Application Of Derivatives

Given: a spherical ball salt, it is dissolving such that the rate of decrease of the volume at any instant is proportional to the surface

To prove: the radius is decreasing at a constant rate

Explanation: Let the radius of the spherical ball of the salt a t any time t be ‘r’.

Let the surface area of the spherical ball be S

Then, S = 4πr²……….(i)

Let V be the volume of the spherical ball,

Then, ……..(ii)

Now as per the given criteria,

Here the negative sign indicates the rate of volume is decreasing.

Or we can write this as

Where k is the proportional constant

Substituting the values from equation (i) and (ii), we get

Now taking out the constant term outside on LHS, we get

Applying the derivatives with respect to t, we get

Cancelling the like terms, we get

Hence the radius of the spherical ball is decreasing at a constant rate.

Hence Proved

Given: circle where its area is increasing at a uniform rate

To prove perimeter varies inversely as the radius

Explanation: Let the radius of the circle be ‘r’.

Let A be the area of the circle,

Then A = πr²……..(i)

As per the given criteria the area is increasing at a uniform rate, then

Now substituting the value from equation (i) in above equation, we get

Now differentiating with respect to t we get

Now let P be the perimeter of the circle, then

P = 2πr

Now differentiating perimeter with respect to t, we get

Applying the derivatives, we get

Now substituting value from equation (ii) in the above equation we get

Cancelling the like terms we get

Converting this to proportionality, we get

Hence the perimeter of the circle with given condition varies inversely as the radius.

Hence Proved

Given: a boy of n height 1.5 m is flying a kite at a height of 151.5 m. The kite is moving with a speed of 10m/s. And the kite is 250 m away from the boy.

To find the speed at which the string is let out

Explanation: the below figure shows the above situation,

From the above figure,

Height of the kite, H = AD = 151.5 m

Height of the boy, b = BC = 1.5 m

Distance between kite and boy, x = CD = BE = 250 m

And BA is the length of the string = y

So, we need to find out the rate of increase of the string

From figure, h = AE

= AD-ED

= 151.5-1.5

= 150m

From figure it is clear that ΔABE forms right-angled triangle

Now applying the Pythagoras theorem, we get

AB² = BE²+AE²

Or y² = x²+h²………..(i)

Now substituting the corresponding values, we get

y² = (250)²+(150)²

⇒ y² = 62500+22500

⇒ y² = 85000

⇒ y = 291.5m………..(ii)

Now differentiate equation (i) with respect to time, we get

Applying the sum rule of the differentiation, we get

Now the height is not increasing so it is constant, so

Applying the derivative with respect to t, we get

Now given the kite is moving with a speed of 10m/s, so

Now substituting corresponding value in equation (iii), we get

Hence the string is let out at a rate of 8.6m/s.

Given: two men A and B start with velocities v at the same time from the junction of the two roads inclined at 45° to each other

To find the rate at which they are being separated

Explanation:

Let A and B move a distance of x on different roads as shown above, there distance at any time t will be same as they have same velocity.

Hence

Now consider ΔAOB, applying the cosine rule, we get

y² = x²+x²-2x.x.cos 45°

Now multiplying and dividing by √2, we get

Now applying the derivative with respect to t, we get

Taking out the constant terms, we get

Substituting the value from equation (i), we get

Hence this is the rate at which the two roads are being separated

Given: a condition

To find the angle θ such that it increases twice as fast as its sine.

Explanation: Let x = sin θ

On differentiating with respect to t, we get

Applying the derivative, we get

But it is given that

Substituting this value in equation (i), we get

Now cancelling the like terms, we get

1 = 2cos θ

But given , this is possible only when

Hence the angle θ is .

Given (1.999)⁵

But the integer nearest to 1.999 is 2,

So, 1.999 = 2-0.001

∴, a = 2 and h = -0.001

Hence, (1.999)⁵ = (2+(-0.001))⁵

Let the function becomes,

f(x) = x⁵………(i)

Now applying first derivative, we get

f’(x) = 5x⁴……….(ii)

Now let f(a+h) = (1.999)⁵

Now we know,

f(a+h) = f(a)+hf’(a)

Now substituting the function from (i) and (ii), we get

f(a+h) = a⁵+h(5a⁴)

Substituting the values of a and h, we get

f(2+(-0.001)) = 2⁵+( -0.001) (5(2⁴))

⇒ f(1.999) = 32+(-0.001)(5(16))

⇒ (1.999)⁵ = 32+(-0.001)(80)

⇒ (1.999)⁵ = 32-0.08

⇒ (1.999)⁵ = 31.92

Hence the approximate value of (1.999)⁵ = 31.92.

Given: a hollow spherical shell with internal radii 3cm and external radii 3.0005 cm

To find: the approximate volume of the metal in the hollow spherical shell

Explanation: Let the internal and external radii of the hollow spherical shell be r and R, respectively.

So it is given,

R = 3.0005 and r = 3

And let the volume of the hollow spherical shell be V.

Then we know,

Now substituting the values of R and r, we get

Now using the differentiation to get the approximate value of (3.0005)³.

But the integer nearest to 3.0005 is 3,

So 3.0005 = 3+0.0005

So let a = 3 and h = 0.0005

Hence, (3.0005)³ = (3+0.0005)³

Let the function becomes,

f(x) = x³………(ii)

Now applying first derivative, we get

f’(x) = 2x²……….(iii)

Now let f(a+h) = (3.0005)³

Now we know,

f(a+h) = f(a)+hf’(a)

Now substituting the function from (ii) and (iii), we get

f(a+h) = a³+h(3a²)

Substituting the values of a and h, we get

f(3+0.0005) = 3³+(0.0005) (3(3²))

⇒ f(3.0005) = 27+(0.0005)(3(9))

⇒ (3.0005)³ = 27+(0.0005)(27)

⇒ (3.0005)³ = 27+0.0135

⇒ (3.0005)³ = 27.0135

Hence the approximate value of (3.0005)³ = 27.0135.

Now substituting this in equation (i), we get

V = 4π(0.0045)

V = 0.018π cm³

Hence the approximate volume of the metal in the hollow spherical shell is 0.018π cm³.

Given: a 2m tall man walks at the rate of m/s towards a m tall street light.

To find: the rate at which the tip of the shadow is moving and also to find the rate at which the length of the shadow changing when he is m from the base of the light.

Explanation:

Here the street light is AB =

And man is DC = 2m

Let BC = x m and CE = y m

And given man walks towards the streetlight at a rate of , and it will negative because the man is moving towards the street light

Hence ………(i)

Now consider ΔABE and ΔDCE

∠DEC = ∠AEB (same angle)

∠DCE = ∠ABD = 90°

Hence by AA similarity,

ΔABE≅ΔDCE

Hence by CPCT,

Now substituting the values from the figure, we get

⇒ 16y = 6(x+y)

⇒ 16y = 6x+6y

⇒ 16y-6y = 6x

⇒ 10y = 6x

Now applying first derivative with respect to t, we get

Now substituting the value from equation (i) in above equation, we get

Hence the rate at which the tip of the shadow is moving is -1m/s, i.e., the length of the shadow is decreasing at the rate of 1m/s.

Let BE = z

So from fig,

z = x+y

Applying the first derivative on above equation with respect to t, we get

Now substituting the corresponding values, we get

Hence the tip of the shadow is moving at the rate of towards the light source.

Given: L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200 (10 – t)²

To find: the rate at which the water is running out of the pool at the end of 5s. And also to find the average rate at which the water flows out during the first 5s.

Explanation: Let the rate at which the water us running out be given by

Given L = 200(10-t)²

Now differentiating the above equation with respect to t, we get

Taking out the constant term, we get

Applying the power rule of differentiation, we get

Now we need to find how fast is the water running out at the end of 5 sec, so finding the value of equation (i) at t = 5, we get the answer.

Hence the rate at which the water is running out of the pool at the end of 5s is 2000 L/s

Now to find the initial rate we will substitute t = 0 in equation (i), we get

……(iii)

So equation (ii) is the final rate and equation (iii) is the initial rate.

Hence the average rate during 5s is

Substituting the corresponding values, we ge

Hence the average rate at which the water flows out during the first 5s is 3000L/s.

Given: a volume of cube increasing at a constant rate

To prove: the increase in its surface area varies inversely as the length of the side

Explanation: Let the length of the side of the cube be ‘a’.

Let V be the volume of the cube,

Then V = a³……..(i)

As per the given criteria the volume is increasing at a uniform rate, then

Now substituting the value from equation (i) in above equation, we get

Now differentiating with respect to t we get

Now let S be the surface area of the cube, then

S = 6a²

Now differentiating surface area with respect to t, we get

Applying the derivatives, we get

Now substituting value from equation (ii) in the above equation we get

Cancelling the like terms we get

Converting this to proportionality, we get

Hence the surface area of the cube with given condition varies inversely as the length of the side of the cube.

Hence Proved

Given: two squares of sides x and y, such that y = x-x²

To find: the rate of change of the area of the second square with respect to the area of the first square

Explanation: Let the area of the first and the second square be A₁ and A₂ respectively.

Then Area of the first square is

A₁ = x²

Differentiating this with respect to time, we get

And the area of the second square is

A₂ = y²…….(ii)

But given, y = x-x²

Substituting this given value in equation (ii), we get

A₂ = (x-x²)² …….(iii)

Now differentiating equation (iii) with respect to t, we get

Now applying the power rule of differentiation, we get

Now applying the sum rule of differentiation, we get

Applying the derivative, we get

We need to find the rate of change of area of the second square with respect to the area of the first square, i.e.,

Substituting values from equation (i) and (iv), we get

By cancelling the like terms we get

Hence the rate of change of the area of the second square with respect to the area of the first square is 2x²-3x+1

Given: two curves 2x = y² and 2xy = k

To find: the condition that these two curves intersect orthogonally

Explanation: Given 2xy = k

Substituting this value of y in another curve equation i.e., 2x = y², we get

Taking cube root on both sides, we get

Substituting equation (ii) in equation (i), we get

Hence the point of intersection of the two cures is

Now given 2x = y²

Differentiating this with respect to x, we get

Now finding the above differentiation value at the point of intersection i.e., at , we get

Also given 2xy = k

Differentiating this with respect to x, we get

Now applying the product rule of differentiation, we get

Now finding the above differentiation value at the point of intersection i.e., at , we get

But the two curves intersect orthogonally, if

m₁.m₂ = -1

Now substituting the values from equation (iii) and equation (iv), we get

Taking cube on both sides we get

k² = 2³ = 8

⇒ k = 2√2

Hence this is the condition for the given two curves to intersect orthogonally.

Given: two curves x² + y² = 8 and xy = 4

To prove: the two curves touch each other

Explanation:

Now given x² + y² = 8

Differentiating this with respect to x, we get

Applying sum rule of differentiation, we get

Also given xy = 4

Differentiating this with respect to x, we get

Now applying the product rule of differentiation, we get

But the two curves touch each other, if

m₁ = m₂

Now substituting the values from equation (ii) and equation (ii), we get

⇒ y² = x²

⇒ x = y…….(iii)

Now substituting x = y in x² + y² = 8, we get

y² + y² = 8

⇒ 2 y² = 8

⇒ y² = 4

⇒ y = ±2

When y = 2,

xy = 4 becomes

x(2) = 4⇒ x = 2

when y = -2,

xy = 4 becomes

x(-2) = 4⇒ x = -2

Hence the point of intersection of the two curve is (2,2) and (-2, -2)

Substituting these points of intersection equation (i) and equation (ii), we get

For (2,2),

∴ m₁ = m₂

For (-2,-2),

∴ m₁ = m₂

Therefore, for both curves to touch each other, the slopes of both the curves should be same.

Hence the two given curves touch each other.

Hence proved

Given: curve √x+√y = 4

To find: the co-ordinates of the point on the curve at which tangent is equally inclined to the axes

Explanation: given √x+√y = 4

Now differentiating this with respect to x, we get

Applying the sum rule of differentiation, we get

Applying the differentiation, we get

This is the tangent to the given curve.

Now it is given that the tangent is equally inclined to the axes,

∴ y = x……….(ii)

Substituting equation (ii) in the curve equation, we get

√y+√y = 4

2√y = 4

√y = 2

⇒ y = 4

When y = 4, then x = 4 from equation (ii)

So the co-ordinates of the point on the curve √x+√y = 4 at which tangent is equally inclined to the axes is (4,4).

Given: the curves y = 4–x² and y = x²

To find: the angle of intersection of the two curves

Explanation: consider first curve

y = 4–x²

Differentiating the above curve with respect to x we get

Consider second curve

y = x²

Differentiating the above curve with respect to x we get

Given y = x²

Substituting this in other curve equation, we get

x² = 4-x²

⇒ 2x² = 4

⇒ x² = 2

⇒ x = ±√2

When x = √2, we get

y = (√2)²⇒ y = 2

When x = -√2, we get

y = (-√2)²⇒ y = 2

Thus the points of intersection are (√2, 2) and (-√2, 2)

We know angle of intersection can be found by following formula,

i.e.,

Substituting the values from equation (i) and equation (ii), we get

For (√2, 2), the above equation becomes,

Hence the angle of intersection of the curves y = 4–x² and y = x² is

Given: two curves y² = 4x and x² + y² – 6x + 1 = 0

To prove: the two curves touch each other at point (1,2)

Explanation:

Now given x² + y² – 6x + 1 = 0

Differentiating this with respect to x, we get

Applying sum rule of differentiation, we get

Solving the above equation at point (1,2), we get

Also given y² = 4x

Differentiating this with respect to x, we get

Now applying the product rule of differentiation, we get

Solving the above equation at point (1,2), we get

From equation (i) and (ii),

∴ m₁ = m₂

Therefore, both curves touch each other at point (1,2).

Hence proved

Given: equation of the curve 3x²–y² = 8, equation of line x+3y = 4

To find: the equation of the normal lines to the given curve which are parallel to the given line

Explanation:

Now given equation of curve as 3x² -y² = 8

Differentiating this with respect to x, we get

Now let slope of the normal to the curve be m₂ is given by

Substituting value from equation (i), we get

The given equation of the line is

x+3y = 4

⇒ 3y = 4-x

the slope of this line is

Since, slope of normal to the curve should be equal to the slope of the line which is parallel to the curve,

∴ m₂ = m₃

Substituting values from equation (ii) and (iii), we get

⇒ 3y = 3x

⇒ y = x……(iv)

Now substituting y = x in the equation of the curve, we get

3x² -y² = 8

⇒ 3x² -x² = 8

⇒ 2x² = 8

⇒ x² = 4

⇒ x = ±2

But from equation (iv)

y = x

⇒ y = ±2

Thus the points at which normal to the given curve is parallel to the given line are (2, 2) and (-2, -2)

Now the equations of the normal are given by

y-2 = m₂(x-2) and y+2 = m₂(x+2)

and

3y-6 = -x+2 and 3y+6 = -x-2

3y+x = 8 and 3y+x = -8

Hence the equation of the normal lines to the curve 3x² – y² = 8 which are parallel to the line x + 3y = 4 are 3y+y = ±8.

Given: equation of a curve x² + y² – 2x – 4y + 1 = 0

To find: the points on the curve x² + y² – 2x – 4y + 1 = 0, the tangents are parallel to the y-axis

Explanation:

Now given equation of curve as x² + y² – 2x – 4y + 1 = 0

Differentiating this with respect to x, we get

Applying the sum rule of differentiation, we get

As it is given the tangents are parallel to the y-axis,

Hence

Substituting the value from equation (i), we get

⇒ y-2 = 0

⇒ y = 2

Now substituting y = 2 in curve equation, we get

x² + y² – 2x – 4y + 1 = 0

⇒ x² + 2² – 2x – 4(2) + 1 = 0

⇒ x² + 4 – 2x – 8 + 1 = 0

⇒ x²– 2x-3 = 0

Now splitting the middle term, we get

⇒ x²-3x+x-3 = 0

⇒ x(x-3)+1(x-3) = 0

⇒ (x+1)(x-3) = 0

⇒ x+1 = 0 or x-3 = 0

⇒ x = -1 or x = 3

So the required points are (-1, 2) and (3, 2).

Hence the points on the curve x² + y² – 2x – 4y + 1 = 0, the tangents are parallel to the y-axis are (-1, 2) and (3, 2).

Given: equation of line , the curve intersects the y-axis

To show: the line touches the curve at the point where the curve intersects the axis of y

Explanation: given the curve intersects the y-axis, i.e., at x = 0

Now differentiate the given curve equation with respect to x, i.e.,

Taking out the constant term,

Now differentiating it we get

Now substitute x = 0, we get

Now consider line equation,

We will differentiate this with respect to x, we get

Taking out the constant terms, we get

Line touches the curve if there slopes are equal.

From equation (i) and (ii), we see that

m₁ = m₂

Hence the line touches the curve at the point where the curve intersects the axis of y.

Given:

To show: the given function is increasing in R.

Explanation: Given

Applying first derivative with respect to x, we get

But the derivative of 2x is 2, so

But the derivative of , so

Applying the sum rule to last part we get

cancelling the like terms, we get

Now adding by taking the LCM, we get

Now for any real value of x, the above value of f(x) is greater than or equal to zero

Hence

And we know, if f’(x)≥0, then f(x) is increasing function.

Hence the given function is increasing function in R.

Given: f (x) = √3 sin X – cos X – 2ax + b

To show: the given function is decreasing in R.

Explanation: Given f (x) = √3 sin X – cos X – 2ax + b

Applying first derivative with respect to x, we get

Applying the sum rule of differentiation, we get

Taking the constant terms out we get

But the derivative of sin X = cos x and that of cos x = -sin x, so

f' (x) = √3cos x-(-sin x)-2a

f' (x) = √3cos X+sin X-2a

Multiplying and dividing RHS by 2, we get

But and , substituting these values in above equation, we get

But cos(A-B) = cos Acos B+sin A.sin B, substituting this in above equation, we get

Now, we know cos x always belong to [-1, 1] for a≥1

∴,

And we know, if f’(x)≤0, then f(x) is decreasing function.

Hence the given function is decreasing function in R.

Given: f (x) = tan^–1(sin X + cos X)

To show: the given function is increasing in .

Explanation: Given f (x) = tan^–1(sin X + cos X)

Applying first derivative with respect to x, we get

Applying the differentiation rule for tan^-1, we get

Applying the sum rule of differentiation, we get

But the derivative of sin X = cos x and that of cos x = -sin x, so

Expanding (sin x+cos x )², we get

But sin²x+cos²x = 1 and 2sin Xcos X = sin2x, so the above equation becomes,

Now for f(x) to be decreasing function,

f’(x)≥0

⇒ cos x- sin x≥ 0

⇒ cos x≥ sin x

But this is possible only if

Hence the given function is increasing function in .

Given: y = –x³+3x²+9x–27

To find: at what point the slope of the curve is maximum, and to also find the maximum value of the slope

Explanation: given y = –x³+3x²+9x–27

By finding the first derivative of the given equation of the curve, we get the slope of the curve.

So slope of the curve is

Now applying the derivative, we get

Now to find the critical point we need to find the derivative of the slope, so

Applying the derivative, we get

Now critical point is found by equating the second derivative to 0, i.e.,

⇒ -6x+6 = 0

⇒ 6x = 6

⇒ x = 1…….(i)

Now we will find the third derivative of the given curve,

i.e.,

Applying the derivative, we get

As the third derivative is less than 0, so the maximum slope of the given curve is at x = 1.

To find the value of the maximum slope we will find first derivative at x = 1, i.e.,

Hence the slope of the curve is maximum at x = 1, and the maximum value of the slope is 12.

Given: f(x) = sin X + √3 cos X

To prove: the given function has maximum value at

Explanation: given f(x) = sin X + √3 cos X

We will find the first derivative of the given function, we get

Now applying the derivative, we get

f'(x) = cos x-√3sin x

Now critical point is found by equating the first derivative to 0, i.e.,

f’(x) = 0

⇒ cos x-√3sin x = 0

⇒ √3sin x = cos x

This is possible only when

Now the second derivative of the function is

Applying the derivative, we get

f’’(x) = -sin x-√3cos x

Now we will substitute in the above equation, we get

f’’(x) = -sin x-√3cos x

Substituting the corresponding value, we get

Hence f(x) has maximum value at .

Hence proved

Given: a right-angled triangle, such that sum of the lengths of its hypotenuse and side is given

To show: the area of the triangle is maximum when the angle between them is

Explanation:

Let ΔABC be the right-angled triangle,

Let hypotenuse, AC = y,

side, BC = x, AB = h

Now sum of the side and hypotenuse is given,

⇒ x+y = k, where k is any constant value

⇒ y = k-x………..(i)

Now let A be the area of the triangle, then we know

Now applying the Pythagoras theorem, we get

y² = x²+h²

Now substituting equation (i) in above equation, we get

Now substituting the value from equation (iii) into equation (ii), we get

Now differentiating above equation with respect to x, we get

Now taking out the constant term we get

Now applying the product rule, we get

Applying the power rule of differentiation on second part in above equation, we get

Now critical point is found by equating the first derivative to 0, i.e.,

⇒k²-2kx = kx

⇒k² = 2kx+kx

⇒ k² = 3kx

⇒ k = 3x

Again, differentiating equation (iv) with respect to x, we get

Taking out the constant term and taking the LCM, we get

Applying the product rule of differentiation, we get

Now applying the power rule of differentiation, we get

Substituting , in above equation, we get

Hence the maximum value of A is at

We know,

Now from figure,

Substituting the value of y = k-x from equation (i), we get

Substituting the value of , we get

This is possible only when

Hence the area of the triangle is maximum only when the angle between them is

Given: function f (x) = x⁵ – 5x⁴ + 5x³ – 1

To find: the points of local maxima, local minima and the points of inflection of f(x) and also to find the corresponding local maximum and local minimum values.

Explanation: given f (x) = x⁵ – 5x⁴ + 5x³ – 1

We will find the first derivative of f(x), i.e.,

Applying the sum rule of differentiation and taking out the constant term, we get

f' (x) = 5x⁴-5.4x³+5.3x²-0

⇒ f' (x) = 5x⁴-20x³+15x²

Now to find the critical point we need to equate the first derivative to 0, i.e.,

f'(x) = 0

⇒ 5x⁴-20x³+15x² = 0

⇒ 5x²(x²-4x+3) = 0

⇒ 5x²(x²-4x+3) = 0

Now splitting the middle term, we get

⇒ 5x²(x²-3x-x+3) = 0

⇒ 5x²[ x(x-3)-1(x-3)] = 0

⇒ 5x²(x-1)(x-3) = 0

⇒ 5x² = 0 or (x-1) = 0 or (x-3) = 0

⇒ x = 0 or x = 1 or x = 3

Now we will find the corresponding y value by substituting the different values of x in given function f(x) = x⁵ – 5x⁴ + 5x³ – 1,

When x = 0, f(0) = 0⁵ – 5(0)⁴ + 5(0)³ – 1 = -1

Hence the point is (0,-1)

When x = 1, f(1) = 1⁵ – 5(1)⁴ + 5(1)³ – 1 = 1-5+5-1 = 0

Hence the point is (0,0)

When x = 3, f(3) = 3⁵ – 5(3)⁴ + 5(3)³ – 1 = 243-405+135-1 = -28

Hence the point is (0,-28)

Therefore, we see that

At x = 3, y has minimum value = -28. Hence x = 3 is point of local minima.

At x = 1, y has maximum value = 0. Hence x = 1 is point of local minima.

And at x = 0, y has neither maximum nor minimum value, hence this is point of inflection.

Given: a telephone company in a town has 500 subscribers and collects fixed charges of Rs 300/- per subscriber per year, company increase the annual subscription and for every increase of Re 1/- one subscriber will discontinue the service

To find: the increase at which the company will get maximum profit

Explanation: company has 500 subscribers, and collects 300 per subscriber per year.

So total revenue of the company will be

R = 500×300 = 150000\-

Let the increase in annual subscription by the company be x.

Then as per the given criteria, x subscribers will discontinue the service.

Therefore the total revenue of the company after the increment is given by

R(x) = (500-x)(300+x)

⇒ R(x) = 500(300+x)-x(300+x)

⇒ R(x) = 150000+500x-300x-x²

⇒ R(x) = 150000+200x-x²

Now we will find the first derivative of the above equation, we get

R'(x) = 0+200-2x…..(i)

Now to find the critical points we will equate the first derivative to 0, i.e.,

R’(x) = 0

⇒ 200-2x = 0

⇒2x = 200

⇒ x = 100……(ii)

Now we will find the second derivative of the total revenue function, i.e., again differentiate equation (i), i.e.,

R’’(x) = 0-2<0

Hence R’’(100) is also less than 0,

Therefore, R(x) is maximum at x = 0, i.e.,

The company should the increase the subscription fee by 100 so that it will get maximum profit.

Given: equation of straight line x cosα + y sinα = p, equation of curve and the straight line touches the curve

To prove: a² cos²α + b² sin²α = p²

Explanation: given equation of the line is

x cosα + y sinα = p

⇒ y sinα = p-x cosα

Comparing this with the equation y = mx+c we see that the slope and intercept of the given line is

and

We know that, if a line y = mx+c touches the eclipse, then required condition is

c² = a²m²+b²

Now substituting the corresponding values, we get

Cancelling the like terms we get

p² = b²sin²α+a²cos²α

Hence proved

Given: an open box with square base is made out of a cardboard of c² area

To show: the maximum volume of the box is cubic units.

Explanation:

Let the side of the square be x cm and

Let the height the box be y cm.

Then area of the card board used is

A = area of square base + 4× area of rectangle

⇒ A = x²+4xy

But it is given this is equal to c², hence

c² = x²+4xy

⇒ 4xy = c²-x²

Then as per the given criteria the volume of the box with square base will be,

V = base×height

Here base is square, so volume becomes

V = x²y……(ii)

Now substituting equation (i) in equation (ii), we get

Now finding the first derivative of the volume, we get

Taking out the constant terms, we get

Applying the sum rule of differentiation, we get

Taking out the constant terms, we get

Applying the differentiation, we get

Now we will apply second derivative test to find out the maximum value of x, so for that let V’ = 0, so equating above equation with 0, we get

⇒ c²-3x² = 0

⇒ 3x² = c²

Differentiating equation (iii) again with respect to x, we get

Taking out the constant terms, we get

Applying differentiation rule of sum, we get

At , the above equation becomes,

Thus the volume (V) is maximum at

∴ Maximum volume of the box is

Hence the maximum volume of the box is cubic units.

Hence proved

Given: rectangle of perimeter 36cm

To find: the dimensions of the rectangle which will sweep out a volume as large as possible, when revolved about one of its sides. Also to find the maximum volume

Explanation: Let the length and the breadth of the rectangle be x and y.

Then it is given perimeter of the rectangle is 36cm,

⇒ 2x+2y = 36

⇒ x+y = 18

⇒ y = 18-x………(i)

Now when the rectangle revolve about side y it will form a cylinder with y as the height and x as the radius, then if the volume of the cylinder is V, then we know

V = πx²y

Substituting value from equation (i) in above equation we get

V = πx²(18-x)

⇒ V = π(18x²-x³)

Now we will find first derivative of the above equation, we get

Taking out the constant terms and applying the sum rule of differentiation, we get

V' = π[18(2x)-3x² ]

V' = π[36x-3x² ]……(ii)

Now to find the critical point we will equate the first derivative to 0, i.e.,

V’ = 0

⇒ π(36x-3x²) = 0

⇒ 36x-3x² = 0

⇒ 36x = 3x²

⇒ 3x = 36

⇒ x = 12……..(iii)

Now we will find the second derivative of the volume equation, this can be done by again differentiating equation (ii), we get

Taking out the constant terms and applying the sum rule of differentiation, we get

V’’ = π[36-3(2x) ]

V’’ = π[36-6x]

Now substituting x = 12 (from equation (iii)), we get

V’’_{x = 12} = π[36-6(12)]

V’’_{x = 12} = π[36-72]

V’’_{x = 12} = -36π<0

Hence at x = 12, V will have maximum value.

The maximum value of V can be found by substituting x = 12 in V = π(18x²-x³), i.e.,

V_{x = 12} = π (18(12)²-(12)³)

V_{x = 12} = π (18(144)-(1728))

V_{x = 12} = π (2592-(1728))

V_{x = 12} = 864π cm³

Hence the dimensions of the rectangle which will sweep out a volume as large as possible, when revolved about one of its sides equal to 12cm.

And the maximum volume is 864π cm³.

Given: sum of the surface areas of cube and a sphere is constant

To find: the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum

Explanation: Let the side of the cube be ‘a’

Then surface area of the cube = 6a²….(i)

Let the radius of the sphere be ‘r’

Then the surface area of the sphere = 4πr²…(ii)

Now given the sum of the surface areas of cube and a sphere is constant, hence adding equation (i) and (ii), we get

6a²+4πr² = k (where k is the constant)

⇒ 6a² = k-4πr²

Now we know the volume of the cube is

V_c = a³

And volume of the sphere is

So the sum of the volumes of cube and sphere is

Now substituting equation (iii) in above equation, we get

Now we will find the first derivative of the volume, we get

Applying the sum rule of differentiation and taking out the constant terms, we get

Applying the power rule of differentiation, we get

Now to find the critical point we will equate the first derivative to 0, i.e.,

V’ = 0

⇒r = 0 or (k-4πr²) = 24r²

⇒r = 0 or k = 4πr²+24r²

⇒r = 0 or (4π+24)r² = k

Now we know, r≠0

Hence

Now we will find the second derivative of the volume equation, this can be done by again differentiating equation (ii), we get

Taking out the constant terms and applying the sum rule of differentiation, we get

Applying the product rule of differentiation, we get

Applying the power rule of differentiation, we get

Applying the differentiation, we get

Hence for , the sum of their volumes is minimum

Now substituting , in equation (iii), we get

Now we will find the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum, i.e.,

a:2r

Hence the required ratio is

a:2r = 1:1

Given: a circle with AB as diameter and C is any point on the circle

To show: area of Δ ABC is maximum, when it is isosceles

Explanation: Let r be the radius of the circle, then the diameter will be

AB = 2r

Now we know, any angle in the semi-circle is always 90°.

Hence ∠ACB = 90°

Now let AC = x and BC = y

Then by applying the Pythagoras theorem to the right-angled triangle ABC, we get

(2r)² = x²+y²

⇒ y² = 4r²-x²

Now we know area of ΔABC is

Now substituting value from equation (i), we get

Now we will find the first derivative of the area, we get

Applying the product rule of differentiation and taking out the constant terms, we get

Applying the power rule of differentiation, we get

Now to find the critical point we will equate the first derivative to 0, i.e.,

A’ = 0

⇒ 2r²-x² = 0

⇒2r² = x²

Or

Now we know, r≠0

Hence and x = r√2

Now we will find the second derivative of the area equation, this can be done by again differentiating equation (iii), we get

Taking out the constant terms and applying the product rule of differentiation, we get

Applying the power rule of differentiation, we get

For in above equation, we get

Hence for , the area of ΔABC is maximum

Now we will find the maximum value by substituting in equation (i), we get

⇒ y = r√2 = x

i.e., the two sides of the ΔABC are equal

Hence the area of Δ ABC is maximum, when it is isosceles

Hence proved

Given: A metal box with a square base and vertical sides is to contain 1024 cm³, the material for the top and bottom costs Rs 5/cm² and the material for the sides costs Rs 2.50/cm².

To find: the least cost of the box

Let the side of the square be x cm and

Let the vertical side of the metal box be y cm.

Then as per the given criteria the volume of the metal box with square base will be,

V=base × height

Here base is square, so volume becomes

V=x²y

This is equal to 1024 cm³. Hence volume becomes

1024cm³=x²y

Now, we will find the total area of the metal box.

As it is given the base is square, that means top is also a square, so

Area of top and bottom = 2x²cm²

Given the material for the top and bottom costs Rs 5/cm², so the material cost for top and bottom becomes

Cost of top and bottom =Rs. 5(2x²)

Now the vertical side of metal box is y cm, so

Area of one side of the metal box = xy cm

Now the metal box has 4 sides, so

Area of all the sides of the metal box = 4xy

Given the material for the sides costs Rs 2.50/cm²

∴ Cost of all the sides of the metal box =Rs. 2.50(4xy)

So the total area of the metal box is

A=2x²+4xy

And the total cost of the metal box is

C=5(2x²)+ 2.50(4xy)

Substituting the value of y from equation (i) in the above equation, we get

Now differentiating both sides with respect to x, we get

Applying differentiation rule of sum, we get

Now applying the derivative, we get

Now we will apply second derivative test to find out the minimum value of x, so for that let C=0, so equating above equation with 0, we get

⇒x³=512

Solving this we get

⇒ x=8

Differentiating equation (ii) again with respect to x, we get

Applying differentiation rule of sum, we get

At x=8, the above equation becomes,

Now at x=8, C’(8)=0 and C’’(8)>0, so as per the second derivative test, x is a point of local minima and C(8) will be minimum value of C.

Hence least cost becomes

⇒C_x=8=640+1280=Rs.1920

Hence the least cost of the metal box is Rs. 1920

Given: a rectangular parallelepiped with sides x, 2x and . The sum of the surface areas of a rectangular parallelepiped and a sphere is given to be constant.

To prove: the sum of their volumes is minimum, if x is equal to three times the radius of the sphere and also find the minimum value of the sum of their volumes

Now the surface area of the rectangular parallelepiped is

Now let the radius of the sphere be r cm, then the surface area of the sphere is,

S_s=4πr²

Now sum of the surface areas of a rectangular parallelepiped and a sphere becomes,

Now given the sum of the surface areas of a rectangular parallelepiped and a sphere is constant, then

So differentiating equation (i) with respect to r, we get

Applying differentiation rule of sum, we get

Taking out the constant terms, we get

Applying the derivative, we get

Let V denotes the sum of volume of the rectangular parallelepiped and a sphere, then

For maxima or minima, the first derivative of volume should be equal to 0, i.e.,

So differentiating equation (iii) with respect to r, we get

Applying differentiation rule of sum, we get

Taking out the constant terms, we get

Applying the derivative, we get

Now substituting the value of from equation (ii), we get

i.e., x is three times the radius of the sphere when sum of the volumes of rectangular parallelepiped and sphere

Hence proved

Now to find the minimum volume of the sum of their volumes we need to find out the second derivative value at x=2r.

Hence applying derivative with respect to r to equation (iv), we get

Applying differentiation rule of sum, we get

Taking out the constant terms, we get

Applying differentiation rule of product to the first part, we get

Now substituting the value of from equation (ii), we get

Now substituting x=3r, we get

This is positive; hence V is minimum when x=3r or , and the minimum value of volume can be obtained by substituting in equation (iii), we get

Is the minimum value of the sum of their volumes.

Let the side of the equilateral triangle be x cm, then the area of the equilateral triangle is

So as per the question rate of side increasing at instant of time t is

Now differentiating area with respect to time t, we get

Taking out the constants we get,

Now applying the derivative, we get

Now substituting the given value of , we get

So when side x=10cm, the above equation becomes,

Hence, the rate at which the area increases, when side is 10 cm is 10√3 cm²/s.

So the correct option is option C.

Let the length of the ladder = 5m=500cm be the hypotenuse of the right triangle so formed as shown in the above figure.

Now let β be the angle between the ladder and the floor, so from the figure we can write that

Now differentiating both sides with respect to time t, we get

Applying the derivatives, we get

But given the top of the ladder slides downwards at the rate of 10 cm/sec, i.e.,

So the above equation becomes,

Now from figure,

Now substituting equation (iii) in equation (ii), we get

Now when the lower end of the ladder is 2m=200cm from the wall, i.e., y=200cm, the above equation becomes,

Hence the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is

So the correct option is option B.

Given

Differentiating both sides with respect to x, we get

Applying the power rule of derivation, we get

Now at (0,0), the above equation becomes

So the given curve at (0,0) has a vertical tangent, which is parallel to Y-axis.

Hence the correct option is option A.

Given the equation of the line is 3x² – y² = 8

Now differentiating both sides with respect to x, we get

Now applying the sum rule of differentiation and differentiation of constant =0, so we get

Taking out the constants we get

Applying the power rule of differentiation we get

So this is the slope of the given curve.

We know the slope of the normal to the curve is

Now given the equation of the line x + 3y = 8

⇒ 3y=8-x

Differentiation with respect to x, we get

So the slope of the line is

Now as the normal to the curve is parallel to this line, hence the slope of the line should be equal to the slope of the normal to the given curve,

⇒ 3y=3x

⇒ y=x

On substituting this value of the given equation of the curve, we get

3x² – y² = 8

⇒ 3x² – (x)² = 8

⇒ 2x² = 8

⇒ x² = 4

⇒ x=± 2

When x=2, the equation of the curve becomes,

3x² – y² = 8

⇒ 3(2)² – y² = 8

⇒ 3(4) – y² = 8

⇒ 12-8= y²

⇒ y² = 4

⇒ y=±2

When x=-2, the equation of the curve becomes,

3x² – y² = 8

⇒ 3(-2)² – y² = 8

⇒ 3(4) – y² = 8

⇒ 12-8= y²

⇒ y² = 4

⇒ y=±2

So, the points at which normal is parallel to the given line are (±2, ±2).

And required equation of the normal to the curve at (±2, ±2) is

⇒ 3(y-(±2))=-(x-(±2))

⇒ 3y-(±6)=-x+(±2)

⇒ x+3y-(±6)-(±2)=0

⇒ x+3y+(± 8)=0

Hence the equation of normal to the curve 3x² – y² = 8 which is parallel to the line x + 3y = 8 is x+3y+(± 8)=0.

So the correct option is option C.

Given the curve ay + x² = 7 and x³ = y, cut orthogonally at (1, 1),

ay + x² = 7

Differentiating on both sides with respect to x, we get

Applying the sum rule of differentiation and also the derivative of the constant is 0, so we get

Applying the power rule we get

The value of above derivative at (1,1), becomes

x³ = y

Differentiating on both sides with respect to x, we get

Applying the power rule we get

The value of above derivative at (1,1), becomes

Now as these two curves cut orthogonally at (1,1), so

so from equation (i) and (ii), we get

⇒ a=6

Hence f the curve ay + x² = 7 and x³ = y, cut orthogonally at (1, 1), then the value of a is 6.

So the correct option is option D.

Given y = x⁴ – 10

Differentiating on both sides with respect to x, we get

Applying the power rule we get

It is also given that the value of x changes from from 2 to 1.99, so the change in x is

Δx=2-1.99=0.01

So the change in y becomes,

Substituting the corresponding values we get

⇒ Δy=4x³×(0.01)

Now at x=2, the change in y becomes

⇒ Δy_x=2=4(2)³×(0.01)

⇒ Δy_x=2=4× 8×(0.01)

⇒ Δy_x=2=0.32

Hence the change in y is 0.32.

Given the equation of the curve is

y (1 + x²) = 2 – x

Differentiating on both sides with respect to x, we get

Applying the power rule we get

We know derivative of a constant is 0, so above equation becomes

Applying the power rule we get

As the given curve passes through the x-axis, i.e., y=0,

So the equation on given curve becomes,

y(1+x²)=2-x

⇒ 0(1+x²)=2-x

⇒ 0=2-x

⇒ x=2

So the given curve passes through the point (2,0)

So the equation (i) at point (2,0) is,

Hence, the slope of tangent to the curve is

Therefore, the equation of tangent of the curve passing through (2,0) is given by

⇒ 5y=-x+2

⇒ x+5y=2

So the equation of tangent to the curve y (1 + x²) = 2 – x, where it crosses x-axis is x+5y=2.

Therefore the correct option is option A.

Given the equation of the curve is

y = x³ – 12x + 18

Differentiating on both sides with respect to x, we get

Applying the sum rule of differentiation, we get

We know derivative of a constant is 0, so above equation becomes

Applying the power rule we get

So, the slope of line parallel to the x-axis is given by

So equating equation (i) to 0, we get

3x²-12=0

⇒ 3x²=12

⇒ x²=4

⇒ x=±2

When x=2, the given equation of curve becomes,

y = x³ – 12x + 18

⇒ y = (2)³ – 12(2) + 18

⇒ y = 8– 24 + 18

⇒ y = 2

When x=-2, the given equation of curve becomes,

y = x³ – 12x + 18

⇒ y = (-2)³ – 12(-2) + 18

⇒ y = -8+24 + 18

⇒ y = 34

Hence the points at which the tangents to the curve y = x³ – 12x + 18 are parallel to x-axis are (2, 2) and (-2, 34).

So the correct option is option D.

Given the equation of the curve is

y = e^2x

Differentiating on both sides with respect to x, we get

Applying the exponential rule of differentiation, we get

As it is given the curve has tangent at (0,1), so the curve passes through the point (0,1), so above equation at (0,1), becomes

So, the slope of the tangent to the curve at point (0,1) is 2

Hence the equation of the tangent is given by

y-1=2(x-0)

⇒ y-1=2x

⇒ y=2x+1

It is given that the tangent to the curve y=e^2x at the point (0,1) meet x-axis i.e., y=0

So the equation on tangent becomes,

⇒ y=2x+1

⇒ 0=2x+1

⇒ 2x=-1

Hence the required point is

Therefore the tangent to the curve y = e^2x at the point (0, 1) meets x-axis at

So the correct option is option B.

Given equation of curve is x = t² + 3t – 8, y = 2t² – 2t – 5

x = t² + 3t – 8

Differentiating on both sides with respect to t, we get

Applying the sum rule of differentiation, we get

We know derivative of a constant is 0, so above equation becomes

Applying the power rule we get

y = 2t² – 2t – 5

Differentiating on both sides with respect to t, we get

Applying the sum rule of differentiation, we get

We know derivative of a constant is 0, so above equation becomes

Applying the power rule we get

Now we know,

Now substituting the values from equation (i) and (ii), we get

As given the curve passes through the point (2,-1), substituting these values in given curve equation we get

x = t² + 3t – 8

⇒ 2= t² + 3t – 8

⇒ t² + 3t – 8-2=0

⇒ t² + 3t – 10=0

Splitting the middle term we get

⇒ t² + 5t-2t – 10=0

⇒ t(t+ 5) -2(t+5)=0

⇒ (t+ 5) (t-2)=0

⇒ t+5=0 or t-2=0

⇒ t=-5or t=2……….(iii)

y = 2t² – 2t – 5

⇒ -1=2t² – 2t – 5

⇒ 2t² – 2t – 5+1=0

⇒ 2t² – 2t – 4=0

Taking 2 common we get

⇒ t² – t – 2=0

Splitting the middle term we get

⇒ t² – 2t +t– 2=0

⇒ t(t– 2)+1(t– 2)=0

⇒ (t– 2)(t+1)=0

⇒ (t– 2)=0 or (t+1)=0

⇒ t=2 or t=-1……..(iv)

So from equation (iii) and (iv), we can see that 2 is common

So t=2

So the slope of the tangent at t=2 is given by

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

So the correct option is option B.

Given the curve x³ – 3xy² + 2 = 0 and 3x²y – y³ – 2 = 0

x³ – 3xy² + 2 = 0

Differentiating on both sides with respect to x, we get

Applying the sum rule of differentiation and also the derivative of the constant is 0, so we get

Applying the power rule we get

Now applying the product rule of differentiation, we get

Let this be equal to m₁

3x²y – y³ – 2 = 0

Differentiating on both sides with respect to x, we get

Applying the sum rule of differentiation and also the derivative of the constant is 0, so we get

Applying the power rule we get

Now applying the product rule of differentiation, we get

Let this be equal to m₂

Multiplying equation (i) and (ii), we get

⇒ m₁.m₂=-1

As the product of the slopes is -1, hence both the given curves are intersecting at right angle i.., they are making angle with each other.

So the correct option is option C

Given f (x) = 2x³ + 9x² + 12x – 1

Applying the first derivative we ge

Applying the sum rule of differentiation and also the derivative of the constant is 0, so we get

Applying the power rule we get

⇒ f’(x)=6x²+18x+12-0

⇒ f’(x)=6(x²+3x+2)

By splitting the middle term, we get

⇒ f’(x)=6(x²+2x+x+2)

⇒ f’(x)=6(x(x+2)+1(x+2))

⇒ f’(x)=6((x+2) (x+1))

Now f’(x)=0 gives us

x=-1, -2

These points divide the real number line into three intervals

(-∞, -2), [-2,-1] and (-1,∞)

(i) in the interval (-∞, -2), f’(x)>0

∴ f(x) is increasing in (-∞,-2)

(ii) in the interval [-2,-1], f’(x)≤0

∴ f(x) is decreasing in [-2, -1]

(iii) in the interval (-1, ∞), f’(x)>0

∴ f(x) is increasing in (-1, ∞)

Hence the interval on which the function f (x) = 2x³ + 9x² + 12x – 1 is decreasing is [-2, -1].

So the correct option is option B.

Given f (x) = 2x + cosx if f:R→R

Applying the first derivative we ge

Applying the sum rule of differentiation and also the derivative of the constant is 0, so we get

Applying the power rule we get

⇒ f’(x)=2-sin x

Now we know, the maximum value of sin x is 1.

So f’(x)>0, ∀ x

Hence the function f is increasing function.

So the correct option is option D.

Given y = x (x – 3)²

⇒ y=x(x²-6x+9)

⇒ y=x³-6x²+9x

Applying the first derivative we get

Applying the sum rule of differentiation, so we get

Applying the power rule we get

By splitting the middle term, we get

Now gives us

x=1, 3

These points divide the real number line into three intervals

(-∞, 1), (1,3) and (3,∞)

(i) in the interval (-∞, 1), f’(x)>0

∴ f(x) is increasing in (-∞,1)

(ii) in the interval (1,3), f’(x)≤0

∴ f(x) is decreasing in (1,3)

(iii) in the interval (3, ∞), f’(x)>0

∴ f(x) is increasing in (3, ∞)

Hence the interval on which the function y = x (x – 3)² is decreasing is (1,3) i.e., 1<x<3

So the correct option is option A.

Given f (x) = 4 sin³x – 6 sin2x + 12 sin x + 100

Applying the first derivative we get

Applying the sum rule of differentiation, so we get

Applying the power rule we get

Applying the derivative,

⇒ f' (x)=12 sin²x (cos x)-12sin x (cos x)+12(cos x)

⇒ f' (x)=12 sin²x cos x-12sin x cos x +12cos⁡x

⇒ f' (x)=12 cos x(sin²x -sin x +1)

Now 1-sin x≥0 and sin²x≥0

Hence sin²x -sin x +1≥0

Therefore, f’(x)>0, when cos x>0, i.e.,

So f(x) is increasing when

and f’(x)<0, when cos x<0, i.e.,

Hence f(x) is decreasing when

Now

Therefore, f(x) is decreasing in

So the correct option is option B

(i) Let f(x)=sin 2x

Applying the first derivative we get

f’(x)=2cos 2x

Putting f’(x)=0, we get

2cos 2x =0

⇒ cos 2x=0

Is possible when

0≤x≤2π

Thus sin 2x is neither decreasing nor increasing on

(ii) Let f(x)=tan x

Applying the first derivative we get

f’(x)=sec² x

As square of any number is always positive,

So f’(x)>0 ∀

Hence tan x is increasing function in

(iii) Let f(x)=cos x

Applying the first derivative we get

f’(x)=-sin x

But we know, sin x>0 for

And –sin x<0 for

Hence f’(x)<0 for

⇒ cos x is strictly decreasing on

(iv) Let f(x)=cos 3x

Applying the first derivative we get

f’(x)=-3sin 3x

Putting f’(x)=0, we get

-3sin 3x=0

⇒ sin 3x=0

As sin θ=0 if θ=0, π, 2π, 3π

⇒ 3x=0,π, 2π, 3π

Now

Since so we write it on number line as

Now the point divide the interval into 2 disjoint interval.

i.e.,

case 1: for

Now

0<3x<π

So when

We know that

sin θ>0 for θ∈ (0,π)

sin 3x>0 for 3x∈ (0,π)

From equation (a), we get

sin 3x>0 for

-sin 3x<0 for

⇒ f’(x)<0 for

⇒ f(x) is strictly decreasing on

case 2: for

Now

So when

We know that

Sin θ<0 in 3^rd quadrant

sin θ<0 for θ∈ (0,π)

sin θ <0 for

sin 3x<0 for

From equation (b), we get

sin 3x<0 for

-sin 3x>0 for

⇒ f’(x)<0 for

⇒ f(x) is strictly increasing on

Thus cos 3x is neither decreasing nor increasing on

So the correct option is option C, i.e., cos x is decreasing in

Given f (x) = tanx – x

Applying the first derivative we get

Applying the sum rule of differentiation, so we get

Applying the derivative,

⇒ f' (x)=sec²x-1

As square of any number is always positive,

So f’(x)>0 ∀ x∈R

Hence tan x –x is always increases.

So the correct option is option A.

Let f(x)= x² – 8x + 17

Applying the first derivative we get

Applying the sum rule of differentiation, so we get

Applying the derivative,

⇒ f' (x)=2x-8

Putting f’(x)=0,we get

2x-8=0

⇒ 2x=8

⇒ x=4

Therefore the minimum value of f(x) at x=4 is given by

f(x)= x² – 8x + 17

f(4)=4²-8(4)+17

⇒ f(4)=16-32+17

⇒ f(4)=1

Hence if x is real, the minimum value of x² – 8x + 17 is 1

So the correct option is option C.

Let f(x)= x³ – 18x² + 96x

Applying the first derivative we get

Applying the sum rule of differentiation, so we get

Applying the derivative,

⇒ f' (x)=3x²-36x+96

Putting f’(x)=0,we get critical points

3x²-36x+96=0

⇒ 3(x²-12x+32)=0

⇒ x²-12x+32=0

Splitting the middle term, we get

⇒ x²-8x-4x+32=0

⇒ x(x-8)-4(x-8)=0

⇒ (x-8)(x-4)=0

⇒ x-8=0 or x-4=0

⇒ x=8 or x=4

⇒ x∈[0,9]

Now we will find the value of f(x) at x=0, 4, 8, 9

f(x)= x³ – 18x² + 96x

f(0)= 0³ – 18(0)² + 96(0)=0

f(4)= 4³ – 18(4)² + 96(4)=64-288+384=160

f(8)= 8³ – 18(8)² + 96(8)=512-1152+768=128

f(9)= 9³ – 18(9)² + 96(9)=729-1458+864=135

Hence we find that the absolute minimum value of f(x) in [0,9] is 0 at x=0.

So the correct option is option B.

Let f(x)= 2x³ – 3x² – 12x + 4

Applying the first derivative we get

Applying the sum rule of differentiation, so we get

Applying the derivative,

⇒ f' (x)=6x²-6x-12

Putting f’(x)=0,we get critical points as

6x²-6x-12=0

⇒ 6(x²-x-2)=80

⇒ x²-x-2=0

Splitting the middle term we get

⇒ x²-2x+x-2=0

⇒ x(x-2)+1(x-2)=0

⇒ (x-2)(x+1)=0

⇒ x-2=0 or x+1=0

⇒ x=2 or x=-1

Now we will find the value of f(x) at x=-1, 2

f(x)= 2x³ – 3x² – 12x + 4

f(-1)= 2(-1)³ – 3(-1)² – 12(-1) + 4=-2-3+12+4=11

f(2)= 2(2)³ – 3(2)² – 12(2) + 4 =16-12-24+4=-16

Hence from above we find that x=-1 is point of local maxima and the maximum value of f(x) is 11.

Whereas x=2 is point of local minima and the minimum value of f(x) is -16.

So the correct option is option C.

Hence the given function 2x³ – 3x² – 12x + 4has one maxima and one minima

Let f(x)= sin x cos x

But we know sin2x=2sin x cos x

Applying the first derivative we get

Applying the derivative,

⇒ f’(x)=cos2x……(i)

Putting f’(x)=0,we get critical points as

cos2x=0

Now equating the angles, we get

Now we will find out the second derivative by deriving equation (i), we get

Applying the derivative,

⇒ f’’ (x)=-sin 2x.2

⇒ f’’(x)=-2sin2x

Now we will find the value of f’’(x) at , we get

But , so above equation becomes

Therefore at , f(x) is maximum and is the point of maxima.

Now we will find the maximum value of sin x cos x by substituting , in f(x), we get

f(x)= sin x cos x

Hence the maximum value of sin x cos x is

So the correct option is option B.

Given f (x) = 2 sin3x + 3 cos3x

Applying the first derivative we get

Applying the sum rule of differentiation and taking out the constant terms, we get

Applying the derivative,

⇒ f' (x)=2.cos3x.3-3.sin3x.3

⇒ f’(x)=6cos3x-9sin3x……(i)

Now we will find the value of f’(x) at , we get

Now split

Now we know cos(2π+θ)=cosθ and sin(2π+θ)=sinθ

Now we know and

And we find that f’(x) at is not equal to 0.

So cannot be point of maxima or minima.

Hence, f (x) = 2 sin3x + 3 cos3x at is neither maxima nor minima.

So the correct option is option D.

Given equation of curve is y = –x³ + 3x² + 9x – 27

Now applying first derivative, we get

Now applying the sum rule of differentiation and the differentiation of the constant term is 0 we get

Now applying the power rule of differentiation we get

This is the slope of the given curve.

Now we will differentiate equation (i) once again to find out the second derivative of the given curve,

Now applying the sum rule of differentiation and the differentiation of the constant term is 0 we get

Now applying the power rule of differentiation we get

Now we will find the critical point by equating the second derivative to 0, we get

-6(x-1) =0

⇒ x-1=0

⇒ x=1

Now we will differentiate equation (ii) once again to find out the third derivative of the given curve,

Now applying the sum rule of differentiation and the differentiation of the constant term is 0 we get

Now applying the power rule of differentiation we get

So the maximum slope of the given curve is at x=1

Now we will substitute x=1 in equation (i), we get

Hence the maximum slope of the curve y = –x³ + 3x² + 9x – 27 is 12.

So the correct option is option B.

Given equation is f (x) = x^x

Let y= x^x………(i)

Taking logarithm on both sides we get

log y=log (x^x)

⇒ log y=x log x

Now applying first derivative, we get

Now applying the product rule of differentiation we get

Now applying the deravative we get

Substituting the value of y from equation (i), we get

Now we will find the critical point by equating equation (i) to 0, we get

x^x (1+log x)=0

⇒ 1+log x =0 as x^x cannot be equal to 0

⇒ log x=-1

But -1=log e^-1

⇒ log x=log e^-1

Equating the terms we get

x= e^-1

Hence f(x) has a stationary point at

So the correct option is option B.

Let ………(i)

Taking logarithm on both sides we get

Now applying first derivative, we get

Now applying the product rule of differentiation we get

Now applying the derivative we get

Substituting the value of y from equation (i), we get

Now we will find the critical point by equating equation (i) to 0, we get

as cannot be equal to 0

But 1=log e¹

Equating the terms we get

Hence f(x) has a stationary point at .

i.e the maximum value of

So the correct option is option C.

Given the first curve is y = 4x² + 2x – 8

Now applying first derivative, we get

Now applying the sum rule of differentiation and the differentiation of the constant term is 0 we get

Now applying the power rule of differentiation we get

This is the slope of the first curve; let this be equal to m₁.

⇒ m₁=8x+2……(i)

Given the second curve is y = x³ – x + 13

Now applying first derivative, we get

Now applying the sum rule of differentiation and the differentiation of the constant term is 0 we get

Now applying the power rule of differentiation we get

This is the slope of the second curve; let this be equal to m₂.

⇒ m₂=3x²-1……(ii)

Now when the curve touch each other, there slope must be equal, i.e.,

m₁=m₂

⇒ 8x+2=3x²-1

⇒ 3x²-8x-1-2=0

⇒ 3x²-8x-3=0

Splitting the middle term, we get

⇒ 3x²+x-9x -3=0

⇒ x(3x+1)-3(3x+1)=0

⇒ (3x+1)(x-3)=0

⇒ (3x+1)=0 or (x-3)=0

Substituting in both the curves equation we get

For first curve, y = 4x² + 2x – 8

For second curve, y = x³ – x + 13

Thus at both the curves do not touch

Substituting x=3 in both the curves equation we get

For first curve, y = 4x² + 2x – 8

⇒y=4(3)²+2(3)-8

⇒y=4(9)+6-8=34

For second curve, y = x³ – x + 13

⇒y= (3)³ – (3) + 13

⇒y=27-3+13=37

Thus at x=3 both the curves do not touch

So the curves y = 4x² + 2x – 8 and y = x³ – x + 13 do not touch each other.

Given curve is y = tan x

Now applying first derivative, we get

This is the slope of the tangent

Substituting (0,0) in the slope of the curve, we get

Therefore the slope of the normal to the curve at (0,0) is -1

Hence the equation of the normal to the curve y=tanx at (0,0) is

y-0=-1(x-0)

⇒ y=-x

or x+y=0

Given f (x) = sinx – ax + b

Now apply the derivative we get

Now applying the sum rule of differentiation and the differentiation of the constant term is 0 we get

Now applying the derivative, we get

f'(x)=cos x-a

Now given f(x) increases on R

⇒ f’(x)≥0 ∀ x

⇒ cos x-a≥0 ∀ x

⇒ cos x≥a ∀ x

This is possible when a≤-1

Hence a∈(-∞, -1]

The values of a for which the function f (x) = sinx – ax + b increases on R are (-∞, -1].

Given

Now apply the derivative we get

Now applying the quotient rule of differentiation and the differentiation of the constant term is 0 we get

We will equate this with 0 to get critical points,

f’(x)=0

⇒ x²-1=0

⇒ x²=1

⇒ x=±1

The intervals formed by these two critical numbers are (-∞, -1), (-1, 0), (0, 1) and (1, ∞)

(i) in the interval (-∞, -1), f’(x)>0

∴ f(x) is increasing in (-∞,-1)

(ii) in the interval (-1, 0), f’(x)<0

∴ f(x) is decreasing in(-1,0)

(iii) in the interval (0, 1), f’(x)>0

∴ f(x) is increasing in (1, ∞)

(iii) in the interval (1, ∞), f’(x)<0

∴ f(x) is decreasing in (1, ∞)

Hence the function decreases in the interval (1, ∞).

Given (a > 0, b > 0, x > 0)

Now apply the derivative we get

Now applying the sum rule of differentiation we get

Now applying the quotient rule of differentiation on second part we get

We will equate this with 0 to get critical points,

f’(x)=0

Now second derivative gives,

Now apply the derivative we get

Now applying the sum rule of differentiation we get

Now applying the quotient rule of differentiation on second part we get

We will equate this with we get

Thus the least value of f(x) is

Multiply and divide by , we get

Hence the least value of the function (a > 0, b > 0, x > 0) is