Observe that the given curve is neither a circle nor an ellipse. For your reference the curve is shown in Figure.

Now,  \(\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} } \)
= -\(\frac{6 cos2t}{6sin3t} = -\frac{cos2t}{sin3t} \).
Therefore, the tangent at any point is
\(y-3sin2t= -\frac{cos2t}{sin3t}(x-2cos3t)\)
That is, x cos 2t + y sin 3t = 3sin 2t sin 3t + 2cos 2t cos 3t.
The slope of the normal is the negative of the reciprocal of the tangent which in this case is \(\frac{sin3t}{cos2t}\). Hence, the equation of the normal is \(y-3sin2t=\frac{sin3t}{cos2t}(x-2cos3t)\).
That is, x sin 3t - y cos 2t = 2sin 3t cos3t 3sin 2t cos 2t = sin 6t - \(\frac{3}{2}\) sin 4t.

The slope of the line y = x is 1. The tangent to the given curve will be parallel to the line, if the slope of the tangent to the curve at a point is also 1. Hence,
\(\frac{dy}{dx}=3x^{2}-6x+1=1\)
which gives \( 3x^{2}-6x=0\)
Hence, x = 0 and x = 2.
Therefore, at (0, –2) and (2, –4) the tangent is parallel to the line y = x.

The distance from the point (1, 1) to any point (x, y) is d =\(\sqrt { { \left( x-1 \right) }^{ 2 }+{ \left( y-1 \right) }^{ 2 } } \). Instead of extremising d, for convenience we extremise D = d² = (x−1)²+ (y− 1)² subject to the condition,  x² + y² = 1 Now, \(\frac { dD }{ dx } \) = 2( x −1)+2(y −1) \(\times\) \(\frac { dy }{ dx } \) where the \(\frac { dy }{ dx } \) will be computed by differentiating x² y² + = 1 with respect to x . Therefore we get, 2x + 2y \(\frac { dy }{ dx } \) = 0 which gives \(\frac { dy }{ dx } \) = -\(\frac { x }{ y } \) 
Substituting this, we get \(\frac { dD }{ dx } \) = 2(x −1) +2(y −1)\(\left(- \frac { x }{ y } \right) \)
= \(\frac { 2\left[ xy-y-xy+x \right] }{ y } \)
Substituting this, we get \(\frac { dD }{ dx } \) =2\(\left[ \frac { x-y }{ y } \right] \)=0
⇒x = y
Since (x, y) lie on the circle x² + y² + =1 we get, 2x² = 1 gives x = 土\(\frac { 1 }{ \sqrt { 2 } } \).  
Hence the points at which the extremum distance occur are \(\left( \frac { 1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) \), \(\left( -\frac { 1 }{ \sqrt { 2 } } ,-\frac { 1 }{ \sqrt { 2 } } \right) \) 
To find the extrema, we apply second derivative test. So,
\(\frac { { d }^{ 2 }D }{ { dx }^{ 2 } } =2\frac { { y }^{ 2 }+{ x }^{ 2 } }{ { y }^{ 3 } } \)
The value of \({ \left( \frac { { d }^{ 2 }D }{ { dx }^{ 2 } } \right) }_{ \left( \frac { 1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) }>0;{ \left( \frac { { d }^{ 2 }D }{ { dx }^{ 2 } } \right) }_{ \left( -\frac { 1 }{ \sqrt { 2 } } ,-\frac { 1 }{ \sqrt { 2 } } \right) }<0\) 
This implies the nearest and farthest points are \(\left( \frac { 1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } \right) \) and \(\left( -\frac { 1 }{ \sqrt { 2 } } ,-\frac { 1 }{ \sqrt { 2 } } \right) \)
Therefore, the nearest and the farthest distances are respectively \(\sqrt { 2 } \)-1 and \(\sqrt { 2 } \)+1

Let x, y be the sides of the rectangle. Hence the area of the rectangle is xy = k (given). The perimeter of the rectangle P is 2(x+ y). So the problem is to minimize 2(x+ y) suject to the condition xy = k. Let \(P(x)=2\left( x+\frac { k }{ x } \right) \)
\(P'(x)=2\left( 1-\frac { k }{ { x }^{ 2 } } \right) \)
P'(x) = 0 gives \(\left( 1-\frac { k }{ { x }^{ 2 } } \right) =0\)
As x, y are sides of the rectangle, \(x=\sqrt { k } \) is a critical number.
Now, P''(x) = \(\frac{4k}{x^3}\) and P''(\(​​\sqrt k\)) >0 \(\Rightarrow\) p(x) and has its minimum value at \(​​\sqrt k\)
Substituting \(x=\sqrt { k } \)  in xy = k we get \(y=\sqrt { k } \) . Therefore the minimum perimeter rectangle of a given area is a square.

The given function is defined and is differentiable at all \(x \in(0, \infty)\) 
f(x) = x log x + 3x.
Therefore f'(x) = log x+1+3 = 4 + log x.
The stationary points are given by 4 + log x = 0
That is x = e^-4
Hence the intervals of monotonicity are (0, e^-4) and \((e^{-4}, \infty)\) 
At \(x=e^{-5}\in(0,e^{-4})\), f'(e^-5) = -1<0 and hence in the interval (0, e^-4) -the function is strictly decreasing.
At \(x=e^{-3}\in(0,e^{-4})\), f'(e^-3) = 1>0 and hence strictly increasing in the interval \((e^{-4}, \infty)\). 
Since f′(x) changes from negative to positive when passing through x = e⁻⁴, the first derivative test tells us there is a local minimum at x = e^-4 and it is f(e-⁴) = -e^-4.

We have,
\(f(x)=log(1+x)-\frac{x}{1+x}\)
Therefore, \(f'(x)=\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\)
= \(\frac{x}{(1+x)^{2}}\).
Hence, f′(x) is \(\begin{cases} <0 \ when-10 \ whenx>0 \end{cases}\)
Therefore f (x) is strictly increasing for x > 0 and strictly decreasing for x < 0. Since f′(x) changes from negative to positive when passing through x = 0, the first derivative test tells us there is a local minimum at x = 0 which is f (0) = 0. Further, for x > 0, f(x) > f (0) = 0 gives
\(log(1+x)-\frac{x}{1+x}>0 \Rightarrow log(1+x)>\frac{x}{1+x}\).

 We have , f(x) = \(x^\frac{2}{3}\) then \(f'(x)=\frac{2}{3}x^{-\frac{1}{3}}\)\(​​​​=\frac{2}{3x^{\frac{1}{3}}}, f'(x)\ne 0 \forall x \in R\) and f'(x) does not exist at x = 0.
Therefore, there are no stationary points but there is a critical point at x = 0.

Because f'(x) changes its sign from negative to positive when passing through x = 0 for the function it has a local minimum at x = 0. The local minimum value is f(0) = 0. Note that here the local minimum occurs at a critical point which is not a stationary point.

Let f (x) = tan x, then the Mclaurin series of f (x) is 
\(f(x)=\sum^{n=\infty}_{n=0} a_{n}x^{n}\), where , \(a_{n}=\frac{f^{(n)}(0)}{n!}\).
Various derivative’s of the function f (x) evaluated at x = 0 is given below :
Now,
\(f'(x)=\frac{d}{dx}(tanx)=sec^{2}(x)\)
\(f''(x)=\frac{d}{dx}sec^{2}(x))=2sec x.sec x.tan x= 2sec^{2}x.tanx\)
\(f'''(x)=\frac{d}{dx}(2sec^{2}(x).tan x)= 2sec^{2}(x).sec^{2}x+tan x.4secx.secx.tanx\)
=\(2sec^{4}x+4sec^{2}x.tan^{2}x\)
\(f^{(iv)}(x)=8sec^{3}(x).secx+tan x+4sec^{2}x.2tanx sec^{2}x+8secx.secx.tanx.tan^{2}x\)
=\(16sec^{4}xtanx+8sec^{2}x.tan^{3}x\)
\(f^{(v)}(x)=16sec^{4}x.sec^{2}x+64sec^{3}x.sec x.tan x.tan x+8 sec^{2}x.3tan^{2}x.sec^{2}x+16secx.secx.tanx.tan^{3}x\)
=\(16sec^{6}x+88sec^{4}x.tan^{2}x+16sec^{2}x.tan^{4}x\).

Substituting the values and on simplification we get the required expansion of the function as
\(tan x=x+\frac{1}{3}x^{3}+\frac{2}{15}x^{5}+...; -\frac{\pi}{2}

Let f(x) = log(1+x) then the Maclaurin series of f (x) is f (x) \(\sum _{ n=0 }^{ n=\infty }{ { a }_{ n }{ x }^{ n } } \) where, \(a_{n}=\frac{f^{n}(0)}{n!}\) f(x) various derivatives of the function f(x) evaluated at x = 0 are given below:

Substituting the values and on simplification we get the required expansion of the function given by,
\((log(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+... -1\)

We have,
f(x) = (x-2)², then
\(f'(x)=2(x-2)=0 \) gives x = 2.
The intervals of monotonicity are \((-\infty,2)\) and \((2,\infty)\)
Since  \(f'(x)<0, \forall x \in (-\infty,2)\) for \((-\infty,2)\) the f(x) is strictly decreasing on \((2,\infty)\)
As \(f'(x)>0, \) for \(x \in(2, \infty)\) the function f(x) is strictly increasing on \((2,\infty)\)
Becasue f'(x) changes its sign from negative to positive when passing through x = 2 for the function f(x) it has a local minimum at x = 2
The local minimum value is f(2) = 0.