Application Of Derivatives

(a) We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to its radius is given by,

When r = 3cm

Then

Therefore, the area of the circle is changing at the rate of 6πcm²/s when its radius is 3cm.

(b) We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to its radius is given by,

When r = 4cm

Then

Therefore, the area of the circle is changing at the rate of 8πcm²/s when its radius is 4cm.

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

Then, V = a³ and S = 6a² where a is the function of time t.

Now, It is given that

Then, by using the chain rule, we get,

………………(1)

So, when a = 12cm, then,

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

We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to time is given by,

It is given that

Then,

Thus, when r = 10cm

Then

Therefore, 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

…….. by chain rule

It is given that

Then,

Thus, when x = 10cm

Then

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

We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to time (t) is given by,

……………….. by chain rule

It is given that

Then,

Thus, when r = 8cm

Then

Therefore, when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing is 80π cm²/s.

We know that circumference of a circle (C) is C = 2πr

Then, Rate of change of circumference with respect to time (t) is given by,

……………….. by chain rule

It is given that

Then,

Therefore, the rate of increase of its circumference is 1.4π cm/s.

(a) It is given that 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, then we have,

The perimeter (P) of a rectangle is given by:

P = 2( x + y )

= 2(-5 +4) = -2 cm/min

Therefore, the rate of change of the perimeter is -2cm/min.

(b) It is given that 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, then we have,

Now, the area (A) of the rectangle is given by

A = x × y

= -5y + 4x

So, when x = 8cm and y = 6cm, then,

Therefore, the rate of change of the area of the rectangle is 2cm²/min.

Let V be the volume of the sphere, then

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

…….. by chain rule

It is given that

Then,

Thus, when r = 15cm

Then,

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

Let V be the volume of the sphere, then

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

…….. by chain rule

Thus, when r = 10cm

Then,

Therefore, the rate at which its volume is increasing with the radius when the later is 10 cm is 400π cm³/s.

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

we know that, by Pythagoras theorem,

x² + y² = 25

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

Now, it is given that

Therefore,

And when x= 4 m, then

=

Therefore, the height of the ladder on the wall is decreasing at the rate of m/s.

It is given that equation of the curve is 6y = x³ + 2.

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

………….. (1)

When the y- coordinate of the particle changes 8 times as fast as the x – coordinate

then, we have,

So, putting the value in (1), we get,

So, when x = +4, then,

And

So, when x = -4, then, y = -62/6 = -31/2

Therefore, the points required on the curve are (4,11) are (-4, -31/2).

As, the air bubble is in the shape of a sphere.

Therefore, V be the volume of the sphere, then

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

…….. by chain rule

It is given that

when, r = 1cm

Then,

Therefore, the rate is the volume of the bubble increasing when the radius is 1 cm is 2πcm³/s.

Let V be the volume of the sphere, then

It is given that, Diameter =

Now, the rate of change of its volume with respect to x is as

Therefore, the rate of change of its volume with respect to x is cm³/s.

Let V be the volume of the cone, then

It is given that, h=

⇒ r = 6h

Therefore,

Now, the rate of change of its volume with respect to time (t) is given by:

Now, the rate of change of its volume with respect to x is as

……… by chain rule

…………….(1)

Now, it is also given that and h = 4cm

So, putting the value in equation (1), we get

Therefore, the height of the sand cone increasing when the height is 4 cm is cm/s.

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

Then,

= 0.021x² – 0.006x +15

So, when x = 17 then,

MC = 0.021(17²) – 0.006(17) +15

= 0.021(289) – 0.102 + 15

= 20.967

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

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

Then,

So, when x = 7 then,

MR = 26(7) + 26 = 208

Therefore, the marginal revenue when x = 7 is Rs. 208.

We know that area of a circle (A) is A = πr²

Then, Rate of change of the area with respect to its radius is given by,

When r = 6cm

Then

Therefore, the area of the circle is changing at the rate of 12πcm²/s when its radius is 6cm.

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

Then,

So, when x = 15 then,

MR = 6(15) + 36 = 126

Therefore, the marginal revenue when x = 15 is Rs. 126.

Let x₁ and x₂ be any two numbers in R.

Then, we have,

x₁ < x₂

⇒ 3x₁ < 3x₂

⇒ 3x₁ +17 < 3x₂ +17

⇒ f(x₁) < f(x₂)

Therefore, f is strictly increasing on R.

Let x₁ and x₂ be any two numbers in R.

The, we have,

x₁ < x₂

⇒ 2x₁ < 2x₂

⇒ e^2x₁ < e^2x₂

⇒ f(x₁) < f(x₂)

Therefore, f is strictly increasing on R.

(a) The function is f (x) = sin x

Then, f’(x) = cos x

Since for each x ϵ , cos x > 0, we have f’(x) > 0

Therefore, f’ is strictly increasing in.

(b) The function is f (x) = sin x

Then, f’(x) = cos x

Since for each , cos x < 0, we have f’(x) < 0

Therefore, f’ is strictly decreasing in.

(c) The function is f (x) = sin x

Then, f’(x) = cos x

Since for each x ϵ , cos x > 0, we have f’(x) >0

Therefore, f’ is strictly increasing in……………….(1)

Now, The function is f (x) = sin x

Then, f’(x) = cos x

Since, for each x ϵ, cos x < 0, we have f’(x) < 0

Therefore, f’ is strictly decreasing in …………(2)

From (1) and (2),

It is clear that f is neither increasing nor decreasing in (0, π).

(a) It is given that function f(x) = 2x² – 3x

⇒ f’(x) = 4x – 3

If f’(x) = 0, then we get,

So, the points divides the real line into two disjoint intervals, and

So, in interval, f’(x) = 4x -3 >0

Therefore, the given function (f) is strictly increasing in interval.

(b) It is given that function f(x) = 2x² – 3x

⇒ f’(x) = 4x – 3

If f’(x) = 0, then we get,

So, the points divides the real line into two disjoint intervals, and

So, in interval f’(x) = 4x -3 < 0

Therefore, the given function (f) is strictly decreasing in interval.

(a) It is given that function f(x) = 2x³ – 3x² – 36x + 7

⇒ f’(x) = 6x² – 6x + 36

⇒ f’(x) = 6(x² – x + 6)

⇒ f’(x) = 6(x + 2)(x – 3)

If f’(x) = 0, then we get,

⇒ x = -2, 3

So, the points x = -2 and x = 3 divides the real line into two disjoint intervals, (-∞,2), (-2,3) and (3,∞)

So, in interval

f’(x) = 6(x + 2)(x – 3) >0

Therefore, the given function (f) is strictly increasing in interval .

(b) It is given that function f(x) = 2x³ – 3x² – 36x + 7

⇒ f’(x) = 6x² – 6x + 36

⇒ f’(x) = 6(x² – x + 6)

⇒ f’(x) = 6(x + 2)(x – 3)

If f’(x) = 0, then we get,

⇒ x = -2, 3

So, the points x = -2 and x = 3 divides the real line into two disjoint intervals, (-∞,2), (-2,3) and (3,∞)

So, in interval

f’(x) = 6(x + 2)(x – 3) < 0

Therefore, the given function (f) is strictly decreasing in interval.

It is given that function f(x) = x² + 2x – 5

f’(x) = 2x + 2

If f’(x) = 0, then we get,

⇒ x = -1

So, the point x = -1 divides the real line into two disjoint intervals, (-∞,-1) and (1,∞)

So, in interval (-∞,-1)

f’(x) = 2x + 2 < 0

Therefore, the given function (f) is strictly decreasing in interval (-∞,-1).

And in interval (1,∞)

f’(x) = 2x + 2 > 0

Therefore, the given function (f) is strictly increasing in interval (1,∞).

Thus, f is strictly increasing for x > -1.

It is given that function f(x) = 10 – 6x – 2x²

f’(x) = -6 – 4x

If f’(x) = 0, then we get,

⇒ x =

So, the point x = divides the real line into two disjoint intervals,

So, in interval

f’(x) = -6 – 4x > 0

Therefore, the given function (f) is strictly increasing in interval.

And in interval

f’(x) = -6 – 4x < 0

Therefore, the given function (f) is strictly decreasing in interval .

Thus, f is strictly decreasing for x > .

It is given that function f(x) = –2x³ – 9x² – 12x + 1

⇒ f’(x) = -6x² – 18x + 12

⇒ f’(x) = -6(x² +3x + 6)

⇒ f’(x) = -6(x + 1)(x + 2)

If f’(x) = 0, then we get,

⇒ x = -1 and -2

So, the points x = -1 and x = -2 divides the real line into two disjoint intervals,

(-∞,-2), (-2,-1) and (-1,∞)

So, in interval (-∞,-2),(-1,∞)

f’(x) = -6(x + 1) (x +2) < 0

Therefore, the given function (f) is strictly decreasing for x < -2 and x>-1.

So, in interval (-2.-1)

f’(x) = -6(x + 1)(x+2) > 0

Therefore, the given function (f) is strictly increasing for -2 < x < -1.

It is given that function f(x) = 6 – 9x – x²

f’(x) = -9 – 2x

If f’(x) = 0, then we get,

⇒ x =

So, the point x = divides the real line into two disjoint intervals,

So, in interval

f’(x) = -9 – 2x > 0

Therefore, the given function (f) is strictly increasing for x < .

And in interval

f’(x) = -9 – 2x < 0

Therefore, the given function (f) is strictly decreasing for x>.

Thus, f is strictly decreasing for x>.

It is given that function f(x) = –(x + 1)³ (x – 3)³

⇒ f’(x) = 3(x + 1)² (x – 3)³ +3(x + 1)³ (x – 3)²

⇒ f’(x) = 3(x + 1)² (x – 3)²[x – 3 + x + 1]

⇒ f’(x) = 6(x + 1)² (x – 3)²(x-1)

If f’(x) = 0, then we get,

⇒ x = -1,3 and 1

So, the points x = -1, x =1 and x = 3 divides the real line into four disjoint intervals,

So, in interval

f’(x) = 6(x + 1)² (x – 3)²(x-1) < 0

Therefore, the given function (f) is strictly decreasing in intervals .

So, in interval

f’(x) = 6(x + 1)² (x – 3)²(x-1) > 0

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

Let y =

Now, x > -1

⇒ 1 + x >0

Also, for all x> -1,

for x > -1

Therefore, f is an increasing function throughout its domain.

It is given that y = [x(x – 2)]², then,

= 4x(x - 2)(x - 1)

Now if

⇒ x = 0, 1,2

So, the points x = 0, x =1 and x = 2 divides the real line into four disjoint intervals,

(-∞,0), (0,1), (1, 2) and (2,∞).

So, in interval(-∞,0),(1,2)

< 0

Therefore, the given function (f) is strictly decreasing in intervals .

So, in interval (0,1) and (2,∞)

Therefore, the given function (f) is strictly increasing for 0 < x < 1 and x>2

We have, y =

Now,

⟹ 8cosθ + 4 = 4 + cos²θ + 4cosθ

⟹ cos²θ - 4cosθ = 0

⟹ cosθ(cosθ-4) = 0

⟹ cosθ = 0 or cosθ = 4

Since, cosθ≠4, cosθ = 0

⟹ cosθ = 0 ⟹ θ = π/2

Now,

In interval,, we have cos θ > 0. Also, 4 > cos θ

⇒ 4 – cosθ > 0

Therefore, cosθ(4 – cosθ) > 0 and also (2 + cosθ)² > 0

Therefore, y is strictly increasing in interval.

Also, the given function is continuous at x = 0 and x = .

Therefore, y is increasing in interval.

The given function is f(x) = logx

It is clear that for x>0,

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

It is given that function f(x) = x² – x + 1

f’(x) = 2x – 1

If f’(x) = 0, then we get,

⇒ x =

So, the point x = divides the interval (-1,1) into two disjoint intervals,

So, in interval

f’(x) = 2x – 1 < 0

Therefore, the given function (f) is strictly decreasing in interval

So, in interval

f’(x) = 2x -1 > 0

Therefore, the given function (f) is strictly increasing in interval for.

Therefore, f is neither strictly increasing and decreasing in interval (-1,1).

(A) Let f₁(x) = cosx

In interval,

Therefore, f₁(x) = cosx is strictly decreasing in interval.

(B) Let f₂(x) = cos2x

Now, 0 < x <

⇒ 0 < 2x < π

⇒ sin2x > 0

⇒ -2sin2x < 0

Therefore, f₂(x) = cos2x is strictly decreasing in interval.

(C) Let f₃(x) = cos3x

Now,

⇒ sin3x = 0

⇒ 3x = π, as xϵ

⇒ x =

The point x = divides the interval into two distinct intervals.

i.e. and

Now, in interval, ,

f₃'(x) = -3sin3x < 0 as (0 < x < => 0 < 3x < π)

Therefore, f₃ is strictly decreasing in interval.

Now, in interval

f₃'(x)=-3sin3x > 0 as

Therefore, f₃ is strictly increasing in interval.

(D) Let f₄ = tanx

In interval,

Therefore, f4 is strictly increasing in interval .

It is given that f (x) = x¹⁰⁰ + sin x –1

Then, f’(x) = 100x⁹⁹ + cosx

In interval (0,1), cos x >0 and 100x⁹⁹ > 0

⇒ f’(x)>0

Therefore, function f is strictly increasing in interval (0,1).

In interval, cos x < 0 and 100x⁹⁹ > 0.

Also, 100x⁹⁹ > cos x

⇒ f’(x) > 0 in

Therefore, function f is strictly increasing in interval .

In interval , cos x < 0 and 100x⁹⁹ > 0.

Also, 100x⁹⁹ > cos x

⇒ f’(x) > 0 on

Therefore, function f is strictly increasing in interval .

Hence, function f is strictly decreasing on none of the intervals.

It is given that function 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
⇒ a < -2x

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

⇒ a < -2x when x ϵ [1, 2]

Now, 1 ≤ x ≤ 2
⇒ -4 ≤ -2x ≤ -2

Therefore, the least value of a for f to be increasing on [1, 2] is given by

⇒ a = -4

Therefore, the least value of a is -4

It is given that

Now, f’(x) =0

The points x =1 and x = -1 divide the real line in three disjoint intervals

(-∞,-1),(-1,1) and (1,∞)

Now in interval, (-1,1)

it is clear that -1 < x < 1

⇒ x² < 1

Therefore, f’(x) = 1- < 0 (-1,1) ~ {0}

Therefore, f is strictly decreasing on (-1,1) ~ {0}

x < -1 or 1 < x

⇒ x² > 1

Therefore, f’(x) = 1- > 0 (-∞, -1) and (1,∞)

Therefore, f is strictly increasing in interval I disjoint from (-1,1)

Hence Proved.

It is given that f (x) = log sin x

In interval, f’(x) = cot x >0

Therefore, f is strictly increasing in.

In interval, f’(x) = cot x < 0

Therefore, f is strictly decreasing in.

It is given that f (x) = log |cos x|

In interval, f’(x) = -tanx < 0

Therefore, f is strictly decreasing on.

In interval, f’(x) = -tanx > 0

Therefore, f is strictly increasing in.

We have, f (x) = x³ – 3x² + 3x – 100

=> f’(x) = 3x² -6x + 3

= 3(x² -2x + 1)

= 3(x-1)2

For any x ϵ R, (x -1)2 > 0

Thus, f’(x) is always positive in R.

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

it is given that y = x² e^–x

then

Now if

⇒ x = 0 and x =2

The points x = 0 and x= 2 divide the real line into three disjoint intervals ie, (-∞,0), (0,2) and (2,∞).

In interval (-∞,0) and (2,∞),

f’(x) < 0 as e^-x is always positive.

Therefore, f is decreasing on (-∞,0) and (2,∞).

In interval (0,2), f’(x)>0

Therefore, f is strictly increasing in interval (0.2).

The given curve y = 3x⁴ – 4x

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

= 12(64) – 4

= 764

Therefore, the slop of the tangent is 764.

The given curve is y =

Then, the slope of the tangent

Therefore, the slope of the tangent is .

The given curve is y = x³ – x + 1

Then, the slope of the tangent

= 12 -1 = 11

Therefore, the slope of the tangent 11.

The given curve is y = x³ – 3x + 2

Then, the slope of the tangent

= 27 -3 = 24

Therefore, the slope of the tangent 24.

The given curve is y = acos³θ and y = asin³θ

Then, the slope of the tangent is given by:

Then, the slope of the tangent is given by:

Therefore, the slope of the tangent 1.

The given curve is x = 1− a sin θ and y = b cos² θ

Then, the slope of the tangent is given by:

Then, the slope of the tangent is given by:

Therefore, the slope of the tangent.

It is given that the curve y = x³ – 3x² – 9x + 7

We know that the tangent is parallel to the x-axis if the slope of the tangent is zero.

Therefore, 3x² -6x-9 = 0

⇒ x² -2x-3 = 0

⇒ (x-3)(x+1) =0

⇒ x =3 and 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 = 12

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

We know that if a tangent is parallel to the chord joining the points (2,0) and (4,4), then

Slope of the tangent = slope of the curve………….(1)

And, the slope of the curve =

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

Now, from (1) we have,

2(x -2) = 2

⇒ x-2 = 1

⇒ x =3

So, when x = 3 then y = (3-2)² = 1

Therefore, required points are (3,1).

It is given that equation of the curve y = x³ – 11x + 5

At which the tangent is y = x –11

⇒ Slope of the tangent = 1

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

⇒ 3x² -11 = 1

⇒ 3x² = 12

⇒ x² = 4

⇒ x = 2

So, when x = 2 then y = (2)³ -11(2) + 5 = -9

And when x = -2 then y = (-2)³ -11(-2) + 5 = 19

Therefore, required points are (2, -9) and (-2, 19).

It is given that equation of the curve y =

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

Now, if the slope of the tangent is -1, then we get,

⇒ (x-1)² = 1

⇒ (x-1) = 1

⇒ x = 2, 0

So, when x = 2 then y = 1

And when x = 0 then y = 1

Therefore, required points are (0, -1) and (2, 1).

Now, the equation of the tangent (0,1) is given by:

y – (-1) = -1(x-0)

⇒ y + 1 = -x

⇒ y + x+ 1 = 0

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

y – 1 = -1(x-2)

⇒ y - 1 = -x +2

⇒ y + x - 3 = 0

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

It is given that equation of the curve y =

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

Now, if the slope of the tangent is 2, then we get,

⇒ 2(x-3)² = -1

⇒ (x-3)² =

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

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

It is given that equation of the curve y = ,

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

Now, if the slope of the tangent is 0, then we get,

⇒ -2(x-1)=0

⇒ x =1

So, when x = 1 then y =

Now, the equation of the tangent (0,) is given by:

y – = 0(x-1)

⇒ y - = 0

⇒ y =

Therefore, the equations of the required line is y = .

(i) It is given that

Now, differentiating both sides with respect to x, we get

We know that the tangent is parallel to the x –axis if the slope is 0 ie,

, which is possible if x =0

Then, for x =0

⇒ y² = 16

⇒

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

(ii) It is given that

Now, differentiating both sides with respect to x, we get

We know that the tangent is parallel to the y–axis if the slope of the normal is 0 ie,

,

⇒ y = 0

Then, for y =0

⇒

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

It is given that equation of curve is x = cost, y = sint

On differentiating with respect to x, we get

Therefore, the slope of the tangent at ) is -1.

When

Then, the equation of the tangent is is

y – = -1(x – )

⇒ x + y -- = 0

⇒ x + y - = 0

Then, slope of normal at

=

Now, equation of the normal at

y – = 1(x –)

⇒ x = y

It is given that equation of curve is y = x⁴ – 6x³ + 13x² – 10x + 5

On differentiating with respect to x, we get

= 4x³ - 18x² +26x -10

Therefore, the slope of the tangent at (0, 5) is -10.

Then, the equation of the tangent is

y – 5 = -10(x – 0)

⇒ y – 5 = 10x

⇒ 10x +y =5

Also, slope of normal at (0,5)

Now, equation of the normal at (0,5)

⇒ 10y - 50 = x

⇒ x -10y +50 = 0

It is given that equation of curve is y = y = x⁴ – 6x³ + 13x²– 10x + 5

On differentiating with respect to x, we get

= 4x³ - 18x² +26x - 10

Therefore, the slope of the tangent at (1, 3) is 2.

Then, the equation of the tangent is

y – 3 = 2(x – 1)

⇒ y – 3 = 2x - 2

⇒ y = 2x +1

Then, slope of normal at (1,3)

=

Now, equation of the normal at (1,3)

y – 3 = (x – 1)

⇒ 2y -6 =- x + 1

⇒ x + 2y - 7 = 0

It is given that equation of curve is y = x³

On differentiating with respect to x, we get

= 3x²

Therefore, the slope of the tangent at (1, 1) is 3.

Then, the equation of the tangent is

y – 1 = 3(x – 1)

⇒ y = 3x - 2

Then, slope of normal at (1,1)

=

Now, equation of the normal at (1,1)

y – 1 = (x – 1)

⇒ 3y -3 =- x + 1

⇒ x + 3y - 4 = 0

It is given that equation of curve is y = x²

On differentiating with respect to x, we get

= 2x

Therefore, the slope of the tangent at (0, 0) is 0.

Then, the equation of the tangent is

y – 0 = 0(x – 0)

⇒ y = 0

Then, slope of normal at (0,0)

=

Now, equation of the normal at (0,0)

x = 0

(a) It is given that equation of the curve is y = x² – 2x +7

On differentiating with respect to x, we get

The equation of the line is 2x – y + 9 = 0

⇒ y = 2x + 9

⇒ Slope of the line = 2

Now we know that if a tangent is parallel to the line 2x – y + 9 = 0, then

Slope of the tangent = Slope of the line

⇒ 2 = 2x – 2

⇒ 2x = 4

⇒ x = 2

Now, putting x = 2, we get

y =4 -4 + 7 = 7

Then, the equation of the tangent passing through (2,7)

⇒ y – 7 = 2(x – 2)

⇒ y – 2x – 3 = 0

Therefore, 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) It is given that equation of the curve is y = x² – 2x +7

On differentiating with respect to x, we get

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

⇒ y =

⇒ Slope of the line = 3

Now we know that if a tangent is perpendicular to the line 5y – 15x = 13, then

⇒ 2x – 2=

⇒ 2x =

⇒ x =

Now, putting x = , we get

y =

Then, the equation of the tangent passing through

⇒ y – = (x – )

⇒

⇒ 36y – 217 = -2(6x -5)

⇒ 36y+12x – 227 = 0

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

The given curve y = 7x³ + 11

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

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

Therefore, the two tangents are parallel.

The given curve y = x³

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

We know that, when the slope of the tangent is equal to the y- coordinate of the point,

Then 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 = 27

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

The given curve y = 4x³ – 2x⁵

Then, the slope of the tangent at the point (x, y) is 12x² – 10x⁴

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

Y – y = (12x² – 10x⁴)(X – x) ………….(1)

When the tangent passes through the origin (0,0), then X =Y=0

Then equation (1) becomes,

-y = (12x² – 10x⁴)(– x)

⇒ y = (12x³ – 10x⁵)

Also, we have y = 4x³ – 2x⁵

⇒ (12x³ – 10x⁵) = 4x³ – 2x⁵

⇒ 8x⁵ – 8x³ = 0

⇒ x⁵ – 2x³ = 0

⇒ x³(x² – 1) = 0

⇒ x = 0 , 1

When x = 0 then y = 0

When x = 1 then y = 2

And when x = -1 then y = -2

Therefore, the required points are (0, 0), (1,2) and (-1, -2).

It is given that x² + y² – 2x – 3 = 0

Now, differentiating both sides with respect to x, we get

We know that the tangents are parallel to the x –axis if the slope of the tangent is 0 ie,

⇒ 1-x = 0

⇒ x = 1

But, x² + y² – 2x – 3 = 0 for x = 1

⇒ y² = 4

⇒ y = 2

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

It is given that ay² = x³

Now, differentiating both sides with respect to x, we get

Then, the slope of the tangent to the given curve at (am², am³) is

Then, slope of normal at (am², am³)

=

Therefore, equation of the normal at (am², am³) is given by:

y - am³ =

⇒ 3my – 3am⁴ = -2x + 2am²

⇒ 2x + 3my – am²(2 + 3m²) = 0

Therefore, equation of the normal at (am², am³) is 2x + 3my – am²(2 + 3m²) = 0

It is given that the equation of the normal to the curve y = x³ + 2x + 6

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

Then, slope of normal to the given curve at any point (x,y)

=Mediumy =

⇒ Slope of the given line =

We know that 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.

⇒

⇒ 3x2 + 2 = 14

⇒ 3x² = 12

⇒ x² = 4

⇒

So, when x = 2 then y = 18

and x -2 then y = -6

Hence, there are two normals to the given curve with the slope and passing through the points (2, 18) and (-2,-6).

Then, the equation of the normal through (2, 18) is:

y -18 =

⇒ 14y -252 = -x +2

⇒ x +14y-254 = 0

And, the equation of the normal through (-2, -6) is:

y –(-6) =

⇒ y + 6 =

⇒ 14y +84 = -x -2

⇒ x +14y+86 = 0

Therefore, the equation of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0 are x +14y-254 = 0 and x +14y+86 = 0.

The equation of a parabola is y² = 4ax, then,

On differentiating it with respect to x, we get

2y

Then, the slope of the tangent at (at², 2at) is

Then, the equation of the tangent at (at², 2at) is given by,

y – 2at =

⇒ ty -2at² = x –at²

⇒ ty = x + at2

Now, Then, slope of normal at (at², 2at)

=

Then, the equation of the normal at (at², 2at) is given by:

y – 2at = -t(x-at²)

⇒ y – 2at = -tx + at³

⇒ y = -tx + 2at + at³

Therefore, the equation of the normal at (at², 2at) is y = -tx + 2at + at³.

It is given that the curves x = y² and xy = k

Now, putting the value of x in y = k, we get

y³ = k

Then, the point of intersection of the given curves is

On differentiating x = y² with respect to x, we get

Then, the slope of the tangent at xy = k at

is

As we know that two curves intersect at right angles if the tangents to the curve at the point of intersection are perpendicular to each other.

So, we should have the product of the tangent as -1.

Then, the given two curves cut at right angles if the product of the slopes of their respective tangent at is -1.

Cubing both side, we get

⇒ 8k² = 1

Therefore, the given two curves cut at right angles if 8k² = 1.

It is given that the equations of the tangent and normal to the hyperbola , then,

On differentiating it with respect to x, we get,

Therefore, the slope of the tangent at (x₀, y₀) is

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

y - y₀ =

Then, slope of normal at (x₀, y₀)

=

Therefore, the equation of the normal at (x₀, y₀) is

y - y₀ = (x - x₀)

Therefore, the equation of the normal at (x₀, y₀) is.

It is given that

Then, the equation of the tangent at any given point (x, y) is given by,

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

⇒ y = 2x +

⇒ 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 = the slope of the line

When x =,

y =

Then, Equation of the tangent passing through the point is given by:

⇒ 24y – 18 = 48x – 41

⇒ 48x -24y =23

Therefore, the equation of the required tangent is 48x -24y =23

It I given that the slope of the normal to the curve y = 2x² + 3 sin x,

Then, the slope of the tangent at x = 0 is

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

Therefore, the slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is.

It is given that tangent to the curve y² = 4x

Then differentiating with respect to x, we have,

Then, the equation of the tangent at any given point (x,y) is given by,

The given line is y = x + 1

⇒ Slope of the line = 1

The line y = 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.

Then, we have,

⇒ y =2

Now, y = x+1

⇒ x = y -1

⇒ x = 2-1 = 1

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

Consider y = √x.

Let x = 25 and Δx = 0.3. Then, we get

Δy = √25.3 - √25

Δy = √25.3 – 5

Δy = √25.3 = Δy + 5

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

= 0.03

Therefore, the approximate value of is 5.03.

Consider y = √x.

Let x = 49 and Δx = 0.5. Then, we get

Δy =

= √49.5-√49

= √49.5 – 7

= √49.5 = Δy + 7

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

dy =

= 0.035

Therefore, the approximate value of is 7.035.

Consider y = .

Let x = 1 and Δx = -0.4. Then, we get

= √0.6-1

= √0.6 = Δy + 1

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

= -0.2

Therefore, the approximate value of is 0.8.

Consider y =

Let x = 0.008 and Δx = 0.001. Then, we get

Δy = (x+∆x)^1/3 - (x)^1/3

= (0.009)^1/3 - (0.008)^1/3 = (0.009)^1/3 - 0.2

= (0.009)^1/3 = Δy + 0.2

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

= 0.208

Therefore, the approximate value of is 0.208.

Consider y =

Let x = 1 and Δx = -0.001. Then, we get

Δy = (x+∆x)^1/10 - (x)^1/10

Δy = (0.999)^1/10 - 1

(0.999)^1/10 = Δy + 1

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

dy =

= -0.0001

Therefore, the approximate value of is 0.9999.

Consider y =

Let x = 16 and Δx = -1. Then, we get

Δy =

=

⇒ = Δy + 2

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

dy =

= -0.03125

Therefore, the approximate value of is -0.03125.

Consider y =

Let x = 27 and Δx = -1. Then, we get

Δy = (x+∆x)^1/3 - (x)^1/3

= (26)^1/3 - 3

= (26)^1/3 = Δy + 3

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

=

Therefore, the approximate value of is 2.9629.

Consider y =

Let x = 256 and Δx = -1. Then, we get

Δy = (x+∆x)^1/4 - (x)^1/4

= (225)^1/4 - (256)^1/4

= (225)^1/4 - 4

= (225)^1/4 = Δy + 4

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

= -0.0039

Therefore, the approximate value of (255)^1/4 is 3.9961.

Consider y =

Let x = 81 and Δx = 1. Then, we get

Δy =

=

⇒ = Δy + 3

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

dy =

= 0.009

Therefore, the approximate value of is 3.009.

Consider y =

Let x = 400 and Δx = 1. Then, we get

Δy =

=

⇒ = Δy + 20

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

dy =

= 0.025

Therefore, the approximate value of is 20.025.

Consider y =

Let x = 0.0036 and Δx = 0.0001. Then, we get

Δy =

=

⇒ = Δy + 0.06

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

dy =

= 0.00083

Therefore, the approximate value of is 0.06083.

Consider y =

Let x = 27 and Δx = -0.43. Then, we get

Δy =

=

⇒ = Δy + 3

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

dy =

= -0.015

Therefore, the approximate value of is 2.984.

Consider y =

Let x = 81 and Δx = 0.5. Then, we get

Δy =

=

⇒ = Δy + 3

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

dy =

= 0.0046

Therefore, the approximate value of is 3.0046.

Consider y =

Let x = 4 and Δx = -0.032. Then, we get

Δy =

=

⇒ = Δy + 8

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

dy =

= 3(-0.032)

= -0.096

Therefore, the approximate value of is 7.904.

Consider y =

Let x = 32 and Δx = 0.15. Then, we get

Δy =

=

⇒ = Δy + 2

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

dy =

= 0.00187

Therefore, the approximate value of is 2.00187.

Let x = 2 and Δx = 0.01. Then, we get,

f(2.01) = f(x + Δx) = 4(x +Δx)² + 5 ( x + Δx) + 2

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)

= (16 + 10 + 2) + (16 + 5)(0.01)

= 28 + 0.21

= 28.21

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

Let x = 5 and Δx = 0.001. Then, we get,

f(5.001) = f(x + Δx) = (x +Δx)³ - 7 ( x + Δx)² + 15

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)² – 7(5)² + 15] +[3(5)² -14(5)] (0.001)

= (125 – 175 + 15) + (75 - 70)(0.001)

= -35 + 0.005

= -34.995

Therefore, the approximate value of f (5.001) is -34.995.

It is given that the volume of a cube (V) of a side x

⇒ V = x³

= (3x²)Δx

= (3x²)(0.01x)

= 0.03x³

Therefore, the approximate change in the volume V of a cube of side x metres caused by increasing the side by 1% is 0.03x³.

It is given that the surface area of cube (S) of a side x

⇒ S = 6x²

= (12x)Δx

= (12x)(0.01x)

= 0.12x²

Therefore, the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1% is 0.12x².

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

Now, it is given that r = 7m and Δr = 0.02m

We know that volume of sphere (V) =

Now,

= (4πr²)Δr

= 4π (7²)(0.02)m³

= 3.92π m³

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

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

Now, it is given that r = 9m and Δr = 0.03m

We know that surface area of sphere (S) =

Now,

= (8πr)Δr

= 8π (9)(0.03)m²

= 2.16π m³

Therefore, the approximate error in calculating its surface area is 2.16π m³.

Let x = 3 and Δx = 0.02. Then, we get,

f(3.02) = f(x + Δx) = 3(x +Δx)² + 15( x + Δx) + 5

Now, Δy = f (x + Δx) – f(x)

⇒ f (x + Δx) = f(x) + Δy

≈ f(x) + f’(x).Δx (as dx = Δx)

⇒ f(3.02) ≈ (3x² + 15x + 5) + (6x + 15) Δx

= [3(3)² + 15(3) + 5] +[6(3) + 15] (0.02)

= (27 + 45 + 5) + (18 + 15)(0.02)

= 77 + 0.66

= 77.66

Therefore, the approximate value of f (3.02) is 77.66.

It is given that the volume of a cube (V) of a side x

⇒ V = x³

⇒ dV =

= (3x²)Δx

= (3x²)(0.03x)

= 0.09x³

Therefore, the approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is 0.09x³.

It is given that f (x) = (2x – 1)² + 3

Now, we can see that (2x – 1)² ≥ 0 for every x ϵ R

⇒ f (x) = (2x – 1)² + 3 ≥ 3 for every x ϵ R

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

2x -1 = 0

Then, Minimum value of

Therefore, function f does not have a maximum value.

It is given that f (x) = 9x² + 12x + 2 = (3x + 2)² - 2

Now, we can see that (3x + 2)² ≥ 0 for every x ϵ R

⇒ f (x) = (3x + 2)² - 2 ≥ -2 for every x ϵ R

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

3x +2 =0

Then, Minimum value of

Therefore, function f does not have a maximum value.

It is given that f (x) = –(x – 1)² + 10

Now, we can see that (x - 1)² ≥ 0 for every x ϵ R

⇒ f (x) = –(x – 1)² + 10 ≤ 10 for every x ϵ R

The minimum value of f is attained when x - 1 = 0

x - 1 = 0

⇒ x = 1

Then, Maximum value of f = f(1) = -(1-1)² + 10 = 10

Therefore, function f does not have a minimum value.

It is given that g(x) = x³ + 1

Now,

x ∈ ℝ

⇒ -∞ ≤ x ≤ ∞

⇒ -∞ ≤ x³ ≤ ∞

⇒ -∞ ≤ x³ + 1 ≤ ∞

The function g neither has a maximum value nor a minimum value.

It is given that f (x) = |x + 2| – 1

Now, we can see that |x + 2| ≥ 0 for every x ϵ R

⇒ f (x) = |x + 2| – 1 ≥ -1 for every x ϵ R

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

|x + 2| =0

⇒ x = -2

Then, Minimum value of f = f(-2) = |-2 + 2| - 1 = -1

Therefore, function f does not have a maximum value.

It is given that g(x) = –|x + 1| + 3

Now, we can see that –|x + 1| ≤ 0 for every x ϵ R

⇒ g(x) = –|x + 1| + 3 ≤ 3 for every x ϵ R

The maximum value of f is attained when |x + 1| = 0

|x + 1| = 0

⇒ x = -1

Then, Maximum value of g = g(-1) = -|-1 + 1| + 3 = 3

Therefore, function f does not have a minimum value.

It is given that h(x) = sin(2x) + 5

Now, we can see that -1 ≤ sin2x ≤ 1

⇒ -1 + 5 ≤ sin2x + 5 ≤ 1 + 5

⇒ 4 ≤ sin2x + 5 ≤ 6

Therefore, the maximum and minimum value of function h are 6 and 4 respectively.

It is given that f(x) = |sin 4x + 3|

Now, we can see that -1 ≤ sin4x ≤ 1

⇒ 2 ≤ sin 4x + 3 ≤ 4

⇒ 2 ≤ |sin 4x + 3| ≤ 4

Therefore, the maximum and minimum value of function h are 4 and 2 respectively.

It is given that h(x) = x + 1, x ∈ (–1, 1)

Now, if a point x₀ is closest to -1, then,

We find for all x₀∈ (–1, 1)

Also, if a point x₁ is closest to 1, then,

We find for all x₁∈ (–1, 1)

Therefore, the function h (x) has neither maximum nor minimum value in (-1, 1).

f(x) = x²

⇒ f’(x) = 2x

Now, f’(x) = 0

⇒ x = 0

⇒ x = 0 is the only critical point which could possibly be the point of local maxima or local minima of f.

⇒ f’’(0) = 2, which is positive.

Then, by second derivative test,

⇒ x = 0 is point of local maxima and local minima of f at x = 0 is f(0) = 0.

g(x) = x³ – 3x

⇒ g’(x) = 3x² - 3

Now, g’(x) = 0

⇒ 3x² - 3 = 0

⇒ 3x² = 3

⇒ x = � 1

g’’(x) = 6x

Now, g’(1) = 6>0

and g’(-1) = -6 < 0

Then, by second derivative test,

⇒ x = 1 is point of local maxima and local minima of g at x = 1 is

g(1) = 1³ – 3 = 1-3 =-2

And,

x = -1 is point of local maxima and local maximum value of g at x = -1 is

g(-1) = (-1)³ – 3(-1) = -1+3 = 2

h(x) = sin x + cos x,

h’(x) = cosx - sinx

Now, h’(x) = 0

⇒ cosx - sinx = 0

⇒ cosx = sinx

⇒ tanx = 1

h’’(x) = -sinx – cosx = -(sinx + cosx)

Then, by second derivative test,

is point of local maxima and local minima of h at is

f (x) = sin x – cos x, 0 < x < 2π

f’(x) = cosx + sinx

Now, f’(x) = 0

⇒ cosx + sinx = 0

⇒ cosx = -sinx

⇒ tanx = -1

’(x) = -sinx + cosx

Then, by second derivative test,

is point of local maxima and the local maximum value of f at is

And,

is point of local minima and the local minimum value of f at is

f (x) = x³ – 6x² + 9x + 15

⇒ f’(x) = 3x² – 12x + 9

Now, f’(x) = 0

⇒ 3x² – 12x + 9 = 0

⇒ 3(x-1)(x-3) = 0

⇒ x = 1,3

g’’(x) = 6x – 12 =6(x-2)

Now, f’(1) = 6(1-2)=-6 < 0

and f’(3) = 6(3-2) = 6 > 0

Then, by second derivative test,

⇒ x = 1 is point of local maxima and local maximum of f at x = 1 is

f(1) = 1³ – 6 +9 +15 = 19

And,

x = 3 is point of local minima and local minimum value of f at x = 3 is

f(3) = 27 – 54 + 27 +15 = 15

Now, g’(x) = 0

=

⇒ x² = 4

⇒ x = �2

Since x > 0, we take x = 2

Then, by second derivative test,

⇒ x = 2 is point of local minima and local minimum of g at x = 2 is

⇒ x = 0

Now, for values close to x = 0 and to the left of 0,

g’(x) > 0.

Also, for values close to x = 0 and to the right of 0,

g’(x) < 0

Then, by first derivative test,

⇒ x = 0 is point of local maxima and local maximum of g at x = 0 is

f’(x) = 0

⇒ 2 – 3x = 0

f (x) = e^x

⇒ f’(x) = e^x

Now, if f’(x) = 0, then e^x = 0.

But, the exponential function can never assume 0 for any value of x.

Therefore, there does not exist c ϵ R such that f’(c) = 0

Hence, function f does not have maxima or minima.

g(x) = logx

Since, log x is defined for a positive number x,

g’(x) > 0 for any x.

Therefore, there does not exist c ϵ R such that f’(c) = 0

Hence, function f does not have maxima or minima.

h(x) = x³ + x² + x +1

⇒ h’(x) = 3x² + 2x +1

h(x) = 0

⇒ 3x² + 2x +1 = 0

Therefore, there does not exist c ϵ R such that h’(c) = 0

Hence, function h does not have maxima or minima.

It is given that f (x) = x³, x ∈ [–2, 2]

⇒ f’(x) = 3x²

Now, f’(x) = 0

⇒ x = 0

Now, we evaluate the value of f 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

Therefore, we have the absolute maximum value of f on [-2, 2] is 8 occurring at x = 2.

And, the absolute minimum value of f on [-2, 2] is -8 occurring at x =-2.

It is given that f (x) = sin x + cos x, x ∈ [0, π]

f’(x) = cosx - sinx

Now, f’(x) = 0

⇒ cosx - sinx = 0

⇒ cosx = sinx

⇒ tanx = 1

⇒ x =

Now, we evaluate the value of f at critical point and at end points of the interval [0, π]

f(0) = sin0 +cos0 = 0+1 = 1

f(π) = sin π + cos π = 0 -1 = -1

Therefore, we have the absolute maximum value of f on [0, π] is √2 occurring at

And, the absolute minimum value of f on [0, π] is -1 occurring at x = π.

It is given that

Now, f’(x) = 0

⇒ x = 4

Now, we evaluate the value of f at critical point x = 0 and at end points of the interval

Therefore, we have the absolute maximum value of f on is 8 occurring at x = 4.

And, the absolute minimum value of f on is -10 occurring at x = -2.

It is given that f (x) = (x − 1)² + 3, x ∈ [−3,1]

⇒ f’(x) = 2(x – 1)

Now, f’(x) = 0

⇒ 2(x-1)

⇒ x = 1

Now, we evaluate the value of f at critical point x = 1 and at end points of the interval [-3, 1].

f(1) = (1 - 1)² + 3 = 0 + 3 = 3

f(-3) = (-3 - 1)² + 3 = 16 + 3 = 19

Therefore, we have the absolute maximum value of f on [-3, 1] is 19 occurring at x =-3.

And, the absolute minimum value of f on [-3,1] is 3 occurring at x = 1.

It is given that the profit function p(x) = 41 – 72x – 18x²

⇒ p’(x) = -72 – 36 x

and p’’(x) = -36

Now, g’(x) = 0

Then, by second derivative test,

x = -2 is point of local maxima of p.

Therefore, Maximum Profit =


= 113

Therefore, the maximum profit that the company can make is 113 units.

Let f (x) = 3x⁴ – 8x³ + 12x² – 48x + 25, x ∈[0,3]

⇒ f’(x) = 12x³ - 24x² + 24x – 48

=12(x³ - 2x² + 2x – 4)

=12[x² (x – 2)+ 2(x – 2)]

=12(x -2)( x²+ 2)

Now, f’(x) = 0

⇒ x =2 or (x²+ 2) = 0 for which there are no real roots.

Therefore, we will only consider x = 2

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

f(2) = 3(2)⁴ – 8(2)³ + 12(2)² – 48(2) + 25

= 3(16) – 8(8) + 12(4) +25

=48 – 64 +48 – 96 + 25

= -39

f(0) = 3(0)⁴ – 8(0)³ + 12(0)² – 48(0) + 25

= 0+ 0 + 0 +25

= 25

f(3) = 3(3)⁴ – 8(3)³ + 12(3)² – 48(3) + 25

= 3(81) – 8(27) + 12(9) +25

=243 – 216 +108 – 144 + 25

= 16

Therefore, we have the absolute maximum value of f on [0,3] is 25 occurring at x =0.

And, the absolute minimum value of f on [0,3] is -39 occurring at x = 2.

It is given that f (x) = sin2x, x ∈ [0, 2π]

f’(x) = 2cos2x

Now, f’(x) = 0

⇒ cos2x = 0

⇒ 2x = 0

⇒ x =

Now, we evaluate the value of f at critical point x = and at end points of the interval [0, 2π]

f’ =

f’ =

f’ =

f’ =

f(0) = sin0, f(2π) = sin2π = 0

Therefore, we have the absolute maximum value of f on [0, 2π] is 1 occurring at

x = and x =.

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

⇒ f’(x) = cosx - sinx

Now, f’(x) = 0

⇒ cosx - sinx = 0

⇒ cosx = sinx

⇒ tanx = 1

Now,

If f’’(x) will be negative when (sinx + cosx) > 0, means both sinx and cosx are positive.

And, we know that sinx and cosx both are positive in the first quadrant.

Then, f’’(x) will be negative when

f’’(x) = -sinx – cosx = -(sinx + cosx)

Now, let us take x =

Then, by second derivative test,

f will be maximum at x =

And, the maximum value of f is

Let f (x) = 2x³ – 24x + 107, x ∈ [1, 3]

⇒ f’(x) = 6x² – 24

=6(x² - 4)

Now, f’(x) = 0

⇒ 6(x² - 4) = 0

⇒ x² = 4

⇒ x = �2

Therefore, we will only consider the interval [1,3]

Now, we evaluate the value of f at critical point x = 2 ϵ [1,3] and at end points of the interval [1,3].

f(2) = 2(2)³ – 24(2) + 107

= 2(8) – 24(2) + 107

= 75

f(1) = 2(1)³ – 24(1)+ 107

= 2 – 24 +107

= 85

f(3) = 2(3)³ – 24(3) + 107

= 2(27) -24(3) +107

= 89

Therefore, we have the absolute maximum value of f on [1,3] is 89 occurring at x =3.

Now, we will only consider the interval [-3, -1]

Now, we evaluate the value of f at critical point x = -2 ϵ [-3, -1] and at end points of the interval [1,3].

f(-3) = 2(-3)³ – 24(-3) + 107

= 2(-27) – 24(-3) + 107

= 125

f(-1) = 2(-1)³ – 24(-1)+ 107

= -2 + 24 +107

= 129

f(-2) = 2(-2)³ – 24(-2) + 107

= 2(-8) -24(-2) +107

= 139

Therefore, we have the absolute maximum value of f on [-3, -1] is 89 occurring at x = -2.

It is given that f(x) = x⁴ – 62x² + ax + 9

Then, f’(x) = 4x³ – 124x+ a

It is given that function f attains its maximum value on the interval [0, 2] at x = 1.

⇒ f’(1) = 0

⇒ 4 – 124 + a = 0

⇒ a = 120

Therefore, the value of a is 120.

It is given that f (x) = x + sin 2x, x ∈ [0, 2π]

f’(x) = 1+ 2cos2x

Now, f’(x) = 0

2x = 2π � , n ϵ Z

⇒ x = nπ � , n ϵ Z

Now, we evaluate the value of f at critical point and at end points of the interval [0, 2π]

f’(0) = 0 + sin 0 = 0

f’(2π) = 2π + sin 4π = 2π + 0 = 2π

Therefore, we have the absolute maximum value of f on [0, 2π] is 2π occurring at

x = 2π and absolute minimum value of f(x) in the interval [0, 2π] is 0 occuring at x = 0.

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

Let P(x) denote the product of the two numbers.

Then, we get,

P(x) = x(24 –x) = 24x – x²

⇒ P’(x) = 24 – 2x

⇒ P’’(x) = -2

Now, P’(x) = 0

⇒ x = 12

And

P’’(12) = -2 <0

Then, by second derivative test,

x = 12 is the point of local maxima of P.

Therefore, the product of the numbers is the maximum when the numbers are 12 and 24 – 12 = 12.

Let the two numbers are x and y such that x + y = 60

⇒ y = 60 –x

Let f(x) = xy³

⇒ f(x) = x(60-x)³

⇒ f’(x) = (60 –x)³ -3x(60 –x)²

= (60-x)²[60 – x – 3x]

= (60-x)²[60 – 4x]

And f’’(x) = -2(60 – x)(60 -4x) -4(60-x)²

= -2(60 – x)[60 -4x + 2(60-x)]

= -2(60 – x)(180 – 6x)

= -12(60 – x)(30 -x)

Now, f’(x) =0

⇒ x = 60 or x =15

When x = 60, f’’(x) = 0.

When x = 15, f’’(x) = -12(60-15)(30-15) = -12×45×15 < 0

Then, by second derivative test, x =15 is a point of local maxima of f.

Then, function xy³ is maximum when x =15 and y = 60 – 15 = 45.

Therefore, required numbers are 15 and 45.

Let one number be x. Then, the other number is y = (35-x)

Let P(x)= x² y⁵

Then, we get,

P(x) = 2x(35 –x)⁵ – 5x²(35-x)⁴

= x(35 –x)⁴ [2(35-x)– 5x]

= x(35 –x)⁴ [70-7x]

= 7x(35 –x)⁴(10-x)

Now, P’’(x) = 7(35 –x)⁴ (10-x)+7x[-(35 –x)⁴ – 4(35-x)³ (10-x)]

= 7(35 –x)⁴ (10-x) - 7x(35 –x)⁴ – 28x(35-x)³ (10-x)]

=7(35 –x)³ [(35-x)(10-x) - x(35 –x) – 4x(10-x)]

=7(35 –x)³ [350-45x+x²-35x+x²-40x+4x²]

=7(35 –x)³ [6x²-120x+350]

Now, P’(x) = 0

⇒ x = 0, 35, 10

When x = 0, 35 This will make the product x2y5 equal to 0.

Therefore, x= 0, 35 cannot be possible values of x.

And when x = 10

Then, we have,

P’’(x) =7(35 –10)³[6(10)²-120(10)+350]

= 7(25)³ [-250]<0

Then, by second derivative test,

x = 10 and y = 35 -10 = 25 is the point of local maxima of P.

Therefore, the required number are 10 and 25.

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

Let S(x) be the sum of these number. Then,

S(x) = x³ + (16-x)³

⇒ S’(x) = 3x² -3(16-x)²

⇒ S’’(x) = 6x + 6(16-x)

Now, S’(x) =0

⇒ 3x² -3(16-x)² = 0

⇒ x² -(16-x)² = 0

⇒ x² – 256 - x² + 32x = 0

⇒ x = 8

Now, S’’(8) = 6(8) + 6(16-8)

= 48 + 48 = 96 > 0

Then, by second derivative test, x = 8 is the point of local minima of S.

Therefore, 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, then, the length and the breadth of the box will be (18 – x) cm each and the height of the box is x cm.

Then, the volume {V(x)} of the box is given by:

V(x) = x(18-x)²

⇒ V’(x) = (18-x)² - 2x(18-x)

= (18 - x)[18- x -2x]

= (18 - x)(18 - 3x)

Now, V’’(x) = (18 - x)(-3) + (18 - 3x)(-1)

= -3(18 - x) - (18 - 3x)

= -54 + 3x - 18 + 3x

= 6x - 72

Now, V’(x) = 0

⇒ x = 18 or 3

If x = 18 then breadth becomes 0 which is not possible

Therefore, x =3

V’’(3) = 6.3 - 72 = -ve

Then, by second derivative test, x= 3 is the point of maxima of V.

Therefore, If we remove a square of side 3cm 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, then, the height of the box is x and the length is 45-2x and the breadth is 24 – 2x.

Then, the volume {V(x)} of the box is given by:

V(x) = x(45-2x)(24-x)

= x(1080-90x-48x+4x²)

= 4x³ – 138x² + 1080x

⇒ V’(x) = 12x² – 276x + 1080

= 12(x² - 23x + 90)

=12(x – 18)(x – 5)

Now, V’’(x) = 24x – 276 = 12(2x-23)

Now, V’(x) = 0

⇒ x = 18 or 5

It is not possible to cut off a square of side 18cm from each corner of the rectangular sheet. So, x cannot be equal to1 8.

Therefore, x =5

V’’(5) = 12(10 – 23) = -156 < 0

Then, by second derivative test, x = 5 is the point of maxima of V.

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

The figure is given below:

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

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

Now, by Pythagoras theorem, we get,

(2a)² = l² + b²

⇒ b² = 4a² – l²

⇒ b =

Therefore, Area of rectangle, A =

.

.

.

Gives 4a² = 2l²

⇒ l = a

⇒ b =

when l =

Then, = -4 < 0

Then, by second derivative test, when l =, then the area of the rectangle is the maximum.

Since, l = b =,

Therefore, the rectangle is square.

Hence proved.

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

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

S = 2πr² + 2πrh

⇒ h

Let V be the volume of the cylinder. Then

V = πr²h

Now,

If

So, when then <0

Then, by second derivative test, the volume is the maximum when

Now, when . then h =

Therefore, the volume is the maximum when the height is the twice the radius or height is equal to diameter.

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

Let V be the volume of the cylinder. Then

V = πr²h = 100(given)

⇒ h =

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

S = 2πr² + 2πrh

Now, , <0

If

So, when then > 0

Then, by second derivative test, the surface area is the minimum when

Now, when then h = cm.

Therefore, the dimensions of the can which has the minimum surface area are and h cm.

Let a piece of length l 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 =

Let r be the radius of the circle,

Then, 2πr = 28 – l

⇒ r =

Therefore, the required area (a) is given by

A = (side of the square)2 + πr2

=

Then,

Now, if

Then,

⇒ (π + 4)l – 112 = 0

⇒

So, when

Then,

Then, by second derivative test, the area (A) is the minimum when .

Therefore, the combined area is the minimum when the length of the wire in making the square is cm while the length of the wire in making the circle is

28 - = cm.

Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.

Let V be the volume of cone.

Then, V=

And height of cone h = R +

Now,

Now, if,

After solving this we get,

So, when, , then < 0

Then, by second derivative test, the volume of the cone is the maximum when

So, when, , h = R +

Therefore, V =

Therefore, the volume of the largest cone that can be inscribed in the sphere is the volume of the sphere.

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

Then, the volume (V) of the cone is given by:

V=

The surface area (S) of the cone

S = πrl

=

Then,

Now, if

So, when then > 0

Then, by second derivative test, the surface area of the cone is the least when .

So when then h =

Therefore, 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.

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

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

It is given that slant height is constant.

Now, r = lsinƟ and h = lcosƟ

Then, the volume of the cone (V)

V =

.

Now, if

sin³Ɵ = 2sinƟcos²Ɵ

⇒ tan²Ɵ = 2

⇒ tanƟ = √2

⇒

Now, when, then tan²Ɵ = 2 or sin²Ɵ = 2cos²Ɵ.

Then, we get

= -4πl³cos³Ɵ < 0 for Ɵ ϵ

Then, by second derivative test, the volume (V) is the maximum when

Therefore, the semi-vertical angle of the cone of the maximum volume and of given slant height is .

Hence Proved.

We know that total surface area of the cone = S = πr(l + r) …(1)

and Volume of the cone(V) =

Then by (1), we get,

P = V²

Now, differentiating P with respect to r, we get,

Now, if, , then

S = 4πr²

Now again differentiating with respect to r, we get

Therefore, P is maximum when S = 4πr²

And V is maximum when S = 4πr²

⇒ πr(l + r) = 4πr²

⇒ l = 3r

SinƟ =

Therefore, semi-vertical angle of right circular cone of given surface area and maximum volume is .

Hence Proved.

It is given that x² = 2y

For each value of x, the position of the point will be

The distance d(x) between the points and (0,5) is given by:

d’(x) = 0

⇒ x³ – 8x = 0

⇒ x(x²-8) =0

⇒ x = 0,

And, .

So, now when x = 0, then d’’(x) = < 0

And when, x = , d’’(x) >0

Then, by second derivative test, d(x) is minimum at

So, when

Therefore, the point on the curve x2 = 2y which is nearest to the point (0,5) is .

Let

Then, f’(x) = 0

⇒ x² = 1

⇒ x = �1

Now,

And,

Also, f’’(-1) = -4 < 0

Then, by second derivative test, f is minimum at x = 1 and the minimum value is given by

Let f(x) =

Now, if f’(x) = 0

Then, we evaluate the value of f at critical point and at the end points of the interval [0, 1].

Therefore, we can conclude that the maximum value of f in the interval [0, 1] is 1.

Let us consider y = x^1/4 and and

Then,

Therefore,

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

dy =

Therefore, the approximate value of is

Let us consider y = and x=32 and

Then,

=

Therefore,

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

Therefore, the approximate value of is

It is given that f(x) =

Then, f’(x) =

Now, f’(x) = 0

⇒ 1 - logx =0

⇒ log x =1

⇒ log x = log e

⇒ x = e

Now, f’’(x) =

Now, f’’(e)=

Therefore, by second derivative test, f is the maximum at x = e.

Let ΔABC be isosceles where BC is the base of fixed length b.

Let the length of the two equal sides of ΔABC be a.

Now, draw a perpendicular AD to BC. Then, we have

In ΔADC, by using Pythagoras theorem,

AD =

Then, Area of triangle (A) =

The rate of change of the area with respect to time (t) is given by:

It is given that the two equal sides of the triangle are decreasing at the rate of 3cm per second.

Then

And when a = b we have,

Therefore, if the two sides are equal to the base, then the area of the triangle is decreasing at the rate of cm²/s.

It is given that curve is y² = 4x.

Differentiating with respect to x, we get,

Now, the slope of the normal at point (1,2) is

Therefore, Equation of the normal at (1,2) is y - 2 = - 1(x – 1)

⇒ y - 2 = - x + 1

⇒ x + y - 3 = 0

We have x = a cosθ + a θ sinθ,

⇒

And y = a sinθ – aθ cosθ

⇒

So,

Then, Slope of the normal at any point θ is .

The equation of the normal at a given point (x,y) is:

y - a sinθ + aθ cosθ = (x - a cosθ - a θ sinθ)

⇒ ysinθ – asin²θ + aθ sinθ cosθ = - x cosθ + acos²θ + aθ sinθ cosθ

⇒ xcosθ + ysinθ – a(sin²θ + cos²θ ) = 0

⇒ xcosθ + ysinθ –a = 0

Now, the perpendicular distance of the normal from the origin is

, which is independent of θ .

Therefore, the perpendicular distance of the normal from the origin is constant.

(i) It is given that f(x) =

Now, if f’(x) =0

⇒ cos x = 0 or cosx = 4

But, cosx = 4 is not possible

Therefore, cosx =0

⇒ x =

Now, x = divides (0,2π) into three disjoints intervals

In the intervals and, f’(x)>0

Therefore, f(x) is increasing for 0< x < and < x < 2π.

In interval, f’(x)<0

Therefore, f(x) is decreasing for < x < .

It is given that f(x) =

Then, f’(x) =0

⇒ 3x⁶ - 3 = 0

⇒ x⁶ = 1

⇒ x = 1

Now, the points x =1 and x = - 1 divide the real line into three disjoint intervals

( - ∞, - 1), ( - 1,1) and (1,∞).

In interval ( - ∞, - 1) and (1,∞) when x < - 1 and x > 1 then f’(x) >0

Therefore, when x < - 1 and x > 1, f is increasing.

And, in interval ( - 1,1) when - 1< x < 1 then f’(x) < 0.

Therefore, when - 1 < x < 1, f is decreasing.

It is given that ellipse

Let the major axis be along the x – axis.1).

Let ABC be the triangle inscribed in the ellipse where vertex C is at (a,0).

Since, the ellipse is symmetrical w.r.t. x - axis and y - axis, we can assume the coordinates of A to be ( - x₁,y₁) and the coordinates of B to be ( - x₁, - y₁).

Now, we have y1 = ±

Therefore, Coordinates of A and the coordinates of B

As the point(x₁,y₁) lies on the ellipse, the area of triangle ABC (A) is given by:

A =

……..(1)

Now,

But, x₁ cannot be equal to a.

⇒ x₁ =

y₁ =

Now,

Also, when x₁ = , then,

< 0

Then, the area is the maximum when x₁ = .

Therefore, Maximum area of the triangle is given by:

A =

Let l, b, and h be the length, breadth and height of the tank respectively.

then, we have h = 2m

Volume of the tank = 8 m³

Volume of the tank = l × b × h

⇒ 8 = l × b × 2

⇒ lb = 4

⇒ b =

Now, area of the base = lb = 4

Area of the 4 walls (A) = 2h(l + b)

Now,

⇒ = 0

⇒ l² = 4

⇒ l = ±2

Since, length cannot be negative therefore l =2.

⇒ b = 2

Now,

When l =2,

Then, by second derivative test, the area is the minimum when l =2.

We have, l =b=h=2

Therefore, Cost of building the base = Rs 70 × (lb) = Rs 70 (4) = Rs 280.

Cost of building the walls = Rs 2h (l + b) × 45 = Rs 90(2)(2 + 2)

= Rs 8(90) = Rs 720.

Required total cost = Rs(280 + 720) = Rs 1000.

Therefore, the total cost of the tank will be Rs 1000.

Let r be the radius of the circle and a be the side of the square.

Then, 2πr + 4a = k (where k is constant)

The sum of the areas of the circle and square (A) is

= πr² + a² = πr² +

Now,

8r = k - 2πr

⇒ (8 + 2π)r = k

⇒ r =

Now, > 0

Therefore, When r = , > 0

⇒ The sum of the area is least when r =

So, when r =

Then a =

Therefore, it is proved that the sum of their areas is least when the side of the square is double the radius of the circle.

Let x and y be the length and breadth of the rectangular window.

Radius of the semi - circular opening =

It is given that the perimeter of the window is 10m.

⇒ x + 2y +

Therefore, Area of the window (A) is given by

=

Now, , then

=0

Then, when x = then < 0.

Therefore, by second derivative test, the area is maximum when length

x = m.

Now, y =

Therefore, the required dimensions of the window to admit maximum light is given by length = m and breadth =m.

Let ΔABC be right - angled at B. Let AB = x and BC = y.

Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and Bc respectively.

Let <C = θ .

Now, we have,

Ac =

Now, PC = b cosecθ

And AP = a secθ

⇒ AC = AP + PC

⇒ AC = a secθ + b cosecθ

Now, if

⇒ asecθ tanθ = bcosecθ cotθ

⇒

⇒ asin³θ =bcos³θ

⇒

⇒

………..(1)

So, it is clear that < 0 when

Therefore, by second derivative test, the length of the hypotenuse is the maximum when

Now, when , we get,

Ac =

Therefore, the maximum length of the hypotenuses is .

It is given that function is 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 - 2)³(x + 1)²[4(x + 1) + 3 (x - 2)]

=(x - 2)³(x + 1)²(7x - 2)

Now, f’(x) =0

⇒ x = - 1 and x = or x = 2

Now, for values of x close to and to the left of

f’(x) > 0.

Also, for values of x close to and to the right of , f’(x) < 0.

Then, x = is the point of local maxima.

Now, for values of x close to 2 and to the left of 2, f’(x) < 0.

Also, for values of x close to 2 and to the right of 2. f’(x) > 0.

Then, x = 2 is the point of local minima.

Now, as the value of x varies through - 1, f’(x) does not changes its sign.

Then, x = - 1 is the point of inflexion.

It is given that f (x) = cos² x + sin x, x ∈ [0, π]

f’(x) = 2cosx( - sinx) + cosx

= - 2sinxcosx + cosx

Now, if f’(x) = 0

⇒ 2sinxcosx = cosx

⇒ cosx(2sinx - 1)=0

⇒ sin x = or cosx = 0

⇒ x =

Now, evaluating the value of f at critical points x = and x = and at the end points of the interval [0,π], (ie, at x = 0 and x =π), we get,

f

f(0)=

f(π)=cos²π + sinπ = ( - 1)² + 0 =1

f

Therefore, the absolute maximum value of f is occurring at x = and the absolute minimum value of f is 1 occuring at x =1, and π.

Let R and h be the radius and the height of the cone respectively.

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

V =

Now, from the right triangle BCD, we get,

BC =

V =

Now, if , then,

Now,

Now, when , it can be shown that < 0.

Therefore, the volume is the maximum when .

When,

Height of the cone = r + .

Therefore, it can be seen that the altitude of the circular cone of maximum volume that can be inscribed in a sphere of radius r is.

Since, f’(x) > 0 on (a,b)

Then, f is a differentiating function (a,b)

Also, every differentiable function is continuous,

Therefore, f is continuous on [a,b]

Let x1, x2 ϵ (a,b) and x2 > x1 then by LMV theorem, there exists c ϵ (a,b) s.t.

f’(c) =

Therefore, f is an increasing function.

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

Now, h =

The volume (V) of the cylinder is given by:

V = πr²h =2 πr²

Now, if

Now,

Now, we can see that at, , < 0.

Therefore, the volume is the maximum when.

When , the height of the cylinder is .

Therefore, the volume of the cylinder is the maximum when the height of the cylinder is .

Hence Proved.

The figure is given below:

Let VAB be a given cone of height h, semi-vertical angle α and let x b ethe radius of the base of the cylinder A'B'DC which is inscribed in the cone VAB.
In triangle VO'A',

VO' = x cotα
OO' = VO - VO' = h - x cotα
Let V be the volume of the cylinder. Then,
V = π(O'B')² (OO')
V = πx²(h - x cotα)
Differentiating with respect to x, we get,

Now, putting dV/dx = 0, for maxima or minima, we get,

Putting the value of x, we get,

Therefore, there is maxima at x = 2h/3 tanα
Hence, putting the value of x, in formula of volume, we get,

Hence, Proved.

Let V be the volume of the cylinder

V = πr²h

= π(10)²h

⇒ V =100πh

Differentiating w.r.t. t we get,

The tank is being filled with wheat at the rate of 314 cubic meter per hour.

Then, we have,

314 =

Therefore, the depth of wheat is increasing at the rate of 1 m/h.

It is given that curve x = t² + 3t – 8, y = 2t² – 2t – 5

Then,

The given points is (2, - 1)

At x = 2, we get

t² + 3t – 8 = 2

⇒ t² + 3t – 10 = 0

⇒ (t – 2)(t + 5) = 0

⇒ t = 2 and - 5

At y = - 1, we get

2t² - 2t – 5 = - 1

⇒ 2t² - 2t – 4 = 0

⇒ 2(t² - t – 2) = 0

⇒ (t – 2)(t + 1) = 0

⇒ t = 2 and - 1

Therefore, the common value of t is 2.

Hence, the slope of the tangent to the given curve at point (2, - 1) is

It is given that the equation of the tangent to the given curve is

y = mx + 1

Now, substituting the value of y in y² = 4x, we get

⇒ (mx + 1)² = 4x

⇒ m²x² + 1 + 2mx - 4x =0

⇒ m²x² + x(2m - 4) + 1 = 0………………..(1)

Since, a tangent touches the curve at one point, the root of equation (1) must be equal.

Thus, we get

Discriminant = 0

(2m - 4)² – 4(m²)(1) = 0

⇒ 4m² + 16 - 16m - 4m² =0

⇒ 16 – 16m = 0

⇒ m =1

Therefore, the required value of m is 1.

It is given that the equation of curve is 2y + x² = 3

Differentiating w.r.t. x, we get,

The slope of the normal to the given curve at point (1,1) is

Then, the equation of the normal to the curve at (1,1) is

⇒ y – 1 =1(x - 1)

⇒ y - 1 = x – 1

⇒ x – y = 0