Explanation
To find where the function is increasing 1) Find its derivatives Given : $$f\left( x \right) =3{ x }^{ 2 }-2x+1$$ $$f^{ 1 }\left( x \right) =6x-2$$ Equate it to zero $$6x-2=0\quad $$ $$6x=2\quad \\ $$ $$x=\dfrac { 1 }{ 3 } $$ (So the values which make derivatives equal to $$0$$ is $$\dfrac { 1 }{ 3 } $$) $$\rightarrow $$ Split and separate the value between intervals $$\left( -\infty ,\infty \right) $$ $$\rightarrow $$ Which gives $$\left( -\infty ,\dfrac { 1 }{ 3 } \right) \left( \dfrac { 1 }{ 3 } ,\infty , \right) $$ Since $$(2,5)$$ and $$ \left( \dfrac { 1 }{ 3 } ,\infty \right)$$ falls under this intervals.Hense both options A and B is the correct answers
Equation of tangent is $$y-2x+1=0$$
It is tangent at $$x=1$$, so for $$x=1$$
$$y-2(1)+1=0\\ \Rightarrow y=1$$
So, its is tangent to the curve at $$(1,1)$$
Slope of tangent $$= -\left (\dfrac{-2}{1}\right)=2$$
$$xy+ax+by=0\\ y+x\dfrac { dy }{ dx } +a+b\dfrac { dy }{ dx } =0$$
Now $$\dfrac { dy }{ dx } =2$$ at $$(1,1)$$
$$1+1(2)+a+b(2)=0\\ \Rightarrow a+2b=-3$$ .....(i)
$$(1,1)$$ also lies on the curve $$xy+ax+by=0$$
$$\Rightarrow 1(1)+a(1)+b(1)=0\\ \Rightarrow a+b=-1$$ .......(ii)
Solving (i) and (ii), we get
$$\Rightarrow a=1,b=-2$$
So, option E is correct.
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