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finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=Inx+1\)

m(tangent) = In x + 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) \(=-\frac{1}{Inx+1}\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

now comparing the slope of normal with the given equation

m(normal) = 1

\(-\frac{1}{Inx+1}=1\)

\(x=\frac{1}{e^2}\)

since this point lies on the curve, we can find y by substituting x

\(y=-\frac{2}{e^2}\)

The equation of normal is given by

\(y+\frac{2}{e^2}=x-\frac{1}{e^2}\)

finding slope of the tangent by differentiating x and y with respect to theta

\(\frac{dx}{d\theta}=1+cos\theta\)

\(\frac{dy}{d\theta}=-sin\theta\)

Dividing both the above equations

\(\frac{dy}{dx}=-\frac{sin\theta}{1+cos\theta}\)

m at theta ( \(\pi/4\) ) = \(-1+\frac{1}{\sqrt{2}}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-1-\frac{1}{\sqrt{2}}\)\(=(-1+\frac{1}{\sqrt{2}})(x-\frac{\pi}{\sqrt{4}}-\frac{1}{\sqrt{2}})\)

We have,

f(x) = x – [x]

∴ f'(x) = 1 > 0

∴ f(x) is an increasing function on (0,1)

Given as f(x) = sin x

f'(x) = (d/dx)(sin x)

⇒ f’(x) = cos x

As given x ∈ (–π/2, π/2).

That is 4^th quadrant, where

⇒ Cos x> 0

⇒ f’(x) > 0

Thus, condition for f(x) to be increasing

Hence, f(x) is increasing on interval (–π/2, π/2).

Given as f(x) = cos² x

f'(x) = (d/dx)(cos² x)

⇒ f’(x) = 2 cos x (–sin x)

⇒ f’(x) = –2 sin (x) cos (x)

⇒ f’(x) = –sin2x

As given x belongs to (0, π/2).

⇒ 2x ∈ (0, π)

⇒ Sin (2x)> 0

⇒ –Sin (2x) < 0

⇒ f’(x) < 0

Thus, condition for f(x) to be decreasing

Hence f(x) is decreasing on interval (0, π/2).

Thus proved

Given as f(x) = x – sin x

f'(x) = (d/dx)(x - sin x)

⇒ f’(x) = 1 – cos x

As given x ϵ R

⇒ –1 < cos x < 1

⇒ –1 > cos x > 0

⇒ f’(x) > 0

Thus, condition for f(x) to be increasing

Hence, f(x) is increasing on interval x ∈ R

Given as f(x) = x³ – 15x² + 75x – 50

f'(x) = (d/dx)(x³ – 15x² + 75x – 50)

⇒ f’(x) = 3x² – 30x + 75

⇒ f’(x) = 3(x² – 10x + 25)

⇒ f’(x) = 3(x – 5)²

As given x ϵ R

⇒ (x – 5)² > 0

⇒ 3(x – 5)² > 0

⇒ f’(x) > 0

Thus, condition for f(x) to be increasing

Hence, f(x) is increasing on interval x ∈ R

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=2x-2\)

m(tangent) = 2x – 2

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

m(tangent) = 2

2x – 2 = 2

x = 2

since this point lies on the curve, we can find y by substituting x

y = 2² – 2 × 2 + 7

y = 7

therefore, the equation of the tangent is

y – 7 = 2(x – 2)

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=\frac{(2x-2)}{(x^2-2x+3)}\)

Now according to question, the slope of all tangents is equal to 0, so

\(-\frac{(2x-2)}{(x^2-2x+3)}=0\)

Therefore the only possible solution is x = 1

since this point lies on the curve, we can find y by substituting x

\(y=\frac{1}{1-2+3}\)

\(y=\frac{1}{2}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-\frac{1}{2}=0(x-1)\)

\(y=\frac{1}{2}\)

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=-\frac{1}{(x-3)^2}\)

Now according to question, the slope of all tangents is equal to 2, so

\(-\frac{1}{(x-3)^2}=2\)

\((x-3)^2=-\frac{1}{2}\)

We can see that LHS is always greater than or equal to 0, while RHS is always negative. 

Hence no tangent is possible.

Given:

The curve is y = 2x² – x + 1and the line y = 3x + 4

First, we will find The Slope of tangent

y = 2x² – x + 1

⇒ \(\frac{dy}{dx}\) = \(\frac{d}{dx}\)(2x²) – \(\frac{d}{dx}\)(x) + \(\frac{d}{dx}\)(1)

⇒ \(\frac{dy}{dx}\) = 4x – 1 ...(1)

y = 3x + 4 is the form of equation of a straight line y = mx + c, where m is the The Slope of the line

so the The Slope of the line is y = 3 (x) + 4

Thus, The Slope = 3. ...(2)

From (1) & (2),we get,

4x – 1 = 3

⇒ 4x = 4

⇒ x = 1

Substituting x = 1in y = 2x² – x + 1,we get,

⇒ y = 2(1)² – (1) + 1

⇒ y = 2 – 1 + 1

⇒ y = 2

Thus, the required point is (1,2)

let x₁, x₂ ∈(0, ∞)

We have, x₁ < x₂

⇒ log_e x₁ < log_e x₂

⇒ f(x₁) < f(x₂)

So, f(x) is increasing in (0, ∞)

finding the slope of the tangent by differentiating the curve

\(m=\frac{dy}{dx}=6x^2-2x\)

m = 4 at (1,4)

normal is perpendicular to tangent so, m₁m₂ = – 1

\(m(normal)=-\frac{1}{4}\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-4=(-\frac{1}{4})(x-1)\)

x + 4y = 17

finding slope of the tangent by differentiating the curve

\(\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}(\frac{dy}{dx})=0\)

\(\frac{dy}{dx}=-\frac{\sqrt{x}}{\sqrt{y}}\)

at (\(\frac{a^2}{4},\frac{a^2}{4}\)) slope m, is – 1

the equation of the tangent is given by y – y₁ = m(x – x₁)

\(y-\frac{a^2}{4}\)\(-1(x-\frac{a^2}{4})\)

\(x+y=\frac{a^2}{2}\)

finding the slope of the tangent by differentiating the curve

\(\frac{2}{3x^{1/3}}+\frac{2}{3y^{1/3}}\frac{dy}{dx}=0\)

\(\frac{dy}{dx}=-\frac{y^{1/3}}{x^{1/3}}\)

m(tangent) at (1,1) = – 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (1,1) = 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 1 = – 1(x – 1)

x + y = 2

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y – 1 = 1(x – 1)

y = x

We have,

f(x) = sin x – bx + c

f'(x) = cos x - b

Given that f(x) is on decreasing function on R

\(\therefore\) f'(x) < 0 for all x ∈ R

\(\Rightarrow\) cos x - b > 0 for all x ∈ R

\(\Rightarrow\) b < cos x for all x ∈ R

But the last value of cos x in 1

\(\therefore\) b ≥ 1

finding the slope of the tangent by differentiating the curve

\(2x=4\frac{dy}{dx}\)

\(\frac{dy}{dx}=\frac{x}{2}\)

m(tangent) at (2,1) = 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (2,1) = – 1

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 1 = 1(x – 2)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y – 1 = – 1(x – 2)

Given as the curve is y = x²

Differentiate the above with respect to x

⇒ y = x²

(dy/dx)_{= 2x^{2 - 1}}  

(dy/dx)_{= 2x ...(1)}

So, (dy/dx)_{= The slope of tangent = tan θ}

The tangent make an angle of 45° with x-axis

(dy/dx)_{= tan(45°) = 1 ...(2)}

Because the tan(45°) = 1

From the equation (1) & (2)

2x = 1

x = 1/2

Substitute x = 1/2 in y = x²

y = (1/2)²

y = 1/4 

Hence, the required point is (1/2,1/4)

We have,

f(x) = x³ – ax

f'(x) = 3x² - a

Given that f(x) is on increasing function

\(\therefore\) f'(x) 0 for all x ∈ R

\(\Rightarrow\) 3x² - a > 0 for all x ∈ R

\(\Rightarrow\) a < 3x² for all x ∈ R

But the last value of 3x² = 0 for x = 0

∴ a ≤ 0

Given:

The curve x² + y² = 13 and the line 2x + 3y = 7

x² + y² = 13

Differentiating the above w.r.t x

⇒ 2x^{2 – 1} + 2y^{2 – 1} \(\frac{dy}{dx}\)= 0

⇒ 2x + 2y\(\frac{dy}{dx}\) = 0

⇒ 2(x + y\(\frac{dy}{dx}\) ) = 0

⇒ (x + y\(\frac{dy}{dx}\) ) = 0

⇒ y\(\frac{dy}{dx}\)= – x

⇒ \(\frac{dy}{dx}\) = \(\frac{-x}{y}\)...(1)

Since, line is 2x + 3y = 7

⇒ 3y = – 2x + 7

⇒ y = \(\frac{-2x+7}{3}\)

⇒ y = \(\frac{-2x}{3}+\frac{7}{3}\)

\(\therefore\) The equation of a straight line is y = mx + c, where m is the The Slope of the line.

Thus, the The Slope of the line is \(\frac{-2}{3}\)...(2)

Since, tangent is parallel to the line,

\(\therefore\) the The Slope of the tangent = The Slope of the normal

\(\frac{-x}{y}\) = \(\frac{-2}{3}\)

⇒ – x = \(\frac{-2y}{3}\)

⇒ x = \(\frac{2y}{3}\)

Substituting x = \(\frac{2y}{3}\) in x² + y² = 13,

⇒ ( \(\frac{2y}{3}\) )² + y² = 13

⇒ ( \(\frac{4y^2}{3}\) ) + y² = 13

⇒ y²( \(\frac{4}{9}+1\) ) = 13

⇒ y²( \(\frac{13}{9}\) ) = 13

⇒ y²( \(\frac{1}{9}\) ) = 1

⇒ y² = 9

⇒ y = \(\pm\)3

Substituting y = \(\pm\)3 in x = \(\frac{2y}{3}\),we get,

x = \(\frac{2x(\pm3)}{3}\)

x = \(\pm\)2

Thus, the required point is (2, 3) & ( – 2, – 3)

Given as the curve is y = x²

y = x²

Differentiate the above with respect to x

(dy/dx)_{= 2x^{2 - 1}}

(dy/dx)_{= 2x ...(1)}

The slope of tangent is equal to the x-coordinate,

(dy/dx)_{= x ...(2)}

From the equation (1) & (2), 

2x = x

⇒ x = 0.

Substitute in y = x²,

y = 0²

⇒ y = 0

Hence, the required point is (0, 0)

Given:

The curve y = x² – 4x + 5 and line is 2y + x = 7

y = x² – 4x + 5

Differentiating the above w.r.t x,

we get the Slope of the tangent,

⇒ \(\frac{dy}{dx}\) = 2x^{2 – 1} – 4 + 0

⇒ \(\frac{dy}{dx}\) = 2x – 4 ...(1)

Since, line is 2y + x = 7

⇒ 2y = – x + 7

⇒ y = \(\frac{-x+7}{2}\)

⇒ y = \(\frac{-x}{2}+\frac{7}{2}\)

\(\therefore\) The equation of a straight line is y = mx + c, where m is the The Slope of the line.

Thus, the The Slope of the line is \(\frac{-1}{2}\) ...(2)

Since, tangent is perpendicular to the line,

\(\therefore\) The Slope of the normal = \(\frac{-1}{\text{The Slope of the tangent}}\)

From (1) & (2),we get

i.e, \(\frac{-1}{2}\) = \(\frac{-1}{2x-4}\)

⇒ 1 = \(\frac{1}{x-2}\)

⇒ x – 2 = 1

⇒ x = 3

Substituting x = 3 in y = x² – 4x + 5,

⇒ y = y = 3² – 4×3 + 5

⇒ y = 9 – 12 + 5

⇒ y = 2

Thus, the required point is (3,2)

slope of given line is 3

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=2x-2\)

m(tangent) = 2x – 2

since both lines are perpendicular to each other

(2x – 2) × 3 = – 1

\(x=\frac{5}{6}\)

since this point lies on the curve, we can find y by substituting x

\(y=\frac{25}{36}-\frac{10}{6}+7=\frac{217}{36}\)

therefore, the equation of the tangent is

\(y-\frac{217}{36}=-\frac{1}{3}(x-\frac{5}{6})\)

Given:

The curve is 2a²y = x³ – 3ax²

Differentiating the above w.r.t x

⇒ 2a² x \(\frac{dy}{dx}\)= 3x^{3 – 1} – 3 x 2ax^{2 – 1}

⇒ 2a² \(\frac{dy}{dx}\)= 3x² – 6ax

⇒ \(\frac{dy}{dx}\) = \(\frac{3x^2-6ax}{2a^2}\)...(1)

\(\therefore\frac{dy}{dx}\)= The Slope of the tangent = tanθ

Since, the tangent is parallel to x – axis

i.e,

⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2)

\(\therefore\) tan(0) = 0

 \(\therefore\frac{dy}{dx}\)= The Slope of the tangent = tanθ

From (1) & (2),we get,

⇒ \(\frac{3x^2-6ax}{2a^2}\) = 0

⇒ 3x² – 6ax = 0

⇒ 3x(x – 2a) = 0

⇒ 3x = 0 or (x – 2a) = 0

⇒ x = 0 or x = 2a

Substituting x = 0 or x = 2a in 2a²y = x³ – 3ax²,

when x = 0

⇒ 2a²y = (0)³ – 3a(0)²

⇒ y = 0

when x = 2

⇒ 2a²y = (2a)³ – 3a(2a)²

⇒ 2a²y = 8a³ – 12a³

⇒ 2a²y = – 4a³

⇒ y = – 2a

Thus, the required point is (0,0) & (2a, – 2a)

Given:

The curve is y = x²

y = x²

Differentiating the above w.r.t x

⇒ \(\frac{dy}{dx}\)= 2x^{2 – 1}

⇒ \(\frac{dy}{dx}\) = 2x ...(1)

Also given the Slope of the tangent is equal to the x – coordinate,

\(\frac{dy}{dx}\) = x ...(2)

From (1) & (2),we get,

i.e,2x = x

⇒ x = 0.

Substituting this in y = x², we get,

y = 0²

⇒ y = 0

Thus, the required point is (0,0)

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=12x^2-3\)

m(tangent) \(=12x^2-3\)

the slope of given line is \(-\frac{1}{9}\), so the slope of line perpendicular to it is 9

12x² -3 = 9

x = 1 or – 1

since this point lies on the curve, we can find y by substituting x

y = 6 or 4

therefore, the equation of the tangent is given by

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 6 = 9(x – 1)

or

y – 4 = 9(x + 1)

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

m(tangent) = 2

\(\frac{3}{2\sqrt{3x-2}}=2\)

\(\frac{9}{16}=3x-2\)

\(x=\frac{41}{48}\)

since this point lies on the curve, we can find y by substituting x

\(y=\sqrt{\frac{41}{16}-2}\)

\(y=\frac{3}{4}\)

therefore, the equation of the tangent is

\(y-\frac{3}{4}=2(x-\frac{41}{48})\)

Given as y = x² + 4x - 16

Differentiate with respect to x, to get the slope of tangent 

dy/dx = 2x + 4

m(tangent) = 2x + 4

The equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

Comparing the slope of tangent with given equation

2x + 4 = 3

x = -1/2

Substitute the value of x in the curve to find y

y = (1/4) - 2 - 16 = - 71/4

So, the equation of tangent is parallel to the given line is 

y + (71/4) = 3(x + (1/2))

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=2x+4\)

m(tangent) = 2x + 4

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

2x + 4 = 3

\(x=-\frac{1}{2}\)

Now substituting the value of x in the curve to find y

\(y=\frac{1}{4}-2-16=-\frac{71}{4}\)

Therefore, the equation of tangent parallel to the given line is

\(y+\frac{71}{4}=3(x+\frac{1}{2})\)

Given:

The curve y = 3x² + 4 and the Slope of the tangent is \(\frac{-1}{6}\)

y = 3x² + 4

Differentiating the above w.r.t x

⇒ \(\frac{dy}{dx}\) = 2 x 3x^{2 – 1} + 0

⇒ \(\frac{dy}{dx}\) = 6x ...(1)

Since, tangent is perpendicular to the line,

\(\therefore\) The Slope of the normal = \(\frac{-1}{\text{The Slope of the tangent}}\)

i.e, \(\frac{-1}{6}\) = \(\frac{-1}{6x}\)

⇒   \(\frac{1}{6}\) = \(\frac{1}{6x}\)

⇒ x = 1

Substituting x = 1 in y = 3x² + 4,

⇒ y = 3(1)² + 4

⇒ y = 3 + 4

⇒ y = 7

Thus, the required point is (1,7).

Given as y² = ax³ + b is y = 4x – 5

Differentiate the given curve, to get the slope of tangent

2y(dy/dx) = 3ax²

dy/dx = 3ax²/2y

m(tangent) at (2, 3) = 2a

The equation of tangent is given by y – y₁ = m (tangent) (x – x₁)

Comparing the slope of a tangent with the given equation

2a = 4

a = 2

(2, 3) lies on the curve, these points must satisfy

3² = 2 × 2³ + b

b = – 7

finding the slope of the tangent by differentiating the curve

\(2ay\frac{dy}{dx}=3x^2\)

\(\frac{dy}{dx}=\frac{3x^2}{2ay}\)

m(tangent) at (am², am³) is \(\frac{3m}{2}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (am², am³) is \(-\frac{2}{3m}\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-am^3=-\frac{2}{3m}(x-am^2)\)

finding the slope of the tangent by differentiating the curve

\(2y\frac{dy}{dx}=3ax^2\)

\(\frac{dy}{dx}=\frac{3ax^2}{2y}\)

m(tangent) at (2,3) = 2a

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

now comparing the slope of a tangent with the given equation

2a = 4

a = 2

now (2,3) lies on the curve, these points must satisfy

3² = 2 × 2³ + b

b = – 7

Given,

f(x) = x³ - 3x² + 4x

\(\therefore\) 3x² - 6x + 4

= 3(x² - 2x + 1) + 1

= 3(x - 1)² + 1 > 0 for all x ∈ R

Hence f(x) is strickly increasing on R

Given f(x) = cos x

\(\therefore\) f'(x) = -sin x

(i) Since for each x ∈ (0, π), sin x > 0

\(\Rightarrow\, \therefore\) f'(x) < 0

So f is strictly decreasing in (0, π)

(ii) Since for each x ∈ (π, 2π), sin x < 0

 \(\Rightarrow\, \therefore\) f'(x) > 0

So f is strictly increasing in (π, 2π)

(iii) Clearly from (1) and (2) above, f is neither increasing nor decreasing in (0, 2π)

finding the slope of the tangent by differentiating the curve

\(c^2(2x+2y\frac{dy}{dx})\)\(=2xy^2+2x^2y\frac{dy}{dx}\)

2xc² - 2xy² = \(2x^2y\frac{dy}{dx}-2x^2y\frac{dy}{dx}\)

\(\frac{dy}{dx}=\frac{xc^2-xy^2}{x^2y-yc^2}\)

m(tangent) at \((\frac{c}{cos\theta},\frac{c}{sin\theta})=-\frac{cos^3\theta}{sin^3\theta}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at = \((\frac{c}{cos\theta},\frac{c}{sin\theta})=-\frac{sin^3\theta}{cos^3\theta}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-\frac{c}{sin\theta}\)\(=-\frac{cos^3\theta}{sin^3\theta}(x-\frac{c}{cos\theta})\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

 \(y-\frac{c}{sin\theta}\)\(=-\frac{sin^3\theta}{cos^3\theta}(x-\frac{c}{cos\theta})\)

Correct Answer is \(\frac{3\pi}{4}\)

Let the principle value be given by x 

Now, let x = \(cot^{-1}(-1)\)

⇒cot x =-1 

⇒cot x= - cot( \(\frac{\pi}{4}\)) ( \(\because cot\left(\frac{\pi}{4}\right)\))=1) 

⇒cot x=cot( \(\pi-\frac{\pi}{4}\)) ( \(\because -cot(\theta)=cot(\pi-\theta\))) 

⇒x =\(\frac{3\pi}{4}\)

finding the slope of the tangent by differentiating the curve

\(\frac{x}{a^2}+\frac{y}{b^2}\frac{dy}{dx}=0\)

\(\frac{dy}{dx}=-\frac{xa^2}{yb^2}\)

m(tangent) at (a cosθ , b sinθ ) = \(-\frac{\cotθ a^2}{b^2}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (a cosθ , b sinθ ) = \(-\frac{\cot\theta a^2}{b^2}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-bsin\theta=-\frac{cot\theta a^2}{b^2}(x-aos\theta)\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-bsin\theta=\frac{b^2}{cot\theta a^2}(x-acos\theta)\)

finding the slope of the tangent by differentiating the curve

\(\frac{x}{a^2}-\frac{y}{b^2}\frac{dy}{dx}=0\)

\(\frac{dy}{dx}=\frac{xb^2}{ya^2}\)

m(tangent) at (a secθ , b tanθ ) = \(\frac{b}{a\sin\theta}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

 m(normal) at (a secθ , b tanθ ) = \(-\frac{a\sin\theta}{b}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-b\tan\theta=\frac{b}{a\sin\theta}(x-a\sec\theta)\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-b\tan\theta=-\frac{a\sin\theta}{b}(x-a\sec\theta)\)

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=2x\)

m(tangent) at (x = 0) = 0

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (x = 0) = \(\frac{1}{0}\)

We can see that the slope of normal is not defined

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y = 0

equation of normal is given by y – y₁ = m(normal)(x – x₁)

x = 0

finding slope of the tangent by differentiating x and y with respect to theta

\(\frac{dx}{d\theta}=-3\sin\theta+3\cos^2\theta\sin\theta\)

\(\frac{dy}{d\theta}=3\cos\theta-3\sin^2\theta\cos\theta\)

Now dividing \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) to obtain the slope of tangent

\(\frac{dy}{dx}\)\(=\frac{3\cos\theta-3sin^2\theta\cos\theta}{-3\sin\theta+3cos^2\theta\sin\theta}\)\(=-\tan^3\theta\)

m(tangent) at theta is\(=-\tan^3\theta\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at theta is cot³θ

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y - 3 sin θ + sin³ θ = -tan³ θ(x - 3cos θ + 3cos³ θ)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

y - 3 sin θ + sin³ θ = cot³ θ(x - 3cos θ + 3cos³ θ)

Given as y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)

Differentiate with respect to x, to get the slope of tangent 

dy/dx = 4x³ - 18x² + 26x - 10

m(tangent) at (0,5) = -10

m(normal) at (0,5) = 1/10

The equation of tangent is given by y - y₁ = m(tangent)(x - x₁)

y - 5 = -10x

y + 10x = 5

The equation of normal is given by y - y₁ = m(normal)(x - x₁)

y - 5 = (1/10)x

\(\cos \begin{Bmatrix}\cos^{-1}\left(\pi-\frac{\pi}{6}+\frac{\pi}{6}\right)\end{Bmatrix}.\)

\(=\cos\{\pi\}\)

\(=\cos\left(\frac{\pi}{2}+\frac{\pi}{2}\right)\)

= -1

(i) Given as tan^-1 (1/√3)

As we know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

Therefore, tan^-1 (1/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)

we know that the value is equal to π/6

So, tan^-1 (1/√3) = π/6

Hence the principal value of tan^-1 (1/√3) = π/6

(ii) Given as tan^-1 (-1/√3)

As we know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

Therefore, tan^-1 (1-/√3) = an angle in (-π/2, π/2) whose tangent is (1/√3)

But we know that the value is equal to -π/6

So, tan^-1 (-1/√3) = -π/6

Hence the principal value of tan^-1 (-1/√3) = – π/6

(iii) Given tan^-1 (cos(π/2))

we know that cos(π/2) = 0

As we know that for any x ∈ R, tan^-1 represents an angle in (-π/2, π/2) whose tangent is x.

So, tan^-1 (0) = 0

Hence the principal value of tan^-1 (cos(π/2) is 0.

(iv) Given tan^-1 (2 cos(2π/3))

we know that cos π/3 = -1

So, tan^-1 (2 cos(2π/3)) = tan^-1 (2 × – ½)

= tan^-1(-1)

= – π/4

Thus, the principal value of tan^-1 (2 cos(2π/3)) is – π/4

finding slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=4x^3-18x^2+26x-10\)

m(tangent) at (x = 1) = 2

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (x = 1) = \(-\frac{1}{2}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

y – 3 = 2(x – 1)

y = 2x + 1

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-3=-\frac{1}{2}(x-1)\)

2y = 7 - x

We have,

cos^-1 [3/5 cos x + 4/5 sin x], x ∈ [-3π/4, π/4]

Now, let cos α = 3/5

So, sin α = 4/5 and tan α = 4/3

cos^-1 [3/5 cos x + 4/5 sin x]

⇒ cos^-1 [3/5 cos x + 4/5 sin x] = cos^-1 [cos α cos x + sin α sin x]

= cos^-1 [cos (α – x)]

= α – x

= tan^-1 4/3 – x

It is given that y = (sec⁴ x – tan⁴ x)

By substituting the value of x = π/3

y = (sec⁴ (π/3) – tan⁴ (π/3)) = 2⁴ – (√3)⁴ = 7

By differentiating both sides w.r.t x

dy/dx = 4 sec³ x sec x tan x – 4 tan³ x sec² x

It can be written as

dy/dx = 4 sec² x tan x (sec² x – tan² x) = 4 sec² x tan x

So (dy/dx)_x= π/3 = 4 sec² (π/3) tan (π/3) = 16 √3

Here the equation of tangent at point (π/3, 7)

y – 7 = 16√3(x – π/3)

On further calculation

y – 7 = 16√3x – 16√3 π/3

By taking LCM as 3

y – 7 = (48√3x – 16√3 π)/3

We get

3y – 21 = 48√3x – 16√3 π

We can write it as

48√3x – 3y – 16√3 π + 21 = 0

Taking negative sign as common

3y – 48√3x + 16√3 π – 21 = 0

It is given that

y = x⁴ – 6x³ + 13x² – 10x + 5

By substituting the value of x = 1

y = (1)⁴ – 6(1)³ + 13(1)² – 10(1) + 5

On further calculation

y = 1 – 6 + 13 – 10 + 5 = 3

Point of contact is (1, 3)

Consider y = x⁴ – 6x³ + 13x² – 10x + 5 as the equation of curve

By differentiating w.r.t. x

dy/dx = 4x³ – 18x² + 26x – 10

So we get

(dy/dx)_{(1, 3)} = 4(1)³ – 18(1)² + 26(1) – 10 = 2

Here the required equation of tangent is

y – y₁ = m(x – x₁)

By substituting the values

y – 3 = 2x – 2

On further calculation

2x – y – 2 + 3 = 0

So we get

2x – y + 1 = 0

Here the required equation of normal i

y – y₁ = -1/m(x – x₁)

By substituting the values

y – 3 = -1/2(x – 1)

On further calculation

2y – 6 = – x + 1

So we get

2y – 6 + x – 1 = 0

x + 2y – 7 = 0

We know that if tangent is parallel to x -axis then it’s slope is equal to 0.

m : dy/dx = 4x - 6

4x – 6 = 0

⇒ x = 3/2

At x = 3/2, y = -17/2

So, the required points are (3/2, -17/2).

Consider y = (sin 2x + cot x + 2)² as the equation of the curve

By substituting the value of x = π/2 we get y = 4

On differentiation of both sides w.r.t. x

dy/dx = 2(sin 2x + cot x + 2) (2 cos 2x – cosec² x)

We get

(dy/dx)_x= π/2 = – 12

Here the equation of normal at point (π/2, 4)

y – 4 = 1/12 (x – π/2)

On further calculation

12y – 48 = x – π/2

We get

24y – 96 = 2x – π

It can be written as

24y – 96 – 2x + π = 0

24y – 2x + π – 96 = 0