\(f(t)=\dfrac{t^2-5t-1}{t^3}\)

\(\Rightarrow\) \(f'(t)\) \(=\underbrace{\dfrac{t^3\dfrac{d}{dt}(t^2-5t-1)-(t^2-5t-1)\dfrac{d}{dt}t^3}{(t^3)^2}}_{Use \,Quotient\,Rule}\)

\(=\dfrac{t^3×(2t-5)-(t^2-5t-1)×3t^2}{t^6}\)

\(f'(2)\)= \(\dfrac{8×(4-5)-(4-10-1)×3×4}{2^6}\)

\(=\dfrac{-8+84}{64}\)

\(=\dfrac{76}{64}\)

\(=\dfrac{19}{16}\)

\(y=\dfrac{1+2x}{3-4x}\)

\(\Rightarrow\,\dfrac{dy}{dx}=\dfrac{(3-4x)×2-(1+2x)×-4}{(3-4x)^2}\)

\(=\dfrac{6-8x-4+8x}{(3-4x)^2}\)

\(=\dfrac{2}{(3-4x)^2}\)

\(\Rightarrow\)  \(y'\) at \((1,\,-1)\) \(=\dfrac{2}{(3-4)^2}=2\)

Equation of tangent is  \(y-(-1)=\) \(y'\)\((x-1)\)

\(\Rightarrow\,y+1=2(x-1)\)

\(\Rightarrow\,2x-y-3=0\)

\(y=\dfrac{x+1}{x-1}\)

\(\Rightarrow\,\dfrac{dy}{dx}=\dfrac{(x-1)×1-(x+1)×1}{(x-1)^2}\)

\(\therefore\,\dfrac{dy}{dx}=\dfrac{-2}{(x-1)^2}\)

Now

\(\dfrac{dy}{dx}=-2\)

\(\Rightarrow\,\dfrac{-2}{(x-1)^2}=-2\)

\(\Rightarrow\,(x-1)^2=1\)

\(\Rightarrow\,x-1=\pm1\)

\(\Rightarrow\,x=2\;or\;x=0\)

\(\therefore\,\) required points are \((0,\,-1)\;or\;(2,\,3)\)

\(\left(\dfrac{f}{g}\right)'=\dfrac{gf'-fg'}{g^2}\)

(Quotient Rule)

\(\left(\dfrac{f}{g}\right)'\Bigg|_{x=3}=\dfrac{g(3)f'(3)-f(3)g'(3)}{(g(3))^2}\)

\(=\dfrac{-2×4-5×7}{(-2)^2}\)

\(=\dfrac{-8-35}{4}\)

\(=\dfrac{-43}{4}\)

\(y=3x-2\sqrt x\)

\(\Rightarrow\) \(y'\) \(=3-\dfrac{2}{2\sqrt x}\)

\(\Rightarrow\)\(y'\)\(=3-\dfrac{1}{\sqrt x}\)

\(\Rightarrow\) \(y'\) at \((4,\,8)=3-\dfrac{1}{2}=\dfrac{5}{2}\)

Equation of normal is \(y-8=\dfrac{-1}{y'}(x-4)\)

\(\Rightarrow\,y-8=\dfrac{-1}{\dfrac{5}{2}}(x-4)\)

\(\Rightarrow\,y-8=\dfrac{-2}{5}(x-4)\)

\(\Rightarrow\,5y-40=-2x+8\)

\(\Rightarrow\,2x+5y-48=0\)

For horizontal tangent \(\to \dfrac{dy}{dx}=0\)

\(y=2x^3-15x^2+36x+1\)

\(\Rightarrow\,\dfrac{dy}{dx}=6x^2-30x+36\)

then  \(\dfrac{dy}{dx}=0\)

\(\Rightarrow\,6x^2-30x+36=0\)

\(\Rightarrow\,x^2-5x+6=0\)

\(\Rightarrow\,(x-2)\,(x-3)=0\)

\(\Rightarrow\,x=2\;or\;x=3\)

If two lines are parallel then there slopes are equal. So the required tangent has slope = 4.

(Same as slope of \(y=4x-1\))

\(\therefore\,\dfrac{dy}{dx}=4\)

\(\Rightarrow\,3x^2+1=4\)

\(\Rightarrow\,3x^2=3\)

\(\Rightarrow\,x=\pm1\)

\(\therefore\) The point at which tangent have slope 4 are \((1,\,0)\) and \((-1,\,-4)\).

\(\therefore \) There are two tangents -

(1)  \(y-0=4(x-1)\)

\(\Rightarrow\,4x-y-4=0\)

(2)  \(y+4=4(x+1)\)

\(\Rightarrow\,4x-y=0\)

(There can be more then one tangent to a curve of same slope).

Slope = 9 at \(x=1\)

\(\Rightarrow\,\dfrac{dy}{dx}\Bigg|_{x=1}=9\)

\(\Rightarrow\,2ax+b\Bigg|_{x=1}=9\)

\(\Rightarrow\,2a+b=9\to\)  (1)

Slope = 13 at \(x=2\)

\(\Rightarrow\,\dfrac{dy}{dx}\Bigg|_{x=2}=13\)

\(\Rightarrow\,4a+b=13\to\)  (2)

Passes through \((0,\,3)\)

\(\Rightarrow3=c\)

Solving equation (1) and (2) for \('a'\) and \('b'\)  we get

 \(a=2,\,\;b=5\)

\(\therefore\) the required curve is \(y=2x^2+5x+3\).