cos3θ − 7sin2θ cosθ]

∴By second derivative test, the volume (V) is the maximum when θ = tan-1 √2

Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is tan-1 √2.

 

Let r be the radius, l be the slant height and h be the height of the cone of given surface area,S.

Also, let α be the semi-vertical angle of the cone.

 

Equation of the curve is  x2 = 2y

For each value of x, the position of the point will be(x, x2/2)

Let P (x, x2/2) be any point on the curve (i), then according to question,

Distance between given point (0, 5) and

Therefore, option (A) is correct.

 

Let f =

∴By second derivative test,f is the minimum atx= 1 and the minimum value is given by f(1) = 1-1+1/1+1+1 = 1/3

The correct answer is D.

 

Then, we evaluate the value off at critical point x = 1/2 and at the end points of the interval [0, 1] {i.e., atx = 0 andx = 1}.

Hence, we can conclude that the maximum value off in the interval [0, 1] is 1.

The correct answer is C.